Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Compute $\int \frac {dx} {(a + b \cos x)^2} $ for $a > b.$ Let $a>b$ be real numbers. As the title suggests, I would like to compute $$\int \frac 1 {(a + b \cos x)^2} \, dx.$$
My attempt consisted of converting the cosine to half-angle tangents, substituting the half-angle tangent for $t$, and simplifying to $$\int \frac{2(1+t^2)}{[(a+b)+(a-b)t^2]^2} \, dt.$$
From here, I multiplied and divided the integrand by $(a-b)$, added an $(a+b)-2b$ to the numerator, and simplified it to $$\frac{4b}{a-b}\int {\left(\frac 1 {[(a+b)+(a-b)t^2]^2} + \frac{2t}{a-b} \right)} \, dt.$$
Now, just looking at the integral left, I substituted $t$ for $\sqrt{\frac{a+b}{a-b}} \tan\theta$, did some simplifying, converted the resulting $\cos^2\!\theta$ in the numerator to $1+\cos 2 \theta$, and wrapped up the integral, finishing up with $$\frac{2\tan \frac{x}{2}}{a-b}-\frac{2b}{(a^2-b^2)^{\frac{3}{2}}}{\left(\arctan \sqrt{\frac{a+b}{a-b}}\tan \frac{x}{2}+ \frac{\sqrt{ \frac{a+b}{a-b}}\tan
\frac{x}{2}}{1+{ \frac{a-b}{a+b}\tan^2 \frac{x}{2}}} \right)}$$
This is apparently wrong. Can someone help me with why?
| Here it is another way to approach it for the sake of curiosity.
Let us make the substitution:
$$\sinh(x) = t\sqrt{\frac{a-b}{a+b}} \Rightarrow \mathrm{d}t = \sqrt{\frac{a+b}{a-b}}\cosh(x)\mathrm{d}x$$
Then the denominator of the given expression can be written as
\begin{align*}
((a+b) + (a-b)t^{2})^{2} & = (a+b)^{2}\left[1 + \left(t\sqrt{\frac{a-b}{a+b}}\right)^{2}\right]^{2}\\\\
& = (a+b)^{2}(1 + \sinh^{2}(x))^{2}\\\\
& = (a+b)^{2}\cosh^{4}(x)
\end{align*}
Besides that, we do also have that
\begin{align*}
2(1 + t^{2}) & = 2\left[1 + \left(\frac{a+b}{a-b}\right)\sinh^{2}(x)\right]\\\\
& = 2\left[1 - \frac{a+b}{a-b} + \left(\frac{a+b}{a-b}\right)(1 + \sinh^{2}(x))\right]\\\\
& = 2\left[-\frac{2b}{a-b} + \left(\frac{a+b}{a-b}\right)\cosh^{2}(x)\right]
\end{align*}
Gathering both results, the proposed integral can be expressed as
\begin{align*}
I & = \sqrt{\frac{a+b}{a-b}}\left[-\int\left(\frac{4b}{(a^{2}-b^{2})(a+b)\cosh^{3}(x)}\right)\mathrm{d}x + \int\left(\frac{2}{(a^{2}-b^{2})\cosh(x)}\right)\mathrm{d}x\right]
\end{align*}
Can you take it from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4089731",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
The order of computing composition of permutation when using cycle notation Given two permutations $p,q$, say $p=(234)$ and $q=(123)$, we can compute the product permutation $pq$ by writing out $pq=(234)(123)$, then turn it into disjoint cycles.
In my understanding, we shall do the latter step in the backward direction: first evaluate on $q$, then on $p$. This is because the product permutation really means the composition of functions. And this gives $pq=(13)(24)$.
However, there are several times that I saw people on the internet doing this in the forward manner, which would give $pq=(12)(34)$.
Could anyone please clarify for me if I'm doing this correctly? Thanks in advance!
| One can understand this by writing out the tables explicitly. Note that permutation $pq$ is applying $q$ and then applying $p$.
\begin{array}{|c|c|c|}
\hline
\text{permutation} & \text{acts on} & \text{produces}\\
\hline
pq=(234)(123)&1&3 \\
\hline
pq=(234)(123)&2&4 \\
\hline
pq=(234)(123)&3&1 \\
\hline
pq=(234)(123)&4&2 \\
\hline
\end{array}
The disjoint cycles are now visible:
\begin{array}{|c|c|c|}
\hline
\text{permutation} & \text{acts on} & \text{produces}\\
\hline
pq=(13)(24)&1&3 \\
\hline
pq=(13)(24)&2&4 \\
\hline
pq=(13)(24)&3&1 \\
\hline
pq=(13)(24)&4&2 \\
\hline
\end{array}
We will use the table below when told specifically that we apply permutation p and then q,
\begin{array}{|c|c|c|}
\hline
\text{permutation} & \text{acts on} & \text{produces}\\
\hline
qp=(123)(234)&1&2 \\
\hline
qp=(123)(234)&2&1 \\
\hline
qp=(123)(234)&3&4 \\
\hline
qp=(123)(234)&4&3 \\
\hline
\end{array}
\begin{array}{|c|c|c|}
\hline
\text{permutation} & \text{acts on} & \text{produces}\\
\hline
qp=(12)(34)&1&2 \\
\hline
qp=(12)(34)&2&1 \\
\hline
qp=(12)(34)&3&4 \\
\hline
qp=(12)(34)&4&3 \\
\hline
\end{array}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4100349",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Bounds on $\sum\limits_{k=1}^n \frac{\sin(k)}{k}$ $\sum\limits_{k=1}^n \frac{\sin(k)}{k}$ converges as $n$ increases, to a limit of $\frac12(\pi-1) \approx 1.0708$
Empirically, it seems to be bounded by about $\frac12(\pi-1) \pm \frac{1.043}{n}$, as shown in the chart below.
What is the precise value of this $1.043$ term?
Added: I think the three answers from Oliver Diaz, Gary, and Gabriel Romon, plus the comments, have demonstrated that the bounds are $$\frac12(\pi-1) \pm \frac1{2\sin(1/2)}\frac{1}{n}$$ not only asymptotically but also as tight actual bounds; $\frac1{2\sin(1/2)}\approx 1.0429148214667441$
created in R with
n <- 1:11000
plot(n, cumsum(sin(n)/n), ylim=c(1.07,1.0715), xlim=c(1,10000), pch=46)
abline(h=(pi-1)/2, col="red")
curve((pi-1)/2 + 1.043/x, from=1, to=max(n), add=TRUE, col="red")
curve((pi-1)/2 - 1.043/x, from=1, to=max(n), add=TRUE, col="red")
| Note that $\displaystyle\left|\sum_{k=1}^n \frac{\sin k}k - \frac12(\pi-1)\right| = \left|\sum_{k=n+1}^\infty \frac{\sin k}k \right|$.
We prove the following inequality:
$$
\left|\sum_{k=n+1}^\infty \frac{\sin k}k \right|
\leq
\frac{1}{2 \sin(\frac 12)} \frac 1{n+1}
+
\frac{3}{4\sin^2(\frac 12)} \frac 1{(n+1)(n+2)}
$$
Let $\displaystyle S_n = \sum_{k=1}^n \sin k = \frac{\cos(1/2)-\cos(n+\frac 12)}{2\sin(\frac 12)}$ and
$\displaystyle T_n
= \sum_{k=1}^n S_k
= \frac{n}{2\tan(\frac 12)} - \frac{\sin(n+1)-\sin (1)}{4\sin^2(\frac 12)}$.
Performing summation by parts twice yields
$$\sum_{k=n+1}^\infty \frac{\sin k}k
= \sum_{k=n+1}^\infty \Big(\frac{S_k}{k(k+1)}\Big) - \frac{S_n}{n+1}
=\sum_{k=n+1}^\infty \Big(\frac{2T_k}{k(k+1)(k+2)}\Big) - \frac{T_n}{(n+1)(n+2)} - \frac{S_n}{n+1}.
$$
Note that
$$\sum_{k=n+1}^\infty \frac{2T_k}{k(k+1)(k+2)} = \frac{1}{\tan(\frac 12)} \frac{1}{n+2} - \frac{1}{2 \sin^2(\frac 12)} \sum_{k=n+1}^\infty \frac{\sin(k+1)-\sin(1)}{k(k+1)(k+2)} $$
and
$$\begin{align}
\frac{T_n}{(n+1)(n+2)} + \frac{S_n}{n+1}
= &\phantom{-}\Big(\frac{1}{\tan(\frac 12)}-\frac{\cos(n+\frac 12)}{2 \sin(\frac 12)} \Big)
\frac 1{n+1} \\ &-
\Big(\frac{1}{\tan(\frac 12)} + \frac{\sin(n+1)-\sin (1)}{4\sin^2(\frac 12)} \Big) \frac 1{(n+1)(n+2)}.
\end{align}
$$
Therefore
$$\begin{aligned}
\sum_{k=n+1}^\infty \frac{\sin k}k =
&\phantom{+}\frac{\cos(n+\frac 12)}{2 \sin(\frac 12)} \frac 1{n+1}
+
\frac{\sin(n+1)-\sin (1)}{4\sin^2(\frac 12)} \frac 1{(n+1)(n+2)}
\\
&-\frac{1}{2 \sin^2(\frac 12)} \sum_{k=n+1}^\infty \frac{\sin(k+1)-\sin(1)}{k(k+1)(k+2)}
\end{aligned}
$$
and
$$\begin{aligned}
\left|\sum_{k=n+1}^\infty \frac{\sin k}k \right|
&\leq \frac{|\cos(n+\frac 12)|}{2 \sin(\frac 12)} \frac 1{n+1}
+
\frac{|\sin(n+1)-\sin (1)|+1}{4\sin^2(\frac 12)} \frac 1{(n+1)(n+2)} \\
&\leq
\frac{1}{2 \sin(\frac 12)} \frac 1{n+1}
+
\frac{3}{4\sin^2(\frac 12)} \frac 1{(n+1)(n+2)}
\end{aligned}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4102159",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 4,
"answer_id": 1
} |
Question related to arithmetico-geometric series The question is $$\text{Find the sum of }0.2+0.004+0.00006+0.0000008+\cdots$$
My solution:
Given series can be written in the form $$S=\frac{2}{10}+\frac{4}{10^3}+\frac{6}{10^5}+\frac{8}{10^7}+\cdots(1)$$ Multiplying both sides by $10^{-2}$, we get
$$\frac{S}{10^2}=\frac{2}{10^3}+\frac{4}{10^5}+\frac{6}{10^7}+\cdots(2)$$
Subtracting equation $(2)$ from equation $(1)$, we get $$S-\frac{S}{100}=\frac{2}{10}+\frac{2}{10^3}+\frac{2}{10^5}+\frac{2}{10^7}+\cdots$$
$$\implies\frac{99}{100}S=\frac{\frac{2}{10}}{1-\frac{1}{100}}=\frac{2}{10}\times\frac{100}{99}=\frac{20}{99}$$ $$\implies S=\frac{20}{99}\times\frac{100}{99}=\frac{2000}{9801}$$ But the answer given in the book is $\dfrac{2180}{9801}$. I checked my solution multiple times but can't figure out my mistake. Any help would be appreciated.
| Alternate method, per request:
Let $\displaystyle S = \frac{2}{10}
\left[1 + \frac{1}{10^2} + \frac{1}{10^4} + \cdots\right] =
\frac{2}{10}\times \frac{100}{99}.$
Then the desired sum is
$$ T = S\left[1 + \frac{1}{10^2} + \frac{1}{10^4} + \cdots\right]
= S \times \left(\frac{100}{99}\right) = \frac{2}{10} \times
\left(\frac{100}{99}\right)^2.$$
Addendum
Actually, the underlying problem: how to evaluate
$$
g(x) = \left(1 + 2x + 3x^2 + 4x^3 + \cdots \right)
~:~ |x| < 1 \tag1$$
has 3 methods of attack. The OP and I (in effect) used closely related but distinct methods, each of intermediate difficulty.
The (3rd) easier method is to use Calculus.
Let $\displaystyle f(x) = \left(1 + x + x^2 + x^3 + \cdots\right) = \frac{1}{1-x}.$
Then $~f'(x),~$ which equals $~~g(x),~$ as shown in equation (1) above, may be equivalently expressed as
$$\frac{(-1)}{(1-x)^2} \times (-1).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4104196",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Minimum value of this expression , was my inequality true or of any help? The problem which I encountered was a,b,c>0 and
$$f(a,b,c):= \left\lfloor\frac{a+b}{c}\right\rfloor+ \left\lfloor\frac{b+c}{a}\right\rfloor+ \left\lfloor\frac{c+a}{b}\right\rfloor$$
where $a$, $b$ and $c$ are positive. Find minimum of $f(a,b,c)$. For what values of $a$, $b$ and $c$ is the function minimum?
Is there a inequality like $\left\lfloor x\right\rfloor+\left\lfloor y\right\rfloor\geq \left\lfloor x+y\right\rfloor$ (and its proof) something which can be helpful here?
As such what I tried to get, expression is equivalent to $\newcommand{\lf}{\left\lfloor}\newcommand{\rf}{\right\rfloor}$
$$\lf\frac{a+b+c}{a}\rf + \lf\frac{a+b+c}{b}\rf + \lf\frac{a+b+c}{c}\rf-3$$
I don't know if that is correct. Can anyone help me?
| Since $\lfloor x\rfloor\gt x-1$, we have, by AM-GM inequality,
$$\begin{align}f(a,b,c)&=\left\lfloor\frac{a+b}{c}\right\rfloor+ \left\lfloor\frac{b+c}{a}\right\rfloor+ \left\lfloor\frac{c+a}{b}\right\rfloor
\\\\&\gt \frac{a+b}{c}-1+ \frac{b+c}{a}-1+ \frac{c+a}{b}-1
\\\\&=\bigg(\frac ba+\frac ab\bigg)+\bigg(\frac cb+\frac bc\bigg)+\bigg(\frac ac+\frac ca\bigg)-3
\\\\&\ge 2\sqrt{\frac ba\cdot\frac ab}+2\sqrt{\frac cb\cdot\frac bc}+2\sqrt{\frac ac\cdot\frac ca}-3
\\\\&=3\end{align}$$
from which we have
$$f(a,b,c)\ge 4$$
whose equality is attained when
$$(a,b,c)=(0.9,1,1)$$
Hence, the minimum of $f(a,b,c)$ is $\color{red}4$.
Added :
Under the condition that $0\lt a\le b\le c$, $f(a,b,c)=4$ holds if and only if $(a,b,c)$ satisfies
$$1\lt \frac ba\lt \frac 32\qquad\text{and}\qquad c\lt \min(3a-b,2b-a)$$
Proof :
We have
$$\left\lfloor\frac{b+c}{a}\right\rfloor\ge \left\lfloor\frac{c+a}{b}\right\rfloor\ge\left\lfloor\frac{a+b}{c}\right\rfloor\ge 0,\qquad \left\lfloor\frac{b+c}{a}\right\rfloor\ge 2,\qquad \left\lfloor\frac{c+a}{b}\right\rfloor\ge 1$$
Suppose that $\bigg(\left\lfloor\dfrac{a+b}{c}\right\rfloor,\left\lfloor\dfrac{b+c}{a}\right\rfloor,\left\lfloor\dfrac{c+a}{b}\right\rfloor\bigg)=(0,3,1)$. Then, $$\begin{align}&\dfrac{a+b}{c}\lt 1,\qquad \dfrac{b+c}{a}\lt 4,\qquad \dfrac{c+a}{b}\lt 2
\\\\&\implies a+b\lt c,\qquad c\lt 4a-b,\qquad c\lt 2b-a
\\\\&\implies a+b\lt 4a-b,\qquad a+b\lt 2b-a
\\\\&\implies 2a\lt b\lt \frac 32a\end{align}$$
which is impossible.
Suppose that $\bigg(\left\lfloor\dfrac{a+b}{c}\right\rfloor,\left\lfloor\dfrac{b+c}{a}\right\rfloor,\left\lfloor\dfrac{c+a}{b}\right\rfloor\bigg)=(0,2,2)$. Then, $$\dfrac{a+b}{c}\lt 1,\qquad \dfrac{b+c}{a}\lt 3$$
$$\implies a+b\lt c\lt 3a-b\implies a+b\lt 3a-b\implies b\lt a$$
which is impossible.
So, we have to have $\bigg(\left\lfloor\dfrac{a+b}{c}\right\rfloor,\left\lfloor\dfrac{b+c}{a}\right\rfloor,\left\lfloor\dfrac{c+a}{b}\right\rfloor\bigg)=(1,2,1)$ which is equivalent to
$$\begin{align}&1\le\frac{a+b}{c}\lt 2,\qquad 2\le\frac{b+c}{a}\lt 3,\qquad 1\le\frac{c+a}{b}\lt 2
\\\\&\iff \frac{a+b}{2}\lt c\le a+b,\qquad 2a-b\le c\lt 3a-b,\qquad b-a\le c\lt 2b-a
\\\\&\iff 1\lt \frac ba\lt \frac 32\qquad\text{and}\qquad c\lt \min(3a-b,2b-a)\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4105403",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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maximum of $x^2-3xy-2y^2$ subjects to $x^2+xy+y^2=1$ without Lagrange multiplier What is the maximum of $x^2-3xy-2y^2$ subjects to $x^2+xy+y^2=1$? I wonder there is a precalculus method, without using the Lagrange multiplier.
| The condition implies $xy = 1 - (x^2 + y^2)$, thus $x^2-3xy - 2y^2 = x^2 - 3 + 3(x^2 + y^2) - 2y^2$ $ = 4x^2 + y^2 - 3$.
Then as previously mentioned, the ellipse $x^2+xy+y^2 = 1$ can be parametrised as $(x,y) = \left(\frac{1}{\sqrt{3}}\cos t+\sin t,\frac{1}{\sqrt{3}}\cos t-\sin t\right)$. Substituting this into $4x^2 + y^2 - 3$ gives:
$$4 \left(\frac{1}{3} \cos^2 t + \frac{2}{\sqrt3} \cos t \sin t + \sin^2 t\right) + \frac{1}{3} \cos^2 t - \frac{2}{\sqrt3} \cos t \sin t + \sin^2 t - 3$$
$$= \frac{5}{3} \cos^2 t + \frac{6}{\sqrt3} \cos t \sin t + 5 \sin^2 t - 3 = \frac{10}{3} \sin^2 t + \sqrt{3} \sin(2t) - \frac{4}{3}$$
$$= -\frac{5}{3} \cos(2t) + \sqrt{3} \sin(2t) + \frac{1}{3}$$
and since the maximum value of $A \cos t + B \sin t$ is $\sqrt{A^2+B^2}$, the maximum value of the expression is $\sqrt{(-5/3)^2 + 3} - \frac{5}{6} = \frac{1 + \sqrt{52}}{3}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4105835",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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calculate expected value of total number of offspring in second generation Suppose a plant has $X$ offspring per year with $P(X=k) = \frac{1}{4}$ for $k = 1,2,3,4.$
Independently, each offspring has from one to four offspring in the next year with the same discrete uniform distribution. Let $Y$ denote the total number of offspring in the second generation. How to calculate the values of $E(Y|X = k)$ for any outcomes of $X$.
| Hint: Start with the definition of expected value:
$$ E[X] = \sum_{i = 1}^n {x*P[X = x]} .$$
If we add in the condition, this becomes:
$$ E[Y | X = x] = \sum_{i = 1}^n {y*P[Y = y | X = x]} .$$
Now, start by asking what happens if the plant has $X = 1$ offspring? There are four possibilities: $Y = 1, 2, 3, 4$. Each has probability $\frac{1}{4}$. So, that gives you the first part of your answer:
$$ E[Y | X = 1] = 1*\frac{1}{4} + 2*\frac{1}{4} + 3*\frac{1}{4} + 4*\frac{1}{4} = \frac{5}{2} .$$
Next, consider what happens when $X = 2$. Now, $Y = 1$ is not possible - each of the two plants in the first generation must have a minimum of $1$ offspring. So, okay, what about $Y = 2$? Well, that is possible. It happens if each of the first two has $1$ offspring, which has probability $\left(\frac{1}{4}\right)^2 = \frac{1}{16}$. Now, what about the probability that $Y = 3$? Well, that can happen two ways - the first offspring has $1$ and the second $2$, or vice versa. So, that probability is $2*\left(\frac{1}{4}\right)^2 = \frac{1}{8}$. The probability that $Y = 4$ is $3*\left(\frac{1}{4}\right)^2 = \frac{3}{16}$ - that's from offspring possibilities of $1$ and $3$, $2$ and $2$, or $3$ and $1$. Keep going through all possible values of $Y | X = 2$. Then, you'll be able to calculate
$$ E[Y | X = 2] = 2*\frac{1}{16} + 3*\frac{1}{8} + 4*\frac{3}{16} + ... $$
You can keep going this route and brute force your way through the problem, or look for a pattern, but that's how to start thinking about this problem.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4106290",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$I_n= \int_{0}^{ + \infty} \dfrac{ x^{4n-2} }{ ( 1+x^4)^n}$
*
*$x \geq 0$
*$n \geq 0$
*$f_n(x)= \dfrac{ x^{4n-2} }{ ( 1+x^4)^n}$
*$I_n= \int_{0}^{ + \infty}f_n(x) $
*What is the limit of $I_n$ ?
*What is an equivalent of $I_n$ ?
$
\begin{align*}
u'&= (-n+1) 4x^3 (1+x^4)^{-n-1}\\
u &=x^{4n-1}\\
v &= x^{4n-1} \\
v' &= (4n-1) x^{4n-2} \\
I_{n+1}&= \int_{0}^{ + \infty} \dfrac{-4n x^3 x^{4n-1} }{ 4 (1+x^4)^{n+1} (-n) }\\
I_{n+1}&= \left[ \dfrac{ x^{4n-1}}{-4n (1+x^4)^n } \right] + \int_{0}^{ + \infty} \dfrac{(4n-1) x^{4n-2}}{4n (1+x^4)^n } \\
I_{n+1} &= \dfrac{4n-1}{4n}I_n \\
I_2 &=\dfrac{4-1}{4 \times 1}I_1 \\
I_{n+1} &= \dfrac{\prod_{k=1}^{n} (4k-1) }{4^n n!} I_1 \\
I_1&= \int_{0}^{ + \infty} \dfrac{ x^{2}}{ (1+x^4) } dx \\
\end{align*}
$
Quanto's answer for $I_1$ :
$$ \int_0^{\infty} \dfrac{ x^{2}}{ 1+x^4 } dx
\overset{x\to \frac1x}=\frac12\int_0^{\infty} \dfrac{ 1+x^2}{ 1+x^4 } dx = \frac12 \int_0^{\infty} \dfrac{ 1+\frac1{x^2}}{ x^2+\frac1{x^2} } dx \\
= \frac12 \int_0^{\infty} \dfrac{ d(x-\frac1{x})}{ (x-\frac1{x})^2+2 } dx =\frac\pi{2\sqrt2}
$$
$I_{n+1} = \dfrac{ \prod_{k=1}^n (4k-1) }{ 4^n n!} I_1 =\prod_{k=1}^n \dfrac{4k-1}{4k} I_1=\prod_{k=1}^n (1- \dfrac{1}{4k}) I_1$
What is the limit ?
$\ln( 1 - u) \sim u$ donc $\ln (1 - \dfrac{1}{4k}) \sim - \dfrac{1}{4k}$
and $- \sum \dfrac{1}{k} = - \infty$ so $I_n \to 0$ ?
| Note that
$$
\log (1 - x) < - x
$$
for $0<x<1$. Therefore,
\begin{align*}
0 & < \prod\limits_{k = 1}^n {\left( {1 - \frac{1}{{4k}}} \right)} = \exp \left( {\sum\limits_{k = 1}^n {\log \left( {1 - \frac{1}{{4k}}} \right)} } \right) \\ & < \exp \left( { - \sum\limits_{k = 1}^n {\frac{1}{{4k}}} } \right) < \exp \left( { - \frac{1}{4}\log n} \right) = \frac{1}{{n^{1/4} }}.
\end{align*}
Thus, by the squeeze theorem,
$$
I_{n + 1} = I_1 \prod\limits_{k = 1}^n {\left( {1 - \frac{1}{{4k}}} \right)} \to 0.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4107233",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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$ \int \frac{x^3}{\sqrt{x^2+x}}\, dx$ I'm trying to solve this irrational integral $$ \int \frac{x^3}{\sqrt{x^2+x}}\, dx$$
doing the substitution
$$ x= \frac{t^2}{1-2 t}$$
according to the rule.
So the integral becomes:
$$ \int \frac{-2t^6}{(1-2t)^4}\, dt= \int (-\frac{1}{8}t^2-\frac{1}{4}t-\frac{5}{16}+\frac{1}{16}\frac{-80t^3+90t^2-36t+5}{(1-2t)^4})\, dt=\int (-\frac{1}{8}t^2-\frac{1}{4}t-\frac{5}{16}+\frac{1}{16}(\frac{10}{1-2t}-\frac{15}{2} \frac{1}{(1-2t)^2}+\frac{3}{(1-2t)^3}-\frac{1}{2} \frac{1}{(1-2t)^4}))\, dt=-\frac{1}{24}t^3-\frac{1}{8}t^2-\frac{5}{16}t-\frac{5}{16}\cdot \ln|1-2t|
-\frac{15}{64}\frac{1}{1-2t}+\frac{3}{64} \frac{1}{(1-2t)^2}-\frac{1}{16 \cdot 12} \frac{1}{(1-2t)^3}+cost $$
with $t=-x+ \sqrt{x^2+x}$.
The final result according to my book is instead $(\frac{1}{3}x^2-\frac{5}{12}x+\frac{15}{24})\sqrt{x^2+x}-\frac{5}{16}\ln( x+\frac{1}{2}+ \sqrt{x^2+x})$
And trying to obtain the same solution putting t in the formulas I'm definitely lost in the calculation...
I don't understant why this difference in the complexity of the solution...
Can someone show me where I'm making mistakes?
| Just for your curiosity since you already received @hamam_Abdallah's answer.
You can compute this integral musch faster with a different change of variable.
Because of the denominator, let $x=\sinh^2(t)$
$$\int \frac{x^3}{\sqrt{x^2+x}}\, dx=2 \int \sinh ^6(t)\,dt$$
Now, using the multiple angle formulae
$$\sinh ^6(t)=\frac{15}{32} \cosh (2 t)-\frac{3}{16} \cosh (4 t)+\frac{1}{32} \cosh (6
t)-\frac{5}{16}$$ and everything becomes simple.
| {
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"source": "stackexchange",
"question_score": "5",
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Proof of $ d(n)\leq \sqrt {3n}$. Show that
$$ d(n)\leq \sqrt {3n}$$
and the equality is true if only if $n=12$,
where $d(n)$ is the number of positive divisors of $n$..
Here is a proof Proof of $ d(n)\leq \sqrt {3n}$.
Let $$ n=\prod_{k=1}^m p_k^{\alpha_k},$$
the proof given above considers the case when $\alpha_k\geq 2$,
What I want to know is how to deal with when some $\alpha_k=1$ and some $\alpha_k\geq 2$.
Any help and hint will welcome, or some other method is provided.
Thanks a lot!
| We will use the following $\textbf{three facts}$:
(a) $$\max_{n\in\mathbb{N}^+}\frac{(n+1)^2}{2^n}=\frac94<3,$$
with equality if and only if $n=2$;
(b) Let $p\geq 3$ and $\alpha\geq 2$, then, by Bernoulli's inequality,
$$p^{\frac{\alpha}{2}}=(1+(p-1))^{\frac{\alpha}{2}}
\ge 1+\frac{p-1}{2}\alpha\ge1+\alpha,$$
with equality if and only if $\alpha=2,p=3$;
(c) Let $$n=\prod_{k=1}^m p_k^{\alpha_k},$$
then $$d(n)=\prod_{k=1}^{m}(1+\alpha_k).$$
The discussion is divided into the following cases:
$(1)$ If $n=p^\alpha$, where $p$ is a prime,
then $d(n)=\alpha+1$.
It is easy to see that
$$d(n)=\sqrt{(\alpha+1)^2}\leq\sqrt{\frac94\cdot2^\alpha}
<\sqrt{3\cdot p^\alpha}=\sqrt{3n}.$$
$(2)$ If $n=p_1p_2\cdots p_m$ with $2\leq p_1<p_2<\cdots<p_m,m\geq 2$,
then $d(n)=2^m$. It is easy to see that
$$4^m<3p_1p_2\cdot4^{m-2}\iff16<3p_1p_2.$$
So
$$d(n)=\sqrt{4^m}<\sqrt{3p_1p_2\cdots p_m}=\sqrt{3n}.$$
$(3)$ If $n=2^2 p_1p_2\cdots p_m$
with $3\leq p_1<p_2<\cdots<p_m$,
then
\begin{align*}
d(n)
&=3\cdot 2^m\\
&=\sqrt{3\cdot 3\cdot 4^m}\\
&\leq\sqrt{3\cdot 2^2p_1p_2\cdots p_m}\\
&=\sqrt{3n}.\ (\mbox{equality is true}\ \iff n=12)
\end{align*}
$(4)$ If $n=2^{\alpha} p_1p_2\cdots p_m$
with $3\leq p_1<p_2<\cdots<p_m,\alpha\geq 3$,
then
\begin{align*}
d(n)
&=(1+\alpha)\cdot 2^m\\
&=\sqrt{(1+\alpha)^2\cdot 4^m}\\
&<\sqrt{\frac94\cdot 2^{\alpha}\cdot4^m}\\
&=\sqrt{3\cdot 2^{\alpha}\cdot 3\cdot4^{m-1}}\\
&\leq\sqrt{3\cdot 2^{\alpha}p_1p_2\cdots p_m}\\
&=\sqrt{3n}.
\end{align*}
$(5)$ If $n=2^{\alpha}p_1^{\alpha_1}p_2^{\alpha_2}\cdots p_m^{\alpha_m}$
with $3\leq p_1<p_2<\cdots<p_m,\alpha\geq 3,\alpha_k\geq 2$,
then
\begin{align*}
d(n)
&=(1+\alpha)\prod_{k=1}^{m}(1+\alpha_k)\\
&<\sqrt{3\cdot 2^{\alpha}}\prod_{k=1}^{m}p_k^{\frac{\alpha_k}2}\\
&=\sqrt{3n}.
\end{align*}
$(6)$ If $n=2^{\alpha}p_1\cdots p_l\cdot p_{l+1}^{\alpha_{l+1}}\cdots p_m^{\alpha_m}$
with $3\leq p_1<\cdots<p_l,3\leq p_{l+1}<\cdots<p_m,\alpha_k\geq 2$,
then
\begin{align*}
d(n)
&=(1+\alpha)\cdot2^l\prod_{k=l+1}^{m}(1+\alpha_k)\\
&\leq\sqrt{\frac94\cdot 2^{\alpha}\cdot 4^l}\prod_{k=1}^{m}p_k^{\frac{\alpha_k}2}\\
&=\left(\frac{3\cdot 4^{l-1}}{p_1\cdots p_l}\right)^{\frac12}
\cdot\sqrt{3\cdot2^{\alpha}p_1\cdots p_l\cdot
p_{l+1}^{\alpha_{l+1}}\cdots p_m^{\alpha_m}}\\
&<\sqrt{3n}.
\end{align*}
| {
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"source": "stackexchange",
"question_score": "1",
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Assistance: "A Gomory cut problem" Hope everything is operating smoothly for everyone here. I am working on a simple problem involving Gomory cutting planes and am having some trouble.
Problem
Consider the following LPP:
\begin{alignat*}{4}
\text{Maximize} \quad \quad \quad x_1 + x_2 & {}{} & & {}{} & & {}{} &\\
\text{Subject To} \quad -3x_1 + 2x_2 & {}\leq{} & 2 & {}{} & & {}{} & & {}{} &\\
4x_1 + x_2 & {}\leq{} & 8 & {}{} & & {}{} & & {}{} &\\
x_1, x_2 & {}\in{} & \mathbb{N}_0 & {}{} & & {}{} & & {}{} &\\
\end{alignat*}
Employing the Simplex method to solve the LP's relaxation, we arrive at the following optimal tableau:
$$
\begin{array}{c|cccc}
z & x_1 & x_2 & x_3 & x_4 & \text{RHS}\\
\hline
0 & 0 & 1 & \frac{4}{11} & \frac{3}{11} & \frac{32}{11} \\
0 & 1 & 0 & -\frac{1}{11} & \frac{2}{11} & \frac{14}{11} \\
\hline
-1 & 0 & 0 & -\frac{3}{11} & -\frac{5}{11} & -\frac{46}{11}
\end{array}
$$
Q: Find two Gomory cutting planes that "chop off" the optimal solution to the LP relaxation for this problem, one using the first row and the other using the second row of the tableau.
My Attempt
We know $x_3 = 2 + 3x_1 - 2x_2$ and $x_4 = 8 - 4x_1 + x_2$. Using the first row of the tableau, we see that $$ \text{frac}(\frac{4}{11})x_3 + \text{frac}(\frac{3}{11})x_4 \geq \text{frac}(\frac{32}{11}) \implies \frac{4}{11}x_3 + \frac{3}{11}x_4 \geq \frac{32}{11} - \bigg\lfloor \frac{32}{11} \bigg\rfloor = \frac{10}{11}$$ We can rewrite this inequality equation in terms of $x_4$, so $x_4 = 8 - 4x_1 + x_2 \geq -\frac{4}{3}x_3 + \frac{10}{3}$. Since $x_3= 2 + 3x_1 - 2x_2$, this implies that $x_2 \leq 2$. This is the first cut (not sure if I am doing this correctly). Now we can look at the second row of the tableau and see that $$ \text{frac}(-\frac{1}{11})x_3 + \text{frac}(\frac{2}{11})x_4 \geq \text{frac}(\frac{14}{11}) \implies -\frac{1}{11}x_3 + \frac{2}{11}x_4 \geq \frac{14}{11} - \bigg\lfloor \frac{14}{11} \bigg\rfloor = \frac{3}{11}$$
Proceeding in a similar manner as before, we can rewrite $x_3$ in terms of $x_4$, which gives us $x_3 \geq -3 + 2x_4$. Since $x_3 = 2 + 3x_1 - 2x_2 \geq -3 + 2x_4$ we get that $x_1 \geq 1$, which forms the second cutting plane (again, not sure if what I've done is accurate or right in any case).
So, the first cutting plane is $x_2 \leq 2$ and the second cutting plane is $x_1 \geq 1$.
| So you have:
$$x_1 - \frac{1}{11}x_3 + \frac{2}{11} x_4 = \frac{14}{11}. \qquad (1)$$
Since all variables are nonnegative, we can round down the coefficients and obtain a valid inequality:
$$x_1 - x_3 \leq \frac{14}{11}.$$
Since $x_1$ and $x_3$ are integer, this inequality is the same as:
$$x_1 - x_3 \leq 1.$$
If you subtract this from $(1)$, you obtain:
$$\frac{10}{11}x_3 + \frac{2}{11} x_4 \geq \frac{3}{11}.$$
Since $x_3 = 2+3x_1-2x_2$ and $x_4=8-4x_1-x_2$, this becomes:
$$\frac{10}{11}(2+3x_1-2x_2) + \frac{2}{11} (8-4x_1-x_2) \geq -\frac{3}{11},$$
which simplifies to the cut:
$$2x_1 - 2x_2 \geq -3.$$
| {
"language": "en",
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"source": "stackexchange",
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If $\frac{7}{2^{1/2} + 2^{1/4} + 1} = A + B*2^{1/4} + C*2^{1/2} + D*2^{3/4},$ find $A,B,C,D$
If $\frac{7}{2^{1/2} + 2^{1/4} + 1} = A + B*2^{1/4} + C*2^{1/2} + D*2^{3/4},$ find $A,B,C,D$ .
What I Tried: I wrote :-
$$\rightarrow \frac{7}{2^{1/2} + 2^{1/4} + 1} = \frac{2^2 + 2 + 1}{2^{1/2} + 2^{1/4} + 1}$$
Then I rationalized the denominator to get :-
$$\rightarrow\frac{(2^2 + 2 + 1)(2^{1/2} + 2^{1/4} - 1)}{(2^{1/2} + 2^{1/4} + 1)(2^{1/2} + 2^{1/4} -1)}$$
$$\rightarrow \frac{2^{5/2}+2^{9/4}-2^2+2^{3/2}+2^{5/4}-2+2^{1/2}+2^{1/4}-1}{(2^{1/2} + 2^{1/4})^2 - 1}$$
$$\rightarrow \frac{2^{5/2}+2^{9/4}-2^2+2^{3/2}+2^{5/4}-2+2^{1/2}+2^{1/4}-1}{( 2^{5/4}+2^{1/2} + 1)}$$
I am stuck here. I probably cannot proceed further unless I find a way to factorise the numerator, but I have not find a way to do that for now. Can anyone help me with this problem?
| Hint : $(2^{\frac{1}{4}}-1) \cdot (2^\frac{1}{2} + 2^\frac{1}{4} +1) = 2^\frac{3}{4} -1$
Hint : $7 = 2^3-1$
| {
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$|\sin(\theta)|=\frac{2}{\pi}-\frac{4}{\pi} \cdot\sum_{m=1}^{\infty}\frac{\cos(2m\theta)}{4 m^2-1}\Rightarrow \sum_{m=1}^{\infty}\frac{1}{16m^2-1} =$? By the equation $|\sin(\theta)|=\frac{2}{\pi}-\frac{4}{\pi} \cdot \sum_{m=1}^{\infty}\frac{\cos(2m\theta)}{4m^2-1}$, how can I get the value of $\sum_{m=1}^{\infty}\frac{1}{16m^2-1}$?
What I tried:
If I substitute $\theta=\frac{\pi}{2}$, $|\sin(\frac{\pi}{2})|=\frac{2}{\pi}-\frac{4}{\pi} \cdot \sum_{m=1}^{\infty}\frac{\cos(m.\pi)}{4m^2-1}=1$. For odd $m$, $\cos(m.\pi)=-1$ and, for even $m$, $\cos(m.\pi)=1$.
If I substitute $\theta=0$, $|\sin(0)|=\frac{2}{\pi}-\frac{4}{\pi} \cdot \sum_{m=1}^{\infty}\frac{1}{4m^2-1}=0$.
$|\sin(\frac{\pi}{2})|+|\sin(0)|=\frac{4}{\pi}-\frac{4}{\pi} \cdot \sum_{m=1}^{\infty}\frac{1+\cos(m.\pi)}{4m^2-1}=1\Rightarrow\frac{4}{\pi}.(1-2.\sum_{n=1}^{\infty}\frac{1}{4.(2.n)^2-1})=1$
$\Rightarrow 1-2.\sum_{n=1}^{\infty}\frac{1}{16.n^2-1}=\frac{\pi}{4} \Rightarrow -2.\sum_{n=1}^{\infty}\frac{1}{16.n^2-1}=\frac{\pi-4}{4} \Rightarrow \sum_{n=1}^{\infty}\frac{1}{16.n^2-1}=\frac{4-\pi}{8}$
| Hint. One may observe that
$$
\frac{\cos \left(2 m \times 0\right)+\cos \left(2 m\times\frac \pi 2\right)}{2} =
\begin{cases}
1, & \text{if $m$ is even}, \\
0, & \text{if $m$ is odd}.
\end{cases}
$$ Then putting successively $\theta:=0$ and $\theta:=\frac \pi 2$ in the given identity and averaging may help.
| {
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"url": "https://math.stackexchange.com/questions/4109782",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Length of cardiod $r=1+\sin\left(\theta\right)$ I'm pretty sure this is correct since I'm able to justify each of the steps but I'm just wondering if there's a more efficient way that isn't just skipping steps? Because I had to do 2 substitutions, one of which converted sine into cosine.
\begin{align*}
\text{Length }= & \ 2 \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sqrt{2 + 2\sin\left(\theta\right)} d\theta \\
= & \ 2 \sqrt{2} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sqrt{1 + \sin\left(\theta\right)} d\theta\\
= & \ - 2\sqrt{2} \int_{\pi}^{0} \sqrt{1+ \sin\left(\frac{\pi}{2}-x\right) } dx \\
= & \ - 2\sqrt{2} \int_{\pi}^{0} \sqrt{1+ \cos\left(x\right) } dx \\
= & \ - 2\sqrt{2} \int_{\pi}^{0} \sqrt{2 \cos^2\left(\frac{x}{2}\right)} dx \\
= & \ - 4 \int_{\pi}^{0} \left|\cos\left(\frac{x}{2}\right)\right| dx \\
= & \ 4 \int_{0}^{\pi} \cos\left(\frac{x}{2}\right) dx \\
= & \ 4 \cdot 2 \int_{0}^{\frac{\pi}{2}} \cos\left(u\right) du \\
= & \ 8 \left[\sin\left(u\right)\right]_{u=0}^{u=\frac{\pi}{2}} \\
= & \ 8 \left[\sin\left(\frac{\pi}{2}\right) - \sin\left(0\right) \right] \\
= & \ 8 \\
\end{align*}
edit: admittedly, I do know of other answers but I have no idea how one would have come across such trigonometric identities, or reason out how such an identity would be beneficial.
| Everything is correct including the setup of the integral (using symmetry), but note that $1 + \sin x = (\cos x/2 + \sin x/2)^2$ as mentioned here.
When taking the square root you have $|\cos x/2 + \sin x/2|$, but since this equals $|\sqrt{2} \cos(x/2 - \pi/4)|$ which is non-negative for $-\frac{\pi}{2} ≤ x ≤ \frac{3 \pi}{2}$, $|\cos x/2 + \sin x/2| = \cos x/2 + \sin x/2$ for your integral.
| {
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"source": "stackexchange",
"question_score": "3",
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Evaluate $\int_{-1}^1 \frac{x^2e^{\arctan x}}{\sqrt{{1+x^2}}}\ \mathrm{d}x$ $$\int_{-1}^1 \frac{x^2e^{\arctan x}}{\sqrt{{1+x^2}}}\ \mathrm{d}x$$
My try:
I tried substituting $x= \tan\theta$ and got,
$\displaystyle\int e^\theta \tan^2\theta \sec\theta\ \mathrm{d}\theta$ and cannot move forward. so I thought of finding two integrals($I$ and $J$)one of which is required and the other is taken such that $I+J, I-J$ are easy to find.but couldn't find such $J$ $\left(I=\displaystyle\int\frac{x^2e^{\arctan x}}{1+x^2}\ \mathrm{d}x\right)$
Also, I don't think I could make use of definite integral properties.
WolframAlpha result for the integral is
$\frac{(x-1)\sqrt{1+x^2}e^{arctanx}}{2}+C$
I also want to know how you got the idea for the solution.
| Observe that:
$$(e^{\arctan(x)} \sqrt{1+x^2})' = \frac{e^{\arctan(x)}}{\sqrt{1+x^2}} + \frac{xe^{\arctan(x)}}{\sqrt{1+x^2}}$$
$$x(e^{\arctan(x)}\sqrt{1+x^2})' = \frac{xe^{\arctan(x)}}{\sqrt{1+x^2}} + \frac{x^2 e^{\arctan(x)}}{\sqrt{1+x^2}}$$
Let us integrate the left-hand side. So, we have:
$$J = \int_{-1}^{1} x(e^{\arctan(x)}\sqrt{1+x^2})' \ dx = [x\sqrt{1+x^2}e^{\arctan(x)}]_{-1}^{1} - \int_{-1}^{1} e^{\arctan(x)} \sqrt{1+x^2} \ dx$$
Now, we have:
$$I = \int_{-1}^{1} \frac{xe^{\arctan(x)}}{\sqrt{1+x^2}} \ dx$$
Let $u = e^{\arctan(x)}$ and $dv = \frac{x}{\sqrt{1+x^2}} \ dx$. Then, $du = \frac{e^{\arctan(x)}}{1+x^2} \ dx$ and $v = \sqrt{1+x^2}$. So, we have that:
$$I = [\sqrt{1+x^2}e^{\arctan(x)}]_{-1}^{1} - \int_{-1}^{1} \frac{e^{\arctan(x)}}{\sqrt{1+x^2}} \ dx$$
Let the integral we want be denoted by $K$. Then:
$$K = J-I = 2\sqrt{2}e^{-\frac{\pi}{4}} + \int_{-1}^{1} -\frac{x^2e^{\arctan(x)}}{\sqrt{1+x^2}} \ dx$$
$$2K = 2\sqrt{2}e^{-\frac{\pi}{4}}$$
$$K = \sqrt{2}e^{-\frac{\pi}{4}}$$
as was desired. I initially had the idea of playing around with the product of $e^{\arctan(x)}$ and $\sqrt{1+x^2}$ because I basically observed that $\frac{x^2e^{\arctan(x)}}{\sqrt{1+x^2}} = x \cdot \frac{xe^{\arctan(x)}}{\sqrt{1+x^2}}$ and I had the idea that the second portion could be obtained by differentiating $e^{\arctan(x)}\sqrt{1+x^2}$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Simplify convolution integral $\int_{-\infty}^\infty\frac{1}{(a^2+t^2)(a^2+(x-t)^2) }dt$ I am trying to show that $$\frac{1}{a^2+x^2} \ast \frac{1}{a^2+x^2}=\int_{-\infty}^\infty\frac{1}{(a^2+t^2)(a^2+(x-t)^2) }dt = \frac{2\pi}{a(4a^2+x^2)}$$ I wrote out the integral for the convolution and it becomes a big exercise in partial fractions and integration of rational functions.
$$ \int(\frac{1}{a^2+t^2}) (\frac{1}{a^2+(x-t)^2}) dt=
\int\frac{x + 2 t}{x (4 a^2 + x^2) (a^2 + t^2)} + \frac{3 x - 2 t}{x (4 a^2 + x^2) (a^2 + x^2 - 2 xt + t^2)} dt =$$
$$=\frac{a (log(a^2 + t^2) - log(a^2 + (x - t)^2)) + x \cdot tan^{-1}(\frac{t}{a}) + x \cdot tan^{-1}(\frac{t - x}{a})}
{a x (4 a^2 + x^2)}
$$
Plugging in the bounds of plus/minus infinity makes the logs cancel out, and gives the 2$\pi x$ from the arctan terms.
My question: Is there a way to simplify / make this any easier using the properties of the convolution? Or are these problems just 'plug and chug' like back in my engineering days?
| Less elegant than @Quanto's answer, let us write
$$a^2+(x-t)^2=(t-r)(t-r)$$ with $r=(x+ia)$ and $r=(x-ia)$ and use partial fraction decomposition
$$\frac{1}{(a^2+t^2)(a^2+(x-t)^2) }=\frac{r s+r t+s t-a^2}{\left(a^2+r^2\right) \left(a^2+s^2\right) \left(t^2+a^2\right)}+$$
$$\frac{1}{\left(a^2+r^2\right) (r-s) (t-r)}-\frac{1}{\left(a^2+s^2\right) (r-s) (t-s)}$$ Three simple antiderivatives and then the result.
| {
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For positive integer $n$, why is $\lfloor \log_{10}(2^n)\rfloor + \lfloor \log_{10}(5^n)\rfloor + 2 = n+1$?
For positive integer $n$, why is $\lfloor \log_{10}(2^n)\rfloor + \lfloor \log_{10}(5^n)\rfloor + 2 = n+1$?
This question comes from counting the number of digits of $10^n$ in terms of the number of digits of $2^n$ and $5^n$. Number of digits of $10^n$ is $n+1$ which equals to the sum of the number of digits of $2^n$ and $5^n$. I know that the number of digits of a positive integer $x$ is $\lfloor \log_{10}(x)\rfloor + 1$.
Using programming, I've checked that this is true for $n$ less than $100$.
| Consider $\lfloor n\log_{10}2\rfloor+\lfloor n\log_{10}5\rfloor+1$. Since $\log_{10}2$ and $\log_{10}5$ are irrational we can replace each floor by a ceiling and a $-1$, i.e. it is equal to
$$\lceil n\log_{10}2\rceil+\lceil n\log_{10}5\rceil-1$$
Now we have the following inequalities for any $x,y$:
$$\lfloor x+y\rfloor\le\lfloor x\rfloor+\lfloor y\rfloor+1$$
$$\lceil x\rceil+\lceil y\rceil-1\le\lceil x+y\rceil$$
This bounds $\lfloor n\log_{10}2\rfloor+\lfloor n\log_{10}5\rfloor+1$ from below by $\lfloor n(\log_{10}2+\log_{10}5)\rfloor=n$ and from above by $\lceil n(\log_{10}2+\log_{10}5)\rceil=n$, so
$$\lfloor n\log_{10}2\rfloor+\lfloor n\log_{10}5\rfloor+1=n\ \forall n\in\mathbb Z^+$$
Putting the multipliers $n$ in the logs and adding $1$ to both sides gives the desired identity.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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$25$ people will be divided into $5$ groups ,each group have $5$ individuals . What is the probability that $25$ people will be divided into $5$ groups ,each group have $5$ individuals . What is the probability that
a-) Dennis , John and Jack are in the same group.
b-)Dennis , John and Jack are in different groups.
c-)Dennis and John are in the same group but not Jack
My attempt :
a-) If they are in same group , then there are $5$ ways to select this group. Moreover , we should select $2$ people for the group by $C(22,2)$ . Then $5 \times \frac{C(22,2) \times C(20,5) \times C(15,5) \times C(10,5) \times C(5,5)}{C(25,5) \times C(20,5) \times C(15,5) \times C(10,5) \times C(5,5)}$
b-)If they are in different groups , we can disribute them by $P(5,3)=60$ ways.Then , $60 \times \frac{C(24,4) \times C(19,4) \times C(14,4) \times C(10,5) \times C(5,5)}{C(25,5) \times C(20,5) \times C(15,5) \times C(10,5) \times C(5,5)}$
c-)We can choose $5$ groups for Dennis and John , 4 groups for Jack ,so $P(5,2)=20$ ways. Then ; $20 \times \frac{C(23,3) \times C(19,4) \times C(15,5) \times C(10,5) \times C(5,5)}{C(25,5) \times C(20,5) \times C(15,5) \times C(10,5) \times C(5,5)}$
Is my solution way correct ? If not ,can you help..
| While there is a simpler method as given in one of the answers, even going by your method the working can be simplified quite a bit.
a) Once you fix Dennis's group, there are $4$ more slots in the group to be picked from $24$ people. But if John and Jack are in the same group, you are left with choosing two spots from $22$ people. So desired probability is
$\displaystyle \small {22 \choose 2} / {24 \choose 4} = \frac{1}{46}$
b) The number of arrangements with Dennis, Jack and John being in different groups
$ = \displaystyle {5 \choose 3} \cdot 3! \cdot \frac{22!}{ (5!)^2 (4!)^3}$
Unrestricted arrangements $ = \displaystyle \frac{25!}{ (5!)^5}$
So dividing, you get desired probability of $\displaystyle \frac{25}{46}$
c) Here I will just use result from a) and b). If we add results from a) and b) and subtract from $1$, that gives us probability that any two of them are together while the third is in a different group. But there are $3$ ways to choose which two of them are together and it is equally likely that any two of them are together. So probability that Dennis and John are in the same group but not Jack will be
$ \displaystyle \small \frac{1}{3} \cdot \big(1 - \frac{1}{46} - \frac{25}{46}\big) = \frac{10}{69}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4130868",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solution Verification: $\sum_{n=1}^{\infty}2^{-n}\left(1+\frac{(-1)^n}{n}\right)^{n^2}$ I don't have the 'official' solution to this series so I wanted to verify mine.
\begin{equation*}
(s_n)=\sum_{n=1}^{\infty}2^{-n}\left(1+\frac{(-1)^n}{n}\right)^{n^2}
\end{equation*}
$\underline{\text{Claim:}}$ $(s_n)$ diverges.
$\underline{\text{Proof:}}$ Using root test:
\begin{equation*}
\sqrt[n]{\left|a_n\right|}=\sqrt[n]{\left|2^{-n}\cdot\left(1+\frac{(-1)^n}{n}\right)^{n^2}\right|}=\sqrt[n]{2^{-n}\cdot\left(1+\frac{(-1)^n}{n}\right)^{n^2}}.
\end{equation*}
Further:
\begin{equation*}
\sqrt[n]{2^{-n}\cdot\left(1+\frac{(-1)^n}{n}\right)^{n^2}}=\sqrt[n]{2^{-n}}\cdot\sqrt[n]{\left(1+\frac{(-1)^n}{n}\right)^{n^2}}=\frac{1}{2}\cdot\left(1+\frac{(-1)^n}{n}\right)^n.
\end{equation*}
\begin{equation*}
\lim_{n\rightarrow\infty}\frac{1}{2}\cdot\left(1+\frac{(-1)^n}{n}\right)^n=\begin{cases}\frac{e}{2} & \text{n is even,} \\ \frac{1}{2e} & \text{n is odd.}
\end{cases}
\end{equation*}
Therefore:
\begin{equation*}
C=\limsup\sqrt[n]{\left|a_n\right|}=\frac{e}{2}.
\end{equation*}
Since $C=\frac{e}{2}>1$: $~~~(s_n)$ diverges.
| Yes. This works.
More directly, we note that, for $n\geq 1,$ $$\left(1+\frac 1 n\right)^n\geq 2$$
This follows by noting the first two terms of the binomial expansion of the left side are $1+n\cdot \frac1n=2.$
Then for even $n$:
$$2^{-n}\left(1+\frac{(-1)^n}n\right)^{n^2}\geq 2^{-n}\cdot 2^n=1.$$
So the terms do not converge to $0.$
| {
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"source": "stackexchange",
"question_score": "2",
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Find Two Rank 1 Matrices I'm looking for rank one matrices B and C such that A = B + C and BC = 0. Where
$$A = \begin{bmatrix}
0 & 2 & 2\\
2 & 4 & 2\\
2 & 2 & 0
\end{bmatrix}$$
with eigenvalues 0, -2, 6 and eigenvectors \begin{bmatrix} 1 \\ -1 \\ 1 \end{bmatrix} \begin{bmatrix} -1 \\ 0 \\ 1 \end{bmatrix} \begin{bmatrix} 1 \\ 2 \\ 1 \end{bmatrix}
Sorry, I don't really know how to format
| You can diagonalize $A$ as $D_A=P^{-1}AP$ with:
$$P=
\begin{pmatrix}
1 & -1 & 1 \\
-1 & 0 & 2 \\
1 & 1 & 1 \\
\end{pmatrix} \text{, }
D_A=
\begin{pmatrix}
0 & 0 & 0 \\
0 & -2 & 0 \\
0 & 0 & 6 \\
\end{pmatrix}
=\text{diag}(0,-2,6)
$$
We try to find $B$ and $C$ with the same eigenvectors as $A$ (it's easier, but with other conditions it might not be possible). In that case the matrices are simultaneously diagonizable and the EVs can be added and multiplied componentwhise when the matrices are added an multiplied. So the conditions translate to the eigenvalues $(b_1,b_2,b_3)$ of $B$ and $(c_1,c_2,c_3)$ of $C$ as follows:
*
*From $BC=0$ we get: $b_1*c_1=0,\: b_2*c_2=0, \:b_3*c_3=0$.
*From $A=B+C$ we get: $b_1+c_1=0$, $b_2+c_2=-2$, $b_3+c_3=6$.
*Since $B$, $C$ are rank 1 neither can have EVs $(0,0,0)$
The equations can be resolved for $D_B=\text{diag}(0,0,6) \text{ and } D_C=\text{diag}(0,-2,0)$ and we have $D_A=D_B+D_C$ and $D_B*D_C=0$.
Now your conditions are fulfilled for $B=PD_BP^{-1}$ and $A=PD_AP^{-1}$. That results in:
$$B=\begin{pmatrix}1&2&1\\ 2&4&2\\ 1&2&1\end{pmatrix},\;C=\begin{pmatrix}-1&0&1\\ 0&0&0\\ 1&0&-1\end{pmatrix}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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Evaluating limits using Taylor expansions The limits are
$$\lim_{x\to 0}(\frac{\cos{x}-e^{-x^2/2}}{x^4})$$
$$\lim_{x\to 0}\frac{e^x\cdot\!\sin{x}-x(1+x)}{x^3}$$
Probably wrong things that I've tried
$\lim_{x\to 0}\frac{\cos{x}-e^{-x^2/2}}{x^4}=\lim_{x\to 0}(\frac{1-\frac{x^2}{2!}+\frac{x^4}{4!}+o(x^5)-1+\frac{x^2}{2}+\frac{x^4}{4}+o(x^4)}{x^4})$
$\lim_{x\to 0}\frac{e^x\cdot\!\sin{x}-x(1+x)}{x^3}=\lim_{x\to 0}\frac{(1+x+\frac{x^2}{2}+o(x^2))(x+o(x^2))-x-x^2)}{x^3}=\lim_{x\to 0}\frac{\frac{x^3}{2}+o(x^2)}{x^3}=1/2$
Could you please help me understand how these kinds of limits can be computed I keep getting all of them wrong and if this goes on for a little bit longer I may have a panic attack.
| You have$$e^x\sin(x)=x+x^2+\frac{x^3}3+o(x^3),$$and therefore$$\lim_{x\to0}\frac{e^x\sin(x)-x-x^2}{x^3}=\frac13.$$
| {
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"source": "stackexchange",
"question_score": "4",
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Condition that satisfies inequality involving fractions and the floor function I want to solve the following inequality for $a$ :
$$\left \lfloor \frac{a}{b} \right \rfloor > \frac{a}{b+1}$$
Where both $a$ and $b$ are strictly positive integers.
My end goal with this is, for a given $b$, to be able to find the smallest $a$ of the form $6 \times 10^{7+p}$ ($p \in \mathbb{Z}$)
If you would like to known more about the context :
I stumbled upon this problem while coding a musical file format converter.
In one file format, the BPM of the song is stored as some integer version of $$\frac{6 \times 10^7}{\textrm{BPM}}$$ which I call the tempo value.
My code internally handles decimal BPMs, so it needs to convert back and forth between the two "representations". I wanted this conversion to be both "idempotent" and "minimal" in some way, "idempotent" as in : if I convert a tempo value to a BPM and back, I get the same value, and "minimal" meaning that the intermediate BPM is stored with the minimum number of decimal places needed to recover the exact tempo value if I just floor the "raw" conversion on the way back.
if $v$ is the tempo value, the "raw" BPM can be computed as $$\textrm{BPM}_\textrm{raw} = \frac{6 \times 10^7}{v}$$ I then need to choose some number of decimal places $p$ to truncate it to, which (thanks wikipedia) can be expressed this way : $$\textrm{BPM}_\textrm{trunc} = \textrm{floor(}\textrm{BPM}_\textrm{raw}, p\textrm{)} = \frac{\left \lfloor 10^p \cdot \textrm{BPM}_\textrm{raw} \right \rfloor}{10^p}$$ And then to convert it back to an integer tempo value, I reuse the formula then floor the result again : $$v_\textrm{recovered} = \left \lfloor \frac{6 \times 10^7}{\textrm{BPM}_\textrm{trunc}} \right \rfloor$$ Plugging every formula together, I want to find a condition on p that satisfies : $$ \begin{align} v &= v_\textrm{recovered} \\ v &= \left \lfloor \frac{6 \times 10^7}{\textrm{BPM}_\textrm{trunc}} \right \rfloor \\ v &= \left \lfloor \frac{6 \times 10^7}{\frac{\left \lfloor 10^p \cdot \textrm{BPM}_\textrm{raw} \right \rfloor}{10^p}} \right \rfloor \\ v &= \left \lfloor \frac{6 \times 10^7}{\frac{\left \lfloor 10^p \cdot \frac{6 \times 10^7}{v} \right \rfloor}{10^p}} \right \rfloor \\ v &= \left \lfloor \frac{6 \times 10^{7+p}}{\left \lfloor \frac{6 \times 10^{7+p}}{v} \right \rfloor} \right \rfloor \\ \end{align} $$
If we rename $v$ to $b$ and $6 \times 10^{7+p}$ to $a$ we need to find a condition on $a$ so that $$b = \left \lfloor \frac{a}{\left \lfloor \frac{a}{b}\right \rfloor}\right \rfloor$$
$b$ is equal to the floor of that fraction if and only if the value of that fraction sits between $b$ and $b+1$ exclusive, which translates to $$ b \leq \frac{a}{\left \lfloor \frac{a}{b}\right \rfloor} < b+1$$ The left part is true, this comes from the fact that $\left \lfloor x \right \rfloor \leq x$ for any $x$, plugging in $\frac{a}{b}$ gives $$\left \lfloor \frac{a}{b} \right \rfloor \leq \frac{a}{b} \Leftrightarrow \frac{a}{\frac{a}{b}} \leq \frac{a}{\left \lfloor \frac{a}{b} \right \rfloor} \Leftrightarrow b \leq \frac{a}{\left \lfloor \frac{a}{b} \right \rfloor}$$ So I just need to find a condition for the right side, which can be rewritten a bit : $$\frac{a}{\left \lfloor \frac{a}{b}\right \rfloor} < b+1 \Leftrightarrow \frac{a}{b+1} < \left \lfloor \frac{a}{b}\right \rfloor$$ And here we are ...
| Let $a=bk+r$, where $0\leq r\leq {b-1}.$
Then we have, $\lfloor {\frac ab} \rfloor=k.$
So, $k>\frac {a}{b+1}.$
So that, $bk+k>bk+r.$
Which means, $k>r$. So, the range will be $a>b(r+1).$ So the range of $a$ for which your inequality is satisfied depends upon what $a$ mod $b$ is. But $a>b^2$ will invariably give safe values.
| {
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"source": "stackexchange",
"question_score": "10",
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Indefinite integration of $\int\frac{1}{\cos(x-1)\cos(x-2)\cos(x-3)}\,\textrm dx$
Integrate $$\int\dfrac{1}{\cos(x-1)\cos(x-2)\cos(x-3)}\,\textrm dx$$
My Attempt:
Using, $$\tan A-\tan B=\dfrac{\sin(A-B)}{\cos A\cdot \cos B}$$
The given integral can be transformed as
$$\int\dfrac{\tan(x-1)}{\cos(x-2)}\,\textrm dx - \int\dfrac{\tan(x-3)}{\cos(x-3)}\,\textrm dx$$
The right most integral can be calculated easily by writing $\tan(x-3)$ as $\frac{\sin(x-3)}{\cos(x-3)}$ and then by a substituiton $\cos(x-3)$ as $t$. But I have no clue for the left most integral. How to evaluate that?
| Hint:
I found the expression $$=\dfrac{\sin(x-1)}{\cos(x-1)\cos(x-3)}-\dfrac{\sin(x-2)}{\cos(x-2)\cos(x-3)}$$
Now,
$$\dfrac{\sin(x-1)}{\cos(x-1)\cos(x-3)}$$
$$=\dfrac{\sin(x-1)}{\sin2}\cdot\dfrac{\sin(x-1-(x-3))}{\cos(x-1)\cos(x-3)}$$
$$=\dfrac{1-\cos^2(x-1)}{\sin2\cos(x-1)}-\dfrac{\sin(x-1)\sin(x-3)}{\sin2\cos(x-3)}$$
The first part can be managed easily.
For the second part,
$$\dfrac{\sin(x-1)\sin(x-3)}{\cos(x-3)}=\dfrac{\sin(x-3+2)\sin(x-3)}{\cos(x-3)}=\dfrac{\cos2\sin^2(x-3)}{\cos(x-3)}-\sin2\sin(x-3)$$
Can you take it home from here?
| {
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Proving $\left(a^2+b^2\right)^2\geqslant(a+b+c)(a+b-c)(b+c-a)(c+a-b)$ for positive reals $a$, $b$, $c$ Question $5$ of BMO1 $2008$:
For positive real numbers $\;a,\;b,\;c,\;$ prove that $$\left(a^2+b^2\right)^2\geqslant(a+b+c)(a+b-c)(b+c-a)(c+a-b)$$
I noticed that the right side can be grouped, but did not get further.
| Expanding both sides and moving positive terms to their respective sides, we get:
$a^4 +b^4+\dfrac{c^4}{2}\geqslant a^2 c^2 + b^2 c^2$
By AMGM we have:
$a^4 + \dfrac{c^4}{4}\geqslant a^2 c^2\;$ and $\;b^4 + \dfrac{c^4}{4}\geqslant b^2 c^2$
| {
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Recurrence Relation: $a_{n+2}+a_n=5\cos (n\pi/3)-7\sin(n\pi/4)$ Recurrence Relation: $a_{n+2}+a_n=5\cos \left(\frac{n\pi}{3}\right)-7\sin\left(\frac{n\pi}{4}\right)$
Attempt:
The solution of the associated homogeneous relation is
\begin{align}
a_n^{(h)}&=c_1\left(\cos \left(\frac{n\pi}{2}\right)+i\sin \left(\frac{n\pi}{2}\right)\right)+c_2\left(\cos \left(\frac{n\pi}{2}\right)-i\sin \left(\frac{n\pi}{2}\right)\right)\\ &=k_1\cos \left(\frac{n\pi}{2}\right)+k_2\sin \left(\frac{n\pi}{2}\right)
\end{align}
where $k_1=c_1+c_2$ and $k_2=(c_1-c_2)i$
then
\begin{align*}
a_n^{(p)} = A\cos \left( \frac{n\pi}{3} \right)+B\sin \left( \frac{n\pi}{3} \right) +C \cos \left( \frac{n\pi}{4} \right)+D\sin \left( \frac{n\pi}{4} \right)
\end{align*}
is the particular solution. The four constants $A, B, C, D$ can be calculated by substituting $a_n^{(p)}$ in the given non-homogeneous recurrence relation. But I find it difficult to calculate.
| Here is our road map:
First we find a particular solution $p_n$, then a complementary solution $f_n$. Set
$a_n = f_n + p_n $, we are done.
It is easy to find a particular solution $p_n$.
For example, set $a_0$ = 0, $a_1$= 1 we get, for n = 2, 3, 4
$p_n$" />
To find the complete solution, we find the complementary function
$f_n$ from the solution of
$x^2$ + 1 = 0.
Its solutions are x= -i or i.
Thus we have $f_n$ = $C_1$ cos nπ/2 + $C_2$sin nπ/2
and $a_n$ = $f_n$ + $p_n$ = $C_1 cos nπ/2 + C_2 sin nπ/2 + p_n$
| {
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"timestamp": "2023-03-29T00:00:00",
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Generating functions in discrete mathematics - How to find coefficient from polynomial with multiplication So I encountered a certain problem when studying generating functions that I seem to be unable to fully solve.
The problem is as follows:
Carol is collecting money from her cousins to have a party for her aunt. If eight of the cousins promise to give 2, 3, 4, or 5 dollars each, and two others each give 5 or 10 dollars, what is the total number of ways Carol can collect exactly 40 dollars?
The solution is the coefficient of the $x^{40}$ term in the generating function
$(x^2 + x^3 + x^4 + x^5)^8(x^5+x^{10})^2$
which may be simplified to finding the coefficient of the $x^{14}$ term in
$(1+x+x^2+x^3)^8(1+x^5) = (\frac{1-x^4}{1-x})^8(1+2x^5+x^{10}) = (1-x^4)^8(1-x)^{-8}(1+2x^5+x^{10})$
This is where I am stuck. To find the coefficient I know that I am suppose to find all possible combinations of how $x^{14}$ can be created by evaluating the polynomial. This would be done quite easily if only the first term $(1-x^4)^8(1-x)^{-8}$ would exist but how I am suppose to do this with the additional term $(1+2x^5+x^{10})$ present? I.e how do I evaluate the different combinations $x^{14}$ may be created and the find the coefficient with the second term present?
There is a solution given in the book, but I do not understand the principle behind it.
| We have $1+2x^5+x^{10}$ and also two other complicated factors:
$$(1-x)^{-8}\quad=\quad1+{-8 \choose 1}(-x)+{-8 \choose 2}(-x)^2+\dots+{-8 \choose 4}(-x)^4+{-8 \choose 5}(-x)^5+{-8 \choose 6}(-x)^6+\dots+{-8 \choose 9}(-x)^{9}+{-8 \choose 10}(-x)^{10}+\dots{-8 \choose 14}(-x)^{14}+\dots,$$
$$(1-x^4)^8\quad=\quad1-{8 \choose 1}x^4+{8 \choose 2}x^8-{8 \choose 3}x^{12}+\dots.$$
For $(1-x)^{-8}$, I have shown only the nine factors that are usable for our purpose. For each of these nine, in order:
*take $x^{10}$ from the first factor; take the $x^4$ term from the third factor.
*take $2x^{5}$ from the first factor; take the $x^8$ term from the third factor.
*take $1$ from the first factor; take the $x^{12}$ term from the third factor.
*take $x^{10}$ from the first factor; take the $1$ term from the third factor.
*take $2x^{5}$ from the first factor; take the $x^4$ term from the third factor.
*take $1$ from the first factor; take the $x^8$ term from the third factor.
*take $2x^{5}$ from the first factor; take the $1$ term from the third factor.
*take $1$ from the first factor; take the $x^4$ term from the third factor.
*take $1$ from the first factor; take the $1$ term from the third factor.
| {
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Is the Diophantine equation $y(x^3-y)=z^2+2$ solvable?
Do there exists integers $a,b$ such that their sum is a perfect cube while their product minus two is a perfect square?
Equivalently, do there exist integers $x,y,z$ such that
$$
y(x^3-y)=z^2+2 \quad ?
$$
The source of this problem is the MathOveflow question "What is the smallest unsolved diophantine equation?" that introduces a way to measure size of a Diophantine equation and asks for a non-trivial equation of the smallest size. I was able to solve all equations of size up to 25, but cannot solve this equation of size 26.
Update 21st July 2021: Thank you Servaes for solving the above equation. I also received a solution by e-mail from Prof. Will Sawin. In the comment you ask what is the next smallest open equations. Now all equations with sizes up to 28 has been solved, so the smallest nontrivial ones are of size 29. Examples are
$$
y^2 - xyz + z^2 = x^3-5
$$
and
$$
y(x^2+2) = 2zx+2z^2+1.
$$
The last equation can of course be formulated as a question whether there exist integers $x$ and $z$ such that $x^2+2$ is a divisor of $2zx+2z^2+1$.
Update 24th July 2021: I now was able to solve $y(x^2+2) = 2zx+2z^2+1$, but not yet $y^2 - xyz + z^2 = x^3-5$.
Update 16th August 2021: Thank you Dipramit Majumdar and B. Sury for solving the equation $y^2 - xyz + z^2 = x^3-5$. Please look at my mathoverflow question
https://mathoverflow.net/questions/400714/can-you-solve-the-listed-smallest-open-diophantine-equations
for the list of next smallest open equations.
| Let $x$, $y$ and $z$ be integers such that $y(x^3-y)=z^2+2$. If $z$ is odd then
$$y(x^3-y)\equiv z^2+2\equiv3\pmod{8},$$
and so $y$ and $x^3-y$ are odd, meaning that $x$ is even. Then $x^3\equiv0\pmod{8}$ and so
$$y(x^3-y)\equiv-y^2\equiv7\pmod{8},$$
a contradiction. On the other hand, if $z$ is even then
$$y(x^3-y)\equiv z^2+2\equiv2\pmod{4},$$
and so either $y$ or $x^3-y$ is even, but not both. It follows that $x$ is odd. In the factorization
$$x^6-8=(x^2-2)(x^4+2x^2+4),$$
the gcd of the two factors on the right hand side divides $12$, but $x^2-2$ is odd and $x^2-2\not\equiv0\pmod{3}$ so in fact the two are coprime. We have
$$x^2-2\equiv3\pmod{4},$$
so $x^2-2$ has a prime factor $p\equiv3\pmod{4}$ with odd multiplicity that does not divide $x^4+2x+4$. It follows that $x^6-8$ has a prime factor $p\equiv3\pmod{4}$ with odd multiplicity. It is a classical result that then $x^6-8$ is not a sum of two squares. But this contradicts the identity
$$x^6-8=(2y-x^3)^2+(2z)^2.$$
So there is no pair of integers whose sum is a perfect cube and whose product is two more than a perfect square.
| {
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synthetic divide polynomial $(2x^3+7x^2-13x-3) \div (2x-3)$ I need to figure out how synthetic division works. The problem is $$ (2x^3+7x^2-13X-3) \div (2x-3)$$ I can do the long division.
$$\require{enclose}
\begin{array}{r}
x^2+5x+1 \\
(2x-3) \enclose{longdiv}{2x^3+7x^2-13x-3}\\ \underline{-2x^3+3x^2} \phantom{100000000} \\ 10x-13x \phantom{1000} \\\underline{-10x+15x} \phantom{1000} \\ 2x-3 \\\underline{-2x+3} \\ 0
\end{array}$$
But when I do synthetic division my answer is slightly different;
$$ 3/2 \left[ \begin{array}{r}{2 \phantom{10}7 \phantom{2}-13\phantom{2}-3} \\ \underline{ \phantom{100}3 \phantom{100} 15 \phantom{1000} 3} \\ 2 \phantom{1}10\phantom{1000} 2 \phantom{1000} 0\end{array} \right] = 2x^2 +10x +2$$
I just realized that I could factor out the two from the polynomial after I synthetic divide. The answer then would be; $$2(x^2+5x+1)$$ The only problem I have is the two graphs would be different. The second graph would stretch along the y axis. If I put a 1 in for x then y =14 a
fter synthetic division. Y would = 7 if I put a 1 in for x after I use long division. I am not sure what to do next.
| When I was writing the question I was hoping to get the steps to synthetic division. I thought I'd show the steps that I came up with from the discussion.
Step 1: is to get the divisor in (x-a) format. There cannot be a coefficient. It is done by factoring:
$$(2x-3)=2(x- \frac32)$$
The problem now looks like this $$\frac{2x^3+7x^2-13x-3}{2(x-\frac32)}$$
Step 2: is divide by 2 $$ \frac{\frac22x^3+ \frac72x^2 - \frac{13}{2}x-\frac32}{x- \frac32}$$
Step 3: Now the problem is setup for synthetic division
$$ \frac32 \left[ \begin{array}{l}1 \phantom{10}3.5 -6.5 -1.5 \\ \underline{\phantom{100}1.5 \phantom{100}7.5 \phantom{10}1.5} \\1 \phantom{100}5 \phantom{100}1 \phantom{1000}0\end{array}\right] = x^2 +5x+1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4164545",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
I need help finding the values of the constants of a function $f\left(x,y\right)=ax^2y+bxy+2xy^2+c$
I need to find the values of the constants a, b, c so that the function has a point $P\left(\frac{2}{3},\frac{1}{3}\right)$ and and minimum value $-\frac{1}{9}$
My solution
$f´(x)=2axy+by+2y^2$
$f´(y)=2x^2+bx+4xy$
Evaluate $P\left(\frac{2}{3},\frac{1}{3}\right)$ in $f´(x)=2axy+by+2y^2$
$2a\frac{2}{3}\cdot \frac{1}{3}+b\cdot \frac{1}{3}+2\left(\frac{1}{3}\right)^2=0$
$b=\frac{-4a}{3}-\frac{2}{2}$
Evaluate $P\left(\frac{2}{3},\frac{1}{3}\right)$ and $b=\frac{-4a}{3}-\frac{2}{2}$ in $f´(y)=2x^2+bx+4xy$
$a\left(\frac{2}{3}\right)^2+\frac{-4a}{3}-\frac{2}{3}\cdot \frac{2}{3}+4\cdot \frac{2}{3}\cdot \frac{1}{3}$
$a=\frac{1}{2}$
Evaluate $a=\frac{1}{2}$ in $b=\frac{-4a}{3}-\frac{2}{2}$
$b=\frac{-4}{3}$
The main problem is the constant c because I can't find a way to find that value and I need to know if what I did is right
| Plug in your values of $a$ and $b$ into $f$ to get your function of $x$ and $y$. Then you want the value of the function to be $-1/9$ at the point $P$, so solve the equation $f(P) = -1/9$ for $c$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4172421",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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A curious summation. Months ago I was interested in calculating new infinite summations and was able to make a nice summation and also find its closed form as follows :
$$\sum \limits_{n=0}^{\infty } \frac{(-1)^{0+1+2+..+..+n}}{(2n+1)^2} = \frac{\sqrt{2}\pi ^2}{16}$$
And also that
$$\sum \limits_{n=0}^{\infty } \frac{(-1)^{0+1+2+..+..+n}}{(2n+1)^4}=\frac{11\sqrt{2}\pi ^4}{1536}$$
Now observe that the the summand contains two negatives and two positive entities alternatively after the first term which is $1$.
What now I want to work out for is the sum so that the summand follows the pattern of three negatives followed by three positives alternatively and so on.
Also what will be the closed form for this pattern. I mean that we can deduce that $(-1)^{1+2+..+..n}$ follows the pattern of 2 negatives and 2 positives similarly what single formula can give the pattern for three negatives and three positives ?
Also can we generalize the result that for let's say pattern of 4 negatives and 4 positives.
So far my works didn't yield anything productive except for the two summations already mentioned above. Hence any help will be really appreciated. Thank you !)
| 1. Let us first analyze the behavior of OP's sums. We find that
\begin{align*}
F(s)
&:= \sum_{n=0}^{\infty}\frac{(-1)^{1+2+\cdots+n}}{(2n+1)^s} \\
&= \sum_{n=0}^{\infty}\frac{1}{(8n+1)^s}
-\sum_{n=0}^{\infty}\frac{1}{(8n+3)^s}
-\sum_{n=0}^{\infty}\frac{1}{(8n+5)^s}
+\sum_{n=0}^{\infty}\frac{1}{(8n+7)^s}
\end{align*}
Now using the discrete Fourier transform, we can confirm that
$$ (-1)^{1+2+\cdots+n} = \frac{\cos(\pi n/4) - \cos(3\pi n/4)}{\sqrt{2}}, $$
and so, $F(s)$ further simplifies to
\begin{align*}
F(s)
&= \frac{1}{\sqrt{2}} \left( \sum_{k=1}^{\infty} \frac{\cos(\pi k/4)}{k^s} - \sum_{k=1}^{\infty} \frac{\cos(3\pi k/4)}{k^s} \right).
\end{align*}
If $s$ is a positive even integer, these sums can be computed using the Fourier series for Bernoulli polynomials. For example, if $s = 4$, then
$$ \sum_{k=1}^{\infty} \frac{\cos(\pi k/4)}{k^4} = -\frac{(2\pi)^4}{2 \cdot 4!} B_4(1/8) \qquad\text{and}\qquad \sum_{k=1}^{\infty} \frac{\cos(3\pi k/4)}{k^4} = -\frac{(2\pi)^4}{2 \cdot 4!} B_4(3/8). $$
So, using $B_4(1/8) - B_4(3/8) = -11/256$, we get
$$ F(4) = \frac{11 \pi^4 \sqrt{2}}{1536} $$
2. Note that the above solution works if the sign patterns can be written either as a Fourier cosine series (if $s$ is even) or as a Fourier sine series (if $s$ is odd). For example, if $(a_n)_{n=0}^{\infty}$ is a sequence with a period of $12$ such that
$$ (a_0, a_1, \dots, a_{11}) = (0, 1, 0, 1, 0, 1, 0, -1, 0, -1, 0, -1), $$
then
$$ a_n = \frac{2}{3} \sin \Bigl(\frac{\pi n}{6}\Bigr) + \frac{1}{3} \sin \Bigl(\frac{\pi n}{2}\Bigr) + \frac{2}{3} \sin \Bigl(\frac{5 \pi n}{6}\Bigr), $$
and so, the above technic allows to compute the series
$$ \sum_{k=1}^{\infty} \frac{a_k}{k^s} = \frac{1}{1^s} + \frac{1}{3^s} + \frac{1}{5^s} - \frac{1}{7^s} - \frac{1}{9^s} - \frac{1}{11^s} + \dots $$
for positive odd integer $s$. In the case of $s = 3$, for instance, this gives
$$ \sum_{k=1}^{\infty} \frac{a_k}{k^3}
= \frac{(2\pi)^3}{2\cdot 3!} \left( \frac{2}{3}B_3(1/12) + \frac{1}{3}B_3(3/12) + \frac{2}{3}B_3(5/12) \right) = \frac{29 \pi^3}{864}. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4175530",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Closed form for $ a_n = 1-2^4 +3^4 - 4^4 + ... \pm (-1)^n \cdot n^4 $ using Generating Functions Problem: Find closed form expression for $ a_n = 1-2^4 +3^4 - 4^4 + ... \pm (-1)^n \cdot n^4 $ using Generating Functions.
Hint: Calculate what function you need to multiply in order to get partial sums with alternating signs.
I don't really know... , I found that $ G(x) = \sum_{n=0} (-1)^n(n+1)^4 x^n =\sum_{n=0} (-1)^n \cos(\pi n) x^n $. I get the same results when trying to solve the recurrence relation $ a_{n+1} = a_n + (-1)^n (n+1)^4 $ which describes the sequence above, so I get no progress. I don't really see how the hint helps me, do you have any ideas? Thanks in advance!
Note: I found Closed form for $1-2+3-4+\cdots(-1)^{n-1}n$ and some more questions that relate to the partial sum above, but I can't see how to relate the partial sum to generating functions.
| Introduce $A(x):=\sum_{n\geq 0}a_n x^{n+1}$ for
$$
a_n=1-2^4+3^4-...+(-1)^{n-1}n^4.
$$
Then observe that the relation
$$
a_{n+1}=a_{n}+(-1)^n(n+1)^4
$$
and multiply it by $x^n$
$$
a_{n+1}x^n=a_{n}x^n+(-1)^n(n+1)^4x^n
$$
and sum for $n\geq 0$ to get
$$
\frac{A(x)}x=A(x)+\frac{(1-x) (x^2-10 x+1)}{(x+1)^5}
$$
yielding
$$
A(x)=\frac{x(x^2-10 x+1)}{(x+1)^5}=x-15 x^2+66 x^3-190 x^4+435 x^5-861 x^6+1540 x^7+O(x^8)
$$
and expanding $A(x)$ in partial fractions
$$
A(x)=\frac{x^3}{(x+1)^5}-\frac{10 x^2}{(x+1)^5}+\frac{x}{(x+1)^5}
$$
should provide (after some work) the correct answer
$$
a_n=-\frac{1}{2} (-1)^n n (n+1) \left(n^2+n-1\right).
$$
EDIT: about the identity
$$
\sum_{n\geq 0}(-1)^n(n+1)^4x^n=\frac{(1-x) (x^2-10 x+1)}{(x+1)^5}
$$
we can reason as follows. First observe that for any formal power series $B(x)=\sum_{n\geq 0}b_nx^n$ we have $(x\partial_x+1)B(x)=\sum_{n\geq 0}(n+1)b_nx^n$. Hence
$$
\sum_{n\geq 0}(-1)^n(n+1)^4x^n=(x\partial_x+1)^4\sum_{n\geq 0}(-x)^n=(x\partial_x+1)^4\frac 1{1+x}
$$
and a (tedious) computation of the derivatives gives the answer.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to find a general term of a trinomial? I want to find the coefficient of $x^0$ in the expansion of $(x + 1 + 1/x)^4$.
Without an expansion, I keep getting "nested binomial" terms if I group two terms together:
$$\binom{4}{k}\binom{k}{m} (x)^{k-m}(1/x)^{m}$$
For which $k-2m = 0$. I cannot solve further without guessing and checking a value. What is a better way?
| We can also apply the binomial theorem twice in order to derive the cofficient of $x^0$. We use the coefficient of operator $[x^k]$ to denote the coefficient of $x^k$ of a series. This way we can write for instance
\begin{align*}
[x^k](1+x)^n=\binom{n}{k}
\end{align*}
We obtain
\begin{align*}
\color{blue}{[x^0]}&\color{blue}{\left(\left(x+1\right)+\frac{1}{x}\right)^4}\\
&=[x^0]\sum_{k=0}^4\binom{4}{k}\left(\frac{1}{x}\right)^k(x+1)^{4-k}\tag{1}\\
&=\sum_{k=0}^4\binom{4}{k}[x^k](x+1)^{4-k}\tag{2}\\
&=\sum_{k=0}^4\binom{4}{k}\binom{4-k}{k}\tag{3}\\
&=\binom{4}{0}\binom{4}{0}+\binom{4}{1}\binom{3}{1}+\binom{4}{2}\binom{2}{2}\tag{4}\\
&=1\cdot 1+4\cdot 3+6\cdot 1\\
&\,\,\color{blue}{=19}
\end{align*}
Note that we do not need any guessing in order derive the wanted coefficient.
Comment:
*
*In (1) we apply the binomial theorem the first time.
*In (2) we use the rule $[x^p]x^qA(x)=[x^{p-q}]A(x)$.
*In (3) we select the coefficient of $x^k$ by applying the binomial theorem a second time.
*In (4) we write the terms of the sum explicitely noting that $\binom{4-k}{k}=0$ for $k=3,4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4178022",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
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Proof verification: Is my induction proof correct? Can anyone proof read my induction proof for this particular task?
For $$ n \in \mathbb{N_0} $$
$a_n$ is defined as $$ a_n= (-1)^{(n+1)}+ 1 + (-3n) + 2n^2 $$
and $b_n$ as:
$$ b_n =\begin{cases}
a_n \quad &\text{if} \, 0 ≤ n ≤ 2, \\
b_{n-1}+b_{n-2}-b_{n-3}+8 \quad &\text{if} \, 3 ≤ n \\
\end{cases}
$$
Prove or disprove by induction that $ a_n = b_n $
$$Induction Base: n=0,n=1,n=2,n=3$$
$$b_0 =a_0 =(−1)0+1 +1−3·0+2·02 =−1+1=0$$
$$b1 =a1 =(−1)1+1 +1−3·1+2·12 =1+1−3+2=1$$
$$b_2 =a_2 =(−1)2+1 +1−3·2+2·22 =−1+1−6+8=2
$$
$$b_3 =b_2 +b_1 −b_0 +8=2+1−0+8=11
$$
Induction Hypothesis: For all $k \in \mathbb {N_0}$ and k<n and n≥3 for some random but fixed $n \in \mathbb{N_0} $ a_k = b_k
Induction Step: $bn =b_{n−1} +b_{n−2} −b_{n−3} +8=(−1)^n +1−3(n−1)+2(n−1)$
$+(−1)^n−1 +1−3(n−2)+2(n−2)^2−(−1)^{n−2} +1−3(n−3)+2(n−3)^2+8$
$=(−1)(−1)(−1)^{n−2} +1−3(n−1)+2(n−1)^2 +(−1)^{n−1} +1−3(n−2)+2(n−2)^2$
$−(−1)^{n−2} −1+3(n−3)−2(n−3)^2 +8 = 1(−1)^{n−2} +1−3n+3+2n^2 −2^{n+1}+1(−1)^{n−1} −3n+6+2n^2 −4n+4$
$−(−1)^{n−2} +3n−9−2n^2 −6n+9+8= (11) =(−1)(−1)(−1)^{n−1} +9−3n+2n^2 −4n+2+2n^2 −8n+8−2n^2 +12n−18$
$=(−1)^{n+1} +1−3n+2n^2 =a_n $
| Let me propose a bit cleaner alternative proof by strong induction.
You have already showed that these two sequences agree on first four terms $n=0,1, 2, 3.$ Then directly compute $a_{n-1}+a_{n-2}-a_{n-3}+8$ using the given formula, and make sure it is equal to $a_n$ for all $n\ge 3.$ Thus, both $(a_n), (b_n)$ satisfies the same recurrence relation. Fix an integer $k\ge 3,$ and assume the result: $a_n =b_n$ holds for all $3\le n\le k,$ then
\begin{align}
a_{k+1}-b_{k+1}
& = (a_{k}+a_{k-1}-a_{k-2}+8)-(b_{k}+b_{k-1}-b_{k-2}+8)\\
& = (a_k-b_k)+(a_{k-1}-b_{k-1})-(a_{k-2}-b_{k-2})\\
& = 0
\end{align}
proves the result for $n=k+1.$
| {
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"timestamp": "2023-03-29T00:00:00",
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How to solve this integral: $\int_0^\infty{\frac{ln x}{x^2+2x+4}}\cdot dx$ I have been trying to solve this integral:- $$\int_0^\infty{\frac{\ln x}{x^2+2x+4}}\cdot dx$$ But I have been unsuccessful so far. I have tried integration by parts ( taking $1/(x^2+2x+4)$ as the second function does not work because it ends of as tan-inverse function which is more compicated to integrate.)
taking $\ln x$ as the second function was interesting, I ended up in this dead end:-
$$\begin{align} I&=\int_0^\infty{\frac{\ln x}{x^2+2x+4}}\cdot dx \\ &= \left[ {\frac{x\ln x -x}{x^2+2x+4}}\right]_0^\infty + \int_0^\infty \frac{2(x+1)(x\ln x-x)}{(x^2+2x+4)^2}\cdot dx \\&= 2\int_0^\infty \frac{(x^2+2x+4-(x+4))(\ln x-1)}{(x^2+2x+4)^2}\cdot dx \\ &= 2\int_0^\infty\frac{\ln x}{x^2+2x+4}\cdot dx - 2\int_0^\infty\frac{1}{x^2+2x+4}\cdot dx -2\int_0^\infty \frac{(x+4)(\ln x-1)}{(x^2+2x+4)^2}\cdot dx \\&= 2I - \left[\frac{1}{\sqrt{3}} \tan^{-1}{\frac{x+1}{\sqrt{3}}} \right]_0^\infty - 2\int_0^\infty \frac{(x+4)(\ln x-1)}{(x^2+2x+4)^2}\cdot dx \end{align}$$
the function in the second step limits to $0$ so I didn't mention it in the third step. Now I have quartic polynomial at the bottom and don't know how to proceed. I have also tried trigonometric substitution ($ x=\tan\theta$)
If possible please try to end this process, I am a high school student so please keep that in mind if you show another method :)
| $$I=\int_{0}^{\infty} \frac{\ln x}{x^2+2x+4} dx$$
Take $x=az$, then,
$$I=\int_{0}^{\infty} \frac{a( \ln a + \ln z)dz}{a^2z^2+2az+4}.$$
Choose $a=2$, then,
$$I=\frac{a}{4}\int_{0}^{\infty}\frac{\ln 2~ dz}{z^2+z+1}+\frac{a}{4}\int_{0}^{\infty} \frac{\ln z~ dz}{z^2+z+1}=I_1+I_2$$
Put $z=1/t$ and $a=2$ in $I_1$, to get $I_2=-I_2 \implies I_2=0$.
Noew you can take it from here as $I_1$ is standard.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Calculate the following Limit $\lim_{n\rightarrow \infty} \sqrt{n}\int_{0}^{\infty} \frac{1-\cos(x)}{1+nx^6}dx$ Calculate the Limit
$$\lim_{n\rightarrow \infty} \sqrt{n}\int_{0}^{\infty} \frac{1-\cos(x)}{1+nx^6}dx$$
I've been working on trying to find the limit using Dominated Convergence.
If we let $f_n(x) = \frac{\sqrt{n}(1-cos(x))}{1+nx^6} $
then
$$f_n(x)\leq \frac{2\sqrt{n}}{1+nx^6}=\frac{2}{\sqrt{n}^{-1}+\sqrt{n}x^6} $$
but of course we run into trouble around the origin.
Any help would be appreciated.
| Hint. By using
$$
1-\cos x=2\sin^2\frac x2
$$ and the change of variable
$$
x=n^{-1/6}u,\quad u^6=nx^6,\quad dx =n^{-1/6}du,
$$one gets
$$
\lim_{n\rightarrow \infty} \sqrt{n}\int_{0}^{\infty} \frac{1-\cos(x)}{1+nx^6}dx=\lim_{n\rightarrow \infty} 2n^{1/3}\int_{0}^{\infty} \frac{\sin^2\left(\frac{u}{2n^{1/6}}\right)}{1+u^6}du
$$then using
$$
\left|2n^{1/3}\sin^2\left(\frac{u}{2n^{1/6}}\right)\right|\le \frac{u^2}{2},\qquad \lim_{n\rightarrow \infty}2n^{1/3}\sin^2\left(\frac{u}{2n^{1/6}}\right)=\frac{u^2}{2},
$$ yields
$$
\lim_{n\rightarrow \infty} \sqrt{n}\int_{0}^{\infty} \frac{1-\cos(x)}{1+nx^6}dx=\frac12\int_{0}^{\infty} \frac{u^2}{1+u^6}du=\frac \pi{12}.
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How can I evaluate the following limit $\lim_{x\to+ \infty} x^2\left(e^{\frac{1}{x}} -e^{\frac{1}{x+1}}\right)$? Evaluate this limit :
$$\lim_{x\to+ \infty} x^2\left(e^{\frac{1}{x}} -e^{\frac{1}{x+1}}\right)$$
I tried to simplify the function :
$$x^2e^{\frac{1}{x}}-x^2e^{\frac{1}{x+1}}=\frac{e^{\frac{1}{x}}}{\frac{1}{x^2}}-\frac{e^{\frac{1}{x+1}}}{\frac{1}{x^2}}$$
So by the substitution $X=\frac{1}x$ I'll have :
$$\lim_{X\to 0} \frac{e^X}{X^2} -\frac{e^{\frac{X}{X+1}}}{X^2}$$
Am I in the right path ?
| We have : \begin{aligned}\lim_{x\to +\infty}{x^{2}\left(\operatorname{e}^{\frac{1}{x}}-\operatorname{e}^{\frac{1}{x+1}}\right)}&=\lim_{x\to +\infty}{x^{2}\operatorname{e}^{\frac{1}{x+1}}\left(\operatorname{e}^{\frac{1}{x}-\frac{1}{x+1}}-1\right)}\\ &=\lim_{x\to +\infty}{\frac{x\operatorname{e}^{\frac{1}{x+1}}}{x+1}\times\frac{}{}\frac{\operatorname{e}^{\frac{1}{x\left(x+1\right)}}-1}{\frac{1}{x\left(x+1\right)}}}\\ &=1\times 1\\ \lim_{x\to +\infty}{x^{2}\left(\operatorname{e}^{\frac{1}{x}}-\operatorname{e}^{\frac{1}{x+1}}\right)}&=1\end{aligned}
Because $ \lim\limits_{x\to +\infty}{\frac{\operatorname{e}^{\frac{1}{x\left(x+1\right)}}-1}{\frac{1}{x\left(x+1\right)}}}=\lim\limits_{X\to 0}{\frac{\operatorname{e}^{X}-1}{X}}=1 $.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4192553",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove $\ln x\ln(1-x)\le \ln^2 2$. Define $f(x)=\ln x\ln(1-x)$ where $x \in (0,1)$. Note that $f(x)$ is symmetric with respect to $x=\frac{1}{2}$.Thus we may only study the range of $f(x)$ over $\left(0,\frac{1}{2}\right]$.
Now, by differentiating we obtain $$f'(x)=\frac{(1-x)\ln(1-x)-x\ln x}{x(1-x)}.$$
Obviously, $f'(x)$ has a zero at $x=\frac{1}{2}$. But can we conclude that this is unique？
| A proof by someone posted here
\begin{align*} \ln x\ln(1-x)&=\sum_{n=1}^\infty\frac{1-x^n-(1-x)^n}{n^2}\\ &\le \sum_{n=1}^\infty\frac{1-2\left[\frac{x+(1-x)}{2}\right]^n}{n^2}\color{blue}{~~(\text{Jensen's inequality)}}\\ &=\sum_{n=1}^{\infty}\frac{1-\frac{1}{2^{n-1}}}{n^2}\\ &=\zeta(2)-2\rm{Li_2}\left(\frac{1}{2}\right)\\ &=\frac{\pi^2}{6}-2\left(\frac{\pi^2}{12}-\frac{\ln^2 2}{2}\right)\\ &=\ln^2 2. \end{align*}
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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A question about solving quotients all. I ran into this problem and was wondering where my logic failed. I was solving an absolute value involving a quotient and went about it the following way:
$$
\begin{eqnarray}
\lvert \frac{x+1}{x-2} \rvert &<& 3\\
\Rightarrow \frac{x+1}{x-2} &>& -3 \hspace{1mm}and\hspace{1mm}\frac{x+1}{x-2} < 3\\
x+1&>&-3(x-2)\hspace{2cm}\text{(Solving for the left inequality)}\\
x+1&>&-3x+6\\
4x&>&5\\
x&>&\frac{5}{4}
\end{eqnarray}
$$
which cannot be true, as $x$ can equal $2$ with these restrictions which is obviously not allowed. Doing the left side the correct way:
$$
\begin{eqnarray}
\frac{x+1}{x-2} &>& -3\\
\frac{x+1}{x-2} + 3 &>& 0\\
\frac{x+1}{x-2} + \frac{3x-6}{x-2} &>& 0\\
\frac{4x-5}{x-2} &>& 0\\
\Rightarrow x < \frac{5}{4}\lor x>2\\
\end{eqnarray}
$$
leads to the solution set $(-\infty,\frac{5}{4})\cup(2,\infty)$. My question is, what happened with the first method where I failed to come up with a solution for $(2,\infty)$, (and why the signage for $x>\frac{5}{4}$ is backwards/incorrect in the first example, does the multiplication by -3 reverse the inequality even if I'm not introducing a new negative term to one side?). I assume since you cannot divide by zero, you're not allowed to multiply the denominator to the other side without restrictions, but I am not quite sure. Thank you, I greatly appreciate it.
| We can only "flip the sign" if what is inside the absolute value is negative. And similarly, we "keep the sign", if what is inside the absolute value bracket is positive.
We could write it this way...
$-3<\frac {x+1}{x-2}<0$ or $0\le \frac {x+1}{x-2}<3$
Or say, the left-hand equation only applies when $-1<x<2$
Then we can solve the two sets of inequalities:
$-1<x<\frac 54$ or $x > \frac {7}{2}$ or $x \le -1$
Now we can combine the intervals
$(-\infty, \frac 54)\cup (\frac 72,\infty)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4192982",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Inductive proof of series (sum of 1 / [i*(i+1)] ) I like to show that
$$
\sum_{i=1}^n \frac{1}{i(i+1)} = 1 - \frac{1}{n+1} \quad \forall n \in \mathbb{N}.
$$
This seems to be a possible solution:
Base case:
$$ n_0 = 1 $$
$$ \sum_{i=1}^1 \frac{1}{1*2} = 1 - \frac{1}{1+1}$$
$$ \frac{1}{2} = \frac{1}{2} $$
Induction hypothesis:
$$ \sum_{i=1}^n \frac{1}{i(i+1)} = 1 - \frac{1}{n+1} $$
Induction step:
$$ \sum_{i=1}^{n+1} \frac{1}{i(i+1)} = \sum_{i=1}^{n} \frac{1}{i(i+1)} + \frac{1}{(n+1)(n+2)} $$
with the induction hypothesis
$$ = 1 - \frac{1}{n+1} + \frac{1}{(n+1)(n+2)} \quad (1) $$
$$ = 1 - \frac{(n+2) \color{red}{-1} }{(n+1)(n+2)} \quad (2) $$
$$ = 1 - \frac{n+1}{(n+1)(n+2)} \quad (3) $$
$$ = 1 - \frac{1}{n+2} \quad (4) $$
Now, my problem is that I cannot follow from equation (1) to equation (2). How it comes to "minus 1" in the numerator.
If I do this I come to
$$ 1 - \frac{(n+2) \color{red}{+1} }{(n+1)(n+2)} \quad (2a) $$
Could someone help?
| $$\frac{\color{red}{-1}}{n+1}+\frac{1}{(n+1)(n+2)}=\frac{\color{red}{-1}(n+2)+1}{(n+1)(n+2)}=\color{orange}{-}\frac{(n+2)\color{orange}{-1}}{(n+1)(n+2)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4194201",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Solve $\ell=\lim _{x \rightarrow \frac{\pi}{2}} \frac{(1-\sin x)\left(1-\sin ^{2} x\right) \ldots\left(1-\sin ^{n} x\right)}{\cos ^{2 n} x}$ $\ell=\lim _{x \rightarrow \frac{\pi}{2}} \frac{(1-\sin x)\left(1-\sin ^{2} x\right) \ldots\left(1-\sin ^{n} x\right)}{\cos ^{2 n} x}$
I have tried pinning it between $ \frac{(1-\sin x)^n}{\cos ^{2 n} x}$ and $ \frac{(1-\sin^nx)^n}{\cos ^{2 n} x}$ but the limits differed. Tried substituting ${x}$ for ${y}={x}-\pi/2$, no results. The derivative seems too complicated to use L'Hopital, so I'm out of ideas.
| We take the limit $\lim\limits_{x \to \frac{\pi}{2}} \frac{1 - \sin^n{x}}{\cos^2{x}}$. We can evaluate this by $\lim\limits_{x \to \frac{\pi}{2}} \frac{1 - \sin^n{x}}{\cos^2{x}} = \lim\limits_{x \to \frac{\pi}{2}} \frac{1 - \sin^n{x}}{1 - \sin^2{x}} = \lim\limits_{u \to 1} \frac{1 - u^n}{1 - u^2}$, which we evaluate to $\frac{n}{2}$ by L'Hopital's rule.
So $\lim\limits_{x \to \frac{\pi}{2}} \frac{(1 - \sin{x}) \cdot ... \cdot (1 - \sin^n{x})}{\cos^{2n}{x}} = \lim\limits_{x \to \frac{\pi}{2}} \prod\limits_{j = 1}^n \frac{1 - \sin^j{x}}{\cos^2 {x}} = \prod\limits_{j = 1}^n \lim\limits_{x \to \frac{\pi}{2}} \frac{1 - \sin^j{x}}{\cos^2 {x}} =
\prod\limits_{j = 1}^n \frac{j}{2} = \frac{\prod\limits_{j = 1}^n j}{\prod\limits_{j = 1}^n 2} = \frac{n!}{2^n}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4196748",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove that $\frac{\ 1}{a^2+b^2+ab}+\frac{\ 1}{b^2+c^2+bc}+\ \frac{\ 1}{c^2+a^2+ca}\ge\ \ \frac{\ 9}{\left(a+b+c\right)^2}$.
Let $a,b,c$ be positive real numbers. Prove that $$\frac{\ 1}{a^2+b^2+ab}+\frac{\ 1}{b^2+c^2+bc}+\ \frac{\ 1}{c^2+a^2+ca}\ge\ \ \frac{\ 9}{\left(a+b+c\right)^2}.$$
I want to prove the inequality with elementary inequalities ( Mean inequalities, Cauchy-Schwarz etc.). I tried to use Cauchy-Schwarz, but I got $$\frac{\ 1}{a^2+b^2+ab}+\frac{\ 1}{b^2+c^2+bc}+\ \frac{\ 1}{c^2+a^2+ca}\ge\frac{9}{2(a^2+b^2+c^2)+ab+bc+ca}.$$
Then we have to show that $$2(a^2+b^2+c^2)+ab+bc+ca\leq (a+b+c)^2 $$
which is not true.
Multiplying the numerators of the fractions in LHS by $a^2,b^2,c^2$ also makes the problem more difficult.
So, how to solve the problem with elementary inequalities?
| I give another idea.
WLOG, we may assume that $a \ge b \ge c$. It's easy to see that the original problem follows immediately by adding two following inequalities:
\begin{align} & \frac{1}{a^2+ab+b^2}+\frac{1}{b^2+bc+c^2} \ge \frac{4}{ab+bc+2ca+2b^2}
\ \ \ \ \ \ \ \ (1) \\ &\frac{4}{ab+bc+2ca+2b^2} + \frac{1}{c^2+ca+a^2} \ge \frac{9}{\left(a+b+c \right)^2} \ \ \ \ \ \ \ \ (2)\end{align}
Indeed, by reducing and factoring, the first inequality is equivalent with:
\begin{align} \left(c-a \right)^2 \left(ab+bc+2ca-b^2 \right) \ge 0 \end{align}
which is true because $a \ge b \ge c $.
For the second one, it can be proved by applying Cauchy-Schwarz as follows:
\begin{align} \frac{4}{ab+bc+2ca+2b^2} &+ \frac{1}{c^2+ca+a^2} \ge \frac{ \left(2+1 \right)^2}{a^2+2b^2+c^2+ab+bc+3ca}\\ &= \frac{9}{ \left(a+b+c \right)^2 -\left(a-b \right) \left(b-c \right)} \ge \frac{9}{ \left(a+b+c \right)^2}\end{align}
Hence, our proof is completed.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4197535",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 3
} |
Linearly express a polynomial of a root Consider the equation $x^2-4x+7=0$ and $x_1,x_2$ its roots. Calculate:
$$\frac{x_1^4-2x_1^3+3}{x_1^3+x_1^2}+\frac{x_2^4-2x_2^3+3}{x_2^3+x_2^2}$$
Rules: cannot manually calculate the roots, must use Vieta's formulas.
Amplifying the fractions in this form seems like a bad idea so I thought of a trick to simplify it.
Substitute $x_1$ in the main equation:
$$x_1^2-4x_1+7=0$$
Multiply by $x_1^2$ then add $2x_1^3+3$ and get: $x_1^4-2x_1^3+3=2x_1^3-7x_1^2+3$
For the same equation multiply by $2x_1$ then add $x_1^2+3$: $2x_1^3-7x_1^2+3=x_1^2-14x_1+3$
Now substract $10x_1+4$ so $x_1^2-14x_1+3=-(10x_1+4)$
$\Rightarrow x_1^4-2x_1^3+3=-(10x_1+4)$
Do you notice how in the chain of equations by each iteration we lose 1 degree?
Same thing for $x_1^3+x_1^2=13x_1-35$. Now the expression is easily computable.
My question is - is there an easier way to linearly express a polynomial of a root?
| from $x^2 = 4x-7$ I got $x^3 = 9x - 28$ and $x^4 = 8x-63$
Next, $x = 4 - \frac{7}{x},$ so $\frac{1}{x} = \frac{-x+4}{7}$ and so $\frac{1}{x^2} = \frac{-4x+9}{49}$
Then I wrote $(x+1)^2 = 6(x+1) - 12; \; \;$ then dividing gives $x+1 = 6 - \frac{12}{x+1}$ and
$ \frac{1}{x+1} = \frac{-x + 5}{12} $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4200966",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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} |
Prove that $ \det(A^4 + A^2 B^2 + 2A^2 + I) \geq 0 $ Problem:
Let $ A $ and $ B $ be an $ n \times n $ matrices with real entries. If $ AB = -BA $, prove that
$$ {\det}{\left(A^4 + A^2B^2 + 2A^2 + I\right)} \geq 0. $$
My Approach:
If $ A $ invertible, then
$$ AB = -BA \implies AA^{-1}B = -BA^{-1}A \implies B = -B \implies B = O. $$
So,
$$ {\det}{\left(A^4 + A^2B^2 + 2A^2 + I\right)} = {\det}{\left(A^4 + 2A^2 + I\right)} = {\det}{\left(\left(A^2 + I\right)^{2}\right)} = \left({\det}{\left(A^{2} + I\right)}\right)^2 \geq 0. $$
If $ B $ invertible, then
$$ AB = -BA \implies AB^{-1}B = -BB^{-1}A \implies A = -A \implies A = O. $$
So,
$$ {\det}{\left(A^4 + A^2B^2 + 2A^2 + I\right)} = {\det}{(I)} = 1 \geq 0. $$
My Questions:
*
*Can we 'center' multiply both sides of the equation $ AB = -BA $ with a matrix? For example, in my solutions, $ AB = -BA \implies AA^{-1}B = -BA^{-1}A $.
*How do we prove it when $ A $ and $ B $ not invertible?
Thanks
| Since $AB=-BA$, so we have $A^2B^2=-(AB)^2$. Now consider the matrix
\begin{align*}
\det(A^4+A^2B^2+2A^2+I)&=\det((A^4+2A^2+I)\color{magenta}{-(AB)^2})\\
&=\det((A^2+I)^2-(AB)^2)\\
&=\det((A^2-AB+I)(A^2+AB+I))\\
&=\det((A^2+\color{blue}{BA}+I)(A^2+AB+I))\\
&=\det(A^2+BA+I)\det(A^2+AB+I)\\
&=\det((A+B)A+I)\det(A(A+B)+I)\\
&=\det((A+B)A+I)^2\\
& \geq 0. & (\because \, A,B \in M_{n}(\Bbb{R}))
\end{align*}
For the second to last line use the result given here:
Sylvester's determinant identity
OR you can check out Weinstein–Aronszajn identity
to show that
$$\det((A+B)A+I)=\det(A(A+B)+I)$$
Note:
Based on @Pythagoras's suggestion, I am adding the following proof for the fact that $AB$ commutes with $A^2+I$. This is required for the (third) step in which I did the factorization:
\begin{align*}
(A^2+I)AB&=A^3B+AB\\
&=A^2(\color{blue}{-BA})\color{blue}{-BA}\\
&=-A(AB)A-BA\\
&=A(BA)A-BA\\
&=(AB)A^2-BA\\
&=-BA^3-BA\\
&=-BA(A^2+I)\\
&=AB(A^2+I).
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4201287",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 0
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Prove that if $a,b,c$ are sides of a triangle and $s$ is the semi-perimeter, then $a^2+b^2+c^2 \geq \dfrac{36}{35}\bigg(s^2 + \dfrac{abc}{s}\bigg)$
Prove that, if $a,b,c$ are the sides of a triangle and $s$ is the semi-perimeter, then $a^2+b^2+c^2 \geq \dfrac{36}{35}\bigg(s^2 + \dfrac{abc}{s}\bigg)$.
What I Tried:- Nothing special really came in my mind. I did not find a way to use Triangle Inequality. What I did was, by AM-GM :-
$$a^2 + b^2 + c^2 \geq \frac{36}{35}\bigg(s^2 + \dfrac{abc}{s}\bigg) > \frac{72}{35}(\sqrt{sabc}).$$
But I couldn't proceed from this.
Another Idea I had was :- $a^2 + b^2 + c^2 = (a + b + c)^2 - 2(ab + bc + ca)$.
$\rightarrow a^2 + b^2 + c^2 = 4s^2 - 2(ab + bc + ca).$
But I did not know how to use this here, and would make the calculations a bit messy, especially of the $\dfrac{36}{35}$ part present there.
Can Anyone Help me? Thank You.
| It's true for any positives $a$, $b$ and $c$.
Indeed, we need to prove that:
$$a^2+b^2+c^2\geq\frac{36}{35}\left(\frac{(a+b+c)^2}{4}+\frac{2abc}{a+b+c}\right)$$ or
$$35(a^2+b^2+c^2)(a+b+c)\geq9(a+b+c)^3+72abc$$ or
$$\sum_{cyc}(13a^3+4a^2b+4a^2c-21abc)\geq0,$$ which is true by Muirhead or by AM-GM.
The best result in this type it's the folowing.
Let $a$, $b$ and $c$ be positive numbers. Prove that:
$$a^2+b^2+c^2+\frac{9abc}{2(a+b+c)}\geq\frac{(a+b+c)^2}{2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4203954",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 3
} |
Prove that $2^n + 5^n + 56$ is divisible by $9$, where $n$ is an odd integer
Prove that $9 \mid2^n + 5^n + 56$ where n is odd
I have proved this by using division into cases based on the value of $n\bmod3$ but it seems a little bit clumsy to me and I wonder if there are other ways to prove it, probably by using modular arithmetic or induction? Below is my proof:
$\text{Case 1, }n\bmod3=0,\text{then $n=3k$ for some odd integer k:}$ $$\begin{align}
2^n+5^n+56 & =2^{3k}+5^{3k}+56
\\ & = 8^k+125^k+56
\\ & \equiv (-1)^k+(-1)^k+2\quad&\left(\bmod9\right)
\\ & \equiv 0\quad&\left(\bmod9\right)
\end{align}$$
$\text{Case 2, }n\bmod3=1,\text{then $n=3k+1$ for some even integer k:}$ $$\begin{align}
2^n+5^n+56 & =2^{3k+1}+5^{3k+1}+56
\\ & = 2\cdot8^k+5\cdot125^k+56
\\ & \equiv 2\cdot(-1)^k+5\cdot(-1)^k+2\quad&\left(\bmod9\right)
\\ & \equiv 9\equiv0\quad&\left(\bmod9\right)
\end{align}$$
$\text{Case 3, }n\bmod3=2,\text{then $n=3k+2$ for some odd integer k:}$ $$\begin{align}
2^n+5^n+56 & =2^{3k+2}+5^{3k+2}+56
\\ & = 4\cdot8^k+25\cdot125^k+56
\\ & \equiv 4\cdot(-1)^k+25\cdot(-1)^k+2\quad&\left(\bmod9\right)
\\ & \equiv -27\equiv0\quad&\left(\bmod9\right)
\end{align}$$
| Hint: $ $ let $\ \begin{align}n &= 2k\!+\!1\\ a&=(-2)^k\end{align}. $ $\bmod 9\,$ it's $\:\!f(a)\equiv\, 2a^2\!+\!5a\!+\!2\,\equiv 2\,(a\!-\!1)^2\equiv 0\,$ by $\, 3\mid a\!-\!1\!\!$
Or, conceptually $\,a\equiv 1\,$ is double root of $\,f(a)\,$ by $\,f(1)\equiv 0\equiv f'(1)\,$ by $\,f'(a) \equiv 4a\!+\!5$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4206716",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 8,
"answer_id": 0
} |
Linear algebra-Coordinate vector Given Base $A=(a_1,a_2,a_3) $
$$ B=B=\left(b_{1}=\left(\begin{array}{c}
1\\
-1\\
0
\end{array}\right),b_{2}=\left(\begin{array}{c}
0\\
1\\
0
\end{array}\right),b_{3}=\left(\begin{array}{c}
0\\
-1\\
1
\end{array}\right)\right) $$
Given transformation matrix : $ \left[I\right]_{A}^{B}=\left[\begin{array}{ccc}
c & b & a\\
b & a & c\\
a & c & b
\end{array}\right] $ and also $ \left[\begin{array}{c}
0\\
0\\
1
\end{array}\right]_{A}+\left[\begin{array}{c}
0\\
0\\
1
\end{array}\right]_{B}=\vec{0} $
I have to find a base $A $ that fulfills these conditions.
In the solution they did: $$ \left(\begin{array}{c}
0\\
0\\
1
\end{array}\right)=b_{2}+b_{3}\Rightarrow\left[\begin{array}{c}
0\\
0\\
1
\end{array}\right]_{B}=\left(\begin{array}{c}
0\\
1\\
1
\end{array}\right)\Rightarrow\left[\begin{array}{c}
0\\
0\\
1
\end{array}\right]_{A}=-\left[\begin{array}{c}
0\\
0\\
1
\end{array}\right]_{B}=\left(\begin{array}{c}
0\\
-1\\
-1
\end{array}\right) $$
It doesn't add up to me, as I thought that $ \left[\begin{array}{c}
0\\
0\\
1
\end{array}\right]_{B} $ means taking the linear combination of $$\left[\begin{array}{c}
0\\
0\\
1
\end{array}\right]_{B}= 0b_1+0b_2+1b_3 = b_3 =\left(\begin{array}{c}
0\\
-1\\
1
\end{array}\right) $$
Would appreciate any insight
| I think $\begin{bmatrix}0\\ 0\\ 1\\\end{bmatrix}_{B} $ means the coordinate vector of $\begin{pmatrix}0\\ 0\\ 1\\\end{pmatrix} