Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
How to add constraints to vectors Is it possible to add constraints to vectors, more specifically, planes? For instance:
In this example, is it possible to constrain the red plane with Cartesian equation $z=0$ to only occupy within the area within the area of the yellow lines, such that only the green line intersects ... | So you want to parametrize the plane such that a point on the plane is defined by two parameters $$ \vec{r} = \mathbf{ S}(u,v) $$
and then set limits to $u$ and $v$ to define your bounds. In case of a plane, you need a local coordinate system with two mutually orthogonal directions $\mathbf{e}_1$ and $\mathbf{e}_2$ tha... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3321078",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Calculus - indefinite integration The integral in which I am interested in is
$$\int x(x^3+1)^{33}\mathrm{d}x$$
I tried to solve by substituting $x^2 = t$, but it didn't help. I find a solution by expanding it with the help of binomial expansion. Can anyone help me with any other method like substitution, by parts?
| This is not easier than expanding using the Binomial theorem, but it's a different way to approach it which you may at least find interesting, and even potentially useful (in other situations if not this one).
For any integer $n \ge 0$, let
$$f(n) = \int x(x^3 + 1)^n dx \tag{1}\label{eq1}$$
For $n \ge 1$, using integra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3322474",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Evaluate $\lim_{n\to\infty} \frac{\left(\frac{3}{2}\right)^{2n}}{\left(\frac{2}{3}\right)^{n-1}+\left(\frac{3}{2}\right)^{2n+1}}$ Evaluate $\lim_{n\to\infty} \frac{\left(\frac{3}{2}\right)^{2n}}{\left(\frac{2}{3}\right)^{n-1}+\left(\frac{3}{2}\right)^{2n+1}}$.
My approach is to do $\lim_{n\to\infty} \frac{\left(\frac{3... | A good general rule: divide numerator and denominator by the largest term:
$$\frac{\left(\frac{3}{2}\right)^{2n}}{\left(\frac{2}{3}\right)^{n-1}+\left(\frac{3}{2}\right)^{2n+1}}
=\frac{\frac{2}{3}}{\left(\frac{2}{3}\right)^{3n}+1}$$
and I think you should now be able to see what happens as $n\to\infty$.
Good luck!
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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How to isolate $x$ when $\cos(x/2)\cos(3x)/2−3\sin(x/2)\sin(3x)= 0$ Asked to find all relative and absolute extrema of $f(x) = \sin\left(\frac{1}{2}x\right)\cos(3x)$ on the interval $[0,\pi]$. I've gotten the derivative, $$f'(x)=\frac{\cos(\frac{x}{2})\cos(3x)}{2} - 3\sin(\frac{x}{2})\sin(3x),$$ just fine, but how wo... | Expanding and varying my own comment:
$$\begin{split}
f'(x)&=\tfrac{1}{2}\cos\tfrac{x}{2}\cos 3x - 3\sin\tfrac{x}{2}\sin 3x\\
&=\tfrac{1}{2}\cos\tfrac{x}{2}\left(\cos 3x - 6\frac{\sin\tfrac{x}{2}\sin x}{\cos\tfrac{x}{2}} \frac{\sin 3x}{\sin x}\right)\\
&=\tfrac{1}{2}\cos\tfrac{x}{2}\left(\cos 3x - 6(1-\cos x) \frac{\si... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3323680",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Evaluate $ \int \frac{3x^{2} (\csc{\sqrt{x^{3}-6}})^{2}}{\sqrt{x^{3}-6}} dx $ $$ \int \frac{3x^{2} (\csc{\sqrt{x^{3}-6}})^{2}}{\sqrt{x^{3}-6}} dx = ? $$
Attempt:
$$ \int \frac{3x^{2} (\csc{\sqrt{x^{3}-6}})^{2}}{\sqrt{x^{3}-6}} dx = \int \frac{3x^{2}}{\sqrt{x^{3}-6} \sin^{2}\sqrt{x^{3}-6}} dx $$
$$ U = \sqrt{x^{3}-6} $... | When you substitute $U=\sqrt{x^3-6}$, you made a mistake.
It should be $$\int{\dfrac{2U}{U\sin^2{U}}dU}=\int{2\csc^2{U}dU}$$, so the answer is
$$-2\cot^2{\sqrt{x^3-6}}+C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3323749",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $a+b+c+d = 4m+1$ where $m$ is a positive number and $a , b,c,d \in \mathbb Z$ , then the minimum value of $a^2+b^2+c^2+d ^2$ is $4m^2 +2m +1$. If $a+b+c+d = 4m+1$ where $m$ is a positive number and $a , b,c,d \in \mathbb Z$ , then the minimum value of $a^2+b^2+c^2+d ^2$ is $4m^2 +2m +1$.
How to prove it without usin... | Wlog. $a\le b\le c\le d$. If $d\ge a+2$, then
$$ (a+1)^2+b^2+c^2+(d-1)^2=a^2+b^2+c^2+d^2+2(1+d-a)<a^2+b^2+c^2+d^2,$$
hence for a minimizer $d\le a+1$. If $k$ of the numbers $a,b,c,d$ are $=a+1$, then $4m+1=a+b+c+d=4a+k$. We conclude $k=1$, and then $a=b=c=m$ and $d=m+1$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Verify the following limit using epsilon-delta definition: $ \lim_{(x,y)\to(0,0)}\frac{x^2y^2}{x^2+y^2}=0$ Show that $$ \lim\limits_{(x,y)\to(0,0)}\dfrac{x^2y^2}{x^2+y^2}=0$$
My try:
We know that, $$ x^2\leq x^2+y^2 \implies x^2y^2\leq (x^2+y^2)y^2 \implies x^2y^2\leq (x^2+y^2)^2$$
Then, $$\dfrac{x^2y^2}{x^2+y^2}\leq x... | Tips
$\lim\limits_{\left(x,y\right)\rightarrow\left(0,0\right)} \dfrac{x^2 y^2}{x^2+y^2} = \lim\limits_{\left(x,y\right)\rightarrow\left(0,0\right)} \dfrac{1}{\frac{1}{x^2}+\frac{1}{y^2}}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3325340",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 3
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Find the sum of the series $1^3 + 3\cdot 2^2 + 3^3 + 3\cdot 4^2 + 5^3 + 3\cdot 6^2...$ up to $n$ terms
Find the sum of first $n$ terms of the series $1^3 + 3\cdot 2^2 + 3^3 + 3\cdot 4^2 + 5^3 + 3\cdot 6^2...$
*
*When $n$ is even.
*When $n$ is odd.
This sum can be written as
$$\sum_{1}^n (2k-1)^{3} +3... | HINT
When $n = 2m$ is even, both sums have the same amount of terms, $n/2 = m$ each.
When $n = 2m-1$ is odd, the left sum has one more term than the right, so there must be $m$ terms in the left and $m-1$ in the right.
Also notice that the even $n$ sum and the odd $n$ sum are different by just one last term in the righ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3327581",
"timestamp": "2023-03-29T00:00:00",
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Discrete Mathematics - Combinatorics proof $\\$ I need to prove $\displaystyle\sum\limits_{k=0}^n\left(\frac{ 1^k+(-1)^k}{2}\right)\cdot\left(\begin{array}{c}n\\ k\end{array}\right)4^k=\frac{5^n+(-3)^n}{2}$
my Attempt below
$\displaystyle\sum\limits_{k=0}^n\left(\frac{ 1^k+(-1)^k}{2}\right)\cdot \left(\begin{array}{c}... | $\sum\limits_{k=0}^n 1^{k} 4^{k}\binom {n} {k}$ is the expansion of $(1+4)^{n}$ and $\sum\limits_{k=0}^n (-1)^{k} 4^{k}\binom {n} {k}$ is the expansion of $(1-4)^{n}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3328751",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Is $\frac{\frac{1}{b}-\frac{1}{2}}{b-2}$ the same as $\frac{\frac{2-b}{2b}}{b-2}$? Have I subtracted 2 fractions correctly?
$\frac{\frac{1}{b}-\frac{1}{2}}{b-2}$
Starting with the numerator which is a difference of fractions, the least common denominator is 2b? So:
$\frac{2(1)}{2b}-\frac{b(1)}{2b}$ = $\frac{2}{2b}-\fra... | $\require{cancel}$Yes, it is right. And now:$$\frac{\frac{2-b}{2b}}{b-2}=-\frac{\frac{\cancel{b-2}}{2b}}{\cancel{b-2}}=-\frac1{2b}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3328994",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Why does this work? What is the best exponent to approximate the logarithm? Suppose i want to get a upper bound of $log_53$ without other tools than elementary algebra.
I can do this:
Let $x = log_53$, then $5^x=3$
I raise both sides to $4$, then: $(5^x)^4 = 3^4$ And since $3^4 < 5^3$ it follows that $5^{4x} < 5^3$, t... | There's no "best exponent", because you can always get better and better exponents, but you can use continued fractions to do this systematically. With continued fractions, we will write $x$ in the form
$$
x = a_0 + \cfrac1{a_1 + \cfrac1{a_2 + \cfrac1{a_3 + \cfrac1{a_4 + \dots}}}}.
$$
where $a_0, a_1, a_2, a_3, a_4,... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Finding matrix representation of a linear transformation Find the matrix representation of the linear transformation $\alpha:\mathbb{R}^3 \rightarrow \mathbb{R}^2$ defined by
$$\alpha : \begin{pmatrix} a\\b\\c \end{pmatrix}\rightarrow \begin{pmatrix} a+b+c\\b+c \end{pmatrix}$$
With respect to bases
$\left\{ \begin{pm... | What you have computed is the matrix for $\alpha$, with respect to the given basis on $\Bbb{R}^3$, but with respect to the standard basis on $\Bbb{R}^2$ (as opposed to the one given). If $B_2$ is the basis on $\Bbb{R}^2$, then the matrix you want is:
$$\left[\begin{array}{c|c|c}\left[\alpha \begin{pmatrix}-1 \\ 0 \\ 2\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3330655",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Find the remainder of $\frac{x^{2015}-x^{2014}}{(x-1)^3}$ Find the remainder of $\frac{x^{2015}-x^{2014}}{(x-1)^3}$.
Let $P(x)=x^{2015}-x^{2014}=Q(x)(x-1)^3+ax^2+bx+c.$ If we put $x=1$ in $P(x)$ and $P'(x)$, we get $a+b+c=0$ and $2a+b=1$. Then: $c=a-1$. The second derivative won't help in finding $b$, so, what should I... | $\dfrac{x^{2014}(x-1)}{(x-1)^3}= \dfrac{x^{2014}}{(x-1)^2}$;
$x^{2014}= Q(x)(x-1)^2+ ax+b$;
Binomial expansion:
$x^{2014}=(1+(x-1))^{2014}=$
$\sum_{k=0}^{n}\binom{2014}{k}1^{n-k}(x-1)^k$
Remainder $ax+b$:
$\binom{2014}{0}1+\binom{2014}{1}(x-1)=$
$1+2014x-2014=2014x-2013$.
