Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Truncation Error in differentiation I dont understand how truncation error ie, the error caused because the definition of h tends to zero, but we take the value(during differentiation) as a non zero really small number; gets reduced because of taking centre difference.
That is, take $\frac{f(x+h)−f(x−h)}{2h}$ instead ... | By taking symmetric steps around the centre, we're able to cancel out the first-order term to achieve a second-order formula. Indeed, consider their Taylor expansions:
\begin{align*}
f(x + h) &= f(x) + f'(x)h + \frac{1}{2}f''(x)h^2 + \frac{1}{6}f'''(x)h^3 + \cdots \\
f(x - h) &= f(x) - f'(x)h + \frac{1}{2}f''(x)h^2 - \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2333350",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Derivative of $(2x-1)(x+3)^{\frac{1}{2}}$ Find the derivative of
$(2x-1)(x+3)^{\frac{1}{2}}$
My try -
$(2x-1)(\frac{1}{2} (x+3)^{\frac{-1}{2}} (x+0) + (x+3)^{\frac{1}{2}} (2)$
$ = (x+3)^{\frac{-1}{2}} (\frac{1}{2}(2x-1) + 2(x+3) $
$= \frac{3x+5.5}{2 (x+3)^{\frac{1}{2}}} $
My numerator is wrong and should be
$6x+... | you did wrong in the first step, the derivate of (x+3) is equal to 1.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2333747",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $ax^2+\frac{b}{x}\ge c$ for all positive $x$ where $a>0$ and $b>0,$ then show that $27ab^2\ge4c^3$ If $ax^2+\frac{b}{x}\ge c$ for all positive $x$ where $a>0$ and $b>0,$ then show that $27ab^2\ge4c^3$
I considered the function $f(x)=ax^2+\frac{b}{x}$ and I put $f'(x)=0$ to find $x^3=\frac{b}{2a}$
I am stuck here.
| By AM-GM $$ax^2+\frac{b}{x}=ax^2+2\left(\frac{b}{2x}\right)\geq3\sqrt[3]{ax^2\cdot\left(\frac{b}{2x}\right)^2}=3\sqrt[3]{\frac{ab^2}{4}}.$$
Thus, we need $$3\sqrt[3]{\frac{ab^2}{4}}\geq c,$$
which gives which you wish.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Compute $\Sigma _{n=0}^{\infty} \frac{1}{n^2+a^2}$ with Fourier series of $e^{ax}$ given $a>0$. As exercise, the teacher asks "Let $g(x)$ be the Fourier series of $e^{ax} x\in[-\pi,\pi]$. Look $g(x)$ up and use it to compute $\Sigma _{n=0}^{\infty} \frac{1}{n^2+a^2}$".
Since $g(x)=e^{ax}=\frac{2\sinh (a\pi)}{\pi}(\frac... | The Fourier coefficients of $f(x)=e^{ax}$ are given by:
$$\widehat{f}(n)=\frac{1}{2\pi}\int_0^{2\pi}e^{at}e^{-int}\,dt=\frac{e^{2(a-in)\pi}-1}{2\pi(a-in)}=\frac{e^{2a\pi}-1}{2\pi(a-in)}$$
so
$$|\widehat{f}(n)|^2=\frac{(e^{2a\pi}-1)^2}{4\pi^2}\frac{1}{a^2+n^2}$$
By Parseval's identity:
$$\sum_{-\infty}^{\infty}|\widehat... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2334002",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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No. Of possible triangles If there are n line segments in a plane of lengths (1,2,3,....,N) then find the no. of triangles that can be formed from these.
Attempt: let the sides be X,Y,Z such that X < Y< Z then the cardinality of required set boils down to finding out pairs such that X + Y > Z other two inequalities ar... | If $X<Y<Z$ and $X+Y>Z$,
$$Z+1\le X+Y\le Y-1+Y$$
and hence
$$Y\ge \frac{Z}{2}+1$$
The number of triangles is
\begin{align}
\sum_{Z=4}^N\sum_{Y=\lfloor \frac{Z}{2}\rfloor+1}^{Z-1}\sum_{X=Z-Y+1}^{Y-1}1&=\sum_{Z=4}^N\sum_{Y=\lfloor \frac{Z}{2}\rfloor+1}^{Z-1}(Y-1-Z+Y)\\
&=\sum_{Z=4}^N\sum_{Y=\lfloor \frac{Z}{2}\rfloor+1}^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2334413",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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How to properly simplify and trig substitute this integral? $\int\frac{{\sqrt {25-x^2}}}x\,dx$
I see that it is the form $a-x^2$.
Let $x = 5\sin\theta$
Then substitute this $x$ value in and
get $\int\frac {\sqrt {25-(5 \sin\theta)^2}}{5 \sin\theta}\, d\theta $
Take the root, simplified to $\int\frac { {5-(5 \sin\theta... | Let us begin by $x=5t $ and $dx=5\,dt $.
it becomes
$$\int\frac {\sqrt {25 (1-t^2)}}{5t}5\,dt $$
$$=5\int \frac {\sqrt {1-t^2}}{t}\,dt $$
now put $t=\sin (u) $ and $dt=\cos (u)\,du $.
we get
$$\pm 5\int \frac {\cos(u)}{\sin (u)}\cos (u)\,du $$
$$=\pm 5 \int \frac {\cos^2 (u)}{1-\cos^2 (u)}\sin (u)\,du $$
put $v=\cos (u... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2335930",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Find volume bounded by 3 equations using integration The prompt is to find the volume of the solid bounded by the equations $x^2 + y^2 -2y = 0$, $z = x^2 + y^2$ and $z \ge 0$
Plotting the equations, I get something like this
We get a paraboloid and a cylinder.
How to know if I have to use double or triple integration... | $\textbf{Second method}$ (I presume that this is the method you want):
Show that the volume enclosed by the paraboloid $z=x^2+y^2$, the $xy$-plane $z=0$, and the cylinder $x^2+(y-1)^2=1$ is $\frac{3\pi}{2}$.
This time, use the change of coordinates:
$$
\begin{align*}
x &= r\cos \theta, \\
y &=r\sin\theta, \\
z &=z... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2336489",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to sum an infinite matrix? Given this matrix that stretches to infinity to the right and up:
$$
\begin{matrix}
...&...&...\\
\frac{1}{4}& \frac{1}{8}& \frac{1}{16}&... \\
\frac{1}{2} & \frac{1}{4}& \frac{1}{8}&... \\
1 & \frac{1}{2}& \frac{1}{4}&... \\
\end{matrix}
$$... | The ''triangular array'' is not really an array but a column of values:
$ 2 $
$2-1=1$
$2-1-\frac{1}{2}=\frac{1}{2}$
$2-1-\frac{1}{2}-\frac{1}{4}=\frac{1}{4}$
$2-1-\frac{1}{2}-\frac{1}{4}-\frac{1}{8}=\frac{1}{8}$
$\cdots$
so there is not a diagonal
and the sum of these values is clearly $=4$
In other words, this is n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2338508",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Proving trigonometric identity $\cos^6A+\sin^6A=1-3 \sin^2 A\cos^2A$ Show that
$$\cos^6A+\sin^6A=1-3 \sin^2 A\cos^2A$$
Starting from the left hand side (LHS)
\begin{align}
\text{LHS} &=(\cos^2A)^3+(\sin^2A)^3 \\
&=(\cos^2A+\sin^2A)(\cos^4A-\cos^2A\sin^2A+\sin^4A)\\
&=\cos^4A-\cos^2A\sin^2A+\sin^4A
\end{align}
Can a... | $$=\cos^4A-\cos^2A\sin^2A+\sin^4A$$
$$=\cos^4A-2\cos^2A\sin^2A+\sin^4A + \cos^2a\sin^2a$$
$$=(\cos^2A-\sin^2A)^2 + \cos^2A\sin^2A=1-3\sin^2A\cos^2A$$
$$(\cos^2A-\sin^2A)^2 + 4\cos^2A\sin^2A=1$$
$2\cos x\sin x=\sin2x$
-> $\sin^22x=4\cos^2\sin^2x$
and
$\cos^2x-\sin^2x =\cos2x$
$$\cos^22A + sin^22A=1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2339401",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 3
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Trig Identity Proof $\frac{1 + \sin\theta}{\cos\theta} + \frac{\cos\theta}{1 - \sin\theta} = 2\tan\left(\frac{\theta}{2} + \frac{\pi}{4}\right)$ I've been working on this for like half an hour now and don't seem to be getting anywhere. I've tried using the double angle identity to write the LHS with $\theta/2$ and have... | Set $\theta=\pi/2-x$, so the left-hand side becomes
$$
\frac{1+\cos x}{\sin x}+\frac{\sin x}{1-\cos x}
$$
Now recall that
$$
\tan\frac{x}{2}=\frac{\sin x}{1+\cos x}=
\frac{1-\cos x}{\sin x}
$$
so the left-hand side is actually
$$
2\cot\frac{x}{2}=2\tan\Bigl(\frac{\pi}{2}-\frac{x}{2}\Bigr)=
2\tan\Bigl(\frac{\pi}{2}+\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2341368",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
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If $x$ is real, evaluate $k$ in absolute inequality
If $x$ is real and $$y=\frac{(x^2+1)}{x^2+x+1}$$ then it can be shown that $$\left|y-\frac{4}{3}\right|\leq k$$ Evaluate $k$
My attempt,
\begin{align}\left(y-\frac{4}{3}\right)^2&\leq k^2\\
\sqrt{\left(y-\frac{4}{3}\right)^2}&\leq k\end{align}
I don't know how to pr... | $\left|y-\frac{4}{3}\right|$ has a maximum at $x=1$ which is $k=\frac{2}{3}$
Derive $f(x)=\frac{x^2+1}{x^2+x+1}-\frac{4}{3}$ and set $f'(x)=0$. Verify that it is a maximum (for instance checking second derivative) then substitute in $\left|y-\frac{4}{3}\right|$ to get $k$.
Hope this helps.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2341689",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 3
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Proving right angle triangle A triangle $A$$B$$C$ is integer sided and has inradius $1$.prove that it is right angled. Please help.give some hints please.
| Let $a$, $b$ and $c$ be sides-lengths of the triangle.
Thus, $$1=\frac{2S}{a+b+c}$$ or
$$4(a+b+c)^2=(a+b+c)(a+b-c)(a+c-b)(b+c-a)$$ or
$$4(a+b+c)=(a+b-c)(a+c-b)(b+c-a).$$
Now, it's obvious that $a+b+c$ is an even number.
Thus, $a+b-c$, $a+c-b$ and $b+c-a$ are even positive numbers.
Let $a+b-c=2w$, $a+c-b=2v$ and $b+c-a=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2343094",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solution of differential equation using Laplace transform Can someone please help me identify my mistake as I am not able to find it.
$y''+y'=3x^2$ where $y(0)=0,y'(0)=1$
$L[y'']+L[y']=3L[x^2]$
$L[y'']=p^2L[y]-py(0)-y'(0)=p^2L[y]-1$
$L[y']=pL[y]-y(0)=pL[y]$
$p^2L[y]-1+pL[y]=\frac{6}{p^3}$
$p^2L[y]+pL[y]=\frac{6}{p^3}+1... | $\frac{6}{p^3(p^2+p)}=\frac{6}{p^4} - \frac{6}{p^3} + \frac{6}{p^2} - \frac{6}{p} + \frac{6}{1+p}$ then reverse transform gives you $x^3-3x^2+6x-6+6e^{-x}$ add the rest which is right in your calculation we get $$y=x^3-3x^2+6x-6+6e^{-x}+1-e^{-x}=x^3-3x^2+6x-5+5e^{-x}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2346773",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove the inequality using AM-GM only. By considering "Arithmetic mean $\geq $ Geometric mean" prove the trigonometric inequality:
$$\sin A + \sin B + \sin C \leq \frac{3\sqrt{3}}{2}. $$
where $A+B+C=180°$.
My try: By using transformation formulae, I proved that
$$\sin A + \sin B + \sin C = 4\cos\left(\frac{A}{2}\rig... | $$\sin A+ \sin B + \sin C=\sin A+ \sin B+ \sin (A+B)$$ $$= \sin(A)(1+ \cos (B))+ \sin(B)(1+\cos (A)) = \frac1{\sqrt3}((\sqrt3 \sin(A))(1+\cos(B))+(\sqrt3 \sin(B))(1+\cos(A))$$ Then use A.M.-G.M. inequality, $$\le \frac1{2\sqrt3}(3\sin^2(A)+1+\cos^2(B)+2\cos(B)+3\sin^2(B)+1+\cos^2(A)+2\cos(A))$$
$$=\frac1{2\sqrt3}(2\sin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2347366",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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Fraction problem I have been reading Basic Math and Pre Algebra for Dummies, the question below has me stumped!
David bought a cake for himself and his friends. He cut a piece for himself that was $1/6$ of the total cake. The Sharon cut a piece that was $1/5$ of what was left. Then Armand cut a piece that was $1/2$ of... | $\frac{1}{6}$ is taken off the bat, so we're left with $\frac{5}{6}$ of a cake. Then $\frac{1}{5}$ of that is taken, so $\frac{1}{5}\times\frac{5}{6}=\frac{1}{6}$ of the original cake is taken. Now we have $1-2\times\frac{1}{6}=\frac{2}{3}$ of the original cake left. Finally, half of the remaining cake is taken, so $\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2348462",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Other ways to evaluate the integral $\int_{-\infty}^{\infty} \frac{1}{1+x^{2}} \, dx$?
$$\int_{-\infty}^{\infty}\frac{1}{x^2+1}\,dx=\pi $$
I can do it with the substitution $x= \tan u$ or complex analysis. Are there any other ways to evaluate this?
| It's a bit of overkill, but since $f(x) = \frac{1}{1+x^{2}}$ is a monotonically decreasing function on $(0, \infty)$, the Riemann-like sum $$\frac{1}{n}\sum_{k=0}^{{\color{red}{\infty}}} f \left(\frac{k}{n} \right)= \frac{1}{n} \sum_{k=0}^{\infty} \frac{1}{1+ \left(\frac{k}{n} \right)^{2}} = n\sum_{k=0}^{\infty} \frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2349224",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 9,
"answer_id": 6
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Find the value of $x$ in the triangle $ABC$ A triangle $ABC$ has angle $A=2y$ and angle $C=y$. Furthermore, $AB=x,BC=x+2$, and $AC=5$. Solve for $x$.
I tried this question whole day but could not find solution of it . Please help me
| Let $AD$ be a bisector of $\Delta ABC$ and $AD=y$.
Thus, $AD=DC=y$ and $BD=x+2-y$.
Now, since $\Delta{ABD}\sim\Delta{CBA}$, we obtain
$$\frac{y}{5}=\frac{x}{x+2}=\frac{x+2-y}{x},$$ which gives
$xy=5x-2y$ and $x^2=(x+2)^2-xy-2y$, which is $xy=4x+4-2y$
and from here $4x-2y+4=5x-2y$ or $x=4$ and we are done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2351755",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Laplace's equation to solve a system in a square Solve the system in the square $S = \{0 < x < 1, 0 < y < 1\}$:
$$\begin{cases}
\Delta u = 0 \ \ \ &\text{for} \ \ (x,y)\in S\\
u(x,0) = 0, \ u(x,1) = x \ \ \ &\text{for} \ \ 0 < x < 1\\
u_x(0,y) = 0, \ u_x(1,y) = y^2 \ \ \ &\text{for} \ \ 0 < y < 1
\end{cases}$$
Attempte... | Given that $\Delta u = 0$ this implies that $\nabla^2 u = 0$ or
\begin{equation}
\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial^2 y} = 0
\end{equation}
Let $u = X(x)Y(y)$ then
\begin{equation}
Y \frac{d^2 X}{d^2 x^2} + X \frac{d^2 Y}{d^2 y} = 0
\end{equation}
Now we have two cases for boundary condi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2352744",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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MCQ The nth derivative of $f(x)=\frac{1+x}{1-x}$ Let $f(x)=\dfrac{1+x}{1-x}$ The nth derivative of f is equal to:
*
*$\dfrac{2n}{(1-x)^{n+1}} $
*$\dfrac{2(n!)}{(1-x)^{2n}} $
*$\dfrac{2(n!)}{(1-x)^{n+1}} $
by Leibniz formula
$$ {\displaystyle \left( \dfrac{1+x}{1-x}\right)^{(n)}=\sum _{k=0}^{n}{\binom {n}{k}}\ (... | You can do it with Leibniz's formula (not that it's easier than without it); just consider that
$$
(1+x)^{(k)}=
\begin{cases}
1+x & k=0 \\
1 & k=1 \\
0 & k>1
\end{cases}
$$
so the formula gives
$$
\left( \frac{1+x}{1-x}\right)^{\!(n)}=
(1+x)\left(\dfrac{1}{1-x}\right)^{\!(n)}+
n\left(\frac{1}{1-x}\right)^{\!(n-1)}
$$
N... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2353190",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Divisibility of $n\cdot2^n+1$ by 3. I want to examine and hopefully deduce a formula that generates all $n\geq0$ for which $n\cdot2^n+1$ is divisible by $3$. Let's assume that it is true for all $n$ and that there exist a natural number $k\geq0$ such that $$n\cdot2^n+1=3k,$$
Now I want to proceed with induction, but cl... | It should also be clear that if $n$ is divisible by $3$, then $n\cdot 2^n$ is divisible by $3$ so $n\cdot 2^n + 1$ is not divisible by $3$.
Therefore, the only candidates are $n=3k+1$ and $n=3k+2$.
Now, look at the remainders of $2^n$ when divided by $3$:
$$2^1\equiv 2\mod 3\\
2^2\equiv 1\mod 3\\
2^3\equiv 2\mod 3\\
2^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2355821",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 0
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If $ a=\frac{\sqrt{x+2} + \sqrt {x-2}}{\sqrt{x+2} -\sqrt{x-2}}$ then what is the value of $a^2-ax$ is equal to
If $$ a=\frac{\sqrt{x+2} + \sqrt {x-2}}{\sqrt{x+2} -\sqrt{x-2}}$$ then
the value of $a^2-ax$ is equal to:
a)2 b)1 c)0 d)-1
Ans. (d)
My attempt:
Rationalizing $a$ we get,
$ x+ \sqrt {x^2-4}$
$a^2=(x+\sqrt{... | Doing rationalization you get:
$$a=\frac{x+\sqrt{x^2-4}}{2}\to a-x=\frac{\sqrt{x^2-4}-x}{2}$$
so,
$$a(a-x)=\frac{(\sqrt{x^2-4})^2-x^2}{4}=-1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2356126",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 7,
"answer_id": 5
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Proving $a^{-2}bcd+b^{-2}cda+c^{-2}dab+d^{-2}abc\geqslant a+b+c+d$
If $a,b,c,d$ are positive real numbers, then prove that $$\frac{bcd}{a^2}+\frac{cda}{b^2}+\frac{dab}{c^2}+\frac{abc}{d^2}\geqslant a+b+c+d$$
Attempt:
$$\frac{bcd}{a^2}+\frac{cda}{b^2}+\frac{dab}{c^2}+\frac{abc}{d^2} = \frac{abcd}{a^3}+\frac{abcd}{b^3}... | We'll replace $a\rightarrow\frac{1}{a}$ and similar.
Thuse, we need to prove that
$$a^3+b^3+c^3+d^3\geq abc+abd+acd+bcd,$$
which is AM-GM or Muirhead.
For example
$$\sum_{cyc}a^3=\frac{1}{3}\sum_{cyc}(a^3+b^3+c^3)\geq\frac{1}{3}\sum_{cyc}3\sqrt[3]{a^3b^3c^3}=\sum_{cyc}abc$$
and we are done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2356408",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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What can be said if $A^2+B^2+2AB=0$ for some real $2\times2$ matrices $A$ and $B$? Let $A,B\in M_2(\mathbb{R})$ be such that $A^2+B^2+2AB=0$ and $\det A= \det B$. Our goal is to compute $\det(A^2 - B^2)$. According to the chain of comments on Art of Problem Solving, the following statements are true:
*
*$\det(A^2+B... | If $A = \pmatrix {a&b\\c&d}, B = \pmatrix {e&f\\g&h}$
$\det (A+B) + \det (A-B)\\(a+e)(c+h) - (b+f)(c+g) + (a-e)(c-h) - (b-f)(c-g)\\ 2ac+2eh - 2bc-2fg\\2(\det A + \det B)$
$\det (A+B) + \det (A-B) = 2(\det A + \det B)$
In which case the initial premise indeed holds.
I am pretty sure this only holds in the $2\times2$ cas... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2356658",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
How to find $1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{6} + \frac{1}{8} + \frac{1}{9} + \frac{1}{12}+\cdots$? Find the sum of series $$1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{6} + \frac{1}{8} + \frac{1}{9} + \frac{1}{12}+\cdots,$$ where the terms are the reciprocals of positive integers whos... | Hint: factor
$$\sum_{i=0}^\infty \sum_{j=0}^\infty \frac{1}{2^i 3^j}$$
as the product of a sum over $i$ and a sum over $j$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2357580",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
What is the maximum number of distinct positive integer's square that sums up to $2002$?
What is the maximum number of distinct positive integer's square that sums up to $2002$ ?
My tries:
$$\frac{n(n+1)(2n+1)}{6} = 2002$$
$$\implies n\approx 17 $$ but am clueless as to how to proceed any further.
| From $\frac {17(17+1)(2\cdot 17+1)}6 \lt 2002 \lt \frac {18(18+1)(2\cdot 18+1)}6$ you can conclude that at most $17$ distinct squares can be chosen to sum to $2002$ because the smallest $18$ add up to too much. Now you need to show that you can find $17$. If we were asked the same question for $34$ we could conclude ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2358892",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Formula for Natural logarithm of $\pi$ Does any formula or expansion exist that gives $\ln \pi$ ?
The expansion should not just be any formula of $\pi$ with a $\ln$ before it. For example $\ln \pi$ = k + $\sum f(x)$ or something of this type.
| We can also derive that
$$1-\sum_{n=1}^\infty \left(\ln(1-\frac{1}{4n^2})+4n coth^{-1}(2n)-2\right)=ln(\pi)$$
If we use $\int_0^{\frac{\pi}{2}} ln(sin(x)) dx = -\frac{\pi}{2}ln(2)$ and replace $sin(x)$ by it's Euler product, we get
$$\int_0^{\frac{\pi}{2}} ln(x \prod_{n=1}^\infty (1-\frac{x^2}{\pi^2n^2})) dx = -\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2360710",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 5
} |
Find minimum of $a+b+c+\frac1a+\frac1b+\frac1c$ given that: $a+b+c\le \frac32$
Find minimum of $a+b+c+\frac1a+\frac1b+\frac1c$ given that: $a+b+c\le \frac32$ ($a,b,c$ are positive real numbers).
There is a solution, which relies on guessing the minimum case happening at $a=b=c=\frac12$ and then applying AM-GM inequal... | For $a=b=c=\frac{1}{2}$ we get a value $\frac{15}{2}$.
We'll prove that it's a minimal value.
Indeed, we need to prove that
$$\sum_{cyc}\left(a+\frac{1}{a}-\frac{5}{2}\right)\geq0$$ or
$$\sum_{cyc}\frac{(a-2)(2a-1)}{a}\geq0$$ or
$$\sum_{cyc}\left(\frac{(2a-1)(a-2)}{a}+3(2a-1)\right)+6\left(\frac{3}{2}-a-b-c\right)\geq0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2360813",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 1
} |
Rotate the parabola $y=x^2$ clockwise $45^\circ$. I used the rotation matrix to do this and I ended up with the equation: $$x^2+y^2+2xy+\sqrt{2}x-\sqrt{2}y=0$$
I tried to plot this but none of the graphing softwares that I use would allow it.
Is the above the correct equation for a parabola with vertex (0,0) and axis o... | Before rotating (and plotting), you need to parametrize your parabola:
$$\begin{cases}
x(t) = t\\
y(t) = t^2
\end{cases},$$
where $t \in \mathbb{R}.$
You are rotating each point of the parabola, and hence:
$$\begin{bmatrix}X(t)\\Y(t)\end{bmatrix} = \frac{\sqrt{2}}{2}\begin{bmatrix}1 & -1\\1 & 1\end{bmatrix}\cdot\begin{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2363075",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 1
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Find n so that the following converges $\int_1^{+ \infty} \left( \frac{nx^2}{x^3 + 1} - \frac 1 {13x + 1} \right) dx$ Question
Determine $n$ such that the following improper integral is convergent
$$ \int_1^{+ \infty} \left(
\frac{nx^2}{x^3 + 1}
-\frac{1}{13x + 1}
\right) dx
$$
I'm not sure how to go abou... | Note that we have
$$\begin{align}
\lim_{b\to \infty}\left(\frac n3\log\left(b^3+1\right)-\frac1{13}\log\left(13b+1\right)\right) &= -\frac1{13}\log(13)\\\\
&+\lim_{b\to \infty}\left(n\log(b)-\frac1{13}\log(b)\right)\\\\
&+\frac n3 \lim_{b\to \infty}\log\left(1+\frac1{b^3}\right)\\\\&-\lim_{b\to \infty}\log\left(1+\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2365228",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 1
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How many five digit numbers divisible by $3$ can be formed using the digits $0,1,2,3,4,7$ and $8$ if each digit is to be used at most once How many five digit numbers divisible by $3$ can be formed using the digits $0,1,2,3,4,7$ and $8$ if each digit is to be used at most once?
The total number of $5$ digit numbers us... | Solution
We've to make $5$ digit numbers using given $7$ digits excluding $2$ digits every time.
Now,$0+1+2+3+4+7+8=25$.
As $25 \pmod 3 \equiv 1$, duplets to be excluded should have their sum $\mod 3=1$.
It's not much difficult to find such duplets. Without much trouble,
we get :$$(0,1),(0,4),(0,7),(1,3),(2,8),(3,4),(3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2365830",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
What is the area bounded by the graph of $ \lfloor x \rfloor + \lfloor y \rfloor =2 $ with the $x$ and $y$-axis? Question: If $x \ge 0$ and $y \ge 0$, then the area bounded by the graph of $ \lfloor x \rfloor + \lfloor y \rfloor =2 $ with the $x$ and $y$-axis is?
Answer provided: $3$ units$^2$.
My doubt: How do we gra... | Note that
$$
\left\lfloor x \right\rfloor = a\quad \Rightarrow \quad a \leqslant x < a + 1
$$
and therefore
$$
\begin{gathered}
\left\{ \begin{gathered}
\left\lfloor x \right\rfloor + \left\lfloor y \right\rfloor = 2 \hfill \\
0 \leqslant \left\lfloor x \right\rfloor ,\left\lfloor y \right\rfloor \hfill \\
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2370171",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
} |
Sum of $k^\text{th}$ Powers of the First $n$ Natural Numbers is a Perfect Square.
Question:
If for some $k$ and $\forall$ $n>2017, n\in \mathbb{N}$, there exist an $x\in \mathbb{N}$ such that
$$1^k+2^k+3^k+\cdots+n^k=x^2$$
then $k=3$.
Minor clarification:
The question says : $\forall n$. Now, we know that $1^3+2^3+.... | Answer posted by OP (solved by some AoPSer)
$f(n)=1^m+...n^m$, obvious, that $m$ is odd.
$f(n)<n^{m+1}$
Let $p$ -big prime.
$f(p-1)=x^2,f(p)=y^2 \to p^m=y^2-x^2 \to m=a+b;b>a,x=\frac{p^b-p^a}{2}$
$4x^2=(p^b-p^a)^2=4f(p-1)<4(p-1)^{m+1}$
$p^{2b}+p^{2a}-2p^{a+b}<4p^{a+b}(p-1)$
$p^{2b-2a}+1-2p^{b-a}<4p^{b-a}(p-1)$
Let $p^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2372522",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
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Prove that a non-positive definite matrix has real and positive eigenvalues I have a $2\times2$ matrix $J$ of rank $2$ and a $2\times2$ diagonal positive definite matrix $Α$. Denote by $J^+$ a pseudoinverse of $J$.
I can find many counterexamples for which $J^+AJ$ is not positive definite (e.g. $J=\left(\begin{smallma... | $J$ is $2 \times 2$ and has rank $2$, so $J$ is invertible. Consequently, the Moore-Penrose pseudoinverse of $J$, $J^+ = J^{-1}$. Then \begin{align*}
J^+ A J = J^{-1} A J
\end{align*} is similar to $A$ and has the same eigenvalues as $A$. $A$ is a diagonal matrix, so is its own symmetric part. Therefore, since ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2372800",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Evaluate the integral $\int\limits_0^{2\pi} \frac {d\theta}{5 - 3\cos(\theta)}$. Evaluate the integral $\displaystyle \int_0^{2\pi} \frac {d\theta}{5 - 3\cos(\theta)}$.
Hint: put $z = e^{i\theta}$.
Is there a way to solve this without using the Residue Theorem and $\tan(z)$? Is Cauchy's integral formula applicable?
| Another approach.
If $|b|<1$ then we can write:
$$\frac{1}{1-b\cos x}=\sum_{n=0}^{\infty} b^n\cos^n x$$
The constant term of the Fourier expansion of $\cos^n x$ is zero when $n$ odd and $\frac{1}{2^n}\binom{n}{n/2}$ when $n$ is even.
Now, $\frac{1}{2\pi}\int_{0}^{2\pi}\frac{dx}{1-b\cos x}$ is the constant of the Fourie... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2372981",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Length of Radius of Circle
This problem came very straightforward to me. Since the perimeter of the square is $32,$ the sidelengths are $8.$ Since the radius of the circle is $4.$ Constructing a smaller square in the lower right hand corner with side length $4$ I can subtract the radius of the circle $4$ from the cons... | $r=12-8 \sqrt{2}$ so $k+w+f=22$
$$GB=\frac{8\sqrt 2-8}{2}=4\sqrt 2-4\\
GH=\frac{GB}{\sqrt 2}=4- 2\sqrt 2\\
r+r\sqrt 2=4\sqrt 2-4\\
r=\frac{4\sqrt 2-4}{\sqrt 2+1}=12-8\sqrt 2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2373047",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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What is an intuitive approach to solving $\lim_{n\rightarrow\infty}\biggl(\frac{1}{n^2} + \frac{2}{n^2} + \frac{3}{n^2}+\dots+\frac{n}{n^2}\biggr)$?
$$\lim_{n\rightarrow\infty}\biggl(\frac{1}{n^2} + \frac{2}{n^2} + \frac{3}{n^2}+\dots+\frac{n}{n^2}\biggr)$$
I managed to get the answer as $1$ by standard methods of so... | Very intuitively, there are a lot of positive terms inside the brackets, so they can add up to something non-zero. An extremely loose way to get at what is going on is to over estimate by $n$ times the largest term and under estimate $n/2$ times the middle term, so you have
$$
\frac{1}{4} = \frac{n}{2}\cdot \frac{n/2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2373357",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "24",
"answer_count": 11,
"answer_id": 10
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If $ a^{2} + b^{2} = c^{2}$ and $c$ is even, prove that $a$ and $b$ are both even. Question:
If $ a^{2} + b^{2} = c^{2}$ and $c$ is even, prove that $a$ and $b$ are both even.
I am not quite sure how to prove this. My guess is proof by contradiction.
Assume the contrary, that is, $a^{2} + b^{2} = c^{2}$ for $c$ even an... | $c$ is even, thus $c^2=4n$ for some integer $n$.
Now if $a,b$ are both odd, then
$$a^2+b^2=(2k+1)^2 + (2m +1)^2$$
$$=4k^2+ 4k + 1 + 4m^2 + 4m+ 1$$
$$=4n' + 2 \equiv 2 \pmod4$$
Thus $4 \nmid (a^2+b^2)$, while $4 \mid c^2$ - contradiction.
Also it is obvious that if $a$ even and $b$ odd, or $a$ odd and $b$ even; then $a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2375495",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $\cos A+\cos B+\cos C=\frac{3}{2}$, prove $ABC$ is an equilateral triangle If in $\Delta ABC$ $$\cos A+\cos B+\cos C=\frac{3}{2}$$prove that it is an equilateral triangle without using inequalities. I tried using the cosine rule as follows:
$$\frac{b^2+c^2-a^2}{2bc}+\frac{a^2+c^2-b^2}{2ac}+\frac{a^2+b^2-c^2}{2ab}=\... | A trick I often use with these sorts of problems is this. Let $A = P+Q$ and $B=P-Q$, so that we want to show that $Q=0$.
