Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Fast/smart way to write polar curve in cartesian Is there a fast way to write the curve:
$$r=\frac{a}{1-\frac{1}{\sqrt{2}}\cos(\theta)}$$
as a cartesian curve $f(x,y)=0$?
It seems I can take
$$r(1-\frac{1}{\sqrt{2}}\cos(\theta)) = a$$
$$r-\frac{x}{\sqrt{2}}=a$$
$$\sqrt{x^2+y^2}-\frac{x}{\sqrt{2}}=a$$
$$x^2+y^2=(a+\fra... | \begin{align*}
(a + \frac{x}{\sqrt{2}})^2 = a^2 + ax\sqrt{2} + \frac{x^2}{2}
\end{align*}
And then you can rewrite the equation
\begin{equation*}
\frac{x^2}{2} - ax\sqrt{2} + y^2 = a^2
\end{equation*}
as
\begin{equation*}
\left( \frac{x}{\sqrt{2}} - a \right)^2 + y^2 = \left( \frac{x - a\sqrt{2}}{\sqrt{2}} \right... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2389057",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
What is the result of $\lim_{x\to0_+} \frac{e^x - x e^x - 1}{\left(e^x - 1 \right)^2}$ without L'Hôpital's rule. I have the limit
$$
\lim_{x\to0_+} \frac{e^x - x e^x - 1}{\left(e^x - 1 \right)^2}
$$
which I need to compute without L'Hôpital's rule.
(The result is $-\frac{1}{2}$ with L'Hôpital's rule).
Thanks.
| It is also simple to look at Taylor expansions around $0$ up to at least second order
$$e^x = 1 + x + \frac{1}{2}x^2... $$ then $$\frac{e^x - xe^x - 1}{(e^x-1)^2} \approx \frac{1 + x + \frac{1}{2}x^2 - x - x^2 - \frac{1}{2}x^3 - 1} {(x+ \frac{1}{2}x^2)^2} = \frac{ - \frac{1}{2}x^2 - \frac{1}{2}x^3}{x^2 + \frac{1}{4}x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2390227",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Isosceles triangle inscribed in an ellipse.
Find the maximum area of an isosceles triangle, which is inscribed inside an ellipse $\dfrac {x^2} {a^2} + \dfrac {y^2} {b^2} = 1$ with its (unique) vertex lying at one of the ends of the major axis of the ellipse.
The three vertices of the triangle would be $(a,0), (x,y), ... | By AM-GM
$$S_{\Delta}=(a-x)y=(a-x)b\sqrt{1-\frac{x^2}{a^2}}=ab\left(1-\frac{x}{a}\right)\sqrt{1-\frac{x^2}{a^2}}=$$
$$=\frac{ab}{\sqrt3}\cdot\sqrt{\left(1-\frac{x}{a}\right)^3\left(3+\frac{3x}{a}\right)}\leq\frac{ab}{\sqrt3}\cdot\sqrt{\left(\frac{3\left(1-\frac{x}{a}\right)+3+\frac{3x}{a}}{4}\right)^4}=\frac{3\sqrt3ab... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2390641",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
When is this integer a perfect square: $n^2 + 20n + 12$
Let $S_n = n^2 + 20n + 12$, where $n$ a positive integer. What is the sum of all possible values of $n$, for which $S_n$ is a perfect square?
Help me to solve it.
| Use $$(n+4)^2<n^2+20n+12<(n+10)^2.$$
I got $n^2+20n+12=(n+8)^2$, which gives $n=13$ or
$n^2+20n+12=(n+6)^2$, which gives $n=3$.
Id est, the answer is $16$
because in cases $n^2+20n+12=(n+5)^2$, $n^2+20n+12=(n+7)^2$ and $n^2+20n+12=(n+9)^2$ we can not get solutions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2391025",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 0
} |
Given prime $p$, for any integer $a$, exist integer $x,y$ such that $p\mid y^2+x^3+a$. Let $p$ be prime number and $a$ an integer. Show that the equation
$$y^2+x^3+a\equiv 0\pmod p$$has integer solutions.
It seem a famous problem? But I can't prove it.
| The solutions to problem N8 of the 53rd IMO, as mentioned by Gerry Myerson, do indeed apply to this problem. In fact it can be generalised. For any strictly positive integer $n$ and prime power $q=p^m$, then for any element $r$ of the finite field $F_q$ of size $q$, the equation
$$
y^2+x^n=r\qquad{\rm(*)}
$$
has soluti... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2391964",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 1,
"answer_id": 0
} |
$\lim_{(x,y)\rightarrow (0,0)}\frac{x^5-\sin(y^3)}{4x^4+3y^2}$ Are the next computations correct?
$$
0\leq \left|\frac{x^5-\sin(y^3)}{4x^4+3y^2}\right|
\leq\frac{|x^5|}{4x^4}+\frac{|y^3|}{3y^2}
=\frac{|x|}{4}+\frac{|y|}{3}
$$
Then, by squeeze theorem, the limit is zero.
| $$0\leq |\frac{x^5-\sin(y^3)}{4x^4+3y^2}|\leq\frac{|x^5|}{4x^4+3y^2}+\frac{|y^3|}{3y^2+4x^4} \leqslant\frac{|x|}{4}+\frac{|y|}{3} \leqslant \frac{|x|+|y|}{3}=\frac{\sqrt{x^2}+\sqrt{y^2}}{3}\leqslant \frac{2(\sqrt{x^2+y^2})}{3}=\frac{2 \delta}{3}$$
If you take $\delta=\frac{3 \epsilon}{2}$ you have the $\epsilon-\delta$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2392352",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $450^\circ<\alpha<540^\circ$ and $\cot\alpha=-\frac{7}{24},$ calculate $\cos\frac{\alpha}{2}$. Why is my solution wrong? The problem says:
If $450^\circ<\alpha<540^\circ$ and $\cot\alpha=-\frac{7}{24},$ calculate $\cos\frac{\alpha}{2}$
I solved it in the following way: $$\begin{align} -\frac{7}{24}&=-\sqrt{\frac{1... | Hint:
$$\cot\alpha=\frac{1-\tan^2\frac\alpha2}{2\tan\frac\alpha2}=\frac{1-t^2}{2t}$$ or
$$t^2+2\cot\alpha\ t-1=0,$$
or
$$t=-\cot\alpha\pm\sqrt{1-\cot^2\alpha}=-\frac34,\frac43.$$
Then $$\cos\frac\alpha2=\pm\frac1{\sqrt{1+t^2}}=\pm\frac1{\sqrt{2-2\cot\alpha\ t}}=\pm\frac45,\pm\frac35.$$
Then given range of $\alpha$ tell... | {
"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": 2
} |
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.
| Let $x+y=S$
$$S^3+(S-x)^3+33 x (S-x)+x^3$$
has a derivation
$$ \frac{d}{dx}(x^3+(S-x)^3+S^3+33 x (S-x) )= -3 (-11 + S) (S - 2 x) $$
So, sollution $ \frac{d}{dx}=0$ leads to $S=11$ or $ S=2x$, i.e. $x=y$
For $x=y$, $x^3+33 x y+(x+y)^3+y^3=10x^3+33x^2=2662$ has a unique real solution $x(=y)=11/2$
For $y=c-x$, functi... | {
"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": 2
} |
Finding the exterma of $x^2+y^2+z^2-yz-zx-xy$ s.t. $x^2+y^2+z^2-2x+2y+6z+9=0$ using Lagrange's multiplier,
Using Lagrange's multiplier method, obtain the maxima and minima of $$x^2+y^2+z^2-yz-zx-xy$$ subject to the condition $$x^2+y^2+z^2-2x+2y+6z+9=0$$
My attempt:
I formed the expression
$$F=x^2+y^2+z^2-yz-zx-xy+\la... | You solve the system and get
$\left[x= -\dfrac{3-2 \lambda}{2 \lambda+3},\;y= -1,\;z= -\dfrac{3 (2 \lambda+1)}{2 \lambda+3}\right]$
then you use the constraint
$x^2+y^2+z^2-2x+2y+6z+9=0$
and substitute. Simplify
$(2 \lambda-3) (2 \lambda+9)=0$
$\lambda=\dfrac{3}{2};\;\lambda=-\dfrac{9}{2}$
Substitute again in the solut... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2395581",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Evaluate $(\sqrt{3}-3i)^6$
Evaluate $$(\sqrt{3}-3i)^6.$$
So I assume that we should write the following in polar form
$r=\sqrt{(\sqrt{3})^2+(-3)^2}=\sqrt{3+9}=\sqrt{12}=2\sqrt{3},$
$\theta=\arctan{\frac{-3}{\sqrt{3}}}=-\frac{\pi}{3}.$
So $$(\sqrt{3}-3i)^6=[2\sqrt{3}e^{-i\frac{\pi}{3}+2\pi k}]^6=1728e^{-2\pi i+2\pi m}... | You are right. It's
$$27\cdot2^6\cos^6(-60^{\circ})=1728.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2395892",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Find an invertible matrix $S$ and a matrix $J$ in Jordan form, such that $S^{-1}AS = J$ $A = \begin{pmatrix}
0 & -1 & 0 & 0 \\
2 & 0 & -1 & 0 \\
3 & -1 & -2 & -1 \\
-1 & 0 & 1 & 1 \\
\end{pmatrix}$
We need to find an invertible matrix $S$ and a matrix in Jordan form $J$, such th... | Once you have found $J$, finding $S$ is very easy. Just write $S=(s_{ij})$ and solve the linear equations in these coefficients given by the equation
$$
AS=SJ.
$$
There are many solutions for $S$ here, and you have to pick one which satisfies $\det(S)\neq 0$.
Edit: Say, $S=(x_1,x_2,x_3,x_4)$ with column vectors $x_1=(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2396759",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Pattern recognition for $S_n=2+5+13+35+...+n^{th}_{term}$ I want to find sum of the first $n^{th}$ term of this sqequence .
$$2,5,13,35,97,275,793,...\\s_n=2+5+13+35+97+...$$
What is the closed form formula for $s_n$?
| Alternatively, the sequence is:
$$2,3\cdot 2-1,3\cdot 5-2,3\cdot 13-4,3\cdot 35-8,...$$
It is the recurrence relation:
$$a_n=3a_{n-1}-2^{n-2},a_1=2.$$
Divide it by $2^{n}$:
$$\frac{a_n}{2^{n}}=\frac{3a_{n-1}}{2^{n}}-\frac14.$$
Denote: $b_n=\frac{a_n}{2^n}$ to get:
$$b_n=\frac{3}{2}b_{n-1}-\frac{1}{4},b_1=1.$$
Solution ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2397987",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Functional equation : $ f(1)^3 + f(2)^3 + \ldots + f(n)^3 = (f(1) + f(2) + \ldots + f(n))^2$ Find all function $f:\mathbb{N}\rightarrow\mathbb{N}\cup\{0\}$ satisfying
$$ f(1)^3 + f(2)^3 + \ldots + f(n)^3 = (f(1) + f(2) + \ldots + f(n))^2$$
$\forall n \in \mathbb{N}$.
Thank you, Batominovski and Guy Fabrice..
Is my und... | Hint: Prove by induction on $n$ that, if $f(n)\neq 0$, then
(1) we have $f(n)=M_n+1$, where $M_n:=\max\big\{f(1),f(2),\ldots,f(n-1)\big\}$ for $n>1$, and $M_1:=0$;
(2) for $n>1$, each the numbers $1,2,\ldots,M_n$ appears only once among $f(1),f(2),\ldots,f(n-1)$ in an increasing order (i.e., $i$ appears before $i+1$).
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2398994",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
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
$$(x+y+z+1)^2=5(xy+yz+xz)+1$$
| Here's a way to more constructively get the answer user399601 provided:
Note that since $(x + y + z)^2 = x^2 + y^2 + z^2 + 2(xy + yz + xz)$, the equation
\begin{equation}\label{eqn:constraint}\tag{1}
(x + y + z)^2 + 2(x + y + z) = 5(xy + yz + xz)
\end{equation}
is equivalent to
\begin{align*}
x^2 + y^2 + z^2 + 2(x + y... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2399830",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 4
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Determining if a vector is in the row space I am just trying to determine if the vector $[0, 7, 4]$ belongs in the row space of the matrix $$A = \begin{bmatrix}
1 & 2 & 0 \\
3 & -1 & 4 \\
1 & -5 & 4 \\
\end{bmatrix}
$$
What I have done so far is created an augmented matrix like so
$$ \l... | A vector $\vec b$ is in the row space of $A$ if and only if $\vec b$ is in the column space of $A^\top$. Thus, to determine if the vector $\vec b=\left[\begin{array}{r}
0 \\
7 \\
4
\end{array}\right]$ is in the row space of $A = \left[\begin{array}{rrr}
1 & 2 & 0 \\
3 & -1 & 4 \\
1 & -5 & 4
\end{array}\right]$, form th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2400912",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
$\sin(40^\circ)<\sqrt{\frac{3}7}$
Prove without using of calculator, that $\sin40^\circ<\sqrt{\frac{3}7}$.
My attempt.
Since $$\sin(40^\circ)=2\sin(20^\circ)\cos(20^\circ)<2\sin(20^\circ)$$
$$=2\sin(60^\circ-40^\circ)=\sqrt{3} \cos(40^\circ)-\sin(40^\circ),$$
$$2\sin(40^\circ)<\sqrt{3} \cos(40^\circ).$$
Hence, $$4\si... | Applying the identity $\sin 3x=3\sin x-4\sin^3 x$ we get
$$\frac{\sqrt 3}2=\sin 120^\circ=3\sin 40^\circ-4\sin^3 40^\circ$$
Now, consider the polynonial
$$P(x)=8x^3-6x+\sqrt 3$$
We know that $\sin40^\circ$ is a root of $P$. Also, we know that $\sin 40^\circ>\sin30^\circ=1/2$. Differentiating $P$, we see that $P$ is inc... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2401341",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 5,
"answer_id": 1
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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... | You can consider $n=1/x$ and study the limit of the function at $0$ from the right:
$$
\lim_{x\to0^+}
\frac{1}{x}\frac{1}{\sqrt{x}}
\left(\sqrt{\frac{1}{x}+1}-a\sqrt{\frac{1}{x}}+\sqrt{\frac{1}{x}-1}\right)
=
\lim_{x\to0^+}\frac{\sqrt{1+x}-a+\sqrt{1-x}}{x^2}
$$
If $a\ne2$, the limit is infinite, because the limit of th... | {
"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": 3
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Polynomial : $P(a)P(a^2)P(a^3)P(a^4)$ Let $P(x) = x^3 + 2x^2+3x+4$ and $a$ be the root of equation $x^4+x^3+x^2+x+1=0$.
Find the value of $P(a)P(a^2)P(a^3)P(a^4)$
Is my answer correct ?
Since root of equation $x^4+x^3+x^2+x+1=0$ is the $5^{th}$ primitive root of 1,
so $a, a^2, a^3, a^4$ are roots of $x^4+x^3+x^2+x+1=0$... | Here is another way. Call the product that you want Q.
Note that $xP(x)-P(x)=x^4+x^3+x^2+x-4$
As you correctly observe the values you are substituting are all roots of $x^4+x^3+x^2+x+1=0$ and hence for these values you get $(x-1)P(x)=-5$ whence $$(a-1)(a^2-1)(a^3-1)(a^4-1)Q=(-5)^4=625$$
Now pairing the first and last f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2402429",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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take $\int_\lambda f \ ds$ (Stieltjes) where $f(x,y) = \sqrt{x^2+y^2}$ and $\lambda(t) = (a\cos t, b\sin t)$
Calculate $\int_\lambda f \ ds$, where $f:\mathbb{R}^2\to\mathbb{R}$
is given by $f(x,y) = \sqrt{x^2+y^2}$ and $\lambda:[0,
2\pi]\to\mathbb{R^2}$ is the path $\lambda(t) = (a\cos t, b\sin t)$,
which descri... | I studied analysis about twenty years ago, so my answer should be considered not as a rigor proof but as an idea.
$$\int_\lambda f\ ds = \int_0^{2\pi} f(t)\frac{ds}{dt}dt.$$
Let $\lambda(t)=(\lambda_x(t), \lambda_y(t))$. Then
$ds=\sqrt{\lambda_x’^2+\lambda_y’^2}dt=\sqrt{a^2\sin^2t+b^2\cos^2 t}dt$,
$$\int_\lambda f\ d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2404183",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find all values of $x$ at which $P(x)=x^4-4x^3+22x^2-36x+18$ is a perfect square Find all values of positive integer $x$ at which the following expression is perfect square
$$P(x)=x^4-4x^3+22x^2-36x+18$$
I tried to assume $P= (x^2+ax+b)^2$ ; and comparing the cofactors , get that $a= -2 ; b= 9$
, but when ... | If $x$ is an integer, then
$$x^4-4x^3-22x^2-36x+18$$
$\qquad$is congruent to $2$ mod $4$ if $x$ is even,
$\qquad$and is congruent to $5$ mod $8$ if $x$ is odd,
hence can't be a perfect square.
(Even squares are congruent to $0$ mod $4$, and odd squares are congruent to $1$ mod $8$.)
Some details . . .
If $x$ is ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2405043",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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Area of square as a function of $\hat{a}$? Suppose $ABCD $ is square ,and $AM=DN=QB=PC$ so $$A'B'C'D'$$
is a square too.
Can someone help me to find area of $\bf{smaller -square}$(or $\color{red} {\Box A'B'C'D'}
$) as a function of angle $\hat{a}$ .It seems to be easy ,but I get stuck on this problem.Thanks in ad... | Label the points $A, B, C, D, E, F, G, H, I, J, K, L$ as in this diagram:
Now, let $AD = x$ and let $\angle DAF = \alpha$.
We wish to express the area of square $IJKL$ in terms of $x$ and $\alpha$.
First, let $DF = AE = y$. The tangent of $\alpha$ can then be written
$$\tan \alpha = \frac{y}{x} \,\,\implies\,\, y =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2406435",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Finding $a + b + c$ for the $101$st number of the form $3^a + 3^b + 3^c$ when these numbers are listed in increasing order $a,b,c\neq0$ & $a,b,c$ are distinct & $a,b,c\in \mathbb{N}$
numbers in form of
$3^a+3^b+3^c$
If we order them in increasing order, what is the sum of
$a+b+c=?$
for the $101$st such number?
There a... | At the top of your question you say $a,b,c\neq 0$ but then the rest of what you say seems to include the possibility that one of them can be $0$. I'll assume that $0$ is not allowed in what follows. Obviously if you include $3^0$ as a possible power the answer will be different, but the method is the same.
As you say,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2406520",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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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:
$$\frac{1}{a}=\frac{2}{9+\sqrt{45}} \ti... | Since the $ 45 $ inside the radical is expressed a product containing a perfect square ($ 9 $), $$ \sqrt{45} = \sqrt{9 \times 5} = \sqrt{9}\sqrt{5} = 3\sqrt{5} $$
Thus, $$ \frac{1}{a} = \frac{9 - \sqrt{45}}{18} = \frac{9 - 3\sqrt{5}}{18} $$
Dividing the numerator and denominator by $ 3 $ (the greatest common factor of ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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What are the number of solutions of $11^x+13^x+17^x-19^x$? My attempt: $ 11^x+13^x+17^x-19^x=0\implies (\frac{11}{19})^x+(\frac{13}{19})^x+(\frac{17}{19})^x=1 $.
Now taking limit as $x\rightarrow\infty$,we get $0=1$,which is absurd.Hence,the equation has no solution.
Is it correct??
| Let $f(x)= 11^x+13^x+17^x-19^x=19^x[(\frac{11}{19})^x+(\frac{13}{19})^x+(\frac{17}{19})^x-1]$.
Clearly,$g(x)=(\frac{11}{19})^x+(\frac{13}{19})^x+(\frac{17}{19})^x-1$ is a strictly decreasing function.Also,$\lim_{x\to \infty}g(x)=-1 $ & $\lim_{x\to -\infty}g(x)=\infty$.Thus $f(x)=0$ has exactly one solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2408618",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proving two binomial coefficient identities based on the expansion of $(1 + x)^{2n}$ This is a very interesting combinatorics problem that I came across in an old textbook of mine. So I know its got something to do with permutations and combinations, which yields the shortest, simplest proofs, but other than that, the ... | Consider a group of $n$ males and $n$ females from which a committee of $n$ need to be chosen.
There are $n + 1$ possible cases:
Case $1$: $0$ males and $n$ females are chosen- $\binom{n}{0} \cdot \binom{n}{n} = \binom{n}{0}^2 ways$
Case $2$: $1$ male and $n -1$ females are chosen- $\binom{n}{1} \cdot \binom{n}{n - 1} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2414040",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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Determine whether the following series is convergent or divergent, and find its sum if it is convergent. PROBLEM
Is the following complex series convergent, and if so, what is its sum?
$$\sum_{n=0}^{\infty}{\frac{\cos(n\theta)}{3^n}}, \text{ } \theta \in \mathbb{R}$$
MY ATTEMPT #1
Let $$a_n = \frac{\cos(n\theta)}{3^n... | As hinted in the comments, the series
$$\sum_{n=0}^{\infty}{\frac{\cos(n\theta)}{3^n}}$$
is convergent (by the Comparison Test) since $\theta \in \mathbb{R}$ implies that
$$\left|a_n\right| \leq \frac{1}{3^n},$$
where $\left(\frac{1}{3}\right)^n$ is a convergent geometric series.
Now, to compute the sum, we consider th... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove that the expression $5^{2n+1} \cdot 2^{n+2} + 3^{n+2} \cdot 2^{2n+1}$ is divisible by 19.
Prove that the expression
$$5^{2n+1} * 2^{n+2} + 3^{n+2} * 2^{2n+1}$$
is divisible by $19$.
I'll skip the basis step (as I have done last time) but I can conclude that it's only divisible by 19 for integers n ≥ 0 (or ... | Without using induction
$$5^{2n+1}2^{n+2}+3^{n+2}2^{2n+1}=20\cdot50^n+18\cdot12^n$$
$$\equiv1\cdot12^n+(-1)\cdot12^n\pmod{19}$$ as $20\equiv1,18\equiv-1,50\equiv12\pmod{19}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2415192",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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true false combinatorics questions about arrangements I've solved some exercises regarding combinatorics. The questions are true\false questions, but I'm not sure that I've done them right. Please take a look and indicate any corrections I need to make.
1) The number of different arrangements of the string 'AAABBC' is... |
The number of different arrangements of the string AAABBB is larger than the number of different arrangements of the string AABBCC.
Your work is correct.
The number of different arrangements of the string AABBCC is equal to the number of ways of cutting the set $\{1, 2, 3, 4, 5, 6\}$ into three subsets with two elem... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2415267",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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What will be the $n^{th}$ term of this given series? Given series is:
$$1+\frac{1\times x^2}{2\times 4}+\frac{1\times 3\times 5\times x^4}{2\times 4\times 6\times 8}+\frac{1\times 3\times 5\times 7\times 9\times x^6}{2\times 4\times 6\times 8\times 10\times 12}+.....\infty$$
I need to find it's $n^{th}$ term but am hav... | Hint: Your $n$th term (counting the first non-$1$ term as the first term and excepting the $0$th term) looks to be
$$x^{2n}\frac{1\times 3 \times \cdots \times (4n-3)}{2\times 4\times \cdots \times (4n)}.$$
This equals
$$x^{2n}\frac{1\times 2 \times \cdots \times (4n-2)}{\left(2\times 4\times \cdots \times (4n-2)\rig... | {
"language": "en",
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"source": "stackexchange",
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About using the lagrange multiplier. Today I have seen a question like the following;
$x+y+z=5$ and $xy+yz+xz=3$ and $x,y,z \in \mathbb{R}^+$
What is the maximum value $x$ can get?
Now it is pretty obvious that question is solvable using many simple elementary methods like say $(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+xz)=25$
$25... | Your function $f(x,y,z,k)=x+5-(y+z)+k(x+y+z-5)$ is not the right function fot Lagrange. The right function is given by
$f(x,y,z,k)=x+5-(y+z)+k(xy+yz+xz-3)$:
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2416180",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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Find $\tan^{-1} (i\sqrt{2})$. Problem: Find $\tan^{-1} (i\sqrt{2})$.
My attempt: We must find $z \in \mathbb{C}$ such that $\tan z = i\sqrt{2}$.
$$\tan z = \frac{\sin z}{\cos z} = i\frac{e^{-iz} - e^{iz}}{e^{iz} + e^{-iz}}$$
Let $u = e^{iz}$. Then $$i\frac{u^{-1}-u}{u+u^{-1}} = i\sqrt{2}$$ so $$1-u^2 = \sqrt{2}(u^2 + 1... | Let $u=e^{iz}$ just as you did.
$$\begin{array}{rcl}
u^2 &=& \dfrac{1-\sqrt{2}}{1+\sqrt{2}} \\
u^2 &=& \dfrac{(1-\sqrt{2})^2}{(1+\sqrt{2})(1-\sqrt{2})} \\
u^2 &=& \dfrac{(1-\sqrt{2})^2}{-1} \\
e^{2iz} &=& -(1-\sqrt{2})^2 \\
e^{2iz} &=& (1-\sqrt{2})^2 e^{i\pi} \\
2iz &=& \ln[(1-\sqrt{2})^2e^{i\pi}] + 2ni\pi \\
2iz &=& 2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2416766",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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If $8\sin x - \cos x=4$, then find possible values of $x$ I am not understanding what exactly can watch do here. First I thought that if I could square it but it was in vain. Please help me.
| $8\sin x -\cos x =4$
solve for $\cos x$
$\cos x = 8\sin x -4$
plug in the fundamental identity $\sin^2 x + \cos^2 x=1$
$\sin^2 x + \left ( 8\sin x -4\right)^2=1$
$65 \sin ^2 x -64 \sin x +15=0$
$\sin x = \dfrac{64\pm \sqrt{64^2-4\cdot 65 \cdot 15}}{130}$
$\sin x = \dfrac{3}{5} \to x = \arcsin \left(\dfrac{3}{5}\right)+... | {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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Show that ${\sum^{n}_{k=1}} \frac{1}{n+k} \le \frac{3}{4}$
Prove that $\displaystyle{\sum^{n}_{k=1}} \frac{1}{n+k} \le \frac{3}{4}$ for each positive integer $n$.
My work. I think that i have to use induction, but i can't see how... What i did:
$$f(n)=\displaystyle{\sum^{n}_{k=1}} \frac{1}{n+k} \implies k(n)=f(n+1)-f... | $f(x) = 1/(n+x)$ is convex in $x$. So we have that the value for $f(x)$ for any $x \in (0 \; n]$ is less or equal to (line equation between the two interval boundaries) $1/n + x (1/(2n)- 1/n)/n $. Summing the values gives
$$
\sum^{n}_{k=1} \frac{1}{n+k} \le \sum^{n}_{k=1} 1/n + k (1/(2n)- 1/n)/n = \\1 + \frac12 n (n+1)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2418810",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Let $(x, y, z)\in \Bbb R$, prove that $\big(\frac{2x-y}{x-y}\big)^2+\big(\frac{2y-z}{y-z}\big)^2 +\big(\frac{2z-x}{z-x}\big)^2 \geqslant 5$
Olympiad Inequation
Let $x$, $y$ and $c$ be distinct real numbers. Prove that:
$$\left(\frac{2x-y}{x-y}\right)^2+\left(\frac{2y-z}{y-z}\right)^2+\left(\frac{2z-x}{z-x}\right)^2 \g... | This is a special case of the following inequality: $$\sum_{a,b,c}\Big(\dfrac{ma-nb}{a-b}\Big)^2\geq m^2+n^2,$$
where $m,n\in\mathbb{R}$ and $a,b,c$ are distinct reals.
The proof is simply completing the square: start by writing $$\dfrac{ma-nb}{a-b} = m+(m-n)\dfrac{b}{a-b}$$ and $$\dfrac{mc-na}{c-a} = n+(m-n)\dfrac{c}... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "4",
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Logarithmic problem If $\log_{16} 15 =a$ and $\log_{12} 18 =b$, then show that $$\log_{25} 24 = \frac{5-b}{16a-8ab-4b+2}.$$
| Let $l_2= \ln 2$,$l_3= \ln 3$ and $l_5= \ln 5$. Now change base & we have
\begin{eqnarray*}
\frac{5-b}{16a-8ab-4b+2} = \frac{5-\frac{l_2+2l_3}{2l_2+l_3}}{16 \frac{l_3+l_5}{4l_2}-8 \frac{l_3+l_5}{4l_2} \frac{l_2+2l_3}{2l_2+l_3} -4 \frac{l_2+2l_3}{2l_2+l_3} +2} \\
\end{eqnarray*}
\begin{eqnarray*}
= \frac{(5(2l_2+l_3)-(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2420437",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that $\int \frac{x^m dx}{\ln x}=\ln(\ln x)+\frac{(m+1)\ln x}{1}+\frac{(m+1)^2 \ln ^2 x}{1 \times 2^2}+\cdots $ Prove that
$$\int \frac{x^m dx}{\ln x}=\ln(\ln x)+\frac{(m+1)\ln x}{1}+\frac{(m+1)^2 \ln ^2 x}{1 \times 2^2}+\frac{(m+1)^3 \ln^3 x}{1 \times 2 \times 3^2}+\cdots \infty $$
My Try:
I started with $$I=\... | Since differentiating is usually easier than finding primitives, and you are given a candidate, the simplest way is to differentiate the right hand side. Write that as
$$f(x) = \ln (\ln x) + \sum_{n = 1}^{\infty} \frac{(m+1)^n(\ln x)^n}{n\cdot n!}.\tag{1}$$
The series converges locally uniformly on $(0,+\infty)$, so it... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Proof $(1+i)^{n}$ = $2^{n/2}(\cos(nπ/4)+i\sin(nπ/4))$ While binomial problem I struck here
$$(1+i)^{n} = 2^{\frac{n}{2}}\left(\cos(\frac{n\pi}{4})+i\sin(\frac{n\pi}{4})\right).$$
Please proof this equation .
Any help will be appreciated.
And Thanks for help
| This is not a binomial question,
Using Taylor series,
$$\sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots$$
$$\cos x=1-\frac{x^2}{2!} +\frac{x^4}{4!} -\cdots $$
$$e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!} +\cdots$$
$$e^{ix}=1+ix+\frac{(ix)^2}{2!}+\frac{(ix)^3}{3!} +\cdots$$
$$e^{ix}=1+ix-\frac{(x)^2}{2!}+i\frac{(x)^3}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2424072",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Is there a general theorem to manipulate recursive functions? I was messing with some recursive functions and realized it was equivalent to the Fibonacci sequence, but I couldn't figure out why. I then played a little further, and discovered some other interesting patterns. Let's consider functions $A,B$.
Let $A(1) = 0... | (Not a complete answer, still maybe relevant.) The recurrence can be written in matrix form as:
$$
\begin{pmatrix}
B_{n} \\
A_{n}
\end{pmatrix}
=
\begin{pmatrix}
1 & 1 \\
1 & 0
\end{pmatrix}
\cdot
\begin{pmatrix}
B_{n-1} \\
A_{n-1}
\end{pmatrix}
\quad\quad \text{with}\;\;
\begin{pmatrix}
B_{1} \\
A_{1}
\end{p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2427240",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 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... | Using AM-GM:
$$12=xy+yz+xz\ge 3\sqrt[3]{(xyz)^2} \Rightarrow xyz\le 8 \ \ \ \ (1)$$
$$xyz=2+x+y+z\le8 \Rightarrow x+y+z\le 6 \ \ \ \ \ (2)$$
The equality occurs when $x=y=z=2$.
Using C-S:
$$(x^2+y^2+z^2)(y^2+z^2+x^2)\ge (xy+yz+zx)^2 \Rightarrow x^2+y^2+z^2\ge 12 \ \ \ \ (3)$$
The equality occurs when $x=y=z=2$.
From t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2427470",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 5
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What is the probability we need k rolls of a fair sided die to get at least one five and one six? I HAVE checked the questions and none of them seem to specify any method to compensate for at least one 5 AND one 6.
Here is how I tried it.
$$ P(k\ Rolls \ and \ last \ is \ a \ six)= P(\ k -1 \ rolls \ have \ at \ leas... | It's not clear to me if the question is whether $k$ rolls suffice or whether exactly $k$ rolls are required. I try to cover both possibilities below.
Let's say the first roll on which we have at least one 5 and at least one 6 is roll $K$.
Let's also say $A$ is the event of rolling no 5 in $k$ rolls and $B$ is the even... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2429791",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Turkevicius inequality From Zdravko Cvetkovski's Inequalities — Theorems, Techniques and Selected:
Let $a$, $b$, $c$ and $d$ be positive real numbers. Prove that:
$$a^4+b^4+c^4+d^4+2abcd \geq a^2b^2+ a^2c^2+ a^2d^2+ b^2c^2+ b^2d^2+ c^2d^2.$$
Please suggest how to prove this inequality using basic methods. No logarit... | Assume $a\geq b\geq c\geq d.$ Let $f(a) = a^4+b^4+c^4+d^4+2abcd - a^2b^2-a^2c^2-a^2d^2-b^2c^2-b^2d^2-c^2d^2.$ Then, $f'(a) = 4a^3 -2a(b^2+c^2+d^2)+2bcd$ and $f''(a) = 12a^2 - 2b^2-2c^2-2d^2\geq 0.$ Hence, $f'$ is increasing and $f'(b) = 2(b^3-b(c^2+d^2)+bcd) \geq 2b(b-c)(b-d)\geq 0.$ Thus, $f(a)$ is increasing and ther... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2430311",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Sum of terms $1^2-2^2+3^2-...$ Prove that: $$1^2-2^2+3^2-4^2+...+(-1)^{n-1}n^2=\frac{(-1)^{n-1}n(n+1)}{2}$$
I proved it by induction but is there any other way to solve it?
If it was not a proof but rather a question like find the term,how to solve it?
I realized that alternate terms were under same sign but can't unde... | $$\sum_{r=1}^{2n}(-1)^{r-1}r^2=\sum_{r=1}^n\{(2r-1)^2-(2r)^2\}=\sum_{r=1}^n(1-4r)$$
$$=\dfrac n2(1-4+1-4n)=\cdots=(-1)^{2n-1}\dfrac{2n(2n+1)}2$$
Now $$\sum_{r=1}^{2n+1}(-1)^{r-1}r^2=(2n+1)^2+\sum_{r=1}^{2n}(-1)^{r-1}r^2=?$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 3
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How to compute $S_{2016}=\sum\limits_{k=1}^{2016}\left(\sum\limits_{n=k}^{2016}\frac1n\right)^2+\sum\limits_{k=1}^{2016}\frac1k$? I came across a question asking the value of the following sum:
\begin{align}
\left(1+ \frac{1}{2}+\frac{1}{3}+\cdots +\frac 1{2015}+\frac{1}{2016}\right)^2 \\
+\left(\frac{1}{2}+\frac{1}{3... | If you expand all:
$$S=1+2\times \frac{1}{2^2}+3\times \frac{1}{3^2}+\cdots +2016 \cdot \frac{1}{2016^2}+$$
$$2\cdot 1\left(1\cdot \frac12+1\cdot \frac13+\cdots 1\cdot \frac{1}{2016}\right)+$$
$$2\cdot 2\left(\frac12 \cdot \frac13+\frac12 \cdot \frac14+\cdots +\frac12 \cdot \frac{1}{2016}\right)+$$
$$\vdots$$
$$2\cdot ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2431950",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
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If $z = 2 - 3i$, show that $z^2 - 4z + 13 = 0$ and hence find the value of $4z^3 - 3z^2 + 169$ This is an example given in my book, and it seems to involve division by 0, so I'm rather confused
Here is how it's done in the book:
We have $z = 2 - 3i $
So, $z - 2 = -3i$
Now, ${(z-2)}^2 = {(-3i)}^2$
Or, $z^2 + 4 - 4z = 9... | Alt. hint: if $z = 2 - 3i$, then $z+\bar z = 2 \operatorname{Re}(z) = 4$ and $ z \bar z = |z|^2 = 2^2+3^2=13$. It follows by Vieta's formulas that $z$ and $\bar z$ are the roots of $z^2-4z+13=0\,$.
But $z^2-4z+13=0 \iff z^2 = \color{blue}{4z -13}\,$, then multiplying by $z \ne 0\,$ it follows that $z^3 = 4z^2-13 z = 4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2433671",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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Induction proof of $F(n)^2+F(n+1)^2=F(2n+1)$, where $F(n)$ is the $n$th Fibonacci number.
Let $F(n)$ denotes the $n$th number in Fibonacci sequence. Then for all $n\in\mathbb{N}$,
$$F(n)^2+F(n+1)^2=F(2n+1).$$
I know how to prove it by using the formula
$$F(n)=\frac{\left(\frac{1+\sqrt5}{2}\right)^n-\left(\frac{1-... | The base cases $n=0$ and $n=1$ are easy to verify.
Note that
\begin{align*}
F(2n+3)&=F(2n+2)+F(2n+1)=(F(2n)+F(2n+1))+F(2n+1)\\
&=F(2n)+2F(2n+1)\\
&=(F(2n+1)-F(2n-1))+2F(2n+1)\\&=
3F(2n+1)-F(2n-1).
\end{align*}
Hence, by the inductive hypothesis,
\begin{align*}
F(2n+3)&=3F(2n+1)-F(2n-1)=3(F(n)^2+F(n+1)^2)-(F(n-1)^2+F(n)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2434090",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Algebraic solution of complex equation For solving algebraically any complex equation involves two components for the real & imaginary parts. Let the real part be - $a$, imaginary part - $b$.
For the complex equation $$x^3 = 1-i $$
Substituting $x = a +bi$, we get: $$(a+bi)^3 = 1 - i.$$
Expanding the l.h.s. :
$$a^3 -(... | Someone has pointed out that one solution of $x^3 - (1-i) =0$ is $\dfrac{-1-i}{\sqrt[3]2}.$
Therefore $x+\dfrac{1+i}{\sqrt[3]2}$ is a factor of $x^3 - (1-i).$ So we can do long division:
$$
\begin{array}{cccccccccc}
& & & & x^2 & - & \dfrac{1+i}{\sqrt[3]2}x & + & i\sqrt[3]2 \\ \\
x + \dfrac{1+i}{\sqrt[3]2} & \Big) & x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2435350",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 5
} |
Show that $\lim_{n\to\infty}\frac{1}{n^4}\sum_{j=1}^{n}\left((2j-1)\sum_{k=1}^{n+1-j}k\right)=\frac{1}{12}$ Show that$$\lim_{n\rightarrow \infty} \frac{1}{n^4} \left(1\left(\sum_{k=1}^{n}k\right)+ 3\left(\sum_{k=1}^{n-1}k\right)+5\left(\sum_{k=1}^{n-2}k\right)+\cdots+(2n-1)\cdot1\right)=\frac{1}{12}$$
This is getting ... | $$\begin{align}
\sum_{r=1}^n\left[(2r-1)\sum_{k=1}^{n-r+1}k\right]
&=\sum_{r=1}^n\left[(2r-1)\sum_{s=r}^n (n-s+1)\right]
&&(s=n-k+1)\\
&=\sum_{r=1}^n\sum_{s=r}^n\sum_{t=s}^n(2r-1)\\
&=\sum_{t=1}^n\sum_{s=1}^t\sum_{r=1}^s (2r-1)
&&(1\le r\le s\le t\le n)\\
&=\sum_{t=1}^n \sum_{s=1}^t s^2\\
&=\sum_{t=1}^n \sum_{s=1}^t \b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2435490",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 2
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How to calculate $\lim_{n\to\infty}f(n)\sin(\frac{1}{n})$ where $f(x)=\int_{x}^{x^2}(1+\frac{1}{2t})^t\sin{\frac{1}{\sqrt{t}}}dt(x>0)$
$$f(x)=\int_{x}^{x^2}\left(1+\frac{1}{2t}\right)^t\sin\left(\frac{1}{\sqrt{t}}\right)dt\hspace{1cm}(x>0)$$
try to find $$\lim_{n\to\infty}f(n)\sin\left(\frac{1}{n}\right)$$
I found... | First prove that $f(n)\to\infty$ as $n\to\infty$ then apply L'Hopital rule
$$\lim_{n\to\infty}f(n)\sin\left(\frac{1}{n}\right)=\lim_{n\to\infty}\frac{f(n)}{\frac{1}{\sin\left(\frac{1}{n}\right)}}=\lim_{n\to\infty}\frac{f'(n)}{\left(\dfrac{1}{\sin\left(\frac{1}{n}\right)}\right)'}$$
$$f'(n)=2 \left(\frac{1}{2 n^2}+1\rig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2440265",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Let $a$, $b$ and $c$ be integers. Prove that if $4|(a+bc)$ and $6|(b+ac)$, then $2|(a^2-b^2)$. This is my work so far.
How can I isolate the $(a^2-b^2)$?
I am not asking for a full solution, just a hint and next step.
All done! Thanks Michael.
| $4|(a+bc)$ then $a+b c = 4k\to c=\dfrac{4 k-a}{b}$
$6|(b+ac)$ then $b+ac=6h$ plug $c$
$b + a \,\dfrac{4 k-a}{b}=6h$
$b^2+4ak-a^2=6bh$
$a^2-b^2=2 (2 a k-3 b h)$
Which means that $2|(a^2-b^2)$
QED
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2443540",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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A system of two nonlinear algebraic equations Is it possible to determine real solutions $x,y$ for the following system of nonlinear equations in terms of the real constants $a,b,k$?
$$
a=kx(1-y^2)\qquad\mbox{and}\qquad b=ky(1-x^2).
$$
If so, what is (are) the solution(s)?
| Square the first equation $a^2 = k^2x^2(1-y^2)^2$. Now $k^2(1-y^2)^2$ minus this equation gives
\begin{eqnarray*}
k^2(1-y^2)^2-a^2 = k^2(1-y^2)^2(1-x^2)
\end{eqnarray*}
Multiply this by $y$ and substitute for $ky(1-x^2)=b$ and we have
\begin{eqnarray*}
\color{red}{y(k^2(1-y^2)^2-a^2) = kb(1-y^2)^2}.
\end{eqnarray*}
Thi... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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how to solve $\int e^{7x}\cos(2x)dx$ I'm having problems to solve this integral, It looks like it will never end because every time I do integration by parts I need to do it again and again...
$$\int e^{7x}\cos(2x)dx$$
$$u=\cos(2x) \ \ \ dv = e^{7x}$$
$$du = -2\sin(2x)dx \ \ \ v = \frac{1}{7}e^{7x}$$
$$\frac {\cos(2... | You're almost there:
\begin{align}
I &= \int e^{7x}\cos(2x)\,dx \\
&= \frac {\cos(2x)e^{7x}}{7}+\frac{2}{7}\int e^{7x}(-2\sin(2x))\,dx \\
&= \frac {\cos(2x)e^{7x}}{7}+\frac{2}{7}\left(\frac {\sin(2x)e^{7x}}{7} - \frac{2}{7}\int e^{7x}\cos(2x)\,dx\right)\\
&= \frac {\cos(2x)e^{7x}}{7}+\frac {2\sin(2x)e^{7x}}{49} - \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2449019",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
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Find a and b, that there are two roots of $x^3 - 5x^2+ 7x = a$ and these numbers are roots of $x^3 - 8x + b = 0$ Find a and b, that there are two roots of $x^3 - 5x^2+ 7x = a$, that they are roots of $x^3 - 8x + b = 0$
I find that $a+b=5x(-x+3) $
Also I tried to solve second equation to get roots, depending on b, but m... | They are roots of the equation
$$x^2=3x-\frac{a+b}{5},$$ which you got.
Thus, since $x^3-8x+b=0$ has these two roots, we see that the equation
$$x\left(3x-\frac{a+b}{5}\right)-8x+b=0$$ or
$$x^2-\frac{1}{3}\left(\frac{a+b}{5}+8\right)x+\frac{b}{3}=0$$
has these two roots.
Thus, we have the following system
$$\frac{1}{3}... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proof that $\binom{n + 3}{4} = n + 3 \binom{n + 2}{4} - 3 \binom{n + 1}{4} + \binom{n}{4}$. I was trying to count the number of equilateral triangles with vertices in an regular triangular array of points with n rows. After putting the first few rows into OEIS, I saw that this was described by A000332: $\binom{n}{4} = ... | Here is a combinatorial proof, with generalization. Let $m$, $n$, and $r$ be non-negative integers such that $m\leq n$. Consider $(m+r)$-subsets of the set $S=\{1,2,\ldots,n,n+1,n+2,\ldots,n+r\}$ which contain all numbers $n+1$, $n+2$, $\ldots$, $n+r$. Such a subset must be of the form $V\cup \{n+1,n+2,\ldots,n+r\}$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2451199",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 4
} |
Value of $1/e$ with the partial sum of $e$ How to show $$1/e=1-1+\frac1{2!}-\frac1{3!}+\frac1{4!}+\dots$$
I have consider the product of the nth partial sums of the expansions for $e$ and $1/e$ but still can't prove there are equal to 1
with the definition $e=1+1+\frac1{2!}+\frac1{3!}+\dots$
| I assume you know that
$$e = 1 + 1 + \frac{1}{2!} + \frac{1}{3!} + \ldots = \sum_{k=0}^{\infty} \frac{1}{k!}.$$
Now let
$$f = 1 - 1 + \frac{1}{2!} - \frac{1}{3!} + \ldots = \sum_{k=0}^{\infty} \frac{(-1)^k}{k!}.$$
By the Mertens theorem and the binomial formula
$$\begin{align*}
e \cdot f & = \left( \sum_{k=0}^{\infty... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2452764",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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A Geometry Question Concerning Six Equilateral Triangles The question is as follows:
The figure below is built by joining six equilateral triangles ABC, ACD, CDE, DEF, EFG, and FGH, all of whose edges are 1 unit long. It is given that HIJKLMB is straight.
(a) There are five triangles in the figure that are similar to ... | Since $\Delta CMB\sim\Delta AMH$, we obtain:
$$\frac{CM}{AM}=\frac{BC}{AH}$$ or
$$\frac{CM}{AC-CM}=\frac{BC}{AH},$$
which says $$CM=\frac{1}{4}.$$
$EK=\frac{1}{2}$ by symmetry.
Now, $\Delta BLC\sim\Delta HLD.$
Thus, $$\frac{CL}{LD}=\frac{BC}{HD}$$ or
$$\frac{CL}{1-CL}=\frac{1}{2},$$
which gives $$CL=\frac{1}{3}.$$
Now,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2452957",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
An integral of a rational function with high degrees: Evaluate $\int_{-\infty}^{\infty}\frac{(x^4 + 1)^2}{x^{12} + 1}dx$.
Calculate: $$\int_{-\infty}^{\infty}\frac{(x^4 + 1)^2}{x^{12} + 1}dx.$$
What I have tried is to divide both numerator and denominator with $x^4 + 1$ and then get two following integrals, because o... |
Hint: For integrals of rational functions of even polynomials over the real line, the following substitution can come in handy:
$$\begin{align}
\mathcal{I}
&=\int_{-\infty}^{\infty}\mathrm{d}x\,\frac{\left(x^{4}+1\right)^{2}}{x^{12}+1}\\
&=\int_{-\infty}^{\infty}\mathrm{d}x\,\frac{\left(x^{4}+1\right)^{2}}{\left(x^{4}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2454663",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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In how many ways can a student score exactly $100$ points on four $50$ point exams? greater than $100$? A student takes up $4$ exams with $50$ points each. In how many ways can he score exactly $100$? Similarly, in how many ways can he score greater than $100$?
My try:
I cannot seem to fill in the $4$ possibilities. Th... | To score exactly 100, first suppose there are no constraints other than having nonnegative numbers. Then the number of solutions to $a+b+c+d = 100$ is simply $\binom{103}{3}$ by a standard stars and bars argument. Now we need only remove solutions where one of the values is greater than $50.$ Suppose $ a > 50,$ then we... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2455355",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 0
} |
Limit of $\sqrt{x^2+3x}+x$ when $x\to-\infty$ Limit of $ \lim_{x\to -\infty}(\sqrt{x^2+3x}+x)$, I know that the final answer is $-3/2$, my question is about Wolfram Alpha step by step solution:
$$x+\sqrt{x^2+3x}=\frac{(x+\sqrt{x^2+3x})(x-\sqrt{x^2+3x})}{x-\sqrt{x^2+3x}}$$
$$=-\frac{3x}{x-\sqrt{x^2+3x}}$$
$$\lim_{x\to-\... | $$
x+\sqrt{x^2+3x}=\frac{x^2-({x^2+3x})}{x-\sqrt{x^2+3x}}=\frac{-3x}{x-\sqrt{x^2+3x}}=\frac{-3}{1+\sqrt{1+\frac3x}}\stackrel{\tiny x\to-\infty}{\longrightarrow} -\frac 32
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2457183",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
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Find the arclength of the function $ x = \frac{1}{3} y^{\frac{-3}{2}} + \frac{1}{7}y^{\frac{7}{2}}$ for $1\leq y \leq 25$. I'm trying to find the arclength of the function $ x = \frac{1}{3} y^{\frac{-3}{2}} + \frac{1}{7}y^{\frac{7}{2}}$ for $1\leq y \leq 25$.
I can find the equation for the length pretty easily but I'm... | HINT:
$$1+\left(\frac{y^{5/2}-y^{-5/2}}{2}\right)^2=\left(\frac{y^{5/2}+y^{-5/2}}2\right)^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2457600",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
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A difficult integral $I=\int_0^1\sqrt{1+\sqrt{1-x^2}}\frac{dx}{1+x^2}$ How to prove
$$I=\int_0^1\sqrt{1+\sqrt{1-x^2}}\frac{dx}{1+x^2}=\sqrt{\sqrt{2}+1}\arctan\sqrt{\sqrt{2}+1}-\frac{1}{2}\sqrt{\sqrt{2}-1}\ln(1+\sqrt{2}+\sqrt{2+2\sqrt{2}})$$
$$ I=\int_0^{\pi/4}\sqrt{1+\sqrt{1-\tan^2y}}dy=\int_0^{\pi/4}\sqrt{{cosy}+\sqrt... | By enforcing the substitution $x=\sin\theta$ we are left with
$$ I = \int_{0}^{\pi/2}\sqrt{1+\cos\theta}\frac{\cos\theta}{1+\sin^2\theta}\,d\theta = \sqrt{2}\int_{0}^{\pi/2}\frac{\cos\theta\cos\frac{\theta}{2}}{1+\sin^2\theta}\,d\theta$$
and by enforcing the substitution $\theta=2\varphi$ we get:
$$ I = 2 \sqrt{2}\int_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2458865",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Given $\frac{(a-b)(b-c)(a-c)}{(a+b)(b+c)(c+a)}=\frac{1}{11}$. Find $\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}.$
Given: $\{a,b,c\}\subset \Bbb R$, $\frac{(a-b)(b-c)(a-c)}{(a+b)(b+c)(c+a)}=\frac{1}{11}\ \ (1)$. Find the value of $\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}\ \ (2).$
This question appeared in OBM 2005, the... | Set $$X = \frac{a}{a+b} + \frac{b}{b+c} + \frac{c}{a+c}$$ and $$Y = \frac{b}{a+b} + \frac{c}{b+c} + \frac{a}{a+c}.$$ Then evidently $X+Y = 3$ and we may compute that $$X - Y = \frac{a-b}{a+b} + \frac{b-c}{b+c} + \frac{c-a}{a+c} = \frac{(a-b)(b-c)(a-c)}{(a+b)(b+c)(a+c)} = \frac{1}{11}$$ Solving this system of linear equ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2461785",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solve $\frac{\left(x-7\right)}{\sqrt{x-3}+2}+\frac{\left(x-5\right)}{\sqrt{x-4}+1}=\sqrt{10}$ The answer in wolfram is 13. Any easier technique to solve this equation?
| $$\frac{x-7}{\sqrt{x-3}+2}+\frac{x-5}{\sqrt{x-4}+1}=\sqrt{10}$$
Substitute $x=y+4$
$$\frac{y-3}{\sqrt{y+1}+2}+\frac{y-1}{\sqrt{y}+1}=\sqrt{10}$$
$$\frac{y-3}{\sqrt{y+1}+2}+\frac{(\sqrt{y}+1)(\sqrt{y}-1)}{\sqrt{y}+1}=\sqrt{10}$$
$$\frac{y-3}{\sqrt{y+1}+2}+\sqrt{y}-1=\sqrt{10}$$
Substitute $y=z-1$
$$\frac{z-4}{\sqrt{z}+2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2462001",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How do you simplify an expression involving fourth and higher order trigonometric functions? The problem is as follows:
Which value of $K$ has to be in order that $R$ becomes independent from $\alpha$?.
$$R=\sin^6\alpha +\cos^6\alpha +K(\sin^4\alpha +\cos^4\alpha )$$
So far I've only come up with the idea that the solu... | $$1=(\sin^2x+\cos^2x)^3= \sin^6x+ \cos^6x+ 3\sin^2x\cos^2x(\sin^2x+\cos^2x)= \sin^6x+ \cos^6x+ 3\sin^2x\cos^2x$$
$$3\sin^2x\cos^2x=\frac{3}{2}((\sin^2x+\cos^2x)^2 - \sin^4x-\cos^4x)=\frac{3}{2}((1- \sin^4x-\cos^4x))$$
Substitute and compare with the equation you have
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2463811",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
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"answer_id": 4
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Find all natural number(s) $n$ such that $55\mid n^2 + 3n + 1$
Find all natural number(s) $n$ such that $55\mid n^2 + 3n + 1$
I see that $n^2+3n+1 =n^2-2n+1+5n\equiv n^2-2n+1=(n-1)^2 \pmod{5}$
I also see that $n^2+3n+1=n^2+3n-10+11=(n-2)(n+5)+11\equiv(n-2)(n+5) \pmod{11}$
After that what?
| There are infinitely many solutions, in particular you have the identity
$$(55t-119)^2+3(55t-119)+1=55(55t^2-235t+251)$$which shows solutions $x=55t-119$ for each integer $t$ you choose. There are maybe other solutions out of this infinity of them.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2466186",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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surface area of $x^2+y^2+z^2-14z=0$, where $0\leq x^2+y^2\leq3z$ What is the simplest way to compute the surface area of
$$x^2+y^2+z^2-14z=0,$$
where
$$0\leq x^2+y^2\leq3z\tag*{?}$$
I am having problems with various parameterizations.
| Well, $z^2-14z+3z\geq x^2+y^2+z^2-14z\geq 0$ combined with $z\geq 0$ gives $z\geq 11$.
Another form of the equation is $x^2+y^2+(z-7)^2=49$, which is a sphere centered at $(0, 0, 7)$ with radius 7.
So we need to calculate a part of the area of sphere, from $z=11$ to $z=14$, or the area of sphere $x^2+y^2+z^2 = 49$ from... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2468400",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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If $a+b+c=0$ prove that $ (a^2+b^2+c^2) \times (a^2+b^2+c^2) = 2(a^4+b^4+c^4)$
If $a+b+c=0$, for $a,b,c \in\mathbb R$, prove
$$(a^2+b^2+c^2) \times (a^2+b^2+c^2) = 2(a^4+b^4+c^4)$$
What is a good way to do this?
This question came from answering this slightly harder question. Those answers were somewhat hard to unde... | Put $c=-a-b$. Then
$$
(a^2+b^2+c^2)^2=4(a^4 + 2a^3b + 3a^2b^2 + 2ab^3 + b^4)=2(a^4+b^4+c^4).
$$
A very similar calculation was given here.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Asymptotic Analysis for two variables i have a function $ f(x,y)= \sqrt{x^2+2xy+3y^2}$ $\space$ , and $g(x,y)= \sqrt{x^2+y^2}$
I have to prove that $f$ is bounded both above and below by $g$ asymptotically $\Rightarrow f(x,y)=\Theta(g(x,y)) \Rightarrow \text{there is c and C $\in {R_{>0}}$ so that $c g(x,y)\leq f(... | $$c\sqrt{x^2+y^2}\leq \sqrt{x^2+2 x y+3 y^2}$$
square both sides
$$c^2 \left(x^2+y^2\right)\leq x^2+2 x y+3 y^2$$
expand and rearrange
$$\left(c^2-1\right) x^2-2 x y+\left(c^2-3\right) y^2\leq 0\quad(*)$$
which is true for any $x,y\in\mathbb{R}$ if
$$\left(1-\left(c^2-1\right) \left(c^2-3\right)\right) y^2<0\land c^2-1... | {
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"timestamp": "2023-03-29T00:00:00",
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Check solution of recurrence relation $a_0 = 3$, $a_1 = 7$, $a_n = 3a_{n-1} - 2a_{n-2}$ for $n \geq 2$ $a_0 = 3, a_1 = 7, a_n = 3a_{n-1} - 2a_{n-2}$ for $n \geq 2$
I know I've got the wrong answer because my outputs don't match. So if someone could show me where I'm going wrong I would be super grateful.
Let $ y = \sum... | If you are interested in solving only this particular relation, then you can do it with induction.
First, it is not difficult to gues some starting values:
$$a_1 = 3 = 2^2-1$$
$$a_2 = 7= 2^3-1$$
$$ a_3 = 3\cdot 7-2\cdot 3 = 15 = 2^4-1$$
$$ a_4 = 3\cdot 15-2\cdot 7 = 31 = 2^5-1$$
So it is reasonably to believe that $a_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2478002",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Let $r$ be a root of the polynomial $p(x) = (\sqrt{5} - 2\sqrt{3})x^3 + \sqrt{3}x - \sqrt{5} + 1$. Find another polynomial $q(x)$ with integer coefficients such that $q(r) = 0$.
I have no clue how to do this question. Can't use rational root theorem and I see no feasible way to get the roots of $p(x)$. Any help would ... | $p(x)=(\sqrt{5}-2\sqrt{3})x^3+\sqrt{3}x-\sqrt{5}+1=\sqrt{5}(x^3-1)+\sqrt{3}(x-2x^3)+1$
Let's set : $\begin{cases}a=x^3-1\\b=x-2x^3\\p(x)=\sqrt{5}a+\sqrt{3}b+1\end{cases}$
For $p(x)=0$ we have $5a^2+3b^2+2\sqrt{15}ab=(-1)^2=1$
And then $60a^2b^2=(1-5a^2-3b^2)^2$
Finally substitute $a,b$ with their values, I find the sam... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2478954",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
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How to Simplify $\sqrt {-X} \times \sqrt {-Y}$ For example : $\sqrt {-18} \times \sqrt {-12}$
Would I start by multiplying the 2 numbers under a square root, In which case the double negatives cancel out?
$ \sqrt {-18\times-12} = \sqrt {216} = 6\sqrt {6}$
Or get the i out of the square roots in which case I get a $i^... | Simply
$$i\sqrt{18}\cdot i\sqrt{12}$$
$$-1\sqrt{12\cdot 18}$$
$$-1\sqrt{216}$$
$$-6\sqrt{6}$$
ALWAYS start with $i$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2481505",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
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How to find the maximum and minimum values of $\frac{8x(x^2-1)}{(x^2+1)^2}$ algebraically? The function is $f(x) = \frac{8x(x^2-1)}{(x^2+1)^2}$.
I have tried using calculus, only to fail.
| $$f(x) = \frac{8x(x^2-1)}{(x^2+1)^2}={8(x^3-x)}{(x^2+1)^{-2}}\\f'(x)=8\left(\left(x^3-x\right)\left(-\dfrac{4x}{\left(x^2+1\right)^3}\right)+\left(3x^2-1\right){(x^2+1)^{-2}}\right)\\=
8\left(-\dfrac{4x^4-4x^2}{\left(x^2+1\right)^3}+\frac{3x^2-1}{(x^2+1)^{2}}\right)
\\=
8\left(\dfrac{4x^2-4x^4}{\left(x^2+1\right)^3}+\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2483031",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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How many solutions does the equation $a + b + c = 10$ if $(1, 4, 5)$ and $(1, 5, 4)$ are not considered distinct?
Question: What if we consider (1,4,5) and (1,5,4) as non-distinct possibilities, then what should we do?
$${{9}\choose{2}}-2\cdot\frac{{9}\choose{2}}{3}$$
| Since solutions obtained by a permutation are considered the same we may assume $1\leq a\leq b\leq c$ to begin with. We therefore put
$$a=1+x_1,\quad b=a+x_2=1+x_1+x_2,\quad c=b+x_3=1+x_1+x_2+x_3\ ,$$
whereby $x_i\geq0$ $\>(1\leq i\leq3)$ and
$a+b+c=3+3x_1+2x_2+x_3=10$,
or $$3x_1+2x_2+x_3=7\ .$$
If $x_1=0$ then $2x_2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2485997",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Let $\alpha $, $\beta $ are the roots of $3x^2+x+5=0$ Let $\alpha $, $\beta $ are the roots of $3x^2+x+5=0$ then find a quadratic equation with roots as $\dfrac {\alpha +1}{\alpha -3}$ and $\dfrac {\beta +1}{\beta -3}$.
I got this solution but didn't understand what procedure is used, what concepts are used and how'... | For a different route, try this.
Let $a=\dfrac {\alpha +1}{\alpha -3}$ and $b=\dfrac {\beta +1}{\beta -3}$.
An equation with roots $a$ and $b$ is $(x-a)(x-b)=0$, or $x^2-(a+b)+ab=0$.
Now
$$
a+b=\frac{2 (\alpha \beta - \alpha - \beta - 3)}{\alpha \beta - 3 \alpha - 3 \beta + 9},
\quad
ab = \frac{\alpha \beta + \alpha + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2487783",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
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Number of sequences that have at least five consecutive positions in which the numbers are in increasing order? A sequence of numbers is formed from the numbers $1, 2, 3, 4, 5, 6, 7$ where all $7!$ permutations are equally likely. What is the probability that anywhere in the sequence there will be, at least, five conse... | This solution does not differ in any essential way from that of N. Shales. I am posting this here so N. Shales can compare our approaches.
Since the sequence contains seven numbers, any block of five consecutive increasing numbers must start in the first, second, or third positions. Let $A_1$, $A_2$, and $A_3$ denote... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2489988",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Does in plane exist $22$ points and $22$ such circles that each circle contains at least $7$ points and each point is on at least $7$ circles.
Does in plane exist $22$ points and $22$ such circles that each circle contains at least $7$ points and each point is on at least $7$ circles.
I have solved this one but now ... | As we already see in the introduction each point is on $7$ circles and each circle contains $7$ points. Also every two circles meet at $2$ points.
Let $A$ be an incident matrix, so $a_{ij} = 1$ if $P_j \in C_i$, else $a_{ij} =0$. Let $M:= A\cdot A^T$, then determinant of
$M$ is a perfect square: $\det (M) =\det (A\cdo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2490876",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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Find $\lim_{x \to 0} (\frac {\sin x}{x})^{\frac{1}{\sin^2x}}$ Is there any other way to compute the problem $$\lim_{x \to 0} (\frac {\sin x}{x})^{\frac{1}{\sin^2(x)}}$$
I try to use L'Hospital rule but it's too complicated.
| In the same spirit as other answers but assuming that you want to go beyond the limit.
$$A=\left(\frac {\sin (x)}{x}\right)^{\frac{1}{\sin^2(x)}}\implies \log(A)={\frac{1}{\sin^2(x)}}\log\left(\frac {\sin (x)}{x}\right)$$ Using Taylor for one more term
$$\frac {\sin (x)}{x}=1-\frac{x^2}{6}+\frac{x^4}{120}+O\left(x^6\ri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2491833",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Evaluate $\lim_{n\rightarrow \infty}4\sqrt{n}\sin (\pi\sqrt{4n^2+\sqrt{n}})$ Evaluate the limit
$$\lim_{n\rightarrow \infty}4\sqrt{n}\sin (\pi\sqrt{4n^2+\sqrt{n}})$$
We know that :
$$\sin (\pi-x)=\sin x$$
So we have :
$$\lim_{n\rightarrow \infty}4\sqrt{n}\sin (\pi-\pi\sqrt{4n^2+\sqrt{n}})$$
Now :
$$(\pi-\pi\sqrt{4n^2+... | I think the following is better.
$$\lim_{n\rightarrow+\infty}4\sqrt{n}\sin (\pi\sqrt{4n^2+\sqrt{n}})=-\lim_{n\rightarrow+\infty}4\sqrt{n}\sin \left(2\pi n-\pi\sqrt{4n^2+\sqrt{n}}\right)=$$
$$=\lim_{n\rightarrow+\infty}4\sqrt{n}\sin\frac{\pi\sqrt{n}}{2 n+\sqrt{4n^2+\sqrt{n}}}=\lim_{n\rightarrow+\infty}4\sqrt{n}\sin\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2493385",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
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Factoring a quartic expression
We have a function $$f(x)=x^{\frac{2}{3}}-\frac{2}{x^{\frac{1}{3}}}+1$$ and let line segment $AB$ be represented by $g(x)$ such that $$g(x)=x-4$$
Find the greatest positive real solution representing the intersection between the two lines.
The first step is obviously to let them equal... | Perhaps it would be easier if you write $t=x^{1\over 3}$. Then the equation you get is $$t^2-{2\over t}+1 = t^3-4 \;\;\;\Longrightarrow \;\;\; t^4-t^3-5 t+2=0$$
Last one can be factor like this $$(t-2)(t^3+t^2+2t-1)=0$$
So $t=2$ and for $t>2$ we get $$t^3+t^2+2t-1 \geq 8+4+4-1 > 0$$
So $t=2$ and $x=8$ is a maximum solu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2494671",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Summation over roots of unity Find the value of $\displaystyle\sum_{r=1}^{4} \frac{1}{2-\alpha^r} $
where $ \alpha^k (k=0,1,2,3,4,5) $ are fifth roots of unity.
My approach:-
As we know that $ \alpha^k (k=0,1,2,3,4,5) $ are fifth roots of unity, then $ \alpha^k - 1$ should be equal to zero. Therefore the final answer ... | Since $\alpha_0,\alpha_1,\alpha_2 \dots \alpha_{4}$ are roots of the equation
$$x^5-1=0 \tag1$$
You can apply Transformation of Roots to find a equation whose roots are$$\frac{1}{2-\alpha_0} , \frac{1}{2-\alpha_1},\dots \frac{1}{2-\alpha_{4}}$$
Let $P(y)$ represent the polynomial whose roots are $\frac{1}{2-\alpha_k}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2497816",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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Solve $\lfloor \sqrt x +\sqrt{x+1}+\sqrt{x+2}\rfloor=x$ $$\lfloor \sqrt x +\sqrt{x+1}+\sqrt{x+2}\rfloor=x$$
I tried to solve this equation.
First thing is $\lfloor \sqrt x +\sqrt{x+1}+\sqrt{x+2}\rfloor \in \mathbb{Z} $ so $x \in \mathbb{Z}$
second $$\sqrt x +\sqrt{x+1}+\sqrt{x+2} \geq \sqrt 0 +\sqrt{0+1}+\sqrt{0+2}... | This is a little plodding, but $x\in\mathbb{Z}$ and $x\ge0$ (required in order for $\sqrt x$ to be real) tells us
$$x\ge\lfloor\sqrt0+\sqrt1+\sqrt2\rfloor=\lfloor2.414\rfloor=2$$
which tells us
$$x\ge\lfloor\sqrt2+\sqrt3+\sqrt4\rfloor=\lfloor5.145\rfloor=5$$
which tells us
$$x\ge\lfloor\sqrt5+\sqrt6+\sqrt7\rfloor=\lflo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2499746",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Prove that $1^2 + 2^2 + \cdots + n^2 = \frac{n(n+1)(2n+1)}{6}$ without using induction. I have to deduce the following formula $$1^2 + 2^2 + \cdots + n^2 = \frac{n(n+1)(2n+1)}{6},$$ while using the given formula $$\binom{k}{0}+\binom{k+1}{1}+\cdots+\binom{k+r}{r}=\binom{k+r+1}{r}$$
I tried to find values for $k$, such ... | Using your identity, we can write
$$\binom{k}{k} + \binom{k+1}{k} + \cdots + \binom{k+r}{k} = \binom{k+r+1}{k+1}$$
and take $k=2$ to give
$$\binom{2}{2} + \binom{3}{2} + \cdots + \binom{r+2}{2} = \binom{r+3}{3}$$ but we know that $\binom{k}{2} = \frac{k(k-1)}{2}$, so
$$\begin{align}\binom{r+3}{3} &= \sum_{i=0}^r \bino... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2500099",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Compute $\zeta(6)$ using Fourier series How can i prove that $\sum_{n=1}^\infty \frac{1}{n^6} = \frac{\pi^6}{945}$ knowing that $f(t)=t^2 = \frac{\pi^2}{3} + 4\sum_{n=1}^\infty \frac{(-1)^n}{n^2}\cos(nt)$ on $(-\pi, \pi)$. I've integrated four times but i don't get anywhere. Thank's for helping.
| Starting with
$$\sum_{n=1}^{\infty} \frac{(-1)^{n}}{n^2} \, \cos(n t) = \frac{t^{2}}{4} - \frac{\zeta(2)}{2}$$
then integrate from $0$ to $t$ to obtain
$$\sum_{n=1}^{\infty} \frac{(-1)^{n}}{n^3} \, \sin(n t) = \frac{t^{3}}{2 \cdot 3!} - \frac{\zeta(2) \, t}{2} + c_{0}.$$
When $t=0$ it is determined that $c_{0} = 0$. In... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2502046",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Prove by induction that $\sum_{i=1}^{n} i \cdot 2^i = (n-1) \cdot 2^{n+1} +2$. Help finding my mistake
Prove by induction:
$$\sum_{i=1}^{n} i \cdot 2^i = (n-1) \cdot 2^{n+1} +2$$
Basis: let $p(n)$ be the predicate.
Let $n=1$ this gives $(1-1) \cdot 2^1+1+2 = 2$ and $1 \cdot 2^1 = 2$ so its true for $p(1)$
Induction... | The following is strange:
when $n=k+1$:
$$\sum_{i=1}^{k+1} i \cdot 2^i = \sum_{i=1}^{k} i \cdot 2^i + \color{red}{((k+1)-1) \cdot 2^{(k+1)+1} +2}$$
$$\sum_{i=1}^{k+1} i \cdot 2^i = \left(\sum_{i=1}^{k} i \cdot 2^i \right)+\color{blue}{(k+1)2^{k+1}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2502457",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Can someone explain the steps of this Partial fraction decomposition?
My thoughts:
$$\frac{(Ax + B)}{(x^2+1)} + \frac{(Cx + D)}{(x^2+4)} = \frac{x}{(x^2+1)(x^2+4)}$$
I combined the left terms
Set the numerator of the combined left term to "x" which is the numerator of the right term
I got
$$4Ax + Cx = x$$
I am not sur... | You have only found the equation for $x$. The full set of equations is derived as follows:
$$(Ax+B)(x^2+4)+(Cx+D)(x^2+1)\equiv x$$
So compare coefficients of each of the powers of $x$:
$$(A+C)x^3\equiv 0\\(B+D)x^2\equiv 0\\(4A+C)x\equiv x\\(4B+D)1\equiv 0$$
Solving this gives $A=-C,B=-D$. So $$4A-A=1\implies A=\frac13... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2503220",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Representing a unit speed curve on a sphere in terms of its Frenet Frame Let $\alpha$ be a unit speed curve with positive curvature $\kappa \gt 0$ and non-zero torsion $\tau \ne 0$, lying on a sphere of radius $r$ centred at $c \in \Bbb{R}^3$.
Show that $\alpha - c = -\frac{1}{\kappa}N - (\frac{1}{\kappa})'\frac{1}{\ta... | Well, we have
$(\alpha - c) \cdot (\alpha - c) = r^2, \tag 1$
since $\alpha$ lies on the sphere of radius $r$ centered at $c$; this is just what equation (1) affirms. If we differentiate (1) with respect to $s$, the arc-length along $\alpha$, we obtain
$2\alpha' \cdot (\alpha - c) = \alpha' \cdot (\alpha - c) + (\alp... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2504918",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
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Equivalent method for solving lyapunov equality I am trying to show that the lyapunov equality:
$A^{T}P + PA = -Q$
Where $Q =
\begin{bmatrix}
1 & 0 \\
0 & 1 \\
\end{bmatrix}
$
is equivalent to solving:
$M
\begin{bmatrix}
P_{11}\\
P_{12}\\
P_{22}\\
\end{bmatrix}
= v
$
The stated solution is:
... | If in your case the Lyapunov equality can be written as
$$
\begin{bmatrix}
-4\,P_{11} - P_{12} - P_{21} & P_{11} - 2\,P_{12} - P_{22} \\
P_{11} - 2\,P_{21} - P_{22} & P_{12} + P_{21}
\end{bmatrix} = -
\begin{bmatrix}
Q_{11} & Q_{12} \\ Q_{21} & Q_{22}
\end{bmatrix} \tag{1}
$$
then by stacking the columns of $(1)$ on to... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2505902",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Arithmetic and geometric progression - 3 terms Find $3$ numbers $x, y, z$ which are consecutive terms of a geometric series, if $xy$, $yz$, $zx$ and $xyz$ are consecutive terms of an arithmetic series.
OK $y=xa$ and $z=xa^2$.
Also $yz=xy+b$
$zx=xy+2b$ and
$xyz=xy+3b$
So by substituting we get: $a=0$ (rejected) and $a=-... | If $x = a$, $y = ar$ and $z = ar^2$, $(a,r \ne 0)$, then
$xy = a^2r$, $yz = a^2r^3$, $zx = a^2 r^2$ and $xyz = a^3r^3$
Since $xy$, $yz$ and $zx$ are in AP.
$$yz = \dfrac{xy + zx}{2} \iff a^2r^3 = \dfrac{a^2r + a^2r^2}{2} \implies 2r^3 = r+ r^2 \implies r(2r^2 - r - 1) = 0$$
So $r = 1, -\dfrac12, 0$.
Also, $$zx=\dfrac{x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2510372",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How do I simplify $\frac{bH}{a\sqrt{a^2+H^2}}$ to $\frac{b}{a\sqrt{a^2H^{-2}+1}}$?
I have trouble following the algebraic simplification between step 2 and 3.
How do I simplify $\frac{bH}{a\sqrt{a^2+H^2}}$ to $\frac{b}{a\sqrt{a^2H^{-2}+1}}$?
| There you go
$$\frac{bH}{a\sqrt{a^2+H^2}}$$
$$=\frac{bH}{a\sqrt{(H^2)(\frac{a^2}{H^2}+1)}}$$
$$=\frac{bH}{aH\sqrt{\frac{a^2}{H^2}+1}}$$
$$=\frac{b}{a\sqrt{a^2H^{-2}+1}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2513749",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Find the number of real roots for $x+\sqrt{a^2-x^2}=b$, $a>0$, $b>0$, as a function of $a$ and $b$
Given: (1) $x+\sqrt{a^2-x^2}=b$, $(a,b,x)\subset \mathbb R$, $a>0$, $b>0$.
Find: number of roots for (1), given possible values for $a$ and $b$.
This is a question from a book for the preparation for math contests.
It s... | As you wrote, we have
$$2x^2-2bx+b^2-a^2=0$$
with$$a^2-x^2\ge 0\quad\text{and}\quad b-x\ge 0,$$
i.e.
$$-a\le x\le a\quad\text{and}\quad x\le b\tag1$$
Here, let us separate it into cases :
*
*Case 1 : If $(a\sqrt 2\gt)\ a\gt b$, then $(1)\iff -a\le x\le b$. We have $\frac{b+\sqrt{2a^2-b^2}}{2}\gt \frac{b+\sqrt{2b^2-... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 4
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Evaluate the following line integral: $\int_l\sqrt{x^2+y^2}dl$ Evaluate the following line integral:
$$\int_l\sqrt{x^2+y^2}dl$$ where $$l:x^2+y^2=ax$$
What I've already done is:
$$x^2+y^2=ax \Rightarrow \left( x-\frac{a}{2} \right)^2+y^2=\left(\frac{a}{2}\right)^2$$
Spherical coordinates:
$$
\begin{cases}
x-\frac{a}{2}... | Your parameterization is making it very hard for you. I would recommend using straight polar coordinates. You should try that on your own before reading this solution.
Letting our path $l=r$, we then have
$$\int_l\sqrt{x^2+y^2}\,dl=\int_rr\,dr$$
Note that the path $x^2+y^2=ax$ becomes
$$r^2=ar\cos\theta$$
which gives ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How can find all solution of $x^2(y-1)+y^2(x-1)=1$ such that $x,y \in \mathbb{Z}$?
How can find all solution of $x^2(y-1)+y^2(x-1)=1$ such that $x,y \in \mathbb{Z}$?
I have no clue to find solution .
I tried to change into quadratic equation $$y^2(x-1)+yx^2-(x^2+1)=0\\y=\frac{-y\pm\sqrt{x^4+4(x-1)(x^2+1)(-1)}}{2(x-... | Write like this:
$$ xy (x+y)-(x+y)^2+2xy =1$$
If we put $a=x+y$ and $b=xy$ we get $$ b ={1+a^2\over a+2}\;\;\;\Longrightarrow \;\;\;a+2\mid a^2+1$$
Now since $a+2\mid a^2-4$ we have $a+2\mid 5$ so $a+2\in\{\pm 1,\pm 5\}$ and thus $a\in\{-3,-1,-7,3\}$
\begin{array}{c|c|c|c}
a & b & x&y\\\hline
-7 & -10 &/&/\\
-3 ... | {
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"timestamp": "2023-03-29T00:00:00",
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Limit of $f(x)$ when $x$ goes to zero Let $f(x) = \frac{1 + \tan x + \sin x - \cos x}{\sin^2 x + x^3}$ . Find value of $\lim_{x \to 0} f(x)$ if it exists . I can solve it using L'Hospital's Rule and Taylor series but I'm looking for another way suing trigonometric identities .
| $$\lim_{x\rightarrow0}\frac{1 + \tan x + \sin x - \cos x}{\sin^2 x + x^3}=\lim_{x\rightarrow0}\frac{2\sin^2\frac{x}{2} +2\sin\frac{x}{2}\cos\frac{x}{2}\left(\frac{1}{\cos{x}}+1\right) }{\sin^2 x + x^3}=$$
$$=\lim_{x\rightarrow0}\left(\frac{\sin\frac{x}{2}}{\frac{x}{2}}\cdot\frac{\sin\frac{x}{2} +\cos\frac{x}{2}\left(\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2520013",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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GCD of $n^3+3n+1$ and $7n^3+18n^2-n-2$ is always $1$ A problem I found yesterday says to prove $\gcd(n^3+3n+1, 7n^3+18n^2-n-2)=1$ for all integers $n\ge 1$. To begin, I used the Euclidean algorithm to observe that $7n^3+18n^2-n-2=7\left(n^3+3n+1\right)+\left(18n^2-22n-9\right)$, so $$\gcd(n^3+3n+1, 7n^3+18n^2-n-2)=\gcd... | HINT: Because both terms are odd, you know that
$$\gcd(n^3+3n+1,18n^2-22n-9)=\gcd(2(n^3+3n+1),18n^2-22n-9).$$
A similar argument works for the prime $3$, and perhaps for $3^2$. Can you take it from here?
| {
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"url": "https://math.stackexchange.com/questions/2520554",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find value of $a_{2012}$ A sequence $\left\{a_n\right\}$ is defined as:
$a_1=1$, $a_2=2$ and
$$a_{n+1}=\frac{2}{a_n}+a_{n-1}$$ $\forall$ $n \ge 2$
Find $a_{2012}$
My Try:
we have
$$a_{n+1}-a_{n-1}=\frac{2}{a_n}$$
$$a_n a_{n+1}-a_{n-1}a_n=2 \tag{1}$$
Replacing $n$ with $n-1$ we get
$$a_{n-1} a_{n}-a_{n-2}a_{n-1}=2 \ta... | By your work we obtain: $$a_{n+2}a_{n+1}=a_{n+1}{a_n}+2,$$
which gives $$a_{n+1}a_n=2+(n-1)2=2n.$$
Thus, $$a_{n}=\frac{2n-2}{\frac{2n-4}{a_{n-2}}}=\frac{n-1}{n-2}a_{n-2},$$
which for even $n$ gives
$$a_n=\frac{(n-1)(n-3)...1}{(n-2)(n-4)...2}a_2=\frac{2(n-1)!!}{(n-2)!!}.$$
Id est, $$a_{2012}=\frac{2\cdot2011!!}{2010!!}.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2524929",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
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Solution to a cubic-like polynomial with fractional powers I wish to solve the following equation for x, but am not certain how:
$\frac{(1+x)^{1/3}}{1+x^{1/3}} = (\frac{3^{1/3}}{.691})(R/a)(p/h)^{-1/3}$
I suspect that there is some way to manipulate this into a cubic equation and solve by the usual methods, but despite... | Set $x^{1/3}=z\to x=z^3$
and $k=(\frac{3^{1/3}}{.691})(R/a)(p/h)^{-1/3}$
the equation becomes
$$\frac{\sqrt[3]{z^3+1}}{z+1}=k$$
$$\frac{z^3+1}{(z+1)^3}=k^3$$
$$\frac{(z+1)(z^2-z+1)}{(z+1)^3}=k^3$$
$$z^2-z+1=k^3(z^2+2z+1)$$
$$z=\frac{-2 k^3-1\pm\sqrt{12 k^3-3}}{2 \left(k^3-1\right)}$$
$$x=z^3\to x=\left(\frac{-2 k^3-1\p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2525294",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Calculating infinite sums of power series. This past week I have been studying power series, first I studied how to determinate the intervals of convergence and I have no problem doing that (usually I just have to apply the root or ratio test of convergence). However, now I am asked to calculate the sums of:
I know so... | \begin{align}
\sum_{n=1}^\infty \frac{n(x+3)^n}{2^n} & = (x+3)\sum_{n=1}^\infty \frac{n(x+3)^{n-1}}{2^n} = (x+3)\sum_{n=1}^\infty \frac d {dx} \, \frac{(x+3)^n}{2^n} \\[10pt]
& = (x+3) \frac d {dx} \sum_{n=1}^\infty \frac{(x+3)^n}{2^n} \qquad \text{See the comment on this step below.} \\[10pt]
& = (x+3) \frac d {dx} \,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2526253",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
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How to simplify an expression I have an expression:
$$\frac{\sqrt{6} + 1}{6} - \frac{\sqrt{10-4\sqrt{6}}}{6}$$
and it seems like it must be equal to $\frac{1}{2}$. How could i simplify this?
| It is : $(\sqrt{6}-2)^2=10-4\sqrt{6}$ so you get :
$$\frac{\sqrt{6} + 1}{6} - \frac{\sqrt{10-4\sqrt{6}}}{6} = \frac{\sqrt{6} + 1}{6}-\frac{\sqrt{(\sqrt6 - 2)^2}}{6}= \frac{\sqrt{6} + 1}{6}- \frac{\sqrt6-2}{6} = \frac{3}{6} =\frac{1}{2}$$
Note that : $\sqrt{6}-2 > 0 $ and that's why you remove the root simply enough.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2526490",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Given $ \frac{1}{r}+\frac{1}{s}=a; \frac{1}{r}\times\frac{1}{s}=b; a+b=r; a\times b=s$, find $a$. (Brazilian Math Olympics, 2016)
Given: $\{a,b,r,s\}\subset \mathbb R$, $a>0$,
$\frac{1}{r}$, $\frac{1}{s}$ are roots for $x^2-ax+b=0$, and
$a$,$~b$ are roots for $x^2-rx+s$.
Find: the numeric value of $a$.
This is qu... | $$\frac{1}{r}+\frac{1}{s}=a\tag1$$
$$\frac{1}{r}\times\frac{1}{s}=b\tag2$$
$$a+b=r\tag3$$
$$a\times b=s\tag4$$
Substituting $(3)(4)$ into $(1)(2)$ gives
$$(1)\implies\frac{1}{a+b}+\frac{1}{ab}=a\implies ab+a+b=a^2b(a+b)=a^3b+a^2b^2\tag5$$
$$(2)\implies \frac{1}{a+b}\times\frac{1}{ab}=b\implies 1=ab^2(a+b)\tag6$$
From $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2527552",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Set of solutions to a sum I want to know the set of integer solutions to the following sum:
$$\sum\limits_{n=1}^{k}\frac{n}{a_n}=1$$
For instance, the series
$a_n=\frac{1}{nk}$ satisfies this for all integer $k$. So far, this is the only solution that I have come up with for a general $k$.
Here are some solutions th... | Yet another generic solution:
since $$\sum_{n=1}^{k-2} \frac{n}{2^{n+1}}=1-\frac{k}{2^{k-1}},$$
we can write the fraction $\dfrac{k}{2^{k-1}}$ in the form
$$\frac{k}{2^{k-1}}=\frac{k-1}{2^{k-1}}+\frac{1}{2^{k-1}} = \color{red}{\frac{k-1}{2^{k-1}}}+\color{darkviolet}{\frac{k}{k\cdot 2^{k-1}}}.$$
Then for given $k$ ($k\g... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2531306",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
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>Does $\{f_n\}$ converge pointwise to a function on $[0,\infty)?$ Let $g_n(x)=\sin^2(x+\frac{1}{n}),x\in (0,\infty)$
and $f_n(x)=\int _0^x g_n(t)\, dt$.
Does $\{f_n\}$ converge pointwise to a function on $[0,\infty)?$
I try to show that $$\int_0^x \sin^2\left(t +\frac1n\right) \, dt = \frac{x}{2} - \frac{1}{4}\sin\l... | You have correctly calculated $f_n$. Now just let $n\to\infty$:
$$f_n(x) = \frac{x}{2} - \frac{1}{4}\sin\left(2x+\frac{2}n\right) + \frac{1}{4} \sin \left(\frac2n\right)= \frac{x}2 - \frac{1}2\cos\left(x+\frac2n\right)\sin x \xrightarrow{n\to\infty} \frac{x}2 - \frac12\cos x\sin x$$
So,
$$\lim_{n\to\infty} f_n(x) = \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2532811",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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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.