Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Find the sum to $n$ terms of the series: $1^2.1+2^2.3+3^2.5+....$ Find the sum to $n$ terms of the series:
$$1^2.1+2^2.3+3^2.5+.....$$
My Attempt:
Here, $n^{th}$ term of $1,2,3,....=n$
$n^{th}$ term of $1^2,2^2,3^2,....=n^2$
Also, $n^{th}$ term of $1,3,5,....=2n-1$
Hence, $n^{th}$ term of the given series is $t_n=n^2(2... | $$\sum\limits_{j=0}^n j^2(2j-1)=2\sum \limits_{j=0}^n j^3-\sum \limits_{j=0}^n j^2=2(\dfrac{n(n+1)}{2})^2-\dfrac{n(n+1)(2n+1)}{6}.$$ You can prove this formulae for $\sum j^3$ and $\sum j^2$ by induction.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Simplifying $\frac{1}{2\cdot\sqrt{e^x}}\cdot e^x$ We have to simplify:
$$\frac{1}{2\cdot\sqrt{e^x}}\cdot e^x$$
I came to the conclusion that the answer would be:
$$\frac{e^x}{2\cdot\sqrt{e^x}}$$
But I was wrong and it was:
$$\frac{\sqrt{e^x}}{2}$$
Where did I go wrong and why?
Thanks
| $$\require{cancel}\frac{1}{2\cdot\sqrt{e^x}}\cdot e^x = \frac{1}{2\cdot\sqrt{e^x}}\cdot e^x\cdot\frac{\sqrt{e^x}}{\sqrt{e^x}} = \frac{1}{2}\cdot\sqrt{e^x}\cdot\frac{e^x}{\sqrt{e^x}\cdot\sqrt{e^x}} = \frac{1}{2}\cdot\sqrt{e^x}\cdot\frac{\cancel{e^x}}{\cancel{e^x}} = \frac{1}{2}\cdot\sqrt{e^x}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Given that $P(x)$ is a polynomial such that $P(x^2+1) = x^4+5x^2+3$, what is $P(x^2-1)$? How would I go about solving this? I can't find a clear relation between $x^2+1$ and $x^4+5x^2+3$ to solve $P(x^2-1)$.
| Hint: let $y=x^2$, then the given condition writes as $P(y+1)=y^2+5y+3$ and it follows that $P(x^2-1)=P\big((y-2)+1\big)= (y-2)^2+5(y-2)+3 = \ldots$
| {
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"timestamp": "2023-03-29T00:00:00",
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What is the integral of $\int\frac{dx}{\sqrt{x^3+a^3}}$?
What is the integral of $$\int\frac{dx}{\sqrt{x^3+a^3}}?$$
I came across this integration in a physics problem. I suspect role of complex numbers here.
'$a$' is a constant
| It's this:
$$\frac{2 \sqrt[6]{-1} \sqrt[3]{a^3} \sqrt{(-1)^{5/6} \left(\frac{\sqrt[3]{-1} x}{\sqrt[3]{a^3}}-1\right)} \sqrt{\frac{(-1)^{2/3} x^2}{\left(a^3\right)^{2/3}}+\frac{\sqrt[3]{-1} x}{\sqrt[3]{a^3}}+1} F\left(\left.\sin ^{-1}\left(\frac{\sqrt{-\frac{(-1)^{5/6} x}{\sqrt[3]{a^3}}-(-1)^{5/6}}}{\sqrt[4]{3}}\right)\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2685265",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Is there a direct proof of the inequality $cyclic \sum\frac {a^2+1}{b+c}\ge 3$ Let $a,b,c>0$ reals.
Prove that
$$cyclic \sum\frac {a^2+1}{b+c}\ge 3$$
I proved it using Nesbitt inequality
$$cyclic \sum \frac {a}{b+c} \ge \frac {3}{2} $$
and the fact that
$$a+\frac {1}{a}\ge 2$$
But i would like to know if there is a str... | By AM-GM and C-S we obtain:
$$\sum_{cyc}\frac{a^2+1}{b+c}=\sum_{cyc}\frac{a^2}{b+c}+\sum_{cyc}\frac{1}{b+c}\geq2\sqrt{\sum_{cyc}\frac{a^2}{b+c}\sum_{cyc}\frac{1}{b+c}}\geq$$
$$\geq2\sqrt{\frac{(a+b+c)^2}{\sum\limits_{cyc}(b+c)}\sum_{cyc}\frac{1}{b+c}}=\sqrt{2(a+b+c)\sum_{cyc}\frac{1}{b+c}}=$$
$$=\sqrt{\sum\limits_{cyc}... | {
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What is the sum of $\sum_{n=1}^\infty\left(\frac{2n+1}{n^4+2n^3+n^2}\right) = $? $$\sum_{n=1}^\infty\left(\frac{2n+1}{n^4+2n^3+n^2}\right)=\sum_{n=1}^\infty\left(\frac{2n+1}{n^2}\frac1{{(n+1)}^2}\right)$$
I assume that I should get a telescoping sum in some way, but I'm couldn't find it yet.
| You'll get a telescoping series, but that is likely just only one of the ingredients for a solution. First, I'll separate the numerator in two parts:
$$
\frac{2n+1}{n^2}\frac{1}{(n+1)^2}=\frac{n+n+1}{n^2(n+1)^2}
$$
$$
\frac{2n+1}{n^2}\frac{1}{(n+1)^2}=\frac{1}{n(n+1)^2}+\frac{1}{n^2(n+1)}
$$
And of course your telescop... | {
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How to deal with $(-1)^{k-1}$ It's a problem on mathematical induction.
$$1^2-2^2+3^2-.....+(-1)^{n-1}n^2=(-1)^{n-1}\frac{n.(n+1)}{2}$$
I have proved it for values of $n=1,2$.
Now I assume for $n=k$
$$P(k):1^2-2^2+3^2-.....+(-1)^{k-1}k^2=(-1)^{k-1}\frac{k.(k+1)}{2}$$.
$$P(k+1):1^2-2^2+3^2-.....+(-1)^{k-1}k^... | You are making a mistake, in that you are assuming that $(-1)^k=1$.
You have
\begin{align}
(-1)^{k-1}\frac{k.(k+1)}{2}+(-1)^k(k+1)^2&=(-1)^{k-1}\frac{(k+1)}{2} [k-(2k+2)]\\ \ \\
&=(-1)^{k-1}\frac{(k+1)}{2} [-(k+2)]\\ \ \\
&=(-1)^{k}\frac{(k+1)(k+2)}{2} \\ \ \\
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Quadratic equation formula help / simplification I have this quadratic equation,
$ x^{2} + \frac{10}{3}x -\frac{80}{3} = 0 $
I use the quadratic formula to solve and simplify
$-10 \pm \frac{\sqrt{100-(4)(3)(-80)}}{6}$ = $ \frac{-10 \pm \sqrt{1060}}{6}$
my book says it should simplify to
$ \frac{1}{3} ( -5 \pm \sqrt{... | $$x=\frac{-10 \pm \sqrt{100-(4)(3)(-80)}}{6} =\frac{-2\times 5 \pm \sqrt{4\times25-(4)(3)(-80)}}{6} =\frac{1}{6}\left(-2\times 5 \pm 2\sqrt{25+240} \right) =\frac{2}{6}\left(-5 \pm \sqrt{265} \right)=\frac{1}{3}\left(-5 \pm \sqrt{265} \right)$$
| {
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"timestamp": "2023-03-29T00:00:00",
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How to find value of $A'$ in eliminating cross product terms Quadratic Curve Rotation? I was studying conics and came around the topic of eliminating cross-product terms when rotating coordinates of a quadratic curve of the form $$A x^2 + B x y + C y^2 + D x + E y + F = 0$$ where
$$\begin{align*}
A x^2 &= A\left(\cos(\... | Just expand and collect the expressions right above. For example, focusing on the $\,\color{red}{{x'}^2}\,$ terms:
$$
\begin{align}
A x^2 + B x y + C y^2 + D x + E y + F &= \color{red}{A}\left(\color{red}{\cos^2(\alpha) x'^2} -2\cos(\alpha)\sin(\alpha) x' y' + \sin^2(\alpha) y'^2\right) \\
&\;\; + \color{red}{B}\left(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2696260",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Inequality $\frac{a}{\sqrt{bc}}\cdot\frac1{a+1}+\frac{b}{\sqrt{ca}}\cdot\frac1{b+1}+\frac{c}{\sqrt{ab}}\cdot\frac1{c+1}\leqslant\sqrt2.$ Let $a,b,c>0$ and $\frac1{a+1}+\frac1{b+1}+\frac1{c+1}=1.\qquad $ Prove $$\frac{a}{\sqrt{bc}}\cdot\frac1{a+1}+\frac{b}{\sqrt{ca}}\cdot\frac1{b+1}+\frac{c}{\sqrt{ab}}\cdot\frac1{c+1}\l... | The hint.
After substitution $a=\frac{y+z}{x}$, $b=\frac{x+z}{y}$ and $c=\frac{x+y}{z}$, where $x$, $y$ and $z$ are positives,
and squaring of the both sides use AM-GM.
Indeed, let $a=\frac{y+z}{x}$ and $b=\frac{x+z}{y}$, where $x$, $y$ and $z$ are positives.
Thus, the condition gives $$\frac{1}{\frac{y+z}{x}+1}+\frac{... | {
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"timestamp": "2023-03-29T00:00:00",
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Show that the equation $y^2 = x^3 +xz^4$ has no solutions in nonzero integers $x,y,z$.
Show that the equation $y^2 = x^3 +xz^4$ has no solutions in nonzero integers $x,y,z$.
There is given hint: Suppose that there is a solution. First show that it can be reduced to a solution satisfying $\gcd(x,z)=1$. Then use the f... | This is not much more than a lengthy remark, but it's worth posting as an answer, I think, because it shows that proving this is at least as hard as proving the strong version of Fermat's Last Theorem for $n=4$. That is, suppose that $a^4+b^4=c^2$ with $abc\not=0$. Then, letting $x=a^2$, $z=b$, and $y=ac$, we first hav... | {
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"url": "https://math.stackexchange.com/questions/2702461",
"timestamp": "2023-03-29T00:00:00",
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How many integer solutions of $x_1+x_2+x_3+x_4=28$ are there with $-10 \leq x_i \leq 20$? I am trying to solve the following question. I have worked through a similar question but the question I have worked through the range was all positive, however this one has a negative side of the range so I am not quite sure how ... | How many integers solutions of $$x_1 + x_2 + x_3 + x_4 = 28$$ are there that satisfy $-10 \leq x_k \leq 20$ for $1 \leq k \leq 4$?
We can convert this to an equivalent problem in the nonnegative integers.
$$x_1 + x_2 + x_3 + x_4 = 28 \tag{1}$$
Let $y_k = x_k + 10$, where $1 \leq k \leq 4$. Then each $y_k$ is a nonnega... | {
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Solve $\sqrt {x^2-3}=x-3$ in $\mathbb R$
Solve $\sqrt {x^2-3}=x-3$ in $\mathbb R$
My attempt:
$|x^2-3|=(x-3)^2$
So $-(x^2-3)=(x-3)^2$ or $(x^2-3)=(x-3)^2$
If $-(x^2-3)=(x-3)^2=x^2+9-6x$
So no solutions in $\mathbb R$
And if $(x^2-3)=(x-3)^2$
So $x^2-3=x^2+9-6x$
Now, can I delete $x^2$ with $x^2$ ? Like this
$x^2-x... | What you solved is correct but you must notice that $$\sqrt {x^2-3}\ge 0\Rightarrow x-3\ge 0\Rightarrow x\ge 3$$
And hence I'd you get any solution less than $3$ then you have to reject it.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2704837",
"timestamp": "2023-03-29T00:00:00",
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Finding value of $\frac{49b^2-33bc+9c^2}{a^2}$
If $a,b,c$ are positive real numbers such that
$a^2+ab+b^2=9, b^2+bc+c^2=52,$
$c^2+ac+a^2=49$. Then finding $\displaystyle \frac{49b^2-33bc+9c^2}{a^2}$ is
Try: let $O$ be a point inside the Triangle $ABC$ such that angle $OAB$ and angle $OBC$ and $OCA$ is $120^\circ$. So... | We can treat the equations as quadratic with variables $a,b,c$, and thus we solve for each of their values using the quadratic formula:
$$\begin{align}
a\to& \frac{1}{2} \left(\sqrt{36-3 b^2}-b\right)\tag{1}\\
b\to& \frac{1}{2} \left(\sqrt{208-3 c^2}-c\right)\tag{2}\\
c\to& \frac{1}{2} \left(\sqrt{196-3 a^2}-a\right)\t... | {
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Spivak's Calculus: Chapter 1, Problem 18b (Quadratic determinant less than zero) The problem in question is as follows:
18b Suppose that $b^2 -4c \lt 0$. Show that there are no numbers $x$ that satisfy $x^2 + bx + c = 0$; in fact, $x^2 + bx + c \gt 0$ for all $x$. Hint: complete the square.
Trying to apply the hint, I ... | From here: $(x + \frac{b}{2})^2 = \frac{b^2 - 4c}{4}$ we get $b^2-4ac \geq 0$ and not $>$. So if $b^2-4c<0$ there is no real solution. So else from that $>$ your conclusion is correct.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Calculating pm problem: $5$ man and $5$ women running in a race Let's say there are $5$ man and $5$ women running in a race. Let the variable X that indicates the highest position in ranking for a woman. So, for example, if $X = 1$, that means that a woman took first place(any woman) and if $X = 6$, that means that wom... | The women are assumed to be indistinguishable. There are ${10 \choose 5}$ ways to place them in $10$ empty slots. You're correct that $6$ is highest value $X$ can take on.
If we are assuming every one of the $10$ people are equally fast, then the probability that a woman is in the first position is $\frac{1}{2}$ by sy... | {
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Minimum of product of three trigonometric function
If $x,y,z$ are real numbers with $x\geq y\geq z \geq 15^\circ$ and $x+y+z=90^\circ$, then find the range of $\cos x\cos y\cos z$.
I tried Jensen inequality and AM-GM inequality:
$$3(\cos x\cos y\cos z)^{\frac{1}{3}}\leq \cos x+\cos y+\cos z\leq 3 \cos\bigg(\frac{x+y+... | The least value of function
$$f(x,y) = \cos x\cos y \cos \left(\dfrac\pi2 - x-y\right),$$
or
$$f(x,y) = \cos x\cos y \sin(x+y),$$
can be achieved in the stationary points of $f(x)$ or in the bounds of the domain.
The stationary points can be defined from the system $f'_x = f'_y = 0,$ or
\begin{cases}
-\sin x \cos y \... | {
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"timestamp": "2023-03-29T00:00:00",
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Evaluate : $\sum_{k=0}^{\infty}{1\over 4+8k+6k^2+2k^3}$ I am attempting a calculation for an integral and encounter this series:
$$S=\sum_{k=0}^{\infty}{1\over 4+8k+6k^2+2k^3}$$
but I am stuck and find it difficult to evaluate its sum.
Can someone offer suggestions?
My Attempt:
Factorising:
$$4+8k+6k^2+2k^3 = 2(k+1)(k... | Observe you have
\begin{align}
4+8k+6k^2+2k^3 =2(k+1)(k+1+i)(k+1-i)
\end{align}
which means
\begin{align}
\frac{1}{4+8k+6k^2+2k^3} = \frac{1}{4i(k+1)(k+1-i)}-\frac{1}{4i(k+1).(k+1+i)}
\end{align}
Then we have
\begin{align}
\sum^\infty_{k=0}\frac{1}{4+8k+6k^2+2k^3} =&\ \frac{1}{4}\sum^\infty_{k=1}\left( \frac{-i}{k(k-i)... | {
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"timestamp": "2023-03-29T00:00:00",
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How to find the sum of: $a^2+(a+d)^2+(a+2d)^2+(a+3d)^2+\cdots+[a+(n-1)d]^2?$ Given the arithmetic series with a common difference, d, first term, a, and n is the nth terms.
Deriving the formula for $S_n$
$$S_n=a+(a+d)+(a+2d)+(a+3d)+\cdots+[a+(n-1)d]\tag1$$
Reverse the sum
$$S_n=[a+(n-1)d]+\cdots+(a+d)+a\tag2$$
$(1)+(2)... | The only method that I can think of to do it is by using the results that
$$
1 + 2 + \dots + n = \frac{n(n + 1)}{2}
$$
and
$$
1^2 + 2^2 + \dots + n^2 = \frac{n(n + 1)(2n + 1)}{6}.
$$
We then have that
$$ \begin{align}
T_n & = a^2 + (a + d)^2 + \dots + (a + (n - 1)d)^2 \\
& = a^2 + (a^2 + 2ad + d^2) + (a^2 + 4ad ... | {
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"timestamp": "2023-03-29T00:00:00",
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We throw a die $8$ times. What is the probability of obtaining exactly two $3$s, three $1$s, three $6$s? We throw a die $8$ times. What is the probability of obtaining exactly two $3$s, three $1$s, three $6$s?
My work:
The sample space $ S:$ "The set of all solutions of throwing the die [eight times]" and $|S|:6^8$... | To find the number of ways to get two 3s, three 1s and three 6s can also be seen as a simple permutation problem of $8$ digits.
Imagine the result of the $8$ throws is recorded as a sequence of $8$ numbers: $d_1 d_2 \ldots d_8$.
Now you want to know in how many different ways can occur the numbers $3,3,1,1,1,6,6,6$ in ... | {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Why am I getting a negative sum? $\frac{1}{1\cdot2} - \frac{1}{2\cdot3}+\frac{1}{3\cdot4}-\frac{1}{4\cdot5}+ \cdots$ The infinite sum I want to find.
$$\frac{1}{1\cdot2} - \frac{1}{2\cdot3}+\frac{1}{3\cdot4}-\frac{1}{4\cdot5}+ \cdots$$
My attempt at it
$$\begin{align} &\;\sum_{n=1}^{\infty} \frac{1}{(2n-1)(2n)} - \su... | Yes, the sum is positive. Note that instead of
$$= \sum_{n=1}^{\infty} \left(\frac{1}{2n-1} + \frac{1}{2n+1}\right) - 2 \sum_{n=1}^{\infty} \frac{1}{2n}$$
which is indefinite since the series are divergent, you should have the limit as $N$ goes to $\infty$ of
$$\sum_{n=1}^{N} \left(\frac{1}{2n-1} + \frac{1}{2n+1}\righ... | {
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"timestamp": "2023-03-29T00:00:00",
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Find the number of real solutions to the system of equations $x=\frac{2z^2}{1+z^2},y=\frac{2x^2}{1+x^2},z=\frac{2y^2}{1+y^2}$ My approach is naive:
Given
$x=\frac{2z^2}{1+z^2},y=\frac{2x^2}{1+x^2},z=\frac{2y^2}{1+y^2}$,
$[\frac{1}{x}+\frac{1}{y}+\frac{1}{z}]=\frac{1}{2}\cdot[\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}]+... | We have $x\ge0$, $y\ge0$ and $z\ge0$.
One possible solution is $x=y=z=0$.
If one of $x,y,z$ is non-zero, the other two are also non-zero.
Now suppose that all of them are non-zero.
$$xyz=\frac{8x^2y^2z^2}{(1+x^2)(1+y^2)(1+z^2)}\le\frac{8x^2y^2z^2}{(2x)(2y)(2z)}=xyz$$
The equality holds if and only if $x^2=y^2=z^2=1$ an... | {
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Prove that $1+\frac{1}{2^3}+\frac{1}{3^3}+\cdots+\frac{1}{2018^3}<\frac{5}{4}$ Question: Prove that $1+\frac{1}{2^3}+\frac{1}{3^3}+\cdots+\frac{1}{2018^3}<\frac{5}{4}$
Attempt: When I tried solving this problem, I paired up $1$ with $\frac{1}{2018^3}$, $\frac{1}{2^3}$ with $\frac{1}{2017^3}$ etc. but it wasn't useful.
| Hint. Note that
$$1+\frac{1}{2^3}+\frac{1}{3^3}+\cdots+\frac{1}{2018^3}<1+\sum_{n=2}^{2018}\frac{1}{n^3-n}$$
and the sum on the right is easy to be found (it's telescopic).
| {
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"answer_id": 1
} |
Solve system of equations: $2x^2-4y^2-\frac{3}{2}x+y=0 \land 3x^2-6y^2-2x+2y=\frac{1}{2}$ $$
\begin{cases}
2x^2-4y^2-\frac{3}{2}x+y=0 \\
3x^2-6y^2-2x+2y=\frac{1}{2}
\end{cases}
$$
I multiplied the first with $-6$ and the second with $4$ and get two easier equations:
$9x-6y=0 \land -8x+8y=2 $ and out of them I get th... | If $2x^2-4y^2-\frac32x+y=0$ and $3x^2-6y^2-2x+2y=\frac12$, then$$3\times(2x^2-4y^2-\frac32x+y)-2\times(3x^2-6y^2-2x+2y)=-1;$$in other words, $-\frac x2-y=-1$. So, replace $y$ with $1-\frac x2$ in the first equation, and you'll get a quadratic equation whose roots are $-3$ and $1$. So, the solutions of the system are $\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2725579",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Simple approaches to prove that $\lim\limits_{x\to 0}\left(\frac{1}{x^2}-\frac{1}{\sin^2 x}\right)=-\frac13\ $?
Find $\lim_{x\to 0}(\frac{1}{x^2}-\frac{1}{\sin^2 x})$
My attempt:
$\lim_{x\to 0}(\frac{1}{x^2}-\frac{1}{\sin^2 x})$
=$\lim_{x\to 0}(\frac{\sin^2 x -x^2}{x^2 \sin^2 x})$ ($\frac{0}{0}$ form)
Applying L'Ho... | $$=\left(\frac{x}{\sin x}\right)^2\left(\frac{\sin x}{x}+1\right)\left(\frac{\sin x -x}{x^3}\right)\to 1^2(1+1)\left(-\frac{1}{6}\right)=-\frac{1}{3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2731206",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Ternary strings with 3 or more consecutive 2's I'm trying to figure out how to find the number of ternary strings of length $n$ that have 3 or more consecutive 2's.
So far I've been able to establish that there is $n(2^{n-1})$ with a single 2.
And I think (but am not certain) that this can be extrapolated to the number... |
We consider the alphabet $V=\{0,1,2\}$ and we are looking for the number $c_n$ of strings of length $n$ having runs of $2$ less than three. The wanted number is $$a_n=3^n-c_n$$
We derive a generating function $C(z)=\sum_{n=0}^\infty c_nz^n$ from which the number $a_n$ can be obtained easily since
\begin{align*}
a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2732562",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
The integer part of $ \sum_{k=2}^{9999} \frac{1}{\sqrt{k}} $ For $k>0 $
$ \frac{1}{2\cdot \sqrt{k+1}}<\sqrt{k+1}-\sqrt{k}<\frac{1}{2\cdot \sqrt{k}} $,
the integer part of $ \sum_{k=2}^{9999} \frac{1}{\sqrt{k}} $
My attempt here was not complying with any of the options given as my attempt was to find the min. value fo... | It follows from
$$ \frac{1}{2\sqrt{k+1}} \le \sqrt{k+1} - \sqrt{k} \le \frac{1}{2\sqrt{k}}$$
that
\begin{equation}
\sum_{k=1}^{9998}\frac{1}{2\sqrt{k+1}} = \sum_{k=2}^{9999}\frac{1}{2\sqrt{k}}
\le \sqrt{9999}-1
\end{equation}
and
\begin{equation}
\sqrt{10000}-\sqrt{2} \le \sum_{k=2}^{9999}\frac{1}{2\sqrt{k}}.
\end{equ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2732776",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Integral of a piecewise function
Check whether the integral $\int_{-1}^2{f(x)[2x+1]}dx$ exists where
$$f(x)=
\begin{cases}
x+2, &\text{$x<0$}\\
1, & \text{$x=0$}\\
xe^{x^2},& \text{$x>0$}
\end{cases}$$
and $[x]$ denotes the greatest integer function. If it exists then calculate the integral.
I'm clueless about h... | Because of the fact that the discontinuities of $f(x)$ are of measure zero, the function is Riemann integrable. A proof of this exists here. This means we can validly take the integral of $f(x)$ on either side of the discontinuity and sum those:
$$\int_{-1}^{2} f(x)[2x+1]dx = \int_{-1}^{0}(x+2)[2x+1]dx + \int_{0}^{2}xe... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2735412",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
On the series $\sum \limits_{n=1}^{\infty} \left ( \frac{1}{n} - \frac{1}{2n-1} - \frac{1}{2n+1} \right )$ I managed to prove through complex analysis that
$$\sum_{n=1}^{\infty} \left ( \frac{1}{n} - \frac{1}{2n-1} - \frac{1}{2n+1} \right ) = 1 -2 \log 2$$
However, I'm having a difficult time proving this result with r... | Notice
$$\sum_{n=1}^p \left(\frac1n - \frac{1}{2n-1} - \frac{1}{2n+1}\right)
= \sum_{n=1}^p \left[\frac2n - \left( \frac{1}{2n-1} + \frac{2}{2n} + \frac{1}{2n+1} \right)\right]\\
= 2H_p - \left(1 + \sum_{n=2}^{2p} \frac{2}{n} + \frac{1}{2p+1}\right)
= 2(H_p - H_{2p}) + 1 - \frac{1}{2p+1}$$
where $H_p = \sum_{n=1}^p \fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2736040",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
} |
Why is $P(X=2.5)=0?$ The discrete random variable X has cdf
$$\left\{
\begin{array}{rcr}
0 & = & \text{for} \ x<1 \\
1/4 & = & \text{for} \ 1\le x<2 \\
3/4 & = & \text{for} \ 2\le x<3 \\
1 & = & \text{for} \ x\ge 3
\end{array}
\right.$$
Find a) $P(X=1)$, b) $P(X=2)$, c) $P(X=2.5)$, d) $P(x\leq2... | It might help to express this as a probability mass function.
$$P_X(x)=\left\{
\begin{array}{rcr}
1/4 & x=1 \\
2/4 & x=2 \\
1/4 & x=3 \\
0 & \text{otherwise}
\end{array}
\right.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2737138",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Using Lagrange Multipliers to Find Extrema of $f$ I have been trying to solve this lagrange problem for an hour with no result. I can't seem to isolate any variable.
$$f(x,y) = y^2 -4xy +4x^2$$
With constraint:
$$x^2 + y^2 = 1$$
I have found $x$ to = $-\left(\frac{λ-4}{2}\right)^2 + 1$. Using the constraint $x$ is als... | You are given $f(x,y)= y^2- 4xy+ 4x^2$ with constraint $g(x, y)= x^2+ y^2= 1$. Then $\nabla f= <-4y+ 8x, 2y- 4x>$ and $\nabla g= <2x, 2y>$. The basic idea of the Lagrange multiplier method is that those two vectors are in the same direction- one is a multiple of the other: $<-4y+ 8x, 2y- 4x>= \lambda<2x, 2y>$. Equati... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2743962",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
When has weighted sum of two specific unimodal functions two peaks? Consider the functions
$f(x) = x e^{-x}$ and $g(x) = xe^{-kx}$ for $x \geq 0$, where $e$ is the Euler constant and $k > 1$. It is easy to check that $f(x)$ is unimodal and has its global maximum at $x_f^* = 1$, while $g(x)$ is also unimodal and has its... | $\def\e{\mathrm{e}}$Note that between any two adjacent local maximum points, there exists exactly one local minimum point, and $f(x) = (1 - λ)x\e^{-x} + λx\e^{-kx}$ is increasing on $\left[ 0, \dfrac{1}{k} \right]$ and decreasing on $[1, +\infty]$, thus$$
f \text{ has two peaks} \Longleftrightarrow f' \text{ has three ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2744572",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Prove that $6$ is a divisor of $n^3 - n$ for all natural numbers. How would you approach such a problem? Induction perhaps? I have been studying proof by induction, but so far I have only solved problems of this nature:
$$1 + 4 + 7 +\dots+ (3n-2) = \frac{n(3n-1)}{2}.$$
| Gibbs has already presented a perfect proof. However, there is another one, that does not involve decomposition of $n^3-n$ and is relatively straightforward.
To say $n^3-n$ is divisible by $6$ is the same as saying $n^3-n \equiv 0 \pmod 6$. Thus, once we compute $n^3-n \pmod 6$ for all possible $n$, we are done. By gen... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2745984",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
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An inequality with condition I have a new inequality this is the following :
Let $x,y,z$ be real strictly positive number such as :
$$-2 = - x y z + x + y + z $$
Then we have :
$$\sqrt{\frac{1}{x}}+\sqrt{\frac{1}{y}}+\sqrt{\frac{1}{z}}\geq \frac{9}{\sqrt{15+((x+1)(y+1)(z+1))^{(\frac{1}{3})}}}$$
My try :
I put t... | The condition gives $\sum\limits_{cyc}\frac{1}{x+1}=1.$
Let $\frac{1}{x+1}=\frac{a}{3},$ $\frac{1}{y+1}=\frac{b}{3}$ and $\frac{1}{z+1}=\frac{c}{3}.$
Hence, $a+b+c=3$ and we need to prove that
$$\sum_{cyc}\sqrt{\frac{a}{b+c}}\geq\frac{9}{\sqrt{15+\frac{3}{\sqrt[3]{abc}}}}$$ or
$$\sum_{cyc}\sqrt{a(a+b)(a+c)}\geq\sqrt{\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2746073",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Finding density function of random variable $Y=X^2$
Suppose $X$ is continuous r.v with c.d.f
$$ F(x) = \begin{cases} 1 - \left( \frac{2}{x} \right)^2 \; \; \; \;
\; \; x>2 \\ 0, \; \; \; \; \text{otherwise} \end{cases} $$
Put $Y=X^2$. Please find density function of $Y$.
Attempt
By definition $f_Y(y) = \frac{d}{dy}... | You were already given the CDF of $X$. Integrating again is not necessary. We simply write $$F_Y(y) = F_X(\sqrt{y}) = 1 - \left(\frac{2}{\sqrt{y}}\right)^2 = 1 - \frac{4}{y}, \quad y > 4,$$ thus $$f_Y(y) = \frac{4}{y^2}$$ as claimed.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2746942",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Finding $P(X + 10/X > 7 )$ where $X$ is uniform
Let $X$ be a continuous r.v. with a continuous uniform distribution on
$[0,10]$. What is $P \left( X + \frac{10}{X} > 7 \right)$?
Attempt
We notice that $X+ \frac{10}{X} > 7$ can be rewritten as $X^2 + 10 - 7X > 0$ which is equivalent to $(X-5)(X-2) > 0$. Thus,
$$ P \le... | Note that we get
$$P[(X-2)(X-5)>0] = P[\{X<2\}\cup\{X>5\}]=1-P[2<X<5],$$
because the parabola $(x-2)(x-5)$ is positive for $x<2$ and $x>5$. Taking the complement yields the last step. Since $X$ is uniformly distributed on $[0,10]$, we know $P[2<X<5] = \frac{3}{10}$, so
$$P\left(X+\frac{10}{X}>7\right) = 1-\frac{3}{10} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2748446",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Integrate $\int \ln(x^2 +1)\ dx$ $$\int \ln(x^2 +1)\ dx$$
I done it using integration by parts where
$\int u\ dv = uv - \int v\ du$
Let $u$ = $\ln(x^2 +1)$
$du = \frac{2x}{x^2+1} dx $
Let $dv = dx$ so $v=x$
$\int \ln(x^2 +1)\ dx = x \ln (x^2 +1) - \int \frac{2x^2}{x^2+1} $
I integrate $2\int \frac{x^2}{x^2+1} $ sepa... | $2\int \frac{x^2}{x^2+1} $
Does not equal the expression with just an x in the numerator after the substitution .
You can add subtract 1 in numerator to integrate $2\int \frac{x^2}{x^2+1} $
This is wrong :- $2\int \frac{x^2}{x^2+1} $
This is what you get after applying by-parts :- $2\int \frac{x^2}{x^2+1} dx $
Since d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2748688",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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If $\frac{\cos^4 \alpha}{x}+\frac{\sin^4 \alpha}{y}=\frac{1}{x+y}$,prove that $\frac{dy}{dx}=\tan^2\alpha$ If $\frac{\cos^4 \alpha}{x}+\frac{\sin^4 \alpha}{y}=\frac{1}{x+y}$,prove that $\frac{dy}{dx}=\tan^2\alpha$
It is very long to direct differentiate it.Can someone help me?
| $\dfrac{c^4}x+\dfrac{s^4}y=\dfrac 1{x+y}\iff(c^4y+s^4x)(x+y)=xy\iff s^4x^2+\underbrace{(c^4+s^4-1)}_{-2s^2c^2}xy+c^4y^2=0$
Thus $(s^2x-c^2y)^2=0\iff y=\tan(\alpha)^2x$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2748907",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Limit of $ (x_{n}) = ( 1 + \frac{1}{3} + \frac{1}{5} + \cdots + \frac{1}{2n-1} ) - a ( 1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n} ) $ Let $ a \in \mathbb{R} $ so that the sequence $ (x_{n})_{n\geq 1} $
$ (x_{n}) = ( 1 + \frac{1}{3} + \frac{1}{5} + \cdots + \frac{1}{2n-1} ) - a ( 1 + \frac{1}{2} + \frac{1}{3} ... | Let me give you a sketch of the argument.
When you increase $n$ by $1$, what happens? In net,
$$
x_{n+1}-x_n=\frac{1}{2(n+1)-1}-a\left(\frac{1}{n+1}\right)=\frac{1}{2n+1}-\frac{a}{n+1}.
$$
Using this, you can show that it must be the case that $a=\frac{1}{2}$. Why? If $a=\frac{1}{2}+\delta$ where $\delta>0$, then
$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2750460",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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How can I show that $(1+2\cos 2\theta)^3=7+2(6\cos 2\theta+3\cos4\theta+\cos6\theta)$ using the tensor product and the Clebsch Gordan Theorem? How can I show that $$(1+2\cos2\theta)^3=7+2(6\cos2\theta+3\cos4\theta+\cos6\theta)$$
using the tensor product and the Clebsch Gordan Theorem?
| We have $$(1+2\cos(2\theta))^3=1+3\cdot 2\cos(2\theta)+3(2\cos(2\theta))^2+(2\cos(2\theta))^3$$ and this is (simplified) $$-1+12\, \left( \cos \left( \theta \right) \right) ^{2}-48\, \left(
\cos \left( \theta \right) \right) ^{4}+64\, \left( \cos \left(
\theta \right) \right) ^{6}
$$
expanding the right-Hand side ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2751247",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
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Find the $x$ value for when the expression is takes the minimum value
Let $E(x)=\sqrt{x^2-2x+5}+\sqrt{x^2-8x+25}.$ Find the $x$ value for when this expression take it's minimum value.
Using the derivative it's pretty hard... and I think not indicated for this problem... I thought about using the inequality of means. ... | Using Minkowski's inequality:
$$\sqrt{(x-1)^2+4}+\sqrt{(x-4)^2+9}=\\
\sqrt{(x-1)^2+2^2}+\sqrt{(4-x)^2+3^2}\ge \\
\sqrt{(x-1+4-x)^2+(2+3)^2}=\sqrt{3^2+5^2}=\sqrt{34}.$$
The equality occurs when $(x-1,2)||(4-x,3)$, that is:
$$\frac{3}{2}(x-1)=4-x \Rightarrow x=\frac{11}{5}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2754592",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
EDL2 : $(2x+1)y''+(4x-2)y'-8y=0$
Solve this equation :
$(2x+1)y''+(4x-2)y'-8y=0 \qquad $ on $ \left]\dfrac{1}{2},+\infty\right[$ and on $\mathbb{R}$
Attempt :
First step : $\varphi_1$
The first step was to find a solution in that form : $\varphi_1(x)=ax^2+bx+c$
I found $\varphi_1(x)= 4x^2+1$
Second step : $\varphi_... | For the integral consider
$$\left(\frac {e^{-2x}}{4x^2+1}\right)'=-2e^{-2x}\frac {(2x+1)^2}{(4x^2+1)^2}$$
$$\left(\frac {e^{-2x}}{4x^2+1}\right)=-2\int e^{-2x}\frac {(2x+1)^2}{(4x^2+1)^2}dx$$
$$\phi_2=-\frac 12 e^{-2x} $$
or
$$\phi_2=e^{-2x} $$
Another look at the differential equation
$$(2x+1)y''+(4x-2)y'-8y=0 \qquad$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2757949",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Find recurrence for $I_n$. Let $I_n=\int_{0}^{1/2} \frac {x^n}{\sqrt{1-x^2}}dx.$ I must find a recurrence for this so I just started using interation by parts:
Let $$f'(x)=x^n\to f(x)=\frac{x^{n+1}}{n+1}$$ and
$$g(x)=\frac 1{\sqrt{1-x^2}}\to g'(x)=\frac x{\sqrt{(1-x^2)^3}}$$
Therefore:
$$I_n=\frac {x^{n+1}}{(n+1)\sqrt{... | $$I_n-I_{n+2}=\int_0^{1/2}\frac{x^n(1-x^2)}{\sqrt{1-x^2}}\,dx=\int_0^{1/2}x^n\sqrt{1-x^2}\,dx\\$$Now do integration by parts on this, by integrating $x^n$ and differentiating the other term.
This gives:
$$I_n-I_{n+2}=\left[\frac{1}{n+1} x^{n+1}\sqrt{1-x^2}\right]_0^{1/2}+\frac{1}{n+1}\int_0^{1/2}\frac{x^{n+2}}{\sqrt{1-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2758516",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Calculate Smith normal form, cyclic group decomposition Can someone please check my working for the following problem?
Let $A$ be the abelian group generated by elements $x,y,z$ with relations $7x+5y+2z=0, 3x+3y=0, 13x+11y+2z=0$. Decompose $A$ as a direct sum of cyclic groups.
$$
M = \begin{pmatrix}
7 & 5 & 2 \\
3 & ... | Since you want to check your answer, there is a direct way to obtain Smith normal form; define inductively
$$d_1d_2\cdots d_k=\mbox{gcd of $k\times k$ minors of matrix under consideration}.$$
Then the Smith normal form is
$$\begin{bmatrix} d_1 & & \\ & d_2 & \\ & & d_3\end{bmatrix}.$$
So $d_1=1$ because it is $gcd(7,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2760082",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Show the next inequality Show that:
$$
\frac{3}{8}\leq \int_{0}^{1/2}\sqrt{\frac{1-x}{1+x}}dx\leq \frac{\sqrt{3}}{4}.
$$
I have
$$m(b-a)\leq\int_{a}^{b}f(x)dx\leq M(b-a),$$
but the function does not present critical points in its derivative.
And how can I show the following?
$$\frac{|a+b|-|a-b|}{ab}=\frac{2}{\max\{|a|,... | If you have
written it correctly,
$\frac{3}{8}
\leq \int_{0}^{1/2}\sqrt{\frac{1-x}{1+2}}dx
\leq \frac{\sqrt{3}}{4}
$
is the same as
$\frac{3}{8}\sqrt{3}
\leq \int_{0}^{1/2}\sqrt{1-x}dx
\leq \frac{\sqrt{3}}{4}\sqrt{3}
$
or
$\frac{3\sqrt{3}}{8}
\leq \int_{0}^{1/2}\sqrt{1-x}dx
\leq \frac{3}{4}
$.
$\sqrt{1-x}$
is monotonic... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2760972",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Generating function to find the sum of digits How many integers between $30,000$ and $100,000$ have a sum of $15$ or less?
I was approaching this problem with a generating function:
$$g(x) = (x^3+x^4+x^5+x^6+x^7+x^8+x^9)(1+x+...+x^9)^4$$
First I'm going to pull out $x^3$ then I would need the coefficients of $x$ to t... | Assuming you are including $30\,000$ and excluding $100\,000$.
Write:
$$\sum_{k=3}^{9}x^k=\frac{x^3-x^{10}}{1-x}\, ,$$
$$\sum_{k=0}^{9}x^k=\frac{1-x^{10}}{1-x}\, .$$
Then
$$g(x)=\left(\sum_{k=3}^{9}x^k\right)\left(\sum_{k=0}^{9}x^k\right)^4=\frac{x^3-x^{10}}{1-x}\left(\frac{1-x^{10}}{1-x}\right)^4$$
Use the operator $(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2762449",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Finding $(x-1)^{-2}$ in Quotient Ring Let $R=\mathbb{Q}[x]$ with $x^3-x^2+x+2=0$. Find $(x-1)^{-2}$. I thought I could use the euclidean algorithm
$$x^3-x^2+x+2=(x-1)(x^2+1)+3$$
but I already have a constant after the first step. Also not sure that finding $(x-1)^{-1}$ would help anyway.
| Well, $(x-1)^{-2}=((x-1)^{-1})^2$, no? As for the rest, you know that $$(x-1)(x^2+1)\equiv -3\pmod{x^3-x^2+x+2}$$ In other words, that $(x-1)\left(-\frac13x^2-\frac13\right)\equiv 1\pmod{x^3-x^2+x+2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2763497",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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How to find $\lim_{n\rightarrow \infty} \frac {n^2}{n^3+n +1} + \frac {n^2}{n^3 +n +2} +.....+\frac {n^2}{n^3 + 2n}$?
How to find
$$
\lim_{n\rightarrow \infty} \frac {n^2}{n^3+n +1} + \frac {n^2}{n^3 +n +2} +.....+\frac {n^2}{n^3 + 2n}?
$$
I was thinking about the Riemann sum, but I am not able to do it.
| \begin{align*}
\sum_{k=1}^{n}\dfrac{n^{2}}{n^{3}+n+k}\leq\sum_{k=1}^{n}\dfrac{n^{2}}{n^{3}+n+1}=\dfrac{n^{3}}{n^{3}+n+1}
\end{align*}
and
\begin{align*}
\sum_{k=1}^{n}\dfrac{n^{2}}{n^{3}+n+k}\geq\sum_{k=1}^{n}\dfrac{n^{2}}{n^{3}+n+n}=\dfrac{n^{3}}{n^{3}+2n},
\end{align*}
now use Squeeze Theorem to conclude.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2764488",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find $a\in\mathbb R$, if it exists, so that $T(2,0,1)=(1,-\frac 52)$ Let $$T:\mathbb R^3\to \mathbb R^2\mid M_{B_1B_2}=\begin{pmatrix}a&1&2\\-1&0&\color{red}1\end{pmatrix}$$ a linear transformation and the basis $$B_1=\{(1,0,0),(0,-3,1),(0,0,-2)\},\qquad B_2=\{(2,0),(0,-1)\}.$$
Find $a\in\mathbb R$, if it exists, so th... | Find the matrix of $T$ with respect to the canonical bases of $\mathbb{R}^{3}$ and $\mathbb{R}^2$, which is
$$
\begin{bmatrix}
2 & 0 \\
0 & -1
\end{bmatrix}
\begin{bmatrix}
a & 1 & 2 \\
-1 & 0 & 1
\end{bmatrix}
\begin{bmatrix}
1 & 0 & 0 \\
0 & -3 & 0 \\
0 & 1 & -2
\end{bmatrix}^{-1}
$$
In order to find the required inv... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2765472",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Bi-nominal expansion of 3 terms Find the coefficient of $x^{17}$ in the expansion of $(3x^7 + 2x^5 -1)^{20}$
I'm stuck in handling this question as I do know how to solve it when it has 2 terms.
But now it has 3.
I have no idea where to begin...
| We can apply the binomial theorem twice in order to find the coeffcient. It is convenient to use the coefficient of operator $[x^k]$ to denote the coefficient of $x^k$ of an expression.
We obtain
\begin{align*}
\color{blue}{[x^{17}]}&\color{blue}{(3x^7+2x^5-1)^{20}}\\
&=[x^{17}]\sum_{k=0}^{20}\binom{20}{k}(3x^7)^k(2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2765979",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Calculate the following convergent series: $\sum _{n=1}^{\infty }\:\frac{1}{n\left(n+3\right)}$ I need to tell if a following series convergent and if so, find it's value:
$$
\sum _{n=1}^{\infty }\:\frac{1}{n\left(n+3\right)}
$$
I've noticed that
$$
\sum _{n=1}^{\infty }\:\frac{1}{n\left(n+3\right)} = \frac{1}{3}\sum _... | Yes, it is telescoping,
\begin{align}
\sum_{n=1}^N \frac{1}{n(n+3)}&= \frac13 \sum_{n=1}^N \left[ \frac1{n}-\frac{1}{n+3}\right]\\
&= \frac13\left[1-\frac{1}{4} +\frac12-\frac1{5}+\frac13-\frac16+\ldots +\frac1{N-2}-\frac1{N+1}+\frac1{N-1}-\frac1{N+2}+\frac1N-\frac1{N+3}\right]\\
&=\frac13\left[1+\frac12+\frac13-\frac1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2768332",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 2
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Is it true that $13|2^n5^{2n+1}+3^{2n+1}7^{n+1}$ for all $n$? Is it true that $13|2^n5^{2n+1}+3^{2n+1}7^{n+1}$ for all $n$?
So what I did was basically simplify the terms on the right $\mod 13$.
$2^n5^{2n+1}+3^{2n+1}7^{n+1} \mod 13$
$= 5\cdot 2^n\cdot 5^n\cdot 5^n+3\cdot 7\cdot 3^{n}\cdot 3^n7^n \mod 13$
$= 5\cdot 50^n... | Induction ...
\begin{eqnarray*}
5 \times 50^{n+1}+21 \times 63^{n+1}=50(\color{red}{5 \times 50^{n}+21 \times 63^{n} }) +\color{red}{13} \times 21\times 63^n.
\end{eqnarray*}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Converting $\frac{(3+3i)^5 (-2+2i)^3}{(\sqrt 3 +i)^{10}}$ to trigonometric form
Convert complex to trig.
$$\frac{(3+3i)^5 (-2+2i)^3}{(\sqrt 3 +i)^{10}}$$
Let us consider
$$(3+3i)^5$$
Here $a = 3$ ,$b=3$
$$\sqrt{a^2+b^2}=\sqrt{3^2+3^2} = \sqrt{18}=3\sqrt{2}$$
$$\theta =\arctan\biggr(\frac{b}{a}\biggr) =\arctan\bigg... | As an alternative by exponential form
*
*$(3+3i)^5=(3\sqrt 2e^{i \pi/4})^5$
*$(-2+2i)^3=(2\sqrt 2e^{i 3\pi/4} )^3$
*$(\sqrt 3 +i)^{10}=(2e^{i \pi/6} )^{10}$
then
$$\frac{(3+3i)^5 (-2+2i)^3}{(\sqrt 3 +i)^{10}}=\frac{3^54\sqrt 2\cdot2^4\sqrt2}{2^{10}}\frac{e^{i 5\pi/4}\cdot e^{i 9\pi/4}}{e^{i 10\pi/6}}=30.375e^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2776335",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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$\vec{u}=\vec{w_{1}}+\vec{w_{2}}, w_1\in W$ and $w_2\in W^\perp$ Let $S[(x_1,y_1,z_1)(x_2,y_2,z_2)]=3x_1x_2+x_2y_1+x_1y_2+y_1y_2+z_1z_2$ be an inner product,
$W=${$(x,y,z)\in\mathbb{R}^3|x+z=0$} a subspace and $\vec u=(1,0,0)$
Find $w_1\in W,$ $w_2\in W^\perp$ (orthogonal in respect of $S$) such that $\vec u=\vec w_1+... | With
$$
\begin{array}{rcl}
\vec{u} & = & (1,0,0)\\
\vec{w}_{1} & = & (w_{11},w_{12},w_{13})\\
\vec{w}_{2} & = & (w_{21},w_{22},w_{23})
\end{array}
$$
The inner product is weighted by
$$
M=\left[\begin{array}{ccc}
3 & 1 & 0\\
1 & 1 & 0\\
0 & 0 & 1
\end{array}\right]
$$
The conditions are
$$
\left\{ \begin{array}{rcl... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Limit of $\frac{\sqrt[n]{(n+1)(n+2)\cdots(2n)}}n$ Compute the limit
$$
\lim_{n \to \infty} \frac{\sqrt[n]{(n+1)(n+2)\cdots(2n)}}n
$$
How can this be done? The best I could do was rewrite the limit as
$$
\lim_{n \to \infty} \left(\frac{n+1}n \right)^{\frac 1n}\left(\frac{n+2}n \right)^{\frac 1n}\cdots\left(\frac{2n}n \r... | If you accept using the following limit that has been asked and proved (for example here) many times here on MSE
*
*$\lim_{n\rightarrow \infty}\frac{\sqrt[n]{n!}}{n} = \frac{1}{e}$,
then your limit is easily derived:
$$\frac{\sqrt[n]{(n+1)(n+2)\cdots(2n)}}n = \frac{ (2n!)^{\frac{1}{n}} }{ (n!)^{\frac{1}{n}}\cdot n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2781063",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
} |
probability that exactly three of them shows the same face and remaining three shows different faces Question
$6$ dice are thrown simultaneously ,find the probability that exactly three of them shows the same face and remaining three shows different faces.
My Approach
Total outcome=$$6^{6}$$
Number of ways to select ... | One may see it as follows:
*
*There are 6 options for the equal faces
*$\binom{6}{3}$ choices, which 3 dice show the same face
*$5\cdot 4 \cdot 3$ choices for the mutually different faces of the other three
*all together there are $6^6$ different throws
$$\frac{6\cdot \binom{6}{3}\cdot 5\cdot 4\cdot 3}{6^6}= \fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2781961",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Why $\frac{a^{2m}-1}{a+1}=\frac{a^2-1}{a+1}(a^{2m-2}+a^{2m-4}+\cdots+a^2+1)?$ I know that $\dfrac{a^2-b^2}{a+b} = a-b$, because$$
a^2-b^2 = aa -ab+ab- bb = a(a-b)+(a-b)b = (a-b)(a+b).
$$
Also, I know that$$
\frac{a^n-b^n}{a+b} = a^{n-1}-a^{n-2}b+\cdots-ab^{n-2}+ b^{n-1}.$$
But I do not understand this equality below:
$... | Because$$
a^{2m} - 1 = (a^2)^m - 1 = (a^2 - 1)((a^2)^{m - 1} + (a^2)^{m - 2} + \cdots + 1).
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2788869",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Prove that $\sqrt{11}-1$ is irrational by contradiction I am working on an assignment in discrete structures and I am blocked trying to prove that $\sqrt{11}-1$ is an irrational number using proof by contradiction and prime factorization.
I am perfectly fine doing it with only $\sqrt{11}$, but I am completely thrown of... | from question,
$$\sqrt {11} -1 = \frac {a}{b}$$ where $(a,b)=1$.
From last step,
$10b^2 = a^2 + 2ab$
$LHS $ is multiple of $5$. then $RHS$ is also multiple of $5$.
in $RHS,a^2 + 2ab$ exactly one of $a$ and $b$ is multiple.
if both $a$ and $b$ are multiple of $5$ , it violated $(a,b)=1$.
if $b$ is multiple of $5$... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 3
} |
Evaluate $\lim\limits_{n\rightarrow \infty}\frac{n+n^2+n^3+\cdots +n^n}{1^n+2^n+3^n+\cdots +n^n}.$ Problem
Evaluate $$\lim\limits_{n\rightarrow \infty}\frac{n+n^2+n^3+\cdots +n^n}{1^n+2^n+3^n+\cdots +n^n}.$$
My solution
Notice that $$\lim_{n \to \infty}\frac{n+n^2+n^3+\cdots +n^n}{n^n}=\lim_{n \to \infty}\frac{n(n^n-1... | A Query to @Teddy38's Answer
Yes, we can agree that $$n+n^2+\cdots+n^n=\frac{n}{n-1}(n^n-1)$$ and $$\frac{n^{n+1}}{n+1}=\int_{0}^{n}x^n\ dx<1^n+2^n+\cdots+n^n<\int_{1}^{n+1}x^n\ dx=\frac{(n+1)^{n+1}-1}{n+1}.$$ Thus,we obtain $$\frac{n+1}{n-1}\cdot \frac{n^{n+1}-n}{(n+1)^{n+1}-1}<\dfrac{n+n^2+n^3+\cdots +n^n}{1^n+2^n+3^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2792441",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "26",
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"answer_id": 2
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How to solve $\frac{1}{2^x+3} \geq \frac{1}{2^{x+2}-1}$? I guess it's easy, but I still need help. The inequality is
$$\frac{1}{2^x+3} \geq \frac{1}{2^{x+2}-1}$$
If you set $t=2^x$, then it becomes
$$\frac{1}{t+3} \geq \frac{1}{4t-1}$$
The set of solutions is
$$x \in (-\infty,-2) \cup \{1\}$$
which is not what I get.
... | The given solution is clearly wrong on the positive side. For instance, take $x=2$, and
$$
\frac17=\frac{1}{2^x+3}\geq\frac{1}{2^{x+2}-1}=\frac1{15}.
$$
Since $t=2^x$ you know that $t>0$. So $t+3>0$. Now if $4t-1>0$, everything is positive and you can multiply to get
$$
4t-1\geq t+3,
$$
which simplifies to $3t\geq4$,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2795294",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Probability that $z$ is EVEN satisfying the equation $x + y + z = 10$ is Question
Three randomly chosen non negative integers $x, y \text{ and } z$ are found to satisfy the equation $x + y + z = 10$. Then the probability that $z$ is even, is"?
My Approach
Calculating Sample space -:
Number of possible solution for... | You are missing the fact that a single even value of $z$ represents several different soluitons, not one.
With $z$ equal to 10, there is only one possible solution (0+0+10).
With $z$ equal to 8, there are 3 possible solutions (0+2+8, 1+1+8, 2+0+8).
With $z$ equal to 6, there are 5 possible solutions.
With $z$ equal to ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2795440",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 1
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Finding the Jordan canonical form of A and Choose the correct option Let $$ A = \begin{pmatrix}
0&0&0&-4 \\ 1&0&0&0 \\ 0&1&0&5 \\ 0&0&1&0
\end{pmatrix}$$
Then a Jordan canonical form of A is
Choose the correct option
$a) \begin{pmatrix}
-1&0&0&0 \\ 0&1&0&0 \\ 0&0&2&0 \\ 0&0&0&-2
\end{pmatrix}$
$b) \begin{pmatri... | All the other answers to this question so far overlook the fact that the matrix in option (b) is not in Jordan canonical form in the first place, so you can eliminate that option without doing any work at all. After eliminating (c) and (d) as you’ve done, that leaves only (a).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2796632",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 4
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Evaluate the limit with exponents using L'Hôpital's rule or series expansion Evaluate the limit$$\lim_{x\to 0}\dfrac{\left(\frac{a^x+b^x}{2}\right)^{\frac{1}{x}}
-\sqrt{ab}}{x}$$ It is known that $a>0,b>0$
My Attempt:
I could only fathom that $$\lim_{x\to 0}\left(\frac{a^x+b^x}{2}\right)^{\frac{1}{x}}=\sqrt{ab}$$
| Using the series expansion for $\log(1+x)$, we get
$$
\frac1x\,\log\left(1+mx+nx^2+O\!\left(x^3\right)\right)
=m+\left(n-\frac{m^2}2\right)x+O\!\left(x^2\right)
$$
Using the series expansion for $e^x$, we get
$$
\left(1+mx+nx^2+O\!\left(x^3\right)\right)^{1/x}
=e^m\left(1+\left(n-\frac{m^2}2\right)x+O\!\left(x^2\right)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2799146",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 0
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Integral results in difference of means $\pi(\frac{a+b}{2} - \sqrt{ab})$ $$\int_a^b \left\{ \left(1-\frac{a}{r}\right)\left(\frac{b}{r}-1\right)\right\}^{1/2}dr = \pi\left(\frac{a+b}{2} - \sqrt{ab}\right)$$
What an interesting integral! What strikes me is that the result involves the difference of the arithmetic and ge... | Let
$0\le a\le b,$
then
\begin{align}
&I_1 = \int\limits_a^b \left\{ \left(1-\frac{a}{r}\right)\left(\frac{b}{r}-1\right)\right\}^{-1/2}\,\mathrm dr = \int\limits_a^b \dfrac{r}{\sqrt{(r-a)(b-r)}}\,\mathrm dr\\
&=\dfrac12\int\limits_a^b \dfrac{2r-(a+b)}{\sqrt{(r-a)(b-r)}}\,\mathrm dr+\dfrac{a+b}2\int\limits_a^b \dfrac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2799576",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "24",
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If the last 3 digits of $2012^m$ and $2012^n$ are identical, find the smallest possible value of $m+n$.
Let $m$ and $n$ be positive integers such that $m>n$. If the last 3 digits of $2012^m$ and $2012^n$ are identical, find the smallest possible value of $m+n$.
Since the 100's digit is 0 in both cases, I just did $2... | You have $12^m\equiv 12^n\pmod{10^3}$ hence $12^m\equiv 12^n\pmod{2^3}$ and $12^m\equiv 12^n\pmod{5^3}$.
For the latter, we have $12^{m-n}\equiv 1\pmod{5^3}$, hence $m\equiv n\pmod{100}$ because the multiplicative order of $12$ modulo $5^3$ is $100$, as computed here.
On the other hand, $12^n(12^{m-n}-1)\equiv 0\pmod{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2800523",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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"answer_id": 0
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The quadratic equation is giving me error can you please help me locate where I am wrong Question. Solve $$\log(x-3) + \log (x-4) - \log(x-5)=0.$$
Attempt. I got $$x^2-8x+17=0.$$
$$\log(x-3)(x-4)/(x-5)=0$$
$$\log(x^2-4x-3x+12)/x-5=0$$
$$x^2-7x+12= 10^0 (x-5)$$
$$x^2-7x-x+12+5=0$$
$$x^2-8x+17=0$$
Hi guys update: appare... | What's the problem? We have\begin{align}\log(x-3)+\log(x-4)-\log(x-5)=0&\iff\log\bigl((x-3)(x-4)\bigr)=\log(x-5)\\&\iff\log(x^2-7x+12)=\log(x-5)\\&\iff x^2-7x+12=x-5\\&\iff x^2-8x+17=0.\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2800854",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Recurrence problem: find $a_{1000}$ from $a_{0}$ Hi I'm stuck at yet another question.
$a_{0}=5$. Given $a_{n+1}a_{n} = a_{n}^{2} + 1$ for all $n \ge 0$, determine $\left \lfloor{a_{1000}}\right \rfloor$.
So I got $a_{n+1}=a_{n} + \frac{1}{a_{n}}$ and then:
$a_{1000}=a_{0}+ \frac{1}{a_{0}} + \frac{1}{a_{1}} + \frac... | Squaring the given relation $a_{n+1} = a_{n} + \frac{1}{a_n}$, we get
$$ a_{n+1}^2 = a_n^2 + \frac{1}{a_n^2} + 2 \Rightarrow a_{n+1}^2 > a_n^2 +2 $$
and iterating we get (for $n = 999$ we get the right side)
$$ a_{n+1}^2 > a_0^2 + 2n + 2 \Rightarrow a_{1000} > 2025 = 45^2 $$
Also, using the terms ($\frac{1}{a_i^2} $ t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2803047",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
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Let $w$ be a primitive root of a unit of order 3, prove that $(1-w+w^2)(1+w-w^2)=4$ The title is the statement of the problem.
I did the following:
$(1-w+w^2)(1+w-w^2)=$
$1+w-w^2-w-w^2+w^3+w^2+w^3-w^4=$
$1-w^2+w^3+w^3-w^4=$
$1-w^2+1+1-w^4=4$, * then,by definition of primitive root of order 3*
$w^4+w^2-1=0$ and the prob... | $\,4 - (1 - w + w^2) (1 + w - w^2) = (1 + w + w^2) (3 - 3 w + w^2)\,$
where $\,1 + w + w^2\,$ is the irreducible polynomial for a primitive cube root of $1$.
Another way is to multiply the product by $\,1-w\,$ and expand
$\,(1 - w + w^2) (1 + w - w^2)(1 - w) = 1 - w - w^2 + 3w^3 - 3w^4 + w^5.\,$ Now use the cube root ... | {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Integrating $\frac{\arctan x}{x\sqrt{\smash[b]{1-x^2}}}$ How to evaluate
$$\int_0^1\frac{\arctan x}{x\sqrt{1-x^2}}\,\mathrm dx\text{?}$$
The steps I can think of is integration by parts as
$$\int_0^1\frac{\arctan x}{x\sqrt{1-x^2}}\,\mathrm dx=\int_0^1\frac{\arctan x}{x}\,\mathrm d(\arcsin x)$$
or integration by substit... | Let $\displaystyle I(a)=\int^{1}_{0}\frac{\tan^{-1}(ax)}{x\sqrt{1-x^2}}dx$
Now $\displaystyle I'(a)=\int^{1}_{0}\frac{1}{(1+a^2x^2)\sqrt{1-x^2}}dx$
Put $x=\tan t$. Then $dx=dt$ and changing limits
So $$I'(a)=\int^{\frac{\pi}{2}}_{0}\frac{1}{1+a^2\sin^2 t}dt=\frac{1}{1+a^2}\int^{\frac{\pi}{2}}_{0}\frac{\sec^2 t}{k^2+\ta... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2805010",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
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Solve the equation $3^{\log_4(x)+\frac{1}{2}}+3^{\log_4(x)-\frac{1}{2}}=\sqrt{x}$
Solve $3^{\log_4(x)+\frac{1}{2}}+3^{\log_4(x)-\frac{1}{2}}=\sqrt{x}$.
I am able to reduce the LHS to $\sqrt{x}=3^{\log_4(x)} \cdot \dfrac{4}{3}$. Squaring both sides do not seem to lead to a result. Do you know how to proceed?
| We have
$$
\begin{eqnarray}
\sqrt{x}
&=& 3^{\color{blue}{\log_4(x)} \color{brown}{+ \frac{1}{2}}} + 3^{\color{blue}{\log_4(x)} \color{brown}{- \frac{1}{2}}} \\
&=& \left( \color{brown}{3^{\frac{1}{2}}} + \color{brown}{3^{-\frac{1}{2}}} \right) \cdot \color{blue}{3^{\log_4(x)}} \tag{factoring out $3^{\log_4(x)}$} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2806518",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Given the area, perimeter and an altitude of a triangle, find some products The triangle $ABC$ has perimeter $360$ and area $2100$. The altitudes are named $AD$, $BE$, $CF$.
The altitudes meet at $H$. If $CF=24$, find the values of $HA \times HD$, $HB \times HE$, $HC \times HF$.
I have observed that all of those produc... | Draw circles $(ABDE), (BECF), (CFAD)$. It is clear that $H$ is the radical center of all three circles. Thus, $HA \cdot HD = HB \cdot HE = HC \cdot HF = x$.
Let $BC = a$, $CA = b$. Clearly, $AB = 175$ as $AB\cdot CF = 4200$. It follows that
$$
\begin{align}
\sqrt{a^2 - 24^2} + \sqrt{b^2 - 24} = AB &= 175 \\
a+b+175 &= ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2806896",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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The implication: $x+\frac{1}{x}=1 \implies x^7+\frac{1}{x^7}=1$ $$x+\frac{1}{x}=1 \implies x^7+\frac{1}{x^7}=1$$
The graphs/range: $\quad y_1(x)=x+\frac{1}{x}, \quad y_7(x)=x^7+\frac{1}{x^7} \quad$
and do not touch the line $\quad y=1\quad$. The relation $\quad x+\frac 1 x=1 \quad$ appears to only be true for certain ... | Define two propositions: $p:= x+\frac{1}{x}−1=0$ and $q:=x^7+\frac{1}{x^7}−1=0$. You are trying to find the values of $x$ for which $p \wedge q$ is true. The actual question is 'prove that $p \Rightarrow q$ is true'. These two propositions are different.
If you try to solve $x+\frac{1}{x}=1$ you end up with $x = e^{i\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2808056",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Show that $\lim_{x\to 1}\frac{1}{x+1}=\frac{1}{2}$. Could someone please explain a step in the following proof?
Show that $\lim_{x\to 1}\frac{1}{x+1}=\frac{1}{2}$.
The above limit exists if for every $\varepsilon > 0$, there exists a real number $\delta > 0$ such that if $0<|x-1|<\delta$, then $\left |\frac{1}{x+1}-\... | The condition for $|x-1|<1$ is just a trick, you can set any number for convenience since we want to show that there exists delta.
First, let us observe the following definition of limit
$$\forall \varepsilon > 0, \exists \delta > 0 \ni 0 < |x-1| < \delta \implies \bigg|\frac{1}{x+1}-\frac{1}{2}\bigg|< \varepsilon$$
So... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2810921",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
For any prime number $n >5$, prove the final digit of $n^4$ is $1$ So I am struggling a bit with this question
$n$ is prime
we can ignore $2$ and $5$ as $n>5$
now if $n$ is prime
for the digits: $\{0,1,2,3,4,5,6,7,8,9\}$
$\{0,2,4,6,8\}$ can be discounted as $n$ cannot be even that
$5$ can be discounted as $n$ is not a ... | On the claim I've done in my comment above: for a prime $n > 5$ we have
$$n \equiv 1 \pmod 2 \Rightarrow n^2 \equiv 1 \pmod 8 \Rightarrow n^4 \equiv 1 \pmod{16};$$
$$n^2 \equiv 1 \pmod 3 \Rightarrow n^4 \equiv 1 \pmod 3;\quad n^4 \equiv 1 \pmod 5$$
(the last two are implied by FLT, or may be verified by hand, consider... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2815848",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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a polynomial of degree $4$ such that $P(n) = \frac{120}{n}$ for $n=1,2,3,4,5$
Let $P(x)$ be a polynomial of degree $4$ such that $P(n) = \frac{120}{n}$ for $n=1,2,3,4,5.$ Determine the value of $P(6)$.
Let $P(x) = ax^4 + bx^3 + cx^2 + dx + e$. For $n=1,2,3,4,5$ I have plugged it into this polynomial and got the follo... | The previous solutions calculate the polynomial, and then deduce the value of $P(6)$.
This was done correctly of course (the solution by @Martin R is particularly elegant).
However, it is possible to calculate the value of $P(6)$ directly, by calculating the finite differences and by using the fact that the 5th-order d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2816980",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
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Is there a special case for equations of curves that have parallel oblique asymtotes? I am given the function $f(x,y)=(x-y)^2(x^2+y^2)-10(x-y)x^2+12y^2+2x+y$ and asked whether the curve $f(x,y)=0$ has two parallel oblique asymptotes, two unique asymptotes, or if there is no oblique asymptote at all. Do I manually find... | The algebraic approach here is not too easy, so here's how to blend in some calculus (and some nontrivial computation, perhaps done with the help of a CAS). Of course, using a graphical CAS, you can easily see that there are two asymptotes with slope $1$. My original guess was that $y=x$ was an asymptote because of the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2818589",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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Does $4(x^3+2x^2+x)^3(3x^2+4x+1) = 4x^3(3x+1)(x+1)^7$? Please provide proof
The answer in the back of the book is different to both my calculations and also the online calculator I crosschecked my answer with...
The Question:
"Differentiate with respect to $x$:"
$
(x^3+2x^2+x)^4
$
My Answer: $4(x^3+2x^2+x)^3(3x^2+4x+1... | $4(x^3+2x^2+x)^3 . (3x^2+4x+1)$
$4x^3 (x^2+2x+1)^3 (3x^2+4x+1)$
$4x^3 (x+1)^6 (3x^2+4x+1)$
$4x^3 (x+1)^6 (3x+1)(x+1)$
$4x^3 (x+1)^7 (3x+1)$
So the book resolved your result in more factors.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2823614",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
Self-intersections of a Lissajous curve Hei,
I want to find the self intersections of a Lissajous curve, for instance:
$$x(t)=\sin 2t$$
$$y(t)=\sin 3t$$
I have been trying for a couple of hours but I really don't get how to compute all the solutions. Basically I was trying to write the relation such that
$$\sin 2u=\sin... | Start with a rational parametrization, using the tangent half angle formula:
$$c=\cos u=\frac{1-a^2}{1+a^2}\qquad s=\sin u=\frac{2a}{1+a^2}$$
Use Wikipedia to find formulas for double and triple angle.
$$\sin 2u=2sc=\frac{(1-a^2)2a}{(1+a^2)^2}
\qquad
\sin3u=3s-4s^3=\frac{6a(1+a^2)^2-32a^3}{(1+a^2)^3}$$
Now assume a sec... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2824293",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Show that $\sin (x/2)+2\sin (x/4)\leq3\sqrt{3}/2$ for all $x \in [0,2\pi]$
Given that $$f (x)=\sin (x/2)+2\sin (x/4)$$
Show that $f (x)\leq3\sqrt{3}/2$ for all $x \in [0,2\pi]$.
My try
$f (x)=\sin (x/2)+2\sin (x/4)$
$f' (x)=\frac {1}{2}\cos(x/2)+\frac {1}{2}\cos (x/4)$
$f' (x)=\frac {1}{2}\cos(x/2)+\frac {1}{2}... | Hint:
Set $x=8y$ to find $$2\sin2y(1+\cos2y)=8\sin y\cos^3y$$
Now $\sqrt[4]{3\sin^2y\cos^6y}\le\dfrac{3\sin^2y+\cos^2y+\cos^2y+\cos^2y}4=\dfrac34$
Squaring we get $$16\sqrt3\sin y\cos^3y\le9$$
The equality occurs if $3\sin^2y=\cos^2y\iff\dfrac{\sin^2y}1=\dfrac{\cos^2y}3=\dfrac1{1+3}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2824634",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 5
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Show that the Huber-loss based optimization is equivalent to $\ell_1$ norm based. Dear optimization experts,
My apologies for asking probably the well-known relation between the Huber-loss based optimization and $\ell_1$ based optimization. However, I am stuck with a 'first-principles' based proof (without using Moreau... | Consider the proximal operator of the $\ell_1$ norm
$$
z^*(\mathbf{u})
=
\mathrm{argmin}_\mathbf{z}
\
\left[
\frac{1}{2}
\| \mathbf{u}-\mathbf{z} \|^2_2
+
\lambda \| \mathbf{z} \|_1
\right]
=
\mathrm{soft}(\mathbf{u};\lambda)
$$
In your case, (P1) is thus equivalent to
minimize
\begin{eqnarray*}
\phi(\mathbf{x})
&=&
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2825704",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 1
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Prove by induction that $n^4-4n^2$ is divisible by 3 for all integers $n\geq1$. For the induction case, we should show that $(n+1)^4-4(n+1)^2$ is also divisible by 3 assuming that 3 divides $n^4-4n^2$. So,
$$ \begin{align} (n+1)^4-4(n+1)&=(n^4+4n^3+6n^2+4n+1)-4(n^2+2n+1)
\\ &=n^4+4n^3+2n^2-4n-3
\\ &=n^4+2n^2+(-6n^2+6n^... | Well, obviously $6n^2$ is divisible by $3$.
And $4n^3-4n=4(n-1)n(n+1)$. Since $(n-1)n(n+1)$ is the product of three consecutive integers, one of them is divisible by $3$ and therefore their product has that property too.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2828422",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 7,
"answer_id": 0
} |
Find the solutions of $2^{x-1}-3 \sqrt{2^{x-1}}+2=0$ Find the solutions of $2^{x-1}-3 \sqrt{2^{x-1}}+2=0$
It seemed easy for me, but I couldn't do it. By Wolfram Alpha, I got that $x=1,3$, but I don't know how to get there.
Any hints?
Thanks.
| $$2^{x-1}-3 \sqrt{2^{x-1}}+2=0$$
$$2^{x-1}+2=3 \sqrt{2^{x-1}}$$
$$9\cdot 2^{x-1}=2^{(x-1)*2}+2^{x+1}+4$$
$$9\cdot2^x\cdot2^{-1}=2^{2x-2}+2^x\cdot2+4$$
$$9\cdot2^x\cdot\frac{1}{2}=2^{2x}\cdot2^{-2}+2^x\cdot2+4$$
$$\frac{9}{2}\cdot2^x=(2^x)^2\cdot\frac{1}{2^2}+2^x\cdot2+4$$
$$\frac{9}{2}\cdot2^x=(2^x)^2\cdot\frac{1}{4}+2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2828939",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Finding the value of $\int^{1}_{0}(1+x)^{m}(1-x)^{n} \,\mathrm d x$
Find the value of $$\displaystyle \int^{1}_{0} (1+x)^{m}(1-x)^{n} \,\mathrm d x$$ where $m, n \geq 1$ and $m, n \in \mathbb{N}$.
Let
$$\displaystyle I_{m,n} = \int^{1}_{0}(1+x)^m(1-x)^n \,\mathrm d x = \int^{1}_{0}(2-x)^m\cdot x^n \,\mathrm d x$$
Pu... | We first express the integrand as a product of $\sin \theta$ and $\cos \theta $ by
letting $x=\cos 2\theta $, yields
$$I = \int_{\frac{\pi}{4}}^{0}\left(2 \cos ^{2} \theta\right)^{m}\left(2 \sin ^{2} \theta\right)^{n}\left(-2 \sin ^{2} \theta\right) d \theta
=2^{m+n+2} \int_{0}^{\frac{\pi}{4}} \cos ^{2 m+1} \theta \s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2829264",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Algebra and matrices. Q. Solve for $x$, $y$ and $z$ if
$$(x+y)(x+z)=15, $$
$$(y+z)(y+x)=18, $$
$$(z+x)(z+y)=30. $$
Solution:
I expanded each equation above as :
$$x^2+xz+yx+yz=15, \tag{1}$$
$$y^2+xz+yx+yz=18, \tag{2}$$
$$z^2+xz+yx+yz=30. \tag{3}$$
Then I subtracted $(1)-(3)$, $(2)-(1)$ and $(3)-(2)$; so I got the equat... | Take $x+y=a,y+z=b,x+z=c$. Then you get
$ac=15\quad[1]$
$ab=18\quad[2]$
$bc=30\quad[3]$
Multiply to get $(abc)^2=8100$, thus getting $abc=90$.
Divide by $[1],[2],[3]$ separately to get
$a=3$
$b=6$
$c=5$
Now, substitute to get
$x+y=3\quad[4]$
$y+z=6\quad[5]$
$x+z=5\quad[6]$
Add to get $x+y+z=7$
Subtract separately to ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2830167",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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How to get rid of the absolute value function when solving an ODE? Problem: $y'+yx=0, \quad y(0)=-1$
We separate it:
$\frac{dy}{dx}=-yx \Rightarrow \int\frac{-1}{y}dy=\int x dx = \frac{1}{2}x^2 + C_1$
With $\int\frac{-1}{y}dy=\log(\frac{1}{|y|})+C_2$
we get
$\log(\frac{1}{|y|})=\frac{1}{2}x^2 + C$
We solve for $y$:
$|... |
$y'+yx=0$
$$
\frac{dy}{y} = -xdx \Rightarrow \int \frac{dy}{y} = \int -xdx \Rightarrow \ln |y| = -\frac{1}{2}x^2 + C_1.
$$
*
*$y > 0$. we have $\ln y = -\frac{1}{2}x^2 + C_1$, so $y = C_1e^{-\frac{1}{2}x^2}$.
*$y < 0$. we have $\ln -y = -\frac{1}{2}x^2 + C_1$, so $y = C_2e^{-\frac{1}{2}x^2}$.
*$y = 0$. we ha... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2832282",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Evaluating $\int_{\frac{-1}{2}}^{\frac{1}{2}} \int_{\frac{-1}{2}}^{\frac{1}{2}} \frac{x^2}{(x^2+y^2)^2 \log^2(\frac{2}{\sqrt{x^2+y^2}})} \,dx\,dy.$
$$\int_{\frac{-1}{2}}^{\frac{1}{2}} \int_{\frac{-1}{2}}^{\frac{1}{2}} \frac{x^2}{(x^2+y^2)^2 \log^2(\frac{2}{\sqrt{x^2+y^2}})} \,dx\,dy.$$
I encountered this integral whi... | We may tackle the equivalent integral
$$ \mathcal{J}\stackrel{\text{def}}{=}\iint_{(0,1)^2}\frac{dx\,dy}{(x^2+y^2)\log^2\left(\frac{1}{4}\sqrt{x^2+y^2}\right)} $$
by writing it as
$$ \int_{0}^{1}\frac{\frac{\pi}{2}\rho}{\rho^2\log^2\left(\rho/4\right)}\,d\rho +\int_{1}^{\sqrt{2}}\frac{\left(\frac{\pi}{2}-2\arccos\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2833345",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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If $M_A(x)=x^3-x$ So $M_{A^2}(x)=x^2-x$ If $M_A(x)=x^3-x$ So $M_{A^2}(x)=x^2-x$ Where $M$ is the minimal polynomial
$x^3-x=x(x^2-1)=x(x-i)(x+i)$ so $A$ is diagonalizable
$A=P^{-1}\begin{pmatrix}
0 & 0 & 0 \\
0 & i & 0 \\
0 & 0 & -i \\
\end{pmatrix}P$
$A^2=P^{-1}\begin{pmatrix}
0 & 0 & 0 \\
0 & i & 0 \\
0 & 0 & -i \... | First you can plug in $A^2$ to see whether it satisfies the polynomial. Indeed:
$$(A^2)^2 - A^2 = A^4 - A^2 = A A^3 - A^2 = AA - A^2=0$$
Now check that $A^2$ doesn't satisfy the linear factors $x$ and $x-1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2834016",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Question about Riemann-Stieltjes problem with Euler Use Euler summation formula and integration by parts to show that
$$
\int_1^3 \left(\frac{[[x]]}{x^3}\right)dx = \frac{37}{72}
$$
Sorry I'm still learning how to use MathJax or how to put math formulas into the text...
Currently, I'm an undergraduate studying Math Ana... | Use the real function defined by $f(x)=\frac{-1}{2x^2}\Rightarrow f'(x)=\frac{1}{x^3}$
Euler summation formula implies:
$$\sum_{n=1}^3f(n)=\int_1^3f(x)dx+\int_1^3f'(x)(x-[|x|]-\frac{1}{2})dx+\frac{f(1)+f(3)}{2}$$
$$\sum_{n=1}^3(\frac{-1}{2n^2})=\int_1^3\frac{-1}{2x^2}dx+\int_1^3\frac{1}{x^3}(x-[|x|]-\frac{1}{2})dx-\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2837128",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Rotation of ellipsoid(quadric) Consider $$φ(x, y , z) = x^2 + 2y^2 + 4z^2 −xy −2xz −3yz$$
find the coordinate transformation (translation or rotation) to eliminate $xy$, $xz$ and $yz$.
In $\mathbb R²$, with conic sections, I would do this with
$$\begin{pmatrix}x \\ y\end{pmatrix}=\begin{pmatrix}\cos\alpha &-\sin\alpha... | Perhaps this will help:
\begin{align}
\phi(x,y,z) &= x^2 -x(y+2z)+{(y+2z)^2\over 4} -{(y+2z)^2\over 4}+ 2y^2 + 4z^2 −3yz\\
& = (x-{y+2z\over 2})^2+{7\over 4}y^2+3z^2-4yz\\
& = {1\over 4}(2x-y-2z)^2+{1\over 12}(36z^2-48yz+21y^2)\\
& = {1\over 4}(2x-y-2z)^2+{1\over 12}((6z)^2-2 \cdot4y\cdot 6z+16y^2+5y^2)\\
& = {1\over ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2838874",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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What is going wrong in this log expansion? I am getting a weird result here:
Let $p_1 = q_1 + \Delta$ and Let $p_2 = q_2 - \Delta$
I use the expansion $\log(1-x) = -x -x^2/2 -x^3/3 - ...$ in the third step. This expansion is valid for $x<1$ so there should be no problem with that.
\begin{align*}
D(\vec{p}||\vec{q})... | As you wrote, $$\log(1-x)\sim -x-\frac12x^2-\frac13x^3-\dots\ne-x-x^2-x^3-\dots$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2840462",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Ways of $30$ people ordering from $5$ spicy dishes and $45$ normal dishes
A restaurant serves $5$ spicy dishes and $45$ regular dishes. A group of 30 people each orders dishes, with at most $2$ spicy dishes ordered. How many possible ways of ordering are there if
a) each dish has to be different?
b) a dish can be orde... |
A restaurant serves $5$ spicy dishes and $45$ regular dishes. A group of $30$ people each orders dishes, with at most $2$ spicy dishes ordered. How many possible ways of ordering are there if each dish is different?
The number of ways a subset of exactly $k$ of the $5$ spicy dishes and exactly $30 - k$ of the $45$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2841610",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
find: $\lim_{n\rightarrow\infty}\frac{\sin\left(x+\frac{1}{n}\right)-\sin\left(x\right)}{\sin\left(x+\frac{1}{n}\right)}$ Find: $$\lim_{n\rightarrow\infty}\frac{\sin\left(x+\frac{1}{n}\right)-\sin\left(x\right)}{\sin\left(x+\frac{1}{n}\right)}$$ as $x\in\mathbb{R}$
My progress:
$$\lim_{n\rightarrow\infty}\frac{\sin\le... | use $\sin(A + B) = \sin(A)\cos(B) + \cos(A)\sin(B)$
$$\frac{\sin\left(x+\frac{1}{n}\right)-\sin\left(x\right)}{\sin\left(x+\frac{1}{n}\right)}$$
= $$\frac{\sin(x)\cos(\frac{1}{n}) + \cos(x)\sin(\frac{1}{n}) - \sin(x)}{\sin(x)\cos(\frac{1}{n}) + \cos(x)\sin(\frac{1}{n})}$$
for sin(x) not zero
$$\lim_{n\rightarrow\infty}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2841804",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Range of $(a_{1}-a_{2})^2+(a_{2}-a_{3})^2+(a_{3}-a_{4})^2+(a_{4}-a_{1})^2$ is
If $a_{1},a_{2},a_{3},a_{4}\in \mathbb{R}$ and $a_{1}+a_{2}+a_{3}+a_{4}=0$ and $a^2_{1}+a^2_{2}+a^2_{3}+a^2_{4}=1.$
Then Range of $$E =(a_{1}-a_{2})^2+(a_{2}-a_{3})^2+(a_{3}-a_{4})^2+(a_{4}-a_{1})^2$$ is
Try:
From $$(a_{1}-a_{2})^2+(a_{2}-a... | By Cauchy inequality we have:
$$(a_1+a_3)^2 \leq 2(a_1^2+a_3^2)$$
and
$$(a_2+a_4)^2 \leq 2(a_2^2+a_4^2)$$
Since $(a_1+a_3)^2=(a_2+a_4)^2$ we have $$2(a_1+a_3)^2 \leq 2(a_1^2+a_3^2) +2(a_2^2+a_4^2) =2$$
So $$ E \leq 4$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2842697",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
Explain why would someone factorise $-\int_{-2}^{3}(x^2-x-6)\,\mathrm dx$ to get $\frac{1}{6}(3+2)^3$ I am confused by the workflow of my book.
It went from $$-\int_{-2}^{3}(x^2-x-6)\,\mathrm dx$$ to $$-\int_{-2}^{3}(x+2)(x-3)\,\mathrm dx$$ and, finally $$\frac{1}{6}(3+2)^3$$
| We have that more generally
$$\begin{align}
-\int_{-a}^{b}(x+a)(x-b)dx&=-\int_{-a}^{b}(x+a)(x+a-(a+b))dx\\&=-\left[\frac{(x+a)^3}{3}-(a+b)\frac{(x+a)^2}{2}\right]_{-a}^{b}\\&=-\frac{(a+b)^3}{3}+\frac{(a+b)(a+b)^2}{2}=\frac{(a+b)^3}{6}.
\end{align}$$
This is also a particular the area of a parabolic segment where $w=a+b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2844349",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
how to prove this limit is zero? $$\lim_{n\to\infty}(\frac{\sqrt{1\times2}}{n^2+1}+\frac{\sqrt{2\times3}}{n^2+2}+\cdots+\frac{\sqrt{n\times(n+1)}}{n^2+n})=0$$
I tried to use $\sqrt{k(k+1)}\le\frac{2k+1}{2}$ or $k(k+1)=\sqrt{k}\times\sqrt{k}\times(k+1)\le\left(\frac{2\sqrt{k}+k+1}{3}\right)^3$ but nothing worked out. Ca... | If you enjoy harmonic numbers, starting with
$$\sum \limits_{i=1}^{n}\dfrac{i}{n^2+i} \lt \sum \limits_{i=1}^{n}\dfrac{\sqrt{i(i+1)}}{n^2+i}\lt \sum \limits_{i=1}^{n}\dfrac{i+1}{n^2+i}$$ you have
$$S_1=\sum \limits_{i=1}^{n}\dfrac{i}{n^2+i}=n+n^2\left(H_{n^2}- H_{n^2+n}\right)$$
$$S_2=\sum \limits_{i=1}^{n}\dfrac{i+1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2845138",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Find The Zeros Of $z^3-2z^2+\frac{1}{4}=0$ Find The Zeros Of $z^3-2z^2+\frac{1}{4}=0$ a. at $\frac{1}{4}<|z|<1$ b. $|z|>1$
I know that $\mathbb{C}$ is a closed algebraic field so we can write the polynomial has a product of first degree polynomials, so we will have to guess one root and divide and find the others. but ... | Writing $z = (4 \sin(s) +2)/3$, the equation becomes
$$ 0 = 4 \sin(s)^3 - 3 \sin(s) - \frac{37}{64} = - \sin(3s) - \frac{37}{64}$$
This is $0$ for $s = -\arcsin(37/64)/3+2 \pi k/3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2846874",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Solving this quadratic equation by completing the square? $-\frac{2}{3}x^{2} - x +2 = 0$
Here's what I did:
However, the textbook answer is $x = -2.6, x = 1.1$. What did I do wrong?
| \begin{align*}
-\frac{2}{3}x^2 - x + 2 & = 0\\
-\frac{2}{3}\left(x^2 + \frac{3}{2}x\right) + 2 & = 0\\
-\frac{2}{3}\left[x^2 + \frac{3}{2}x + \left(\frac{1}{2} \cdot \frac{3}{2}\right)^2\right] \color{red}{- \left[-\frac{2}{3}\left(\frac{1}{2} \cdot \frac{3}{2}\right)^2\right]} + 2 & = 0 \tag{1}\\
-\frac{2}{3}\left(x^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2849285",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Compute $\int_{0}^{\pi/2}\csc^2(x)\ln\left(\frac{a+b\sin^2 x}{a-b\sin^2 x}\right)dx$
I would like to compute $I$, using a different method, other than below
$$I=\large\int_{0}^{\pi/2}\csc^2(x)\ln\left(\frac{a+b\sin^2 x}{a-b\sin^2 x}\right)\mathrm dx$$
$a\ge b$
Applying integration by parts:
$\large u=\ln\left(\frac{a... | As suggested in comments by @Dahaka
Let $I(b)$ represent the given integral. Hence we have $I(0)=0$.
Now differentiating $I(b)$ with respect to $b$ we get $$I'(b) =\int_0^{\infty} \csc ^2x \cdot \frac {a-b\sin^2x}{a+b\sin^2x}\cdot\left( \frac {\sin^2x(a-b\sin^2x)+\sin^2x(a+b\sin^2x)}{(a-b\sin^2x)^2}\right) dx$$
Simpl... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2849502",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Prove $ \min \left(a+b+\frac1a+\frac1b \right) = 3\sqrt{2}\:$ given $a^2+b^2=1$ Prove that
$$ \min\left(a+b+\frac1a+\frac1b\right) = 3\sqrt{2}$$
Given $$a^2+b^2=1 \quad(a,b \in \mathbb R^+)$$
Without using calculus.
$\mathbf {My Attempt}$
I tried the AM-GM, but this gives $\min = 4 $.
I used Cauchy-Schwarz to get $\q... | We will square the whole inequality
$$\frac{(a+b)^2}{(ab)^2}+(a+b)^2+2\frac{(a+b)^2}{ab}\geq 18$$
simplifying and using that $$a^2+b^2=1$$ we get
$$2(ab)^3-13(ab)^2+4ab+1\geq0$$
this is equivalent to $$(2ab-1)((ab)^2-6ab-1)\geq 0$$
Now we have $$a^2+b^2\geq 2ab$$ this is $$ab\le \frac{1}{2}$$
so both factors $$2ab-1,(a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2850000",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 9,
"answer_id": 1
} |
Minimum value of $\frac{b+1}{a+b-2}$
If $a^2 + b^2= 1 $ and $u$ is the minimum value of the $\dfrac{b+1}{a+b-2}$, then find the value of $u^2$.
Attempt:
Then I tried this way: Let $a= bk$ for some real $k$.
Then I got $f(b)$ in terms of b and k which is minmum when $b = \dfrac{2-k}{2(k+1)}$ ... then again I got a... | From
$$
\frac{b+1}{a+b-2}= u\Rightarrow L\to a = 2-b\frac{b+1}{u}
$$
So $L$ should be tangent to $a^2+b^2=1\;$ then substituting we have the condition
$$
(2b^2-4b+3)u^2+(4+2b-2b^2)u+(b+1)^2 = 0
$$
and solving for $u$
$$
u = \frac{(b+1)^2}{b^2-b\pm\sqrt{(1-b) (b+1)^3}-2}
$$
but tangency implies on $\sqrt{(1-b) (b+1)^3}=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2852399",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 5
} |
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Fractions in Questions and Answers
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