$A+B+C = 180 \Rightarrow C = 180 - (A+B) = 180 - 2P$. The condition becomes $$\cos(P+Q)+\cos(P-Q)+\cos(180°-2P) = \frac{3}{2}$$
Using standard trig identities yields
$$2\cos(P)\cos(Q)-2\cos^2(P)+1 = ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2375579",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Finding the general Taylor Series of a function I have to find the general Taylor Series Expansion, about $0$, of the following function.
$$ \sqrt{x^4 -6x^2+1} $$
I have tried to use the identity
$$ \left(1+t\right)^{1/2} = \sum_{n\ge 0}\frac{(-1)^{n+1}}{4^n (2n-1)} \binom{2n}{n}t^n $$.
and then substitute... | Keep it simple, for $|x|<1$
$\sqrt{1+x}=1+\dfrac{x}{2}-\dfrac{x^2}{8}+\dfrac{x^3}{16}+O(x^4)$
Plug $x\to x^4-6 x^2$ and get
$$\sqrt{1+ x^4-6 x^2}= \dfrac{1}{16} \left(x^4-6 x^2\right)^3-\dfrac{1}{8} \left(x^4-6 x^2\right)^2+\dfrac{1}{2} \left(x^4-6 x^2\right)+1=\\=1 - 3 x^2 - 4 x^4 - 12 x^6+O(x^7)$$
The initial series ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2376117",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 2
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How to simplify an expression involving several square roots without a calculator? $$\frac{5 \sqrt{7}}{4\sqrt{3\sqrt{5}}-4\sqrt{2\sqrt{5}}}- \frac{4 \sqrt{5}}{\sqrt{3\sqrt{5}}-\sqrt{2\sqrt{5}}}$$
This type of questions are common in the university entrance examinations in our country but the calculators are not allowed... | $$\frac{5 \sqrt{7}}{4\sqrt{3\sqrt{5}}-4\sqrt{2\sqrt{5}}}- \frac{4 \sqrt{5}}{\sqrt{3\sqrt{5}}-\sqrt{2\sqrt{5}}}=\frac{5\sqrt7-16\sqrt5}{4(\sqrt{3\sqrt5}-\sqrt{2\sqrt5})}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2377485",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Solving a functional equation and general theory of functional equations $f(x-1)$ + $f(x+1)$ = $kf(x)$
given that $k>1$ then I need to find $f(x)$.
I am clueless about solving functional equations please help me this . Also give some general advice about how to solve functional equations.
| If we write $\displaystyle F(x) = \sum_{n=0}^\infty f(n)x^n$, then we have
$$f(n-1) + f(n+1) = kf(n)$$
$$\sum_{n=0}^\infty f(n-1)x^n + \sum_{n=0}^\infty f(n+1)x^n = kF(n)$$
$$\sum_{n=-1}^\infty f(n)x^{n+1} + \sum_{n=1}^\infty f(n)x^{n-1} = kF(x)$$
$$f(-1)+\sum_{n=0}^\infty f(n)x^{n+1} -f(0)x^{-1}+\sum_{n=0}^\infty f(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2377966",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Is $\displaystyle\sum_{k=1}^n \frac{k \sin^2k}{n^2+k \sin^2k}$ convergent? Let $\displaystyle x_n=\sum_{k=1}^n\frac{k \sin^2k}{n^2+k \sin^2k}$ for all $n>0$. How can we prove that $(x_n)$ is convergent?
| To prove that the limit is $\frac{1}{4}$, we just need to prove that
$\sum_{k=1}^n k \sin ^2 (k) = \frac{n^2}{4} + o(n^2)$. In fact :
$\sum_{k=1}^n k \sin ^2 (k) = \sum_{k=1}^n \frac{k}{2} (1 - \cos(2k)) = \frac{n(n+1)}{4} - \frac{1}{2} \sum_{k=1}^n k\cos(2k)$.
Let now $f$ be the function defined by
$f(x) = \sum_{k=0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2378459",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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If $x^2+x+1$ is a factor of $f (x)= ax^3+bx^2+cx+d$ , then the real root of $f (x)=0$ is If $x^2+x+1$ is a factor of $f (x)= ax^3+bx^2+cx+d$ , then the real root of $f (x)$ is
A) $-d/a$
B) $d/a$
C) $a/d $
D) none of these
My try
I take $f (x) = ( x^2+x+1)(x+1) = x^3+2x^2+2x+1$
Real root $x= -d/a =-1$
And also ... | The only way $x^2+x+1$ can be a factor of $ax^3+\cdots+d$ is if the
complementary factor is $ax+d$. As $x^2+x+1>0$ for all real $x$, the real solutions of $ax^3+\cdots+d=0$ must be the real solutions of $ax+d=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2381145",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
If $\sum_{k=0}^n \frac{A_k}{x+k}=\frac{n!}{x(x+1)(x+2)(x+3)\cdots(x+n)}$ Then find $A_7$ If $$\sum_{k=0}^n \frac{A_k}{x+k}=\frac{n!}{x(x+1)(x+2)(x+3)\cdots(x+n)}$$ Then find $A_7$
My Try:
I have considered a function
$$f(x)=x^{A_0}(x+1)^{A_1}(x+2)^{A_2}\cdots(x+n)^{A_n}$$
taking natural log on both sides and then diff... | \begin{eqnarray*}
\frac{1}{x(x+1)}&=&\frac{1}{x}-\frac{1}{x+1} \\
\frac{2}{x(x+1)(x+2)}&=&\frac{1}{x(x+1)}-\frac{1}{(x+1)(x+2)}=\frac{1}{x}-\frac{2}{x+1}+\frac{1}{x+2} \\
\frac{6}{x(x+1)(x+2)(x+3)}&=&\frac{2}{x(x+1)(x+2)}-\frac{2}{(x+1)(x+2)(x+3)}\\&=&\frac{1}{x}-\frac{3}{x+1}+\frac{3}{x+2}- \frac{1}{x+3}\\
\cdots
\end... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2381927",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
} |
find pairs of real numbers $x, y$ to satisfy this equation equation is: $(x + y)^2 = (x + 3)(y − 3)$
I'm not asking for a solution, but an approach. How do I prove this kind of question? I have tried to arrange it so that it
is $x + y$ = ....
But I still get nothing, nothing intuitive at least. What is a way to tac... | Hint:
By expanding we have
\begin{align*}
(x+y)^2&=(x+3)(y-3)\\
\iff x^2+2xy+y^2&=xy-3x+3y-9\\
\iff x^2+xy+y^2+3x-3y+9&=0\\
\iff \tfrac12(x+y)^2+\tfrac12(x+3)^2+\tfrac12(y-3)^2&=0
\end{align*}
So we must have $$x+y=x+3=y-3=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2382026",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Easy irreducibility over $\mathbb{F}_5$ One of the last days I was looking for help in a problem here and, somewhere, I saw someone asking something that involved the irreducibility of the polynomial $f(x)=x^4+2$ over $\mathbb{F}_5$. In that case the argment was that since $f$ has no roots in $\mathbb{F}_5$ it is enoug... | More general, $x^4\equiv -1\pmod{x^4+1}$, so $x^{4k+r}\equiv (-1)^{k}x^r\pmod{x^4+1}$, so $x^{24}\equiv (-1)^6=1\pmod{x^4+1}$.
The same trick works for dividing $x^{24}-1$ by $x^4+2$:
$$x^4\equiv -2\pmod{x^4+2}\\
x^{24}\equiv (-2)^6\pmod{x^4+2}\\
x^{24}-1\equiv 3\pmod{x^4+2}$$
More generally, over $\mathbb F_q$, $$x^{q... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2382611",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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A nice but somewhat challenging binomial identity
When working on a problem I was faced with the following binomial identity valid for integers $m,n\geq 0$:
\begin{align*}
\color{blue}{\sum_{l=0}^m(-4)^l\binom{m}{l}\binom{2l}{l}^{-1}
\sum_{k=0}^n\frac{(-4)^k}{2k+1}\binom{n}{k}\binom{2k}{k}^{-1}\binom{k+l}{l}
=\frac{... | Firstly let us evaluate the inner sum on the left hand side. Using the beta function identity quoted above along with the identity $\left. \binom{k+l}{l} = d^l/dx^l x^{k+l}/l! \right|_{x=1}$ we have:
\begin{equation}
S^{(n)}_l:=\sum\limits_{k=0}^n \frac{(-4)^k}{2k+1} \binom{n}{k} [\binom{2k}{k}]^{-1} \binom{k+l}{l} = \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2384932",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "38",
"answer_count": 3,
"answer_id": 2
} |
Find $\lvert m\rvert$ given $\sqrt[3]{m+9}=3+\sqrt[3]{m-9}$ I've already found the solution, but I was wondering if there is a faster or alternative method. My solution is found below:
$$\sqrt[3]{m+9}=3+\sqrt[3]{m-9}$$
$$(m+9)^\frac{1}{3}-(m-9)^\frac{1}{3}=3$$
$$\bigl((m+9)^\frac{1}{3}-(m-9)^\frac{1}{3}\bigr)^3=27$$
$$... | Hint:
Let $\sqrt[3]{m+3}=a$ and $\sqrt[3]{m-3}=b$
$a-b=?$
$a^3-b^3=?$
Can you find $ab$ and $a^3+b^3=(a+b)\{(a-b)^2+ab\}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2386952",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Area between arc and line I have been working on a problem for a few days now. This was a challenge problem on a lecture for Trigonometry. I managed to find an equation for the radius, but wasn't able to solve it.
Problem: Line $AB$ is drawn such that $\overline{AB} = 20$. Minor arc $AB$ is drawn with endpoints $AB$ su... | As angryavian commented, the problem reduces to : find the zero of $$f(\theta)=\frac \theta {\sin\left(\frac\theta 2\right)}- 2.1\qquad \text{or}\qquad g(\theta)=\theta-2.1{\sin\left(\frac\theta 2\right)}$$ which can only be solved using numerical methods.
Let $x=\frac\theta 2$ to make the equation $$x=1.05 \sin(x)$$ a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2387122",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Find the real and imaginary part of z let $z=$ $$ \left( \frac{1 + \sin\theta + i\cos\theta}{1 + \sin\theta - i\cos\theta} \right)^n$$
Rationalizing the denominator:
$$\frac{1 + \sin\theta + i\cos\theta}{1 + \sin\theta - i\cos\theta}\cdot\left( \frac{1 + \sin\theta + i\cos\theta}{1 + \sin\theta + i\cos\theta}\right) =... | $$\left( \frac { 1+\sin \theta +i\cos \theta }{ 1+\sin \theta -i\cos \theta } \right) ^{ n }={ \left( 1+\frac { 2i\cos \theta }{ 1+\sin \theta -i\cos \theta } \right) }^{ n }=\\ ={ \left( 1+\frac { 2i\cos \theta \left( 1+\sin \theta +i\cos \theta \right) }{ \left( 1+\sin \theta -i\cos \theta \righ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2387172",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 8,
"answer_id": 5
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Solving $\frac{1}{(x-1)} - \frac{1}{(x-2)} = \frac{1}{(x-3)} - \frac{1}{(x-4)}$. Why is my solution wrong? I'm following all hitherto me known rules for solving equations, but the result is wrong. Please explain why my approach is not correct.
We want to solve:
$$\frac{1}{(x-1)} - \frac{1}{(x-2)} = \frac{1}{(x-3)} - \f... | Here is another way, which also illustrates how easy it is to make a mistake with these kinds of problems. If we take the negative fractions to the other side of the equation we get $$\frac 1{x-1}+\frac 1{x-4}=\frac 1 {x-3}+\frac 1{x-2}$$ which becomes $$\frac {2x-5}{(x-1)(x-4)}=\frac {2x-5}{(x-2)(x-3)}$$Now cancel and... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2388701",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 7,
"answer_id": 3
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Number of partitions into $k$ parts: recursion has characteristic polynomial $(X-1)(X^2-1) \cdots (X^k-1)$? Let $p_k(n)$ denote the number of partitions of $n$ into $k$ parts. For example, $p_3(6)=3$ because $6$ admits the partitions $1+1+4$, $1+2+3$ and $2+2+2$ into $3$ parts.
For $k=1$ we have $p_1(n)=1$ and therefor... | What you call $p_k$ I will call $r_k$, and use $p_k$ to count partitions of $n$ into at most $k$ parts. This gives the obvious recurrence $p_k=r_k+p_{k-1}$ (as functions of $n$). We may regard $X$ as the shift operator, satisfying $(Xf)(n):=f(n+1)$, in which case $(X^k-1)p_k$ equals $X^kp_{k-1}$, i.e.
$$ p_k(n+k)=p_k(n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2390031",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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What is the result of $\lim_{x\to0_+} \frac{e^x - x e^x - 1}{\left(e^x - 1 \right)^2}$ without L'Hôpital's rule. I have the limit
$$
\lim_{x\to0_+} \frac{e^x - x e^x - 1}{\left(e^x - 1 \right)^2}
$$
which I need to compute without L'Hôpital's rule.
(The result is $-\frac{1}{2}$ with L'Hôpital's rule).
Thanks.
| $\lim \limits_{x \to0+}$
$\frac{e^x-xe^x-1}{\left(e^x-1\right)^2}\cdot\frac{x^2}{x^2}$=
$\lim \limits_{x \to0+}$
$\frac{e^x-x+x-xe^x-1}{x^2}\cdot\lim \limits_{x \to0+}$
$\frac{x^2}{\left(e^x-1\right)^2}$
=
$\lim\limits_{x\to 0^{+}}\frac{e^x-x+x-xe^x-1}{x^2}\cdot\lim \limits_{x \to0+}\left[\frac{x}{\left(e^x-1\right)}\r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2390227",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Inequality : $ \sum_{cyc} \frac{a+b}{\sqrt{a+2c}} \geq 2 \sqrt{a+b+c}$
Let $a$, $b$ and $c$ be positive real numbers. Prove that:
$$ \displaystyle\sum_{cyc} \frac{a+b}{\sqrt{a+2c}} \geq 2 \sqrt{a+b+c}.$$
My attempt :
By Holder inequality,
$ \left(\displaystyle\sum_{cyc} \frac{a+b}{\sqrt{a+2c}}\right)^2 \displaysty... | Your idea to use Holder was very good!
By Holder $$\left(\sum_{cyc}\frac{a+b}{\sqrt{a+2c}}\right)^2\sum_{cyc}(a+b)(a+2c)\geq8(a+b+c)^3.$$
Thus, it's enough to prove that
$$8(a+b+c)^3\geq4(a+b+c)\sum_{cyc}(a+b)(a+2c)$$ or
$$2(a+b+c)^2\geq\sum_{cyc}(a+b)(a+2c),$$ which is
$$\sum_{cyc}(a-b)^2\geq0.$$
Done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2391240",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $x$, $y$ and $z$ are distinct positive integers and $x+y+z=11$ then what is the maximum value of $(xyz+xy+yz+zx)$?
If $x$, $y$ and $z$ are distinct positive integers and $x+y+z=11$ then what is the maximum value of $(xyz+xy+yz+zx)$?
We know that product is maximum when difference between $x$, $y$ and $z$ is minimu... | For $x=5$, $y=4$ and $z=2$ we get a value $78$.
We'll prove that it's a maximal value.
Indeed, let $x<y<z$, $x=1+a$, $y=2+a+b$ and $z=3+a+b+c$,
where $a$, $b$ and $c$ are no-negative integer numbers.
Thus, the condition gives
$$1+a+2+a+b+3+a+b+c=11$$ or
$$3a+2b+c=5,$$
which says that $a\in\{0,1\}$.
Now, let $a=0$.
T... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2391573",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
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If $450^\circ<\alpha<540^\circ$ and $\cot\alpha=-\frac{7}{24},$ calculate $\cos\frac{\alpha}{2}$. Why is my solution wrong? The problem says:
If $450^\circ<\alpha<540^\circ$ and $\cot\alpha=-\frac{7}{24},$ calculate $\cos\frac{\alpha}{2}$
I solved it in the following way: $$\begin{align} -\frac{7}{24}&=-\sqrt{\frac{1... | $$\frac {7^2}{24^2}=\frac {1+\cos 2\alpha}{1-\cos 2\alpha} \implies$$ $$\implies 7^2(1-\cos \alpha)=24^2(1+\cos \alpha)\implies$$ $$\implies 7^2- 7^2\cos 2\alpha = 24^2+ 24^2 \cos 2\alpha\implies$$ $$\implies 7^2-24^2= (7^2+24^2)\cos 2\alpha =25^2 \cos 2\alpha\implies$$ $$\implies -527=625\cos 2\alpha .$$
Th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2393005",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 0
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Where the object that moves along the intersection of $x^2+y^2=1$ and $y+z=1$ needs to be if we want the sum $x+2y+z$ to be max/min?
An object is moving along the curve which is derived from the intersection of the cylinder $x^2+y^2=1$ and the plane $y+z=1$. Where does the object need to be located if we want to maxim... |
An object is moving along the curve which is derived from the intersection of the cylinder $x^2+y^2=1$ and the plane $y+z=1$. Where does the object need to be located if we want to maximize/minimize the sum $x+2y+z$? What is the min/max sum?
So you're looking for the extreme values of $f(x,y,z) = x+2y+z$ for points $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2393232",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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Proving $x-\frac{x^3}{6} < \sin(x) < x - \frac{x^3}{6} + \frac{x^5}{120} ~~ \forall x \in \Bbb R^+$ using Taylor's expansion I'm trying to prove
$$x-\frac{x^3}{6} < \sin(x) < x - \frac{x^3}{6} + \frac{x^5}{120} ~~ \forall x \in \Bbb R^{+}$$
I wrote $\sin(x)=x-x^3/3!+x^5/5!-x^7/7!+\dots$ and then the expression for ... | hint
Prove that
$$1-\frac {x^2}{2}\le \cos (x)\le 1-\frac {x^2}{2}+\frac {x^4}{24} $$
and integrate.
to prove this, it is easier to show that
$$-x\le -\sin (x)\le -x+\frac {x^3}{6} $$
or the easiest
$$0\le 1-\cos (x)\le \frac {x^2}{2} $$
then integrate twice between $0$ and $x $.
Taylor expansion give only local infor... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2394867",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
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Find $x\in \Bbb Z$ such that $x=\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}$
Find $x\in \Bbb Z$ such that $x=\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}$
Tried (without sucess) two different approaches: (a) finding $x^3$ by raising the right expression to power 3, but was not able to find something useful in the result t... | If you know the cubic formula for solving $$x^3 + px + q = 0$$ that helps. The formula is $$x = \sqrt[3]{-\frac{q}{2}+\sqrt{\frac{q^2}{4}+\frac{p^3}{27}}}+\sqrt[3]{-\frac{q}{2}-\sqrt{\frac{q^2}{4}+\frac{p^3}{27}}}$$ Given the "$2$"s in your expression, you might be led to consider $q=-4$, so the cubic $x^3 + px -4 = 0$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2395060",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 7,
"answer_id": 5
} |
If $x^3+y^3+(x+y)^3+33 xy=2662$, $x,y\in \Bbb R$, find $S=x+y$.
If $x^3+y^3+(x+y)^3+33 xy=2662$ and $\{x,y\}\subset \Bbb R$, find $S=x+y$.
This question from an olympiad contest. Answer stated: $S=x+y=11$
Tried to develop $(x+y)^3$ to find something useful for the situation, but without success.
| Rewrite our equation in the following form.
$$(x^3+y^3-11^3+3xy)+(x+y)^3-11^3=0$$
Now, we can use $a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-ac-bc).$
Thus, we obtain:
$$(x+y-11)(x^2+y^2+11^2-xy+11x+11y+(x+y)^2+11(x+y)+11^2)=0$$ or
$$(x+y-11)(2x^2+2y^2+xy+22x+22y+242)=0$$
and since
$$2x^2+2y^2+xy+22x+22y+242=(x+11)^2+(y+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2395460",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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maximize volume of triangle in given perimeter question: Find the triangle with perimeter 2a given that, when we rotate it around one of its sides, the solid obtained have the maximum volume.
suppose we have a triangle of sides x,y,z such that $x+y+z=2a$ and $h$ is the height, we rotate it around the side with size z,... | From $Area=\frac12 \sqrt{a (a-x) (a-y) (a-z)}$
we get $h=\dfrac{2 \sqrt{a (a-x) (a-y) (a-z)}}{z}$
Plug in the volume $V=\frac{1}{3} \pi h^2 z$ and get
$V=\dfrac{4 \pi a (a-x) (a-y) (a-z)}{3 z}$
with the constraint $x+y+z-2a=0$
We use the Lagrangian multiplier and consider
$f(x,y,z,k)=\dfrac{4 \pi a (a-x) (a-y) (a-z)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2396644",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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A question of trigonometry on how to find minimum value.
Find The minimum value of $$2^{\sin^2 \alpha} + 2^{\cos^2 \alpha}.$$
I can easily get the maximum value but minimum value is kinda tricky. Please help.
| $f'(\alpha)=2\log 2 \sin \alpha \cos \alpha \left[2^{\sin ^2\alpha}- 2^{\cos ^2\alpha}\right]$
$f'(\alpha)=0 \to 2\sin\alpha\cos\alpha=0$ or $2^{\sin ^2\alpha}- 2^{\cos ^2\alpha}=0$
$ \sin 2\alpha=0\to 2\alpha=k\pi\to\alpha=\dfrac{k\pi}{2}$
$2^{\sin ^2\alpha}- 2^{\cos ^2\alpha}=0\to 2^{\sin ^2\alpha}= 2^{\cos ^2\alpha}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2396947",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
Angle bisector problem In triangle $ABC$, $AY$ is a perpendicular to the bisector $\angle ABC$ and $AX$ is a perpendicular to the bisector of $\angle ACB$. If $AB= 9cm$ , $AC=7cm$ and $BC= 4cm$, then the length of $XY$ is?
Any help would be much appreciated.
Thank you.
| Let $AB=c$, $AC=b$, $BC=a$, $\measuredangle BAC=\alpha$, $\measuredangle ABC=\beta$, $\measuredangle ACB=\gamma$,
$R$ be a radius of circumcircle of $\Delta ABC$, $S$ be an area of the triangle
and let be our bisectors intersects in the point $O$.
Thus, $$\measuredangle XAY=90^{\circ}-\frac{\beta}{2}+90^{\circ}-\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2398555",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How prove infinitely many postive integers triples $(x,y,z)$ such $(x+y+z)^2+2(x+y+z)=5(xy+yz+zx)$
show that there exsit infinitely many postive integers triples $(x,y,z)$
such
$$(x+y+z)^2+2(x+y+z)=5(xy+yz+zx)$$
May try it is clear $(x,y,z)=(1,1,1)$ is one solution，and
$$(x+y+z+1)^2=5(xy+yz+xz)+1$$
| If $(1,y,z)$ is a solution then $(1,z,3z-y+1)$ is also a solution because \begin{align*} &\quad \Big( 1 + z + (3z - y + 1) \Big)^2 + 2 \Big( 1 + z + 3z - y + 1 \Big) - 5 \Big( z + z(3z - y + 1) + 3z - y + 1 \Big) \\ &= (1 + y + z)^2 + 2(1 + y + z) - 5(y + yz + z). \end{align*} You can use this to generate the infinite ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2399830",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
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Integration with logarithmic expression. I'm trying to solve an integration problem from the book which is the $$\int\frac{\sqrt{9-4x^2}}{x}dx$$ using trigonometric substitution. The answer from the book is $$3\ln\left|\frac{3-\sqrt{9-4x^2}}{x}\right|+\sqrt{9-4x^2}+C.$$
I have almost the same solution where there's a ... | Here's what I did.
$$\int\frac{\sqrt{9-4x^2}}{x}dx \\
\begin{align}
& a=3\\
& u=2x\\
\text{since} \quad& u=a\sin\theta \\
& 2x=3\sin\theta \\
& 2dx=3\cos\theta; \quad x=\frac{3}{2}\sin\theta; \quad \sin\theta=\frac{2x}{3}; \quad \theta= \sin^1\frac{2x}{3}\\
& dx = \frac{3}{2}\cos\theta \\ \\
\sqrt{a^2-u^2}& =\sqrt{9-(2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2400006",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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The easiest way to solve the improper integral $\int_0^{+\infty} \frac{\ln(t)}{(1+t^2)^4} {d}t$? What is the easiest way to solve this improper integral?
$$I = \int_0^{+\infty} \frac{\ln(t)}{(1+t^2)^4} {d}t $$
The value of this integral according to wolfram alpha is $-\frac{23\pi}{96}$. I could find the same result but... | $\begin{align}
I&=\int_0^1\ \frac{\ln x}{(1+x^2)^4}dx +\int_1^{\infty}\frac{\ln x}{(1+x^2)^4}\ dx
\end{align}$
Perform the change of variable $y=\frac{1}{x}$ in the second integral,
$\begin{align}
I&=\int_0^1\ \frac{\ln x}{(1+x^2)^4}dx -\int_1^{\infty}\frac{x^6\ln x}{(1+x^2)^4}\ dx\\
&=\int_0^{1}\frac{(1-x^6)\ln x}{(1+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2400537",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Convergence of sequence $\lim_{n \rightarrow \inf }n\sqrt{n}(\sqrt{n+1}-a\sqrt{n}+\sqrt{n-1})$ I am able to show that the sequence converge only if $a=2$, I also did a numeric analysis of the sequence, and I am sure that the value of this limit is $-\frac{1}{4}$. However, I do not know how to prove that
$\lim_{n \righ... | For an analytic approach, we can use a higher order version of MVT.
When $a = 2$, the expression
$$\sqrt{n-1} - 2\sqrt{n} + \sqrt{n+1} = \left. \sum_{k=0}^{2}(-1)^k\binom{2}{k} \sqrt{x+k}\;\right|_{x = n-1}$$
is the second order finite differences of the function $x \mapsto \sqrt{x}$ at $n-1$.
The MVT we need has the ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2402012",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
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How to solve the following identity? $$
\sum_{i=2}^{n-k+1} \frac{{{n-i}\choose{k-1}}}{{n} \choose k} \frac{1}{i -1} = \frac{k}{n}\sum_{i=k +1}^{n} \frac{1}{i - 1}.
$$
I came across this identity while I was solving a probability problem. I was able to show that it is true for $k=1$ and $k=2$. But I haven't yet cracked ... | Multiplication with $\binom{n}{k}$ gives
\begin{align*}
\sum_{i=2}^{n-k+1}\binom{n-i}{k-1}\frac{1}{i-1}&=\binom{n}{k}\frac{k}{n}\sum_{i=k+1}^n\frac{1}{i-1}\tag{1}\\
&=\binom{n-1}{k-1}\left(H_{n-1}-H_{k-1}\right)
\end{align*}
In (1) we apply the binomial identity $\binom{n}{k}=\frac{n}{k}\binom{n-1}{k-1}$ and the defini... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2404287",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 1,
"answer_id": 0
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multiple integral finding limit integration question :
$\iint_D \ln(x^2 + y^2)dxdy$ ,
domain : $4\le x^2+y^2 \le 9$
im having hard times to find the limit for integration because it is in polar coordinates, i got that $2 \le r\le 3$ but i dont know how to find the $\theta$ limit.
and also i dont know the graph act... | You make the change of variables:
$$x=r\cos{\theta},y=r\sin{\theta}.$$
Hence:
$$4\le x^2+y^2 \le 9 \Rightarrow 2^2\le r^2\le 3^2 \Rightarrow 2\le r\le 3.$$
The domain is the region between the cocentric circles with the center at the origin and the radii $2$ and $3$, therefore:
$$0\le \theta \le 2\pi.$$
Note: If the ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Trigonometry pre calculus level good question Here is a trigonometry problem.
Given
$$\frac{\cos(\alpha-3\theta)}{\cos^3(\theta)}=\frac{\sin(\alpha-3\theta)}{\sin^3(\theta)} = m$$
Show that
$$m^2+m\cos(\alpha) = 2.$$
I tried to convert $\sin^3(x)$ into $\sin(3x)$ and similarly to cosine term, but couldn't get ... | From the equation we have following:
$m \sin^3\theta = \sin\alpha \cos3\theta-\cos\alpha \sin3\theta \cdots (1)$
$m \cos^3\theta = \cos\alpha \cos3\theta+\sin\alpha \sin3\theta \cdots (2)$
$\cos3\theta \times (2)-\sin3\theta \times (1) \rightarrow m(\cos^3\theta\cos3\theta-\sin^3\theta\sin3\theta)=\cos\alpha$
$\cos3\th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2405352",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Infinite power summation where no common ratio can be seen Problem in book states to evaluate the sum:
$$\sum_{i=0}^\infty \frac{i^2}{4^i}$$
I can come up with the series S = $\frac{1^2}{4^1} + \frac{2^2}{4^2} + \frac{3^2}{4^3} + \frac{4^2}{4^4} + \frac{5^2}{4^5} + ...$
And 4S = $ 1 + \frac{2^2}{4^1} + \frac{3^2}{4^2} ... | \begin{eqnarray*}
\sum_{i=0}^{\infty} i^2 x^i = \frac{x(1+x)}{(1-x)^3} \mid_{x=\frac{1}{4}} = \frac{\frac{1}{4} \left( 1+ \frac{1}{4} \right)}{\left(1- \frac{1}{4} \right)^3} = \color{red}{ \frac{20}{27}} .
\end{eqnarray*}
EDIT:
You are happy with a geometric sum ?
\begin{eqnarray*}
\sum_{i=0}^{\infty} x^i = \frac{1}{1... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Generalizing the dot product to multivectors I am studying the book Linear and Geometric Algebra (Macdonald), and I've been stuck on a couple related, seemingly-elementary problems for a couple of days.
5.3.4. Suppose that $\mathbf{a} \bot \mathbf{b}$. Show that $\mathbf{a} \cdot (\mathbf{a} \land \mathbf{b}) = |\mathb... | I don't have that book by MacDonald (although I recall liking the preprint of that I read eons ago), and I don't remember the lingo that he uses.
Define a k-vector as a quantity having a single grade.
Examples of a 2-vector:
$$\begin{aligned} &\mathbf{e}_1 \mathbf{e}_2 + \mathbf{e}_3 \mathbf{e}_4 \\ &\mathbf{e}_1 ... | {
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"url": "https://math.stackexchange.com/questions/2406014",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
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Series $\sum_{k=1}^n\frac{\tan\frac x{2^k}}{2^{k-1} \cos\frac x{2^{k-1}}}$ Evaluate
$$\sum_{k=1}^n\frac{\tan\frac x{2^k}}{2^{k-1} \cos\frac x{2^{k-1}}}$$
I am struck with calculation for this question
| Hint:$$2\cot (2x)=\cot x-\tan x\\ \to \tan x=2\cot (2x)-\cot x$$so
$$\frac{\tan\frac x{2^k}}{2^{k-1} \cos\frac x{2^{k-1}}}=\\
\frac{2\cot (2\frac x{2^k})-\cot \frac x{2^k}}{2^{k-1} \cos\frac x{2^{k-1}}}$$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "1",
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Let $a = \frac{9+\sqrt{45}}{2}$. Find $\frac{1}{a}$ I've been wrapping my head around this question lately:
Let
$$a = \frac{9+\sqrt{45}}{2}$$
Find the value of
$$\frac{1}{a}$$
I've done it like this:
$$\frac{1}{a}=\frac{2}{9+\sqrt{45}}$$
I rationalize the denominator like this:
$$\frac{1}{a}=\frac{2}{9+\sqrt{45}} \ti... | The answer is b, which is equivalent to your answer after simplifying.
This is because $ 9 - \sqrt{45} = 9 - \sqrt{9 * 5} = 9 - 3\sqrt{5}$. Then,$ \frac{9 - 3\sqrt{5}}{18} = \frac{3 - \sqrt{5}}{6}. $
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to calculate $\lim_{x\to\infty}\frac{7x^4+x^2 3^x+2}{x^3+x 4^x+1}$?
$$\lim_{x\to\infty}\frac{7x^4+x^2 3^x+2}{x^3+x 4^x+1}$$
I can't seem to find away to get rid of the $3^x$ and $4^x$ and then resolve it.
| Use equivalents:
$$\begin{aligned}
7x^4+x^2\, 3^x+2&\sim_\infty x^2 3^x\\
x^3+x\,4^x+1&\sim_\infty x\,4^x
\end{aligned}\quad\text{hence}\quad\frac{7x^4+x^2\, 3^x+2}{x^3+x\,4^x+1}\sim_\infty\frac{ x^2 3^x}{ x\,4^x}=x\Bigl(\frac34\Bigr)^{\!x}\to 0.$$
| {
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"source": "stackexchange",
"question_score": "5",
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Prove that the function $f(z) = \sum_{n=0}^\infty \dfrac{1}{1+z^n}, \quad |z|>1,$ is holomorphic. I attempted to solve the following problem and would like feedback (corrections, suggestions for improvement, or a better way).
Prove that the series $\sum_{n=0}^{\infty} \dfrac{1}{1+z^n}$ converges for $|z|>1$ and that th... | This is hard work! Take $R>1$ and consider $U=\{z:|z|>R\}$. On $U$,
$$\left|\frac{1}{1+z^n}\right|\le\frac{1}{R^n-1}\le\frac{R^{-n}}{1-1/R}$$
(for $n\ge1$). So on $U$ the series is uniformly convergent, and the sum
is therefore holomorphic on $U$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Product of consecutive natural numbers Let $n,k$ be two natural numbers, $n\ge k+2$ and
$$(n-k-1)(n-k)(n-k+1)(n-k+2)=k(k+1)n(n+1).$$
In the LHS we have product of 4 consecutive numbers and in the RHS we have the product of 2 consecutive numbers on the product of 2 consecutive numbers.
Problem. Prove that $k,... | Let $a:=n-k\ (\ge 2)$.
Then, we have
$$(a-1)a(a+1)(a+2)=k(k+1)(k+a)(k+a+1)$$
This can be written as
$$(2k^2+2ka+2k+a)^2=4a^4+8a^3-3a^2-8a$$
Now, for $a\gt 2$, we have
$$(2a^2+2a-2)^2\lt 4a^4+8a^3-3a^2-8a\lt (2a^2+2a-1)^2$$
from which we have that $4a^4+8a^3-3a^2-8a$ cannot be a square number for $a\gt 2$.
So, we have t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2413453",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Proving/disproving $∀n ∈ \text{positive integers}$, $\left\lceil{\frac{4n^2+1}{n^2}}\right \rceil = 5$? Is my proof getting anywhere for proving/disproving $∀n ∈ \text{postive integers}$, $$\left\lceil{\frac{4n^2+1}{n^2}}\right \rceil = 5\text{?}$$
$$\left\lceil{\frac{4n^2}{n^2}}+\frac{1}{n^2}\right \rceil = 5$$
$$\lef... | Hint: The ceiling of a real number is always an integer. So $0<\lceil x \rceil\leq 1$ is equivalent to $\lceil x\rceil = 1$, or $0<x\leq 1$.
| {
"language": "en",
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"source": "stackexchange",
"question_score": "2",
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Vector Analysis Identity One Question given in class was to prove that:
$$\mathbf{(A \cdot B \times C)(a \cdot b \times c)} $$
is equal to
$$\begin{vmatrix} \mathbf A \cdot a & \mathbf A \cdot b & \mathbf A \cdot c \\
\mathbf B \cdot a & \mathbf B \cdot b & \mathbf B \cdot c \\ \mathbf C \cdot a & \mathbf C \cdot b & \... | Just for reference in case someone want to see the actual algebraic manipulation.
In addition to some symmetry relations among dot and cross products
$$p \times q = -q \times p \quad\text{ and }\quad p\cdot(q \times r) = q \cdot( r \times p) = r \cdot( p \times q)$$
The key identities we need are:
$$\begin{align}
p \ti... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2417245",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $8\sin x - \cos x=4$, then find possible values of $x$ I am not understanding what exactly can watch do here. First I thought that if I could square it but it was in vain. Please help me.
| Put $t = \tan \frac{x}{2}$. Then $\sin x = \frac{2t}{1+t^2}, \cos x = \frac{1-t^2}{1+t^2}$. The equation reduces to
$$3t^2 - 16t + 5 = 0$$ and hence $(3t-1)(t-5) = 0$. Hence the solutions are given by $\tan \frac{x}{2} = 5$ or $\frac{1}{3}$
Let us verify the solutions graphically. The following shows the graphs of $8 \... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Strange inequality over the positive reals Let $a$, $b$ and $c$ be positive real numbers with $2a^3b+2b^3c+2c^3a=a^2b^2+b^2c^2+c^2a^2$. Prove that $$2ab(a-b)^2+2bc(b-c)^2+2ca(c-a)^2 \geqslant (ab+bc+ca)^2$$
| We need to prove that
$$\sum_{cyc}(2a^3b+2a^3c-5a^2b^2-2a^2bc)\geq0$$ or
$$\sum_{cyc}(2a^3b+2a^3c-5a^2b^2-2a^2bc)+3\sum_{cyc}(2a^3b-a^2b^2)\geq0$$ or
$$\sum_{cyc}(4a^3b+a^3c-4a^2b^2-a^2bc)\geq0.$$
Let $a=\min\{a,b,c\}$, $b=a+u$ and $c=a+v$.
Hence,
$$\sum_{cyc}(4a^3b+a^3c-4a^2b^2-a^2bc)=$$
$$=6(u^2-uv+v^2)a^2+(5u^3+3u^2... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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What's the maximum value of $\frac{2}{a^2+1}-\frac{2}{b^2+1}+\frac{3}{c^2+1}$ given that: $abc+a+c=b$
Given that: $abc+a+c=b$. What's the maximum value of $$\frac{2}{a^2+1}-\frac{2}{b^2+1}+\frac{3}{c^2+1}.$$
(DO NOT use trigonometric methods)
| prove that $$\frac{2}{a^2+1}-\frac{2}{b^2+1}+\frac{3}{c^2+1}\le \frac{10}{3}$$ and the equal sign holds if $$a=\frac{2}{3}\left(\frac{1}{2\sqrt{2}}-\sqrt{2}\right),b=-\sqrt{2},c=-\frac{1}{2\sqrt{2}}$$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "2",
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Solve the equation $18x^2-18x \sqrt{x}-17x-8 \sqrt{x}-2=0$
Solve the following equation. $$18x^2-18x \sqrt{x}-17x-8 \sqrt{x}-2=0.$$
Taking $\sqrt{x}=t$ we get equivalent equation $18t^4 -18t^3 - 17t^2-8t-2=0$.
From this point I have tried to factor it , write RHS as sum of two squares and its variants but nothing se... | First, the condition is $x\geq 0$.
Now, as you did, put $t = \sqrt{x}$, then we have the equation.
$$18t^4-18t^3 -17t^2-8t-2 = 0.$$
Obviously, $t=0$ is not the solution of this equation. Then, we can divide two sides by $t^4$, $$18-\frac{18}{t}-\frac{17}{t^2}-\frac{8}{t^3}-\frac{2}{t^4} = 0$$
Let $u = \frac{1}{t}$, we ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Finding derivative of $\frac{x}{x^2+1}$ using only the definition of derivative I think the title is quite self-explanatory. I'm only allowed to use the definition of a derivative to differentiate the above function. Sorry for the formatting though.
Let $f(x) = \frac{x}{x^2+1}$
$$
\begin{align}
f'(x)&= \lim_{h\to 0} \... | \begin{align}
f'(x)
&= \lim_{h\to 0} \frac {f(x+h)-f(x)}{h}\\
&= \lim_{h\to 0} \frac {\frac{x+h}{(x+h)^2+1}-\frac{x}{x^2+1}}{h}\\
&= \lim_{h\to 0} \frac{(x+h)(x^2+1)-x((x+h)^2+1)}{h(x^2+1)((x+h)^2+1)}\\
&= \lim_{h\to 0} \frac{h - h^2x - hx^2}{h(x^2+1)((x+h)^2+1)}
\end{align}
$$= \lim_{h\to 0} \frac{1 - hx - x^2}{(... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solve the following nonlinear system of equations:$ xy+xz+yz=12 , xyz=2+x+y+z$
Solve the following system of equations in $\Bbb R^+$:
$$
\left\{
\begin{array}{l}
xy+yz+xz=12 \\
xyz=2+x+y+z\\
\end{array}
\right.
$$
I did as follows.
First equation yields to: $(x+y+z)^2-x^2-y^2-z^2=24$. Now, using second equat... | I will use the cubic equation with roots $x$, $y$, $z$, inspired by the answer of @dxiv:.
$$X^3 - s X^2 + 12 X - (2+s) =0$$
The discriminant of the equation is $-(2s+13)(2s+15)(s-6)^2$. If $s\ge 0$ then the discriminant is positive only for $s=6$, which gives $a=b=c=2$. Therefore: the system has only one positive solu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2427470",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 3
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For what values of $a$ does the system have infinite solutions? Find the solutions. The system is
$$\left\{
\begin{array}{rcr}
x+ay+z & = & 1 \\
ax+y+z & = & 1+a \\
x-y+z & = & 2+a
\end{array}
\right.$$
After row reducing I got
$$\left\{
\begin{array}{rcr}
x+ay+z & = & 1 \\
-(1+a)y+0 & = & 1+a \\
(1-a)z& = & 2-a + (1... | The only way for this system not to have a unique solution is if the matrix
$$
\left[
\begin{array}{ccc}
1 & a &1 \\
a & 1 & 1 \\
1 & -1 &1
\end{array}
\right]
$$
is singular. Its determinant is $1 - a^2$. So it's singular if $a =1 $ or $a = -1$. If $a = 1$, the first two rows are inconsistent equations. If $a = -... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2427722",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find the total number of coefficients in the expansion of $(x^2-x +1)^{22}$ that are divisible by $6$. Find the total number of coefficients in the expansion of $(x^2-x +1)^{22}$ that are divisible by $6$.
So I am really stuck on this one.. I immediately defeated when I saw that I need to compute $\binom {22} {a,b,c}$... | You can find which coefficients are divisible by $3$ by noting that $x^2-x+1\equiv(x+1)^2\pmod{3}$ so we are seeking the coefficients of $(x+1)^{44}$ which are divisible by $3$. Now notice that:
$$(x+1)^{44}=(x+1)^{27}(x+1)^{9}(x+1)^{6}(x+1)^2\equiv (x^{27}+1)(x^9+1)(x^{6}-x^3+1)(x^2-x+1)$$
Inspection sees the zero ter... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2429982",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 1
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Yet another family of hypergeometric sums that has a closed form solution. Let $m \ge 2$ and $j\ge 0$ be integers. Now, let $0 < z < \frac{(m-1)^{m-1}}{m^m}$ be a real number. Consider a following sum:
\begin{equation}
{\mathfrak S}^{(m,j)}(z) := \sum\limits_{i=0}^\infty \binom{m \cdot i + j}{i}\cdot z^i
\end{equation}... | Hint: The series ${\mathfrak S}^{(m,j)}(z)$ is strongly related with the generalized binomial series $B_m(z)$
\begin{align*}
B_m(z)=\sum_{i=0}^\infty\binom{mi+1}{i}\frac{1}{mi+1}z^i
\end{align*}
defined in (5.58) in Concrete Mathematics
by R.L. Graham, D.E. Knuth and O. Patashnik.
The series $B_m(z)$ satisfies the i... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Sum of terms $1^2-2^2+3^2-...$ Prove that: $$1^2-2^2+3^2-4^2+...+(-1)^{n-1}n^2=\frac{(-1)^{n-1}n(n+1)}{2}$$
I proved it by induction but is there any other way to solve it?
If it was not a proof but rather a question like find the term,how to solve it?
I realized that alternate terms were under same sign but can't unde... | You could use formal manipulations with a truncated power series:
$$\frac{1-x^{n+1}}{1-x} = 1 + x + x^2 + \cdots + x^n \\
\frac{d}{dx} \left( \frac{1-x^{n+1}}{1-x} \right) = 1 + 2x + \cdots + n x^{n-1} \\
x \frac{d}{dx} \left( \frac{1-x^{n+1}}{1-x} \right) = x + 2x^2 + \cdots + n x^n \\
\frac{d}{dx} \left( x \frac{d}{d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2430887",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 8,
"answer_id": 1
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Integrating a Rational Function with a Square Root : $ \int \frac{dx}{\sqrt{x^2-3x-10}} $
Integrate $$ \int \frac{dx}{\sqrt{x^2-3x-10}}. $$
I started off with the substitution $\sqrt{x^2-3x-10} = (x-5)t$. To which I got
$$ x = \frac{2+5t^2}{t^2-1} \implies \sqrt{x^2-3x-10} = \left(\frac{2+5t^2}{t^2-1} - 5\right)t = \... | Use: $\int \frac{dx}{\sqrt{x^2-a^2}}=\log|x+\sqrt{x^2-a^2}|+C$
Hence $$\int \frac{dx}{\sqrt{x^2-3x-10}} = \int \frac{dx}{\sqrt{x^2-3x+\frac{9}{4}-\frac{49}{4}}}=\int \frac{dx}{\sqrt{(x-\frac{3}{2})^2-\frac{49}{4}}} $$
$$=\{x-\frac{3}{2}=t; dx=dt\}= \int \frac{dx}{\sqrt{t^2-\frac{49}{4}}}=\log|t+\sqrt{t^2-\frac{49}{4}}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2431066",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Compute $2\arctan{2}+\arcsin{\frac{4}{5}}.$ Proceeding as you guys have tought me on my previous posts by setting the whole thing equal to $X$ taking the tangent of the whole expression and then using the additon formula for tangent:
$$\tan{X}=\tan{\left(2\arctan{2}+\arcsin{\frac{4}{5}}\right)}=\frac{\tan{(2\arctan{2}... | The last implication is false: $\tan{X} = 0 \implies X = k\pi$ for some $k \in \mathbb{Z}$. To find out which value, it's probably simplest to estimate the terms:
*
*$\arctan{2}>\arctan{1}=\pi/4$, so the sum is certainly bigger than $\pi/2$.
*On the other hand, $0<\arcsin{(4/5)} < \pi/2$ and $\arctan{2}<\pi/2$, so... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2432331",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Algebraic solution of complex equation For solving algebraically any complex equation involves two components for the real & imaginary parts. Let the real part be - $a$, imaginary part - $b$.
For the complex equation $$x^3 = 1-i $$
Substituting $x = a +bi$, we get: $$(a+bi)^3 = 1 - i.$$
Expanding the l.h.s. :
$$a^3 -(... | Since $1-i= \sqrt 2 (\cos315^\circ + i\sin315^\circ),$ the cube roots of $1-i$ must be
$$
2^{1/6} (\cos(105^\circ+n120^\circ) + i\sin(105^\circ+n120^\circ))
$$
where the only values of $n$ we need to consider are $0,$ $1,$ and $2.$
If we can believe the tables on this page, we have
$$
\cos105^\circ = -\frac 1 4 (\sqrt ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 3
} |
Density of multiplication table Is there any easy way to show $$\lim_{N \to \infty} \frac{1}{N^2}\#\{ab : 1 \le a,b \le N\} = 0$$ A quick calculation I did shows that the number of positive integers $\le N^2$ with a prime divisor $p > N$ is at most the order of $(\log 2) \cdot N^2$, so just getting rid of the numbers ... | For $N \ge 1$, let $$E_N := \{n \le N : \omega(n) \in [\frac{9}{10}\log\log N,\frac{11}{10}\log\log N]\},$$ where $\omega(n) := \sum_{p \mid n} 1$. An easy double-counting / [second moment method] argument shows that $|E_N| = N-o(N)$ as $N \to \infty$. So, $$\frac{1}{N^2}\#\{ab : 1\le a,b \le N\} = \frac{1}{N^2}\#\{ab ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2437052",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
} |
For each $a \in \Bbb{Z}$ work out $\gcd(3^{16} \cdot 2a + 10, 3^{17} \cdot a + 66)$ The problem is the following: For each $a \in \Bbb{Z}$ work out $\gcd(3^{16} \cdot 2a + 10, 3^{17} \cdot a + 66)$
This is what I have at the moment:
Let's call $d = \gcd(3^{16} \cdot 2a + 10, 3^{17} \cdot a + 66)$
Then $d \ \vert \ 3^{1... | $$E=\gcd(3^{16}2a+10,3^{17}a+66)=\gcd(b,c)=\gcd(b,c,3c-2b)=\gcd(b,c,102)$$
$$E=\gcd(b \mod 102,b\mod 102)=\gcd(36 a+10,3a+66)=\gcd(d,e)$$
$$E=\gcd(d-12e \mod 102,e)=\gcd(34,e)=\gcd(34,3(a+22))=\gcd(34,a+22)$$
because $34 \mod 3 \neq 0$
so $E=\gcd(a,2)\times \gcd(a+5,17)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2437747",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
How to find values such that the curves $y=\frac{a}{x-1}$ and $y=x^2-2x+1$ intersect at right angles? Problem:
Find all values of $a$ such that the curves $y = \frac{a}{x-1}$ and $y = x^2-2x+1$ intersect at right angles.
My attempt:
First, I set the two curves equal to each other:
$ \frac{a}{x-1} = x^2 - 2x + 1 $
$ \fr... | hint: $\dfrac{a}{x-1} = x^2- 2x+1 \implies a = (x-1)^3 = \dfrac{x-1}{2}$. Can you solve for $x$ and then $a$?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2442718",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Find equation of largest circle passing through $(1,1)$ and $(2,2)$ and which does not cross boundaries of first quadrant.
Find equation of largest circle passing through $(1,1)$ and $(2,2)$ and which does not cross boundaries of first quadrant.
my attempt
I tried writing a family through line $x=y$ and $(1,1)$ and $... | The circles passing through $A$ and $B$ are a linear combination of the equation of the line $AB:x-y=0$ and the circle having diameter $AB$, $\mathscr{C}:x^2+y^2-3x-3y+4=0$
Therefore its general equation is $x^2+y^2-3x-3y+4+k(x-y)=0$
$x^2+y^2-3x-3y+4+k(x-y)=0$ intersect $x-$axis if $y=0$. Substitute in the circle and ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2443053",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Find the solutions to $\left\lfloor\left(\frac{5}{3} \right)^n\right\rfloor = 3^m$
Find all positive integer solutions to $\left\lfloor\left(\dfrac{5}{3}
\right)^n\right\rfloor = 3^m$.
Let $a_n = \left\lfloor\left(\dfrac{5}{3}
\right)^n\right\rfloor$. Then $$a_n = 1,2,4,7,12,21,35,59,99,165,275,459,765,1276,2126,3544... | If we write
$$
\left(\frac{5}{3}\right)^n = \left[ \left(\frac{5}{3} \right)^n \right] + \left\{ \left(\frac{5}{3}\right)^n \right\},
$$
then supposing that
$$
\left[ \left(\frac{5}{3} \right)^n \right]=3^m
$$
implies the inequality
$$
0 < 5^n - 3^{n+m} < 3^n.
$$
Applying lower bounds for linear forms in (two complex)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2443538",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 1,
"answer_id": 0
} |
Showing a sequence converges from the definition of limit I am having some trouble proving that this series converges to zero
$b_n=\dfrac{n+5}{n^2-n-1}$, $n\geq 2$
directly from the definition of limit. My attempts have only left me with a lengthy, complicated form for $n$ in terms on $\epsilon$ so I must be missing so... | Notice that $n^2 - n - 1 = \dfrac{1}{4} \left( 2n - 1 + \sqrt{5} \right) \left( 2n - 1 - \sqrt{5} \right)$ so we can rewritten:
\begin{align}
b_{n} & = \dfrac{4n+20}{\left( 2n - 1 + \sqrt{5} \right) \left( 2n - 1 - \sqrt{5} \right)} \\
& = \dfrac{4n - 2 + 2\sqrt{5} + 22 - 2\sqrt{5}}{\left( 2n - 1 + \sqrt{5} \right) \le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2443784",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Find limit of trigonometric function $$\lim_{x\rightarrow0} \frac{\tan^3(3x)-\sin^3(3x)}{x^5}$$
I think it should be decomposed with $$\lim_{x\to0} \frac{\sin x}{x}=1$$ but I'm always getting indefinity $\frac{0}{0}$.
| Take $3x=a \to x=\frac{a}{3}$
$$\quad{\lim_{x\rightarrow0} \frac{tg^3(3x)-\sin^3(3x)}{x^5}=\\
\lim_{a\rightarrow0} \frac{tg^3(a)-\sin^3(a)}{(\frac a3)^5}=\\3^5\lim_{a\rightarrow0} \frac{tg^3(a)-\sin^3(a)}{(a)^5}=\\
3^5\lim_{a\rightarrow0} \frac{(\tan(a)-\sin(a))(\tan^2(a)+\sin^2(a)+\tan (a).\sin (a))}{(a)^5}=\\
3^5\li... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2444868",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Complex quintic equation Given the equation $x^5=i$, I need to show by both algebraic and trigonometrical approaches that
$$\cos18^{\circ}=\frac{\sqrt{5+2\sqrt5}}{\sqrt[5]{176+80\sqrt5}}$$
$$\sin18^{\circ}=\dfrac1{\sqrt[5]{176+80\sqrt5}}$$
Trying by trigonometric approach,
$x^5$ = i $\;\;\;\;$ -- eqn. (a)
=> x = $... | Trigonometrical way: One has
$$\sin 36^\circ = \cos 54^\circ = 4\cos^3 18^\circ - 3\cos 18^\circ$$
$$2\sin 18^\circ\cos 18^\circ = 4\cos^3 18^\circ - 3\cos 18^\circ$$
$$2\sin 18^\circ = 4\cos 18^\circ -3 = 1-4\sin 18^\circ$$
So, $\sin 18^\circ$ is positive root of equation $4x^2+2x-1=0$, or $\sin 18^\circ = \frac{-1+\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2445926",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.