Originally:
$x^{2014}(x-1)$ is divided by $(x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3331414",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Finding a formula for $(1^3)\cdot(n)+(2^3)\cdot(n-1)+(3^3)\cdot(n-2)+ \cdots + (n^3)\cdot(1)$
I need a formula for this sum:
$$(1^3)\cdot(n)+(2^3)\cdot(n-1)+(3^3)\cdot(n-2)+ \cdots + (n^3)\cdot(1)$$
I have found this formula :
$$1\cdot n +2\cdot(n-1)+3\cdot(n-2)+ \cdots +(n-1)\cdot 2 +n\cdot1= \frac16n(n+1)(n+2)$$
... | Use generating functions.
Write your sum as:
$\begin{align*}
S_{n + 1}
&= \sum_{0 \le k \le n + 1} k^3 (n + 1 - k)
\end{align*}$
So this is a convolution. It is the coefficients of the series:
$\begin{align*}
S(z)
&= \sum_{n \ge 0} S_n z^n \\
&= \left( \sum_{k \ge 0} k^3 z^k \right)
\cdot... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Solve System of equation using elimination? \begin{align}
I:&& ~~ x+\frac12y &= 6
\\[.5em]
II:&& ~~ \frac32x + \frac{3}{2}y &= {17 \over 2}
\end{align}
when $x$ was multiplied by $(-3/2)$ in first equation the $x$ will be canceled and the resulting $y = -2/3$ and $x = 19/3$.
But when fractions were simplified first the... | This can solved in an entirely algorithmic way, calculating the RREF of the augmented matrix. First we multiply the second equation by $2$. Then:
\begin{align}
\begin{bmatrix}1&\frac 12& 6 \\ 3&3&17
\end{bmatrix}&\rightsquigarrow
\begin{bmatrix}1&\frac 12&\phantom{-}6 \\ 0&\frac32 &-1
\end{bmatrix}\rightsquigarrow
\beg... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to determine algebraically whether an equation has an infinite solutions or not? I was learning for the first time about partial fraction decomposition. Whoever explains it, emphasises that the fraction should be proper in order to be able to decompose the fraction. I was curious about knowing what happens if I try... | $$\text{Given: }\frac{(Ax + B)(x - 3) + (Cx + D)(x + 5)}{(x + 5)(x - 3)} = \frac{x^2 - 4}{(x + 5)(x - 3)}$$
$$(Ax^2-3Ax+BX-3B)+(Cx^2+5Cx+Dx+5D)-(x^2-4)=0$$
$$(A+C-1)x^2+(-3A+B+5C+D)x+(-3B+5D+4)=0$$
$$(\mathbf{A+C}-1)x^2+(\mathbf{-3A-3C}+B+\mathbf{5C+3C} -D)x-(3B-5D-4)=0$$
Noting the coefficient of $x^2$: $A+C=1$, we ca... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3335420",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
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"answer_id": 5
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A double integral for $\frac{\pi}{2} \ln 2$.
Show that
\begin{eqnarray*}
I=\int_0^1 \int_0^1 \frac{dx \,dy}{\sqrt{1-x^2y^2}} = \frac{\pi}{2} \ln 2.
\end{eqnarray*}
My try ... from this question here we have
\begin{eqnarray*}
\int_0^1 \frac{ \sin^{-1}(x)}{x} \,dx = \frac{\pi}{2} \ln 2 .
\end{eqnarray*}
And from th... | Why not go head-on? It seems to work here. That is, write the integral as $$\int_0^1\int_0^1\frac {\mathrm d x \mathrm d y}{\sqrt{1-x^2y^2}}=\int_0^1\int_0^1\frac {\mathrm d x \mathrm d y}{y \sqrt{\frac{1}{y^2}-x^2}}.$$ Setting $x=\frac1y\sin\phi$ as usual helps us evaluate the first integral, or you can simply note th... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 3
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Integrating $\int \frac{3}{x^2+12x+45}$ My professor is giving a mastery exam over integrals and one of the sections covered is un-factorable quadratics like this problem that are supposed to be done by completing the square.
$$\int \frac{3}{x^2+12x+45}$$
Completing the square and everything gives $$\int \frac{3}{(x+6)... | Since\begin{align}\int\frac3{x^2+12x+45}\,\mathrm dx&=\int\frac3{(x+6)^2+9}\,\mathrm dx\\&=\int\frac{\frac39}{\frac{(x+6)^2}9+1}\,\mathrm dx\\&=\int\frac{\frac13}{\left(\frac{x+6}3\right)^2+1}\,\mathrm dx,\end{align}the substitution $y=\frac{x+6}3$ leads indeed to the answer $\arctan\left(\frac{x+6}3\right)$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find table values of this $y=(x-4)(x-2)(x+1)(x+4)$ how to find this $$y=(x-4)(x-2)(x+1)(x+4)$$
I know that x - intercept is $$4,2,-1,-4$$
and y- intercept is $$32$$
In table values
$$ x = -5,-3,-1,2,3$$
now What is the values of $y$ ?
Thanks in advance
| Let the function be $f(x)$ for convenience.
Plug in each value:
$f(-5)=(-5-4)(-5-2)(-5+1)(-5+4)=(-9)(-7)(-4)(-1)=252$
$f(-3)=(-3-4)(-3-2)(-3+1)(-3+4)=(-7)(-5)(-2)(1)=-70$
$f(-1)=(-1-4)(-1-2)(-1+1)(-1+4)=(-5)(-3)(0)(3)=0$
$f(2)=(2-4)(2-2)(2+1)(2+4)=(-2)(0)(3)(6)=0$
$f(3)=(3-4)(3-2)(3+1)(3+4)=(-1)(1)(4)(7)=-28$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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How to show that $ \sum_{k=1}^{2^n}\frac{1}{k}\geq 1+\frac{n}{2}$? So in order to prove that the Harmonic series diverge, I want to show that:
$$ \sum\limits_{k=1}^{2^n}\dfrac{1}{k}\geq 1+\dfrac{n}{2}$$
It is clear that if we expand the sum that this inequality holds true,
$$n=3,\ \sum\limits_{k=1}^{2^3=8}\dfrac{1}{k}=... | *
*Use the fact that $\frac{1}{3}+\frac{1}{4} \geq \frac{1}{4}+\frac{1}{4}$ and $\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8} \geq \frac{1}{8}+\frac{1}{8} +\frac{1}{8}+\frac{1}{8}$
*There are also various other proofs which you might be interested to see. Here is a link: http://scipp.ucsc.edu/~haber/archives/physi... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solving an integral with power of $\exp$ Is it possible to get a "closed" form for the following integral
$$\int_{\mathbb{R}}\int_{\mathbb{R}}e^{-\frac{2}{3}\vert x\vert^{\beta}}e^{-\frac{1}{3}\vert y-x\vert^{\beta}}\frac{1}{\vert y\vert^{\gamma}}dxdy,$$
where $1<\beta\le 2\;,0<\gamma<1?$
The case where $\beta=2$ does... | We have that $\frac{2}{3} = a-b$ and $\frac{1}{3} = b$ so we will try to deal with it as generally as possible (since those specific numbers are unlikely to have any symmetry associated with them).
We can convert the integral into polar coordinates and get an integral of the form
$$\int_{\alpha_1}^{\alpha_2} \int_0^\in... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Proving that $5^n - 1$ is divisible by $4$ by mathematical induction. I have done it, but I am not sure that the inductive step is right. Can anybody please clear me about it?
Basic steps as:
Taking $n=1$: $p(1)=5-1=4$.
Inductive hypothesis: Assume the statement is true for $p(k)$. $5^k - 1$ is divisible by $4$.
Ind... | *
*For n=1 the statement is true.
*Assume that $$5^n-1$$ is divisible by 4.
*Let us consider $$5^{(n+1)}-1$$. We have to prove that it is divisible by4 also.
We have that
$$
\begin{gathered}
5^{n + 1} - 1 = 5 \cdot 5^n - 1 = 5 \cdot 5^n - 5 + 5 - 1 = \hfill \\
\hfill \\
= 5\left( {5^n - 1} \right) + 4... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Evaluate $\sum\limits_{n=0}^{\infty} \frac{\cos(nx)}{2^n}$ where $\cos x = \frac15$ Evaluate
$$\sum_{n=0}^{\infty} \dfrac{\cos(nx)}{2^n}$$
where $\cos x = \frac{1}{5}$.
This is a complex number question. But I don’t know where to start. Maybe need to use the DeMoivre’s Theorem?
| If $S=\sum_{r=0}^\infty y^r\cos r x$
Using scale of relation from this or $\#187$ of this(downloadable)
$$(1-2y\cos x+y^2)S$$
$$=1+y\cos x+y^2\cos2x+y^3\cos3x+\cdots$$
$$-2y\cos x-2y^2\cos^2x-2y^3\cos2x\cos x-2y^4\cos3x\cos x-\cdots$$
$$+y^2+y^3\cos x+y^4\cos2x+y^5\cos3x+\cdots$$
$$=1+y(-\cos x)+y^2(\cos2x-2\cos^2x+1)+... | {
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"timestamp": "2023-03-29T00:00:00",
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Proving the inequality $4(a^6+b^6) \ge (a+b)(a^2+b^2)(a^3+b^3)$ Prove that $4(a^6+b^6) \ge (a+b)(a^2+b^2)(a^3+b^3)$. When does the inequality hold?
I really don't know how to prove the inequality and would like to know how.
I mainly tried to factorise the LHS-RHS fully but I could never properly do it:
https://imgur.c... | You can simplify first:
$$4(\underbrace{a^6+b^6}_{(1)}) =4\require{cancel}\cancel{(a^2+b^2)}(a^4-a^2b^2+b^4)\ge (a+b)\cancel{(a^2+b^2)}(\underbrace{a^3+b^3}_{(2)}) \Rightarrow\\
4((a^2-b^2)^2+a^2b^2)\ge (a+b)^2((a-b)^2+ab) \Rightarrow \\
3(a^2-b^2)^2+ab(4ab-(a+b)^2)\ge 0 \Rightarrow \\
3(a-b)^2(a+b)^2-ab(a-b)^2\ge 0 \R... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3342070",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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Solutions to a quadratic given 1 solution in form a+bi I was just really confused as to how I only ended up with 1 of 2 answers for the following question.
Given that $-2+bi$ is a solution of $x^2+ax+(3+a) $ find constants $a$ and $b$ given that they are real.
As soon as I saw that $-2+bi$ was a solution, I immediately... | Let the second root be $c-ib$ (because $2+ib+c-ib=a$ must be real). Then
$$(x+2-ib)(x-c+ib)=x^2+(2-c)x-2c+b^2+i(bc+2b)=x^2+ax+a+3.$$
For the imaginary term to vanish, $$b=0\lor c=-2$$
gives us the three solutions
$$a=7,b=0,c=-5$$ and $$a=4,b=\pm\sqrt3,c=-2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3343042",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Calculate $\int_0^{\pi} \frac{ \cos ( kx ) }{ 1 - 2 \tau \cos (x ) } dx$ I need to calculate the following integral:
$$\int_0^{\pi} \frac{ \cos ( kx ) }{ 1 - 2 \tau \cos (x ) } dx$$
for $k \geq 0$ and $| \tau | < \frac{1}{2}$. For $k=0$, I use the reparameterization $t = \tan (x /2)$, but I have no idea how to do it fo... | Using the fact that for all $r\in (-1,1)$ and all real $x$,
\begin{align*}
\sum_{n=-\infty}^\infty r^{|n|}e^{inx} =\frac{1-r^2}{1+r^2 - 2r\cos x}
\end{align*} (for background, this is the series representation of the Poisson kernel) we get
\begin{align*}
\int_0^{\pi} \frac{\cos(kx)}{1+r^2 -2r\cos x} dx =& \frac 1 2 \in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3343190",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
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Prove that $\left|\left\{\frac{n}{1}\right\} - \left\{\frac{n}{2}\right\} - \cdots - (-1)^n\left\{\frac{n}{n}\right\}\right| \le \sqrt{2n}$.
For all positive integers $n$, prove that $$\large \left|\left\{\frac{n}{1}\right\} - \left\{\frac{n}{2}\right\} + \left\{\frac{n}{3}\right\} - \cdots - (-1)^n\left\{\frac{n}{n}\... | Let $m = \lfloor\sqrt{2n} + 1\rfloor$. We have that $$\left|\left\{\frac{n}{1}\right\} - \left\{\frac{n}{2}\right\} + \left\{\frac{n}{3}\right\} - \cdots - (-1)^n\left\{\frac{n}{n}\right\}\right|$$
equals the absolute value of
$$\underbrace{\left(\left\{\frac{n}{1}\right\} + \left\{\frac{n}{2}\right\} - \left\{\frac{n}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3346484",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to solve $\int^1_0 (1+7x)^{1/3}dx$? I worked through $\int^1_0 (1+7x)^{1/3}dx$ and I got $\frac{3}{4}+C$ for the answer. However, I forgot about the exponent when I found the difference of the sides of the integral.
I am retrying it, but I've realized I don't know how to find a number to the power of $\frac{4}{3}$.... | You got $\dfrac{1}{7}\left[\dfrac{3(1+7)^{4/3}}{4}-\dfrac{3(1+0)^{4/3}}{4}\right].$
Note that $(1+7)^{4/3}=8^{4/3}=(8^{1/3})^4=2^4=16$ and $(1+0)^{4/3}=1^{4/3}=1,$
so you actually got the correct answer: $\dfrac17\left[\dfrac{3\times16}4-\dfrac{3\times1}4\right]=\dfrac17\dfrac{45}4=\dfrac{45}{28}.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3349010",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solve Diophantine equation with three variables part two I want to find all solutions of $$x^2+y^2+z^2-xy-yz-zx-x-y-z=0$$
Solutions need not to be primitive. I found several parametric family. For example
$(m^2, m^2+m , (m+1)^2)$
$(m^2, m^2+m+2 , (m+1)^2)$
$(m^2+1, m^2+m, (m+1)^2+1)$
$(m^2+1, m^2+m+4, (m+1)^2+1)$
$(m^2... | I was wrong about the formula... it looks like this....
$$X^2+Y^2+Z^2=XY+XZ+ZY+X+Y+Z$$
$$X=s(3(k^2-kt+t^2)s-k-t)$$
$$Y=s(3(k^2-kt+t^2)s+2k-t)$$
$$Z=s(3(k^2-kt+t^2)s-k+2t)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3350399",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Finding $x^3 + y^3$ when $y = \frac{3-\sqrt5}{3+\sqrt5} , x = \frac{3+\sqrt5}{3-\sqrt5} $ $y = \frac{3-\sqrt5}{3+\sqrt5} , x = \frac{3+\sqrt5}{3-\sqrt5} $
$x^3 + y^3 =?$
my answer =
$(3 + \sqrt5)^3 = 47 + 32\sqrt5$
$(3 - \sqrt5)^3 = 47 - 32\sqrt5$
$x^3 + y^3 = \frac{47 + 32\sqrt5}{47 - 32\sqrt5} + \frac{47 - 32\sqrt5}{... | You have made mistakes in computing $(3+\sqrt 5 )^{3}$ and $(3-\sqrt 5 )^{3}$. These are $72+32\sqrt 5$ and $72-32\sqrt 5$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3350519",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
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find minimum of maximum of two functions
I need to find the angle $\theta$ so that: $$\max(\cos^2(\theta),1-\cos^2(45-\theta))$$ is minimized.
OK, so I wrote \begin{align*}f(\theta)&=\max(\cos^2(\theta),1-\cos^2(45-\theta))\\
&=\frac{\cos^2(\theta)+1-\cos^2(45-\theta)+|\cos^2(\theta)-1+\cos^2(45-\theta)|}{2}\end{ali... | Is known that
$$2\cos^2a = 1+\cos 2a.$$
So
$$f(\theta) = \dfrac12\max\left(1+\cos2\theta,2-\left(1+\cos\left(\dfrac\pi2-2\theta\right)\right)\right),$$
$$f(\theta) = \dfrac12+\dfrac12\max(\cos2\theta,-\sin2\theta).$$
Since
$$\cos2\theta+\sin2\theta = \sqrt2\left(\sin\dfrac\pi4\cos2\theta+\cos\dfrac\pi4\sin2\theta\rig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3350686",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Prove a trigonometric identity: $\cos^2A+\cos^2B+\cos^2C+2\cos A\cos B\cos C=1$ when $A+B+C=\pi$ There is a trigonometric identity:
$$\cos^2A+\cos^2B+\cos^2C+2\cos A\cos B\cos C\equiv 1\text{ when }A+B+C=\pi$$
It is easy to prove it in an algebraic way, just like that:
$\quad\cos^2A+\cos^2B+\cos^2C+2\cos A\cos B\cos... | A purely geometric way doesn't look likely, because the degrees of the cosine terms (two and three respectively) don't match. For what it's worth, here is an alternative trigonometric derivation.
Writing $2\cos B\cos C$ in the second term as $\cos (B+C)+\cos(B-C)$ transforms our expression to$$\cos^2A+\cos^2B+\cos^2C+[... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3350887",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
"answer_count": 4,
"answer_id": 3
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For positive real numbers $a,b,c$ prove that $ a^4 + b^4 + c^4 \ge abc(a+b+c)$ For positive reals $a,b,c$ prove that $$ a^4+b^4+c^4 \ge abc(a+b+c). $$
I tried to pls around trying to reorganize to get AM-GM but i couldn't
Thanks for the help in advance.
| Hint
$$\frac{a^4+a^4+b^4+c^4}{4}\geq abca \\
\frac{a^4+b^4+b^4+c^4}{4}\geq abcb \\
\frac{a^4+c^4+b^4+c^4}{4}\geq abcc \\$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3353216",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding all the complex roots of $x^3 - x^2 + 1 = 0$ I'm trying to calculate the exact values of the roots of $x^3 - x^2 +1 =0$ (When I say exact value, here I mean find the roots in the form of $a + bi$ for some $a,b \in \mathbb{R}$
My first thought is to get it to a form where I can use Cardano's method and so, since... | Without change of variables, using Cardano method, the raw results are
$$x_1=\frac{1}{3} \left(1-\sqrt[3]{\frac{2}{25-3 \sqrt{69}}}-\sqrt[3]{\frac{1}{2}
\left(25-3 \sqrt{69}\right)}\right)$$
$$x_2=\frac{1}{3}+\frac{1}{6} \left(1+i \sqrt{3}\right) \sqrt[3]{\frac{1}{2} \left(25-3
\sqrt{69}\right)}+\frac{1-i \sqrt{3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3354712",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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For $x$ and $k$ real numbers, for what values of $k$ will the graphs of $f(x)=-2\sqrt{x+1}$ and $g(x)=\sqrt{x-2}+k$ intersect?
For $x$ and $k$ real numbers, for what values of $k$ will the graphs of $f(x)=-2\sqrt{x+1}$ and $g(x)=\sqrt{x-2}+k$ intersect?
I tried to make an equation of them, but I’m stuck with the two ... | Hint: You must solve the equation
$$\sqrt{x-2}+k=-2\sqrt{x+1}$$ for $$x\geq 2$$ Writing this equation in the form
$$-k-\sqrt{x-2}=2\sqrt{x+1}$$ then it must be
$$k^2+2\geq x$$ and you can square it.
After squaring one times we get
$$2k\sqrt{x-2}=3x+6-k^2$$
Squaring again we get
$$-k^4+10 k^2 x+4 k^2-9 x^2-36 x-36=0$$
S... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3355215",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
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Minimizing the error for a second degree interpolating polynomial
Construct the second degree polynomial $q_2(t)$ that approximates $g(t) = \sin(\pi t)$ on the interval [0,1] by minimizing
$$\int_0^1 [g(t) - q_2(t)]^2dt$$
A useful integral: $\int_0^1 (6t^2-6t+1)^2dt=\frac{1}{5}$
My attempt:
We Know $q_2(t)$ will ... | I do not see why the hint was even given since
$$I=\int_0^1 (zt^2+yt+x)^2\,dt=\int_0^1 (x^2+2 x yt+ \left(2 x z+y^2\right)t^2+2 y z t^3+ z^2 t^4)\,dt$$ that is to say
$$I=x^2+x y+\frac{2 x z}{3}+\frac{y^2}{3}+\frac{y z}{2}+\frac{z^2}{5}$$ So, for the total
$$R=\frac{1}{2}-\frac{4x}{\pi}-\frac{2y}{\pi}-\frac{2z(\pi^2-4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3356458",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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On the equation $\sqrt{x^2-9}=\frac{2(x+3)}{(x-3)^2}-x$ I'm trying to solve the equation
$$\sqrt{x^2-9}=\frac{2(x+3)}{(x-3)^2}-x \tag{1}$$
Attempt: I have rewritten the equation as
\begin{align*}
\sqrt{x^2-9} = \frac{2(x+3)}{\left (x-3 \right )^2}-x &\Leftrightarrow \sqrt{x^2-9} = \frac{2(x+3)-x\left ( x-3 \right )^2}... | Difference of two squares identity and its Conjugates can help:
$\sqrt{x^2-9} = \frac{2(x+3)}{(x-3)^2} - x$
$(x-3)^2(\sqrt{x^2-9} + x) = 2(x+3)$
$(x-3)^2(\sqrt{x^2-9} + x)(\sqrt{x^2-9} - x) = 2(x+3)(\sqrt{x^2-9} - x)$
$(x-3)^2(x^2-9 - x^2) = 2(x+3)(\sqrt{x^2-9} - x)$
$-9(x-3)^2 = 2(x+3)(\sqrt{x^2-9} - x)$
$\frac{-9(x-3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3357131",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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What is the probability that a four-digit number formed by randomly selecting four-digits without replacement is greater than $4321$? A four-digit number is formed by randomly selecting four digits, without replacement, from the set $D = \{ 1, 2, 3, 4, 5, 6, 7\}$. What is the probability that the resulting number is gr... | What you have done thus far is correct. That leaves cases in which the first number is $4$ and the second number is $3$. The probability of first obtaining $4$, then obtaining $3$ is $\frac{1}{7} \cdot \frac{1}{6} = \frac{1}{42}$. We now consider two cases: the third digit is larger than $2$ or the third digit is $2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3358132",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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How to calculate $\int_0^1 \frac{\ln(1-x)}{x} dx$? I have tried to let$f(a)=\int_0^1\frac{\ln(1-ax)}{x} dx$, which means
$f'(x)=-a\int_0^1 \frac{1}{x(1-ax)} dx=a\int_0^1 \left[\frac{a}{ax-1}-\frac{1}{x}\right] dx$.
However, I am stuck at here.
| Let us consider
$$ I = \int_{0}^{+\infty}\frac{\arctan(x)}{1+x^2}\,dx =\frac{1}{2}\left[\arctan^2(x)\right]_{0}^{+\infty}=\frac{\pi^2}{8}$$
and the parametric integral
$$ I(a) = \int_{0}^{+\infty}\frac{\arctan(ax)}{1+x^2}\,dx. $$
We have $I(0)=0$ and $I(1)=I=\frac{\pi^2}{8}$. By differentiation under the integral sign... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3358436",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
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Two circles tangent to each other and tangent to a line Two circles with centers $A$ and $B$ are externally tangent at point $C$.
tangent to the two circles. Given that the radii of the two circles are $2cm$ and $3cm$, respectively, find $\frac{DC}{FC}$
| Adjusting Matthew's figure:
$\hspace{2cm}$
Use Cosine theorem:
$$\frac{FC}{DC}=\frac{\sqrt{3^2+3^2-2\cdot 3^2\cdot \cos \beta}}{\sqrt{2^2+2^2-2\cdot 2^2\cdot \cos (180^\circ-\beta)}}=\sqrt{\frac{18-18\cdot \frac15}{8+8\cdot \frac15}}=\sqrt{\frac32}.$$
Note: $\cos \beta =\frac{BE}{AB}=\frac{FB-FE}{AC+BC}=\frac{FE-AD}{AC... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3362914",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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The closed-form of $\sum_{n=0}^{\infty}\frac{(-1)^n H^{(2)}_{n}}{(2n+1)^2} $ How to Prove that
$$ \sum_{n=0}^{\infty}\frac{(-1)^nH^{(2)}_{n}}{(2n+1)^2} \;\;=\;\;\frac{7 \pi \; \zeta(3)}{4}-\frac{\zeta(2)G}{2}+\frac{45\zeta(4)}{8}-\frac{\Psi^{(3)}\big(\frac{1}{4}\big)}{128}$$
where $H_n^{(m)}=\sum_{k=1}^n\frac1{k... | Solution by Cornel Valean.
\begin{align}
S&=\sum_{n=1}^\infty\frac{(-1)^{n-1}H_n^{(2)}}{(2n+1)^2}\\
&=\sum_{n=1}^\infty(-1)^{n-1}H_n^{(2)}\int_0^1-x^{2n}\ln x\ dx\\
&=\int_0^1\ln x\sum_{n=1}^\infty(-x^2)^nH_n^{(2)}\ dx\\
&=\int_0^1\frac{\ln x\operatorname{Li}_2(-x^2)}{1+x^2}\ dx\\
&=\int_0^\infty\frac{\ln x\operatorna... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3364852",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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Prime Number Theory (Dividers) When $N = a^\alpha × b^β × c^γ$ is divided by $a$, the number of dividers decreases by $m$ units; when divided by $b$, decreases by $n$ and, dividing by $c$, decreases by $p$. Determine the exponents $α, β$ and $γ$.
Mine ended up giving in this system here:
$$
(α-a)(β+1)(γ+1)-(γ+1)(β+1) =... | When $N = a^\alpha \times b^{\beta} \times c^{\gamma}$ is divided by $a$, it becomes $\frac{N}{a} = a^{\alpha - 1} \times b^{\beta} \times c^{\gamma}$. Thus, as metamorphy's question comment indicates that if we can assume $a$, $b$ and $c$ are distinct primes, then since the number of divisors decreases by $m$, this me... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3369823",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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how to solve $\lim_{n\to\infty}{\left(\sum_{k=1}^{n}{\frac{1}{\sqrt{n^2+k}}}\right)^{n}}$? $\displaystyle\left(\sum_{k=1}^{n} \frac{1}{\sqrt{n^{2}+1}}\right)^{n}\ge\left(\sum_{k=1}^{n} \frac{1}{\sqrt{n^{2}+k}}\right)^{n}\ge\left(\sum_{k=1}^{n} \frac{1}{\sqrt{n^{2}+n}}\right)^{n}$
left=$\displaystyle\lim_{n\to\infty}{\l... | We have
$$\lim_{n\to\infty}{\left(\sum_{k=1}^{n}{\frac{1}{\sqrt{n^2+k}}}\right)^{n}}=\lim_{n\to\infty} \frac1{n^n}{\left(\sum_{k=1}^{n}{\frac{1}{\sqrt{1+k/n^2}}}\right)^{n}}$$
and
$$\frac{1}{\sqrt{1+k/n^2}}=1-\frac12\frac{k}{n^2}+O\left(\frac{k^2}{n^4}\right)$$
therefore
$$\sum_{k=1}^{n}{\frac{1}{\sqrt{1+k/n^2}}}=n-\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3370438",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 0
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Solving $2\sin(x+30^\circ) = \cos(x+150^\circ)$ for $x$ between $0^\circ$ and $360^\circ$
Solve this equation for $x$ between $0^\circ$ and $360^\circ$:
$$2\sin(x+30^\circ) = \cos(x+150^\circ)$$
Anyway thank you so much. This is my first question.
| $$2\sin(x+30^{\circ}) = \cos(x+150^{\circ})$$
By the Cofunction Identity,
$$2\sin(x+30^{\circ}) = \sin(90^{\circ}-(x+150^{\circ}))$$
$$2\sin(x+30^{\circ}) = \sin(-x-60^{\circ})$$
$$2\sin(x+30^{\circ}) = \sin(-(x+60^{\circ}))$$
By the Negative Angle Identity,
$$2\sin(x+30^{\circ}) = -\sin(x+60^{\circ})$$
$$2\sin(x+30^{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3372946",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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factor $x^9 + 243x^3 + 729$ Factor the polynomial $$x^{9} + 243x^{3} + 729$$
it might be helpful to see it like this $$x^{9}+3^{5}x^{3}+3^{6}$$
I would imagine this being done without a calculator, but I don't see how to do it.
| Observe that $x^9+243x^3+729$ = $(x^3+9)^3 - 27x^6$ = $(x^3+9)^3-(3x^2)^3$.
Now using the formula for difference of cubes we obtain:
$(x^3+9)^3-(3x^2)^3$ = $(x^3+9-3x^2)((x^3+9)^2+(x^3+9)3x^2+(3x^2)^2)$ .
Doing the calculations on the second parenthesis and reordering the terms we get that:
$x^9+243x^3+729 = (x^3 - 3... | {
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"timestamp": "2023-03-29T00:00:00",
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Solve the equation $\sqrt{a(2^{x}-2)+1}=1-2^{x}$ for every value of the parameter $a$. Question : Solve the equation $\sqrt{a(2^{x}-2)+1}=1-2^{x}$ for every value of the parameter a.
I have solved the problem as follows
$\sqrt{a\left(2^{x}-2\right)+1}=1-2^{x}$
$a\left(2^{x}-2\right)+1=\left(1-2^{x}\right)^{2}=2^{2 x}... | By letting $y=1-2^x$, you can simplify the problem to:
$$
\sqrt{a(3-y)+1} = y
$$
Since root is non-negative, $y\ge0$. Since $2^x$ is always positive, $y<1$. Thus we want to find the solution of the equation above with the condition $0\le y<1$.
Since both sides are non-negative, we can square them and keep the equality:... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3376613",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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What is the principal root of $\sqrt{-i}$? Where is the mistake in this solution?
$$\sqrt{-i}=\sqrt {\cos \frac{3 \pi}2+i\sin \frac{3 \pi}2}=\cos \frac{3 \pi}4+i\sin \frac{3 \pi}4=-\frac{\sqrt 2}2+i \frac{\sqrt 2}2$$
WA gives me different result:
$$\frac{\sqrt 2}2-i \frac{\sqrt 2}2$$
Why the principal root must be ... | Since you are asked to find the principal root, you need to solve according to the definition of the principal root below.
Given the complex number $z$,
\begin{equation}
z = r \mathrm{e}^{i \theta}, ~ -\pi < \theta \le \pi.
\end{equation}
its principal square root is defined as,
\begin{equation}
\sqrt{z} = \sqrt{r} \m... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3376738",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 3
} |
Ways to Arrange 7 things given restrictions Let's say you have three $z$'s, two $x$'s, and two $y$'s. How many ways are there to arrange those $7$ variables given that $x$ and $y$ cannot be together?
Ex: $zxzxzyy$ and $xxzzzyy$ are valid while $zyxzyzx$ and $xyzzyxz$ are not.
I tried approaching this problem by findin... | We may select three of the seven positions for the $z$'s, two of the remaining four positions for the $x$'s, and fill the remaining two positions with the two $y$'s in
$$\binom{7}{3}\binom{4}{2}\binom{2}{2} = \binom{7}{3}\binom{4}{2}$$
ways. From these, we must subtract the number of arrangements in which an $x$ and $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3377314",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Interesting function : $f(x)=\frac{\ln (\frac{x+2}{x+1})}{\ln (1+\frac{1}{x})}$ , $x>1$ Question :
Prove this function :
$$f(x)=\dfrac{\ln \left(\dfrac{x+2}{x+1}\right)}{\ln \left(1+\dfrac{1}{x}\right)},\quad x>1$$
is increasing.
I don't know how I solve by my effort is :
Derivative of $f$ is :
$$f'(x)=\dfrac{(x+... | I didn't check all the detail of your derivation, but
$$(x+2)\ln \left(\frac{x+2}{x+1}\right)-x\ln \left(\frac{x+1}{x}\right)=\frac{x+2}{x+1}\ln \left(1+\frac1{x+1}\right)^{x+1}-\ln \left(1+\frac1x\right)^x≥0$$
and $$\left(1+\frac1x\right)^x$$ is an increasing function.
EDIT (aimed to clarify any single step)
We start... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3379519",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Prove that $2^{n+2}$ is a divisor of the number $k^{2^{n}}-1$ Question :
If n is a natural number and k is an odd number
$k\in \mathbb{Z}$ then show that $2^{n+2}$ is a divisor of
the number $k^{2^{n}}-1$
My try as following :
I'm think use induction :
$n=1$ then $2^{3}=8$ since $k$ is odd so : $k^{2}=1\mod 8$
We ... | The trick is $k^{2*m} - 1 = (k^m -1)(k^m+1)$ and if $2^z\mid k^m -1$ then $k^m$ is odd and $k^m + 1$ is even so $2^z|k^m-1$ and $2\mid k^m+1$ so $2^z*2\mid (k^m-1)(k^m+1)$.
So by induction:
Base case: If $k$ is odd the $k = 2m+1$ for some $m$ and $(2m+1)^2-1= 4m^2 + 4m = 4m(m-1)$ either $m$ or $m-1$ is even so $2|m(m-1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3381826",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Find positive integers such that $2n^3 + 5|n^4 +n+1$ $$ 2n^3 + 5 | 2n^4 +2n +2 - 2n^4 - 5n$$
$$= 2n^3+5 | 2-3n$$
$$3n-2≥2n^3 + 5$$
is this correct? Is there a more efficient way?
| You can use polynomial long division instead:
$$
\require{enclose}
\begin{array}{rll}
\frac{1}{2}n && \\[-3pt]
2n^3+5 \enclose{longdiv}{n^4+0n^3+0n^2+n+1}\kern-.2ex \\[-3pt]
\underline{- (n^4 \quad \quad \quad \quad \ \ + \frac{5}{2}n) \ \ \ } \\[-3pt]
- \frac{3}{2}n+ 1 \\[-3pt]
\end{array}
$$
Th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3382310",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
} |
Ramanujan's Nested Radical By noting Ramanujan's Nested Radical, we have
$3 = \sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+\cdots}}}}$
On the other hand, we can manipulate the number $4$ by applying the similar principle. Here we have
$\begin{aligned} 4 & = \sqrt{16} \\ & = \sqrt{1+15} \\ & = \sqrt{1+2 \cdot \dfrac{15}{2}} \\ & ... | In Ramanujan's radical, if you stop after $n$ nested radicals the last term inside the radical will be $\sqrt{1}$ but in your case, if you stop after $n$ nested radicals the last term is a term which is increasing with $n$. Hence the difference.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3382458",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
} |
Solution of a limit of a sequence $\frac{\sqrt{4n^2+1}-2n}{\sqrt{n^2-1}-n}$ I'm trying to solve the limit of this sequence without the use an upper bound o asymptotic methods:
$$\lim_{n\longrightarrow\infty}\frac{\sqrt{4n^2+1}-2n}{\sqrt{n^2-1}-n}=\left(\frac{\infty-\infty}{\infty-\infty}\right)$$
Here there are my diff... | From here
$$\frac{\sqrt{4n^2+1}-2n}{\sqrt{n^2-1}-n}=\frac{\sqrt{4+\dfrac{1}{n^2}}-2}{\sqrt{1-\dfrac{1}{n^2}}-1}$$
we can use that
$$\sqrt{4+\dfrac{1}{n^2}}=2\sqrt{1+\dfrac{1}{4n^2}}\sim 2\left(1+\dfrac{1}{8n^2}\right)=2+\dfrac{1}{4n^2}$$
$$\sqrt{1-\dfrac{1}{n^2}}\sim 1-\dfrac{1}{2n^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3384280",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Find the sum of $\binom n0 - \binom n2 +\binom n4 -\binom n6 \cdots$ Find the sum of $\binom n0 - \binom n2 +\binom n4 -\binom n6 \cdots$
Using Binomial expansion of $(1+x)^n$,
$$ \binom n0 +\binom n1 x^1 + \binom n 2 x^2 + \cdots + \binom n nx^n $$
Substituting $x = i$,
$$(1+i)^n = \binom n0 +\binom n1 i^1 + \binom ... | So you have $$(1+i)^{n}= \binom{n}{0} + \binom{n}{1} i + \binom{n}{2}i^2 + \cdots = \binom{n}{0}-\binom{n}{2} + \binom{n}{4} + i \cdot\left[\binom{n}{1}-\binom{n}{3} + \cdots\right]$$
*
*Write $(1+i)^{n}= (\sqrt{2})^{n}\cdot \left(\frac{1}{\sqrt{2}}+i \cdot\frac{1}{\sqrt{2}}\right)^{n}$ and see if you can find the r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3385119",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
How to solve this inequality : Do you know what is this problem or how to solve it ?
Let $x,y,z\in\mathbb{R}^+$ such that $x+y+z=1$. Proove that
$$\frac{(xy+yz+zx+1)(3x^3+3y^3+3z^3+1)}{9(x+y)(y+z)(z+x)}\ge\left(\frac{x\sqrt{x+1}}{\sqrt[4]{9x^2+3}}+\frac{y\sqrt{y+1}}{\sqrt[4]{9y^2+3}}+\frac{z\sqrt{z+1}}{\sqrt[4]{9z^2+3... | If I understood right you used homogenization and assumption $x+y+z=3$.
Now use C-S:
$$\sqrt[4]{x^2+3}=\sqrt{\frac{1}{2}\sqrt{(1+3)(x^2+3)}}\geq\sqrt{\frac{1}{2}(x+3)}.$$
Also, use Holder: $$9(x^3+y^3+z^3)\geq(x+y+z)^3.$$
After this you'll get something obvious:
$$\sum_{cyc}(x-y)^2(x+y)\geq0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3385718",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Let $\xi$ be the fifth primitive root of $1$ an let $\zeta= \xi + \frac{1}{\xi}$. Prove that $\zeta^2 + \zeta =1$ Let $\xi$ be the fifth primitive root of $1$ an let $\zeta= \xi + \frac{1}{\xi}$. Prove that $\zeta^2 + \zeta =1$
In my attempt I used $\xi = \cos(72k) + i\sin(72k)$ with $k=1,2,3,4$
Where I got $\zeta = \f... | Write $\zeta= \xi + \dfrac{1}{\xi} = \xi + \xi^4$, because $\xi^5=1$. Then
$$\zeta^2 + \zeta = \xi^2 + 2\xi \xi^4 + \xi^8 + \xi + \xi^4 = \xi^2 + 2 + \xi^3 + \xi + \xi^4 = 1 + 1 + \xi + \xi^2 + \xi^3 + \xi^4 = 1
$$
because $1 + \xi + \xi^2 + \xi^3 + \xi^4 = 0$, since $0=\xi^5-1=(\xi-1)(1 + \xi + \xi^2 + \xi^3 + \xi^4)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3386226",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Find the roots of $z^4-3z^2+1=0$ in polar form. Question :
Prove that the solutions of $z^4-3z^2+1=0$ are given by :
$$z=2\cos{36^\circ},2\cos{72^\circ},2\cos{216^\circ},2\cos{252^\circ}$$
My work :
First of all, i want ro find the roots with quadratic formula
$\begin{align}
&(z^2)^2-3z^2+1=0\\
&z^2=\dfrac{3\pm \sqrt{5... | Set $y = z^2$, your equation is then equivalent to
$$y^2 -3y +1=0.$$
This is a quadratic equation and can be solve with the quadratic formula that you mentioned:
$$y_{1,2} = \dfrac{3 \pm \sqrt{(-3)^2 - 4 \cdot 1}}{2}=\dfrac{3 \pm \sqrt{5}}{2}.$$
Now observe that
$$y_1 = \dfrac{3+\sqrt{5}}{2}= \dfrac{\frac{1}{2}+\sqrt{5... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3386733",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
Domain of inequation inside squared root How should I go when restricting the roots of the inequation:
$\sqrt {x^2+5x+6} - \sqrt {x^2-x+1} \lt 1$?
By restricting both the squared roots, I know that:
$x \le 3$ and $x \ge -2$
However when simplifying the whole inequation, I get the two roots:
$\frac{-13-\sqrt{73}}{16}$ a... | It's $$\sqrt{x^2+5x+6}<1+\sqrt{x^2-x+1}$$ or
$$x^2+5x+6<1+x^2-x+1+2\sqrt{x^2-x+1}$$ or
$$\sqrt{x^2-x+1}>3x+2.$$
Now, consider two cases:
*
*$x< -\frac{2}{3},$ which is $x\leq-3$ or $-2\leq x<-\frac{2}{3}$;
*$x\geq-\frac{2}{3}.$
Can you end it now?
I got the following answer:
$$(-\infty,-3]\cup\left[-2,\frac{-13+\sq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3388042",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Showing that $\frac{2^{\sqrt n}}{1+2^{\sqrt n}/x^n}$ is approximated by $2^{\sqrt n}\left(1- \frac{2^{\sqrt n}}{x^n} \right)$ Sorry for this inconvenient question but really I can't see how to show that $$\frac{2^{\sqrt n}}{1+ \frac{2^{\sqrt n}}{x^n}}\quad\text{is approximated by}\quad 2^{\sqrt n } \left(1- \frac{2^{\s... | $$\frac{2^{\sqrt{n}}}{1+\frac{2^{\sqrt{n}}}{x^n}}=\frac{x^n2^{\sqrt{n}}}{x^n+2^{\sqrt{n}}}=2^{\sqrt{n}}\left(\frac{x^n}{x^n+2^{\sqrt{n}}}\right)=2^{\sqrt{n}}\left(\frac{x^n+2^{\sqrt{n}}}{x^n+2^{\sqrt{n}}}-\frac{2^{\sqrt{n}}}{x^n+2^{\sqrt{n}}}\right)=2^{\sqrt{n}}\left(1-\frac{2^{\sqrt{n}}}{x^n+2^{\sqrt{n}}}\right)$$
now... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3388710",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
If $\alpha$, $\beta$, $\gamma$ and $\delta$ be the roots of the polynomial $x^4+px^3+qx^2+rx+s$, prove the following.
If $\alpha$, $\beta$, $\gamma$ and $\delta$ be the roots of the polynomial $x^4+px^3+qx^2+rx+s$, show that $$(1+\alpha^2)(1+\beta^2)(1+\gamma^2)(1+\delta^2)=(1-q+s)^2+(p-r)^2.$$
I have tried putting
\... | Denoting $\alpha=x_1$, $\beta=x_2$, $\gamma=x_3$ and $\delta=x_4$, we have $$\small\prod_{i=1}^4(1+x_i^2)-\prod_{i=1}^4(1-x_i^2)=\small2(x_1x_2x_3+x_1x_2x_4+x_1x_3x_4+x_2x_3x_4)+2(x_1+x_2+x_3+x_4)$$ so taking out a factor of $x_1x_2x_3x_4$ from the first term in brackets, $$\prod_{i=1}^4(1+x_i^2)=2\left(\prod_{i=1}^4x_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3390796",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
} |
$x$ and $y$-intercepts of an absolute value function $f(x)=-3|x-2|-1$ I am to find the $x$ and $y$-intercepts of the function $$f(x)=-3|x-2|-1.$$ The solution is provided in my book as
$(0, -7)$; no $x$ intercepts.
I cannot see how this was arrived at. I attempted to find the x intercepts and arrived at $2-\frac{1}{3... | $$|x-2|=-\frac{1}{3}$$
Two cases:
1) If $x\ge2$, the equation becomes $x-2 = -\frac{1}{3}$, which has no solution.
2) If $x<2$, the equation becomes $-(x-2 )=-\frac{1}{3}$, which has no solution.
Thus, there is no x intercepts. (0,-7) is the y intercept as seen from $f(0)=-7$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3391033",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Taylor series of $\sinh{(x)}$ at $\ln{(2)}$. Determine the Taylor series of $\sinh{(x)}$ about $x = \ln{(2)}$.
Equating each derivative at $x = \ln{(2)}$ gives:
\begin{equation*}
\begin{split}
f'(\ln{(2)}) &= \frac{2+\frac{1}{2}}{2} = \frac{5}{4} \\
f''(\ln{(2)}) &= \frac{2-\frac{1}{2}}{2} = \frac{3}{4} \\
f^{(3)}(\l... | $$\sinh(x)=\dfrac{e^x-e^{-x}}2=\sum_{r=0}^\infty\dfrac{x^{2r+1}}{(2r+1)!}$$
Set $x=\ln2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3395794",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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If $A$ is a rotation matrix, then $||Ax||=||x||$. Attempt:
Let $$
A =
\begin{bmatrix}
\cos\theta & -\sin\theta \\
\sin\theta & \cos\theta
\end{bmatrix},
x = \begin{bmatrix} x_1 \\ x_2 \end{bmatrix}
$$
Then,
$$Ax =
\begin{bmatrix}
\cos\theta & -\sin\theta \\
\sin\theta & \cos\theta
\end{bmatrix} \cdot \begin{bmatri... | You are right and the restriction is not important.
As an alternative we can use that
$$|Ax|^2=(Ax^T)(Ax)=x^TA^TAx=x^Tx=|x|^2$$
since $A$ is orthogonal and thus $A^TA=I$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3401051",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Linear Transformation and Basis $$
\begin{array}{l}{\text { 2.) Consider the linear transformation } T: \mathbb{R}^{2} \rightarrow \mathbb{R}^{2},\left[\begin{array}{l}{x} \\ {y}\end{array}\right] \mapsto\left[\begin{array}{c}{-2 x-5 y} \\ {2 x+4 y}\end{array}\right]} \\ {\text { Find a basis } \mathcal{B} \text { of }... | We have that the matrix from basis $B$ to canonical is
$$u_C=M_Bu_B$$
then
$$T_C(u)=\left[\begin{array}{cc}{-2} & {-5} \\ {2} & {4}\end{array}\right]u_C=\left[\begin{array}{cc}{-2} & {-5} \\ {2} & {4}\end{array}\right]M_Bu_B$$
and then
$$T_B(u)=M_B^{-1}T_C(u)=M_B^{-1}\left[\begin{array}{cc}{-2} & {-5} \\ {2} & {4}\end{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3403899",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Find $x$ so that rational function is an integer Find all rational values of $x$ such that $$\frac{x^2-4x+4}{x^2+x-6}$$ is an integer.
How I attempt to solve this: rewrite as $x^2-4x+4=q(x)(x^2+x-6)+r(x)$. If we require that $r(x)$ be an integer then we can get some values of $r(x)$ by solving $x^2+x-6=0$, so that $x=-... | For $x \neq 2$ is $$\frac{x^2-4x+4}{x^2+x-6}=\frac{x-2}{x+3}=1-\frac{5}{x+3}.$$
To obtain an integer with an integer $x,\;$ $x+3$ must be a divisor of $5,$ or equivalently $x+3 \in \{-1,1,-5,5\}.$
This gives the solutions $-4,-2$ and $8,$ because $x=2$ is excluded.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3406495",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
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Let $x>1,y>1$ and $\left(\log_e x \right)^2+\left(\log_e y \right)^2=\log(x^2)+log(y^2)$, then find the maximum value of $x^{\log_e y}$ Let $x>1,y>1$ and $\left(\log_e x \right)^2+\left(\log_e y \right)^2=\log(x^2)+log(y^2)$, then find the maximum value of $x^{\log_e y}$
My attempt is as follows:
As $x>1,y>1$ , so $\lo... | Here is a way where you do not need any standard inequality like AM-GM or Cauchy-Schwarz.
The fist step is, as already shown in an other solution, to get rid of the logarithms by substituting $x=e^a$, $y=e^b$ with $a,b >0$ which gives:
*
*$a^2+b^2 = 2(a+b)$ and
*maximize $x^{\ln y} =e^{\ln x\cdot \ln y} = e^{ab}$.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3410401",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Show that $(1 - x)^{-k} \le 1 + 2 k x$ when $0 < k x <1/2$ and $k \ge 1$. I think it is true that
$$(1 - x)^{-k} \le 1 + 2 k x$$
when $0 < k x <1/2$ and $k \ge 1$.
This can be proved by using mean-value theorem to show that
$$
(1-x)^{-k} \le 1 + k x / (1-x)^{k+1}
\le
1+ 2kx
$$
under the given condition.
Is there's a m... | $(1 - x)^{-k} \le 1 + 2 k x
$
if
$0 \lt x < \dfrac1{2k}$.
This is
$1
\le (1-x)^k(1+2kx)
$.
If $k=1$ this is
$1
\le (1-x)(1+2x)
=1+x-2x^2
$
or $x \ge 2x^2
$
or
$x \le \frac12$
which is true.
Suppose true for $k$.
Want to show
$1
\le (1-x)^{k+1}(1+2(k+1)x)
$.
Assume
$0 \lt x < \dfrac1{2(k+1)}$.
$\begin{array}\\
(1-x)^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3413754",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Least value of unit vector $|a+b|^2+|b+c|^2+|c+a|^2$ If $a, b, c$ are unit vectors, then least value of $|a+b|^2+|b+c|^2+|c+a|^2$ will be equal to
(1) 1
(2) 3
(3) 9
(4) 12
If am using the concept $a=c=-b$, I am getting the answer 4, but not matching with options provided
| Let $a,b,c$ be three unit vectors then $$(a+b+c)^2\ge 0 \implies a^2+b^2+c^2+2(a,b+b.c+c.a) \ge 0$$ $$ \implies 2(a.b+b.c+c.a) \ge -3 ~~~(1)$$
Then $$|a+b|^2+|b+c|^2+|c+a|^2=2(a^2+b^2+c^2)+2(a.b+b.c+c.a) \ge 6-3=3.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3415387",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Divisibility by 7 of a number consisting of 0 and 1's A decimal number is of arbitrary length consisting of 0 and 1 only i.e. (10,111,10000,11111,1010, number of digits in the number can be upto 100 )
Can this number ever be divisible by 7 if yes, is there any efficient way to list all those numbers
| Note that $10 \equiv 3 \bmod 7$ and the powers of $3$ modulo $7$ are:
$$3^0=1,\ 3^1=3,\ 3^2=2,\ 3^3=6,\ 3^4=4,\ 3^5=5$$
It follows that $10^k \equiv 3^r \bmod 7$ where $r$ is the remainder of $k$ modulo $6$, so you can find the remainder of a number consisting of only $0$s and $1$s modulo $7$ by splitting its decimal e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3417330",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Solving a system of modular equations I am trying to solve this system of equations：
$24x ≡ 12 \hspace{5pt} (\mathrm{mod} \hspace{2pt} 63)$
$x ≡ 2\hspace{5pt} (\mathrm{mod} \hspace{2pt} 27)$
So I know that the second equation is equivalent to $x=27y+2$, but when I plug it into $24x ≡ 12\hspace{5pt} (\mathrm{mod} \hsp... | Do you know about modular inverse ?
Here you have $\quad 2y\equiv 3 \pmod 7$
Notice that when we multiply by $4$ we get $\quad 8y\equiv 12\pmod 7$
And since $8\equiv 1\pmod 7$ and $12\equiv 5\pmod 7$ we get in the end $\quad y\equiv 5\pmod 7$
We say that $4$ is the inverse of $2$, since $2\times 4\equiv 1\pmod 7$
Now... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3428639",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
$x,y,z > 0$, prove that $ 3xyz + x^{3}+ y^{3} + z^{3} \ge 2 \left[ (xy)^{3/2} + (yz)^{3/2} + (xz)^{3/2} \right] $ $x,y,z > 0$, prove that
$$ 3xyz + x^{3}+ y^{3} + z^{3} \ge 2 \left[ (xy)^{3/2} + (yz)^{3/2} + (xz)^{3/2} \right] $$
Without using Schur's inequality,
Attempt:
By $C.S$:
$$ (xy)^{3/2} + (yz)^{3/2} + (xz)^{... | Let $x=e^a$, $y=e^b$, $z=e^c$ $\;$
and $a\ge b \ge c$ (WLOG)
Our inequality is equivalent to
$$3e^{a+b+c} + e^{3a}+ e^{3b} + e^{3c} \ge 2 \left( e^{3(a+c)/2} + e^{3(a+c)/2} + e^{3(a+b)/2} \right)$$
If $a+c\le 2b$ ,
This inequality is true by Karamata (Majorization Inequality) with convex function $f(x)=e^x$ for all... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3435099",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
If $f(x)$ is a polynomial of degree $4$ and $f(n)=n+1$ for $n=1,2,3,4$. Find $f(5)$ Question: If $f(x)$ is a polynomial of degree $4$ and $f(n)=n+1$ for $n=1,2,3,4$. Find $f(5)$
If we construct $g(x)=f(x)-(x-2)(x-3)(x-4)(x-5)$, then is it possible to find f(5)?
| Note that, $f(x)=\lambda(x-1)(x-2)(x-3)(x-4)+x+1$. Here $\lambda$ is a non-zero scalar. Hence $f(5)=\lambda 4\cdot 3\cdot 2\cdot 1+6=6+24\lambda$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3435200",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Evaluating integrals involving products of exponential and Bessel functions over the interval $(0,\infty)$ While trying to solve a challenging system of dual integral equations resulting from a mixed boundary value problem arising in a fluid mechanical problem, the four following non-trivial convergent improper integra... | I believe these integrals have a simple analytical form. I will demonstrate for $I_1$ and I hope you can see how to do the others similarly.
I write $I_1$ out as originally stated:
$$I_1 = \int_0^{\infty} d\lambda \, \lambda^{1/2} \, e^{-\lambda} \, J_1(\lambda r) \, J_{1/2}(\lambda t)$$
Note that
$$J_{1/2}(\lambda t)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3436474",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Find the smallest value of the following expression $\sqrt{(x-9)^2 +4} + \sqrt{x^2+y^2}+ \sqrt{(y-3)^2 +9}$
Find the smallest value of the following expression:
$$\sqrt{(x-9)^2 +4} + \sqrt{x^2+y^2} + \sqrt{(y-3)^2 +9}$$
I tried to derive the expression with respect to $x$ and $y$ and then equal the derivative to ze... | Use the Minkowski inequality:
$$
\begin{aligned}
&\sqrt{(x-9)^{2}+4}+\sqrt{x^{2}+y^{2}}+\sqrt{(y-3)^{2}+9}\\
=&\sqrt{(9-x)^{2}+2^2}+\sqrt{x^{2}+y^{2}}+\sqrt{3^2+(3-y)^{2}}\\
\geqslant&\sqrt{(9-x+x)^2+(2+y)^2}+\sqrt{3^2+(3-y)^{2}}\\
\geqslant&\sqrt{(9-x+x+3)^2+(2+y+3-y)^2}=13
\end{aligned}
$$
"$=$" holds iff $(9-x,2)$ a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3437202",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 1
} |
General formula for the power series of $\dfrac{1}{(1+x)^3}$ I wish to find the general formula for the following power series:
$\dfrac{1}{(1+x)^3}=1-3x+6x^2-10x^3+15x^4-21x^5+28x^6...$
The difference between the first and second term is $2$
The difference between the second and third term is $3$
The difference between... | $$\frac {1}{1+x} = 1-x+x^2-x^3+x^4-....$$
Differentiate and you get
$$ \frac {-1}{(1+x)^2} = -1+2x-3x^2+4x^3-....$$
Differentiate again and you get
$$ \frac {2}{(1+x)^3} = 2-6x+12x^2-....$$
$$ \frac {1}{(1+x)^3} = 1-3x+6x^2-....$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3437309",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Why is the graph of $8x^2+8xy+2y^2+2x+y+5=0$ empty? For the equation
$$8x^2+8xy+2y^2+2x+y+5=0$$
Comparing it with standard equation of conic section we get that
$h^2-ab=0$ and
$$\Delta= \left|\begin{array}{ccc}
a & h & g \\
h & b & f \\
g & f & c \end{array}\right| = 0$$
So as per I know it should represent a pair of... | $$ 8 \left(x + \frac{y}{2} + \frac{1}{8} \right)^2 + \frac{39}{8} $$
$$ P^T H P = D $$
$$\left(
\begin{array}{rrr}
1 & 0 & 0 \\
- \frac{ 1 }{ 8 } & 0 & 1 \\
- \frac{ 1 }{ 2 } & 1 & 0 \\
\end{array}
\right)
\left(
\begin{array}{rrr}
16 & 8 & 2 \\
8 & 4 & 1 \\
2 & 1 & 10 \\
\end{array}
\right)
\left... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3445951",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Prove $S=\{a+b\sqrt[3]{2} +c\sqrt[3]{4}:a,b,c \in \mathbb{Q}\}$ is a ring. Prove $S=\{a+b\sqrt[3]{2} +c\sqrt[3]{4}:a,b,c \in \mathbb{Q}\}$ is a ring.
My professor has gone over how to prove a subring, but I am not sure how to prove a ring.
This is what I have so far, but I am not sure if it is similar to proving a sub... | This is a subset of the real numbers with the same ring structure, you just have to see that it is closed under adition and multiplication as distribution, identity and associativity are free. Clearly, it is closed under addition. To prove that it is closed under adition we only have to check that products of (additive... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3446309",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
If $n\geq 3$, $9^n \equiv a (\mod 100)$ and $9^{n+1} \equiv b (\mod 100)$, then $a+b=90$. I noticed a pattern in the powers of 9 modulo 100.
$9^1 \equiv 9 \pmod{100}$
$9^2 \equiv 81 \pmod{100}$
$9^3 \equiv 29 \pmod{100}$
$9^4 \equiv 61 \pmod{100}$
.
.
.
and conjectured the following:
If $n\geq 3$ is an odd integer wher... | From the binomial theorem, it follows that,
for $n$ odd, $9^n=(10-1)^n\equiv10n-1\bmod100$,
and $9^{n+1}=(10-1)^{n+1}\equiv-10(n+1)+1\bmod100$.
Therefore, for $n$ odd, $9^n+9^{n+1}\equiv-10\equiv90\bmod100$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3446410",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Is $G = (\Bbb Q^*,a*b=\frac{a+b}{2})$ a group, monoid, semigroup or none of these?
Is $G = (\Bbb Q^*,a*b=\frac{a+b}{2})$ a group, monoid, semigroup or none of these?
I tried to solve it by assuming $ a,b,c \in G $ such that $a*(b*c)=(a*b)*c$. Then$$\frac{a+\frac{b+c}{2}}{2} = \frac{\frac{a+b}{2}+c}{2}$$
and by taking... | Let $a=2$ and $b=c=1$. This should be a counterexample to associativity.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3448020",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 2
} |
Computing the cycle decomposition of the composition of permutations. Let $\sigma,\tau\in S_5$ be given by
$$
\sigma = \begin{pmatrix}1&3&5\end{pmatrix}\begin{pmatrix}2&4\end{pmatrix},\quad \tau = \begin{pmatrix}1&5 \end{pmatrix}\begin{pmatrix}2&3\end{pmatrix}.
$$
I want to find the cycle decomposition of $\tau\sigma$.... | $4$ is taken to $2$ and $2$ is taken to $3$, so it's $(1243)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3448678",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Triangle sides $a,b,c$ are in arithmetic progression. Show $\sin^2(A/2)\csc2A$, $\sin^2(B/2)\csc2B$, $\sin^2(C/2)\csc2C$ are in harmonic progression
If sides $a,b,c$ of the triangle ABC are in arithmetic progression, prove that $\sin^2\frac{A}{2}\mathrm{cosec}(2A),\sin^2\frac{B}{2}\mathrm{cosec}(2B),\sin^2\frac{C}{2}\... | $$f(x)=\dfrac{2\sin2x}{1-\cos x}=\dfrac{4\sin x(1-(1-\cos x))}{1-\cos x}=\dfrac{4\sin x}{1-\cos x}-4\sin x=4\cot\dfrac x2-4\sin x$$
Using https://en.m.wikibooks.org/wiki/Trigonometry/Solving_triangles_by_half-angle_formulae and
https://en.m.wikipedia.org/wiki/Law_of_sines
$f(A)=\dfrac{4s(s-a)}\triangle -\dfrac{4a}{2R}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3450752",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Defining a function at some given point to make it continuous So I have a function
$$ f(x,y) = \frac{xy(x^2-y^2)}{x^2+y^2} $$
I have to define the function at $(0,0)$ so that it is continuous at origin. Is there any general approach we follow in this type of problems? I am not sure but I think Sandwich Theorem has to... | Please check out the new answer
$$ |xy(x^2-y^2)| <= |xy||(x^2-y^2)| <= |x||y||x^2+y^2|$$
Now,
$$ |x||y||(x^2+y^2)| = \sqrt(x^2)\sqrt(y^2)|x^2+y^2| <= (\sqrt{(x^2+y^2)})(\sqrt{(x^2+y^2)})|x^2+y^2| $$
The last expression is nothing but $(x^2+y^2)^2$ . Therefore
$$ \frac{ |xy(x^2-y^2)| }{(x^2+y^2)}<=\frac{(x^2+y^2)^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3451854",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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Smallest value of $x^2+5y^2+8z^2$ given $xy+yz+xz=-1$. The question is:
Find the smallest value of $x^2+5y^2+8z^2$ given $xy+yz+xz=-1$.
Here's what I've tried so far. Dividing by $xyz$, I get $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{-xyz}$. Squaring both sizes:
$$\left( \frac{1}{x}+\frac{1}{y}+\frac{1}{z} \right... | Let $k$ be a minimal value.
Thus, $k>0$ and $$x^2+5y^2+8z^2\geq k$$ or
$$x^2+5y^2+8z^2+k(xy+xz+yz)\geq0$$ or
$$x^2+k(y+z)x+5y^2+8z^2+kyz\geq0,$$ for which we need
$$k^2(y+z)^2-4(5y^2+8z^2+kyz)\leq0$$ or
$$(20-k^2)y^2+(4k-2k^2)yz+(32-k^2)z^2\geq0,$$ for which we need
$$20-k^2>0$$ and $$(2k-k^2)^2-(20-k^2)(32-k^2)=0$$ o... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3452821",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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If there exists a positive integer $n$ such that any element $x$ of a ring $R$ satisfies $x^{4^n+2} = x$, then every element $x$ in $R$ is idempotent Let $R = (A,+,\cdot)$ be a ring. If there exists a positive integer $n$ such that any element $x$ of a ring $R$ satisfies $x^{4^n+2} = x$, then every element $x$ in $R$ i... | Let $N$ be $4^n+2$ for short. It is an even number. Then from
$$
x = x^N=(-x)^N=-x
$$
we obtain $2x=0$ for all $x\in R$. So we work "in the characteristic two". In particular, $(x+y)^{4^n}=x^{4^n}+y^{4^n}$ for all $x,y$ in the ring. This implies:
$$
\begin{aligned}
x+1&=(x+1)^N
\\
&=(x+1)^{4^n}(x+1)^2
\\
&=\left(x^{4^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3454587",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
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Minimise $2x+y$ subject to $x^2+y^2=1$ using KKT I'm doing this exercise in preparing for the final exam in optimization:
$$\begin{align*}
\text{min} &\quad 2x+y \\
\text{s.t} & \quad x^2+y^2=1
\end{align*}$$
Could you please verify if I correctly understanding the KKT's theorem? Thank you so much for your help!
My... | Your final solution is correct. But this problem doesn't use inequalities in the constraint so KKT is not needed. Just basic Lagrangian multipliers. You should consider
$$\begin{align*}
\text{max} &\quad xy \\
\text{s.t} & \quad x+y^2\leq2\\
&\quad x,y\geq 0
\end{align*}$$
Solution can be found here for your practice:... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3456783",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Split into partial fractions $\frac{3x^2+2x+1}{(x+2)(x^2+x+1)^2}$ This is from "Calculus Made Easy", Exercises 10, Question 15 (page 147). I've worked this one over and over and still haven't made progress.
This is my initial setup:
$$\frac{3x^2+2x+1}{(x+2)(x^2+x+1)^2} = \frac{A}{(x+2)} + \frac{Bx+C}{(x^2+x+1)} + \fra... | Now that you know $A=1$ set $A$ equal to $1$. You get
$$3x^2+2x+1 = (x^2+x+1)^2 + (Bx+C)(x+2)(x^2+x+1) + (Dx+E)(x+2)$$
$$3x^2+2x+1 - (x^4+2x^3+3x^2+2x+1) = (Bx+C)(x+2)(x^2+x+1) + (Dx+E)(x+2)$$
$$-x^4-2x^3 = (Bx+C)(x+2)(x^2+x+1) + (Dx+E)(x+2)$$
$$-x^3(x+2) = (Bx+C)(x+2)(x^2+x+1) + (Dx+E)(x+2)$$
$$-x^3 = (Bx+C)(x^2+x+1) ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3459213",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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How to show $B^2=B$ when $rank(B) + rank(I_d - B) = d$? I have a question on this specific question from the past entrance examination of a university.
https://www.ism.ac.jp/senkou/kakomon/math_20190820.pdf
Related post: How to calculate eigenvalues of a matrix $A = I_d - a_1a_1^T - a_2a_2^T$
Problem
$≥3$, $_$ is an id... | Let $B'=I_d-B$. Observe that if $v\in\ker B\cap \ker B'$ then $v=Bv+B'v=0$, hence $\dim(\ker B\cap \ker B')=0$. Thus
$$
\dim(\ker B+\ker B')=\dim \ker B+\dim \ker B'=(d-\textrm{rank} B)+(d-\textrm{rank} B')=d.
$$
On the other hand since $B$ and $B'$ commute, every $v\in \ker B+\ker B'$ satisfies $BB'v=0$. Thus $BB'=0$,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3460891",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Classic Complex Numbers - Given $z+\frac 1z=2\cos 3^\circ$, find least integer greater than $z^{2000}+\frac 1{z^{2000}}$ Given that $z$ is a complex number such that $z+\frac 1z=2\cos 3^\circ,$ find the least integer that is greater than $z^{2000}+\frac 1{z^{2000}}.$
Solution: We have $z=e^{i\theta}$, so $e^{i\theta}+\... | Note that, given $z+\frac1z = 2\cos a$ for any $a$,
$$z^n+\frac1{z^n}=2\cos (na)$$
holds true, which is shown recursively below,
$$z^n+\frac1{z^n}= \left(z+\frac1z\right)\left(z^{n-1}+\frac1{z^{n-1}}\right)
-\left(z^{n-2}+\frac1{z^{n-2}}\right)$$
$$=2\cos a \cdot 2\cos [(n-1)a] - 2\cos [(n-2)a] $$
$$=2[\cos (na) + \c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3463278",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Determining the value of $\sum_{n=2}^\infty \frac{1}{n^n-1}$ $$\displaystyle\sum_{n=1}^\infty\sum_{m=2}^\infty \frac{1}{m^{mn}}=\sum_{n=1}^\infty\left(\frac{1}{2^{2n}}+\frac{1}{3^{3n}}+\frac{1}{4^{4n}}+\cdots\right) \tag{$\star$}$$
I have a strong suspicion that the above summation converges, although I'm not sure how ... | If you want more figures $$\displaystyle \sum_{n=2}^\infty \frac{1}{n^n-1}=0.37605925334160927467565605197313357724916639675$$ All the digits are obtained summing up to $31$.
An amazing approximation of it is
$$\frac{76 \pi ^2+651 \pi-563 }{308 \pi ^2+481 \pi+1385}$$ which is in a relative error of $2.56\times 10^{-18}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3464100",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
How can I prove analytically that the golden ratio is less then $\frac{\pi^2}{6}$. Or stated in other terms, prove that
$$\frac{1+\sqrt{5}}{2} < \sum_{n=1}^{\infty}\frac{1}{n^2}$$
| \begin{align}
\frac{1+\sqrt{5}}{2}\lt\frac{\pi^2}{6}
\iff&3+3\sqrt{5}\lt\pi^2\\
\iff&3\sqrt{5}\lt\pi^2-3\\
\iff&(3\sqrt{5})^2\lt(\pi^2-3)^2\quad(\text{both sides are positive})\\
\iff&45\lt\pi^4-6\pi^2+9\\
\iff&\pi^4-6\pi^2-36\gt0\\
\end{align}
Using, for example, $3.14\lt\pi\lt3.15$ we have that
$$\pi^4-6\pi^2-36\gt3.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3464638",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
What is the probability of the office being deemed functional? An IT office consists of 10 computers of which exactly 4 are working. To check if the office is functional, officials inspect four of the computers picked at random (without replacement). The office is deemed functional if at least three of the four compute... | The two answers are numerically exactly the same. Yours is $\frac{576}{5040}+\frac{24}{5040}=\frac{600}{5040}$ and the provided solution is $\frac{24}{210}+\frac{1}{210}=\frac{25}{210}$. Both are $\frac{5}{42} \approx 0.1190476$.
The factor of $4!=24$ in the numerator and denominator is simply the number of ways of o... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3466047",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Using Fermat's Little Theorem to Show Divisibility I was asked to prove, using Fermat's Little Theorem, that $11|5^{10n+8}-4$ for $n\ge0$. I proved it but I was wondering whether there's an easier way (still using Fermat's). Here is my proof:
\begin{alignat}{3}
11|5^{10n+8}-4&\iff5^{10n+8}-4&&\equiv0 &&&\mod11\\
\quad&... | Much easier way!
By FLT $5^{10} \equiv 1 \pmod{11}$ so $5^{10n+8}\equiv 5^8$ and $5^{10n +8} -4 \equiv 5^8 -4\pmod {11}$.
So you just have to show that one case the $5^8 \equiv 4 \pmod {11}$. Then every case will be $5^{10n + 8} - 4\equiv 0 \pmod{11}$
Admittedly that requirse calculations but there are 3 ways, eac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3467319",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 2
} |
The differential equation$ \frac{d^2 x}{dt^2}=f(x)$ We are supposed to solve the diff. eq.
$$x''=\frac{1}{x^3}$$
for $x(0)=1$ and $x'(0)=1$ using a particular approach. I've gotten so far:
Using the interval $I=(0,\infty)$ and $a \in I$ we define $U:I\to \mathbb{R}$:
$$U(x):=-\int_a^x\frac{1}{t^3}dt = \frac{1}{2x^2}-\... | $$\frac{d^2 x}{dt^2}=\frac{1}{x^3}$$
Multiply by $2 \frac{dx}{dt}$ on both sides to get
$$2 \frac{dx}{dt} \frac{d^2x}{dt^2}=2 \frac{dx}{dt} \frac{1}{x^3} \implies \frac{d}{dt} \left( \frac{dx}{dt}\right)^2=\frac{2}{x^3}\frac{dx}{dt}.$$
Next, integrate w.r.t.t both sides
$$\int \frac{d}{dt} \left( \frac{dx}{dt}\right)^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3468341",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
What is the probability that $THTH$ occurs before $HTHH$ in an infinite sequence of coin flips?
What is the probability that $THTH$ occurs before $HTHH$ in an infinite sequence of coin flips?
The expected number of flips until you first see $THTH$ is $6$, while the expected number until you first see $HTHH$ is $10$.... | We have states $\{H,T,HT,TH,THT,HTH,THTH,HTHH\}$ which are the prefixes of your strings. We can write down a Markov transition matrix for these states:
$$A = \frac{1}{2} \begin{pmatrix} 1 & 0 & 1 & 0 & 0 & 0 & 0 & 0\\
0 & 1 & 0 & 1 & 0 & 0 & 0 & 0\\
0 & 1 & 0 & 0 & 0 & 1 & 0 & 0\\
1 & 0 & 0 & 0 & 1 & 0 & 0 & 0\\
0 & 1 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3469403",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Limit of $\lim_{x \to 0} \frac{(1+x^5)^{10} -1}{(\sqrt{1+x^3}-1)(\sqrt[5]{1+x^2}-1) }$ I tried using symbolab to get the limit of
$$\lim_{x \to 0} \frac{(1+x^5)^{10} -1}{(\sqrt{1+x^3}-1)(\sqrt[5]{1+x^2}-1) }$$
but it couldn't solve it. With WolframAlpha I got $100$ for the limit.
Can someone show me how it's done "per ... | \begin{align*}
&\dfrac{(1+x^{5})^{10}-1}{(\sqrt{1+x^{3}}-1)(\sqrt[5]{1+x^{2}}-1)}\\
&=\dfrac{(\sqrt{1+x^{3}}+1)((1+x^{5})-1)((1+x^{5})^{9}+\cdots+1)((1+x^{2})^{4/5}+\cdots+1)}{((1+x^{3})-1)((1+x^{2})-1)}\\
&=(\sqrt{1+x^{3}}+1)((1+x^{5})^{9}+\cdots+1)((1+x^{2})^{4/5}+\cdots+1)\\
&\rightarrow 2\cdot10\cdot 5\\
&=100.
\en... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3469782",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
How to Solve this Simultaneous DE? Using laplace transform How to solve this Simultaneous DE using laplace transform
$x''+ y''+3x = 15e^{-t}$
$y''-4x'+3y = 15\sin(2t)$
with initial values $x(0)=35$, $x'(0)= -48$ and $y(0)=27$?
| Notice that:
\begin{align}
x''+ y''+3x = 15e^{-t} & \Rightarrow_{\mathcal{L}} s^2 X - sx(0) - x'(0) + s^2 Y - sy(0) + y'(0) + 3X= 15 \frac{1}{s+1}\\
& \Rightarrow s^2 X - 35s + 48 +s^2 Y - 27s + a + 3X = \frac{15}{s+1}.
\end{align}
and
\begin{align}
y''-4x'+3y = 15\sin(2t) & \Rightarrow_{\mathcal{L}} s^2 Y - sy(0) + y'... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3470923",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
What is the eccentricity of the conic given by $r = \frac{2}{1 + \cos \theta - \sin \theta}$? A very simple question. Usually, Conic section in polar graph given as,
$$r = \frac{ke}{1 - e \cos\theta}$$
or $\sin\theta$, or something along this way.
However, given
$$r = \frac{2}{1 + \cos \theta - \sin \theta}$$
how... | You have
\begin{align}
\cos\theta-\sin\theta&=-\sqrt2\,\left(-\frac1 {\sqrt2}\cos\theta+\frac1{\sqrt2}\sin\theta\right)\\ \ \\&=-\sqrt2\,\left(\cos\frac{3\pi}4\cos\theta+\sin\frac{3\pi}4\sin\theta\right)\\ \ \\ &
=-\sqrt2\,\cos\left(\theta-\frac{3\pi}4\right)
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3475764",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to solve the recursion: $f(n)=3-\frac{1}{f(n-1)}$? I have some problems with the recursion: $f(n) = 3 - \frac{1}{f(n-1)} $, and the initial is $f(1)= \frac{5}{2}$.
Can anyone give me some hints of this?Thanks!!!
| The given equation is
$$f(n)f(n-1)-3f(n-1)+1=0 \implies f(n-1)(f(n)-3)+1=0.$$
Let $$f(n)-3=\frac{g(n-1)}{g(n)},$$ then
we get $$g(n)+3g(n-1)+g(n-2)=0.$$
Now we take $g(n)= x^n$, then
$$x^2+3x+1=0 \implies x= a,b =\frac{-3 \pm \sqrt{5}}{2}$$
so $g(n)=p a^n+q b^n$. So $$f(n)=\frac{p a^{n-1}+ q b^{n-1}}{pa^n+qb^n}+3= \f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3476948",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Limit of $\frac { 2^{\sqrt{ (\ln n)^2+ \ln n^2}}}{n^2+1}$ as $n\to\infty$ I've tried to solve the limit
$$ \lim_{n \to \infty} \frac { 2^{\sqrt{ (\ln n)^2+ \ln n^2}}}{n^2+1}$$
but I'm not sure.
$$ \frac { 2^{\sqrt{ (\ln n)^2+ \ln n^2}}}{n^2+1} = \frac { 2^{\sqrt{ (\ln n)^2+ 2\ln n}}}{n^2+1} = \frac { 2^{\ln n \sqr... | Your intuition is correct, but the part where you employ the symbol $\sim$ lacks a rigorous justification. To fully go into detail, you have the following expressions and estimates:
$$\frac{2^{\ln{n}\sqrt{1+\frac{2}{\ln{n}}}}}{n^2+1}=\frac{n^{{\ln{2}} \sqrt{1+\frac{2}{\ln{n}}}}}{n^2+1} \leqslant \frac{n^{\sqrt{3}\ln{2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3477804",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Find value of $a_1$ such that $a_{101}=5075$ Let $\{a_n\}$ be a sequence of real numbers where $$a_{n+1}=n^2-a_n,\, n=\{1,2,3,...\}$$
Find value of $a_1$ such that $a_{101}=5075$.
I have
$$a_2=1^2-a_1$$
$$a_3=2^2-a_2=2^2-1^2+a_1$$
$$a_4=3^2-2^2+1^2-a_1$$
$$a_5=4^2-3^2+2^2-1^2+a_1$$
$$\vdots$$
$$a_{101}=100^2-99^2+98^... | Yes, your solution is correct. Another method of solution is to note that if $$a_{n+1} = n^2 - a_n,$$ we want to find some (possibly constant) function of $n$ such that $$a_{n+1} - f(n+1) = -(a_n - f(n)).$$ This of course implies $$f(n+1) + f(n) = n^2.$$ A quadratic polynomial should do the trick: suppose $$f(n) = ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3482467",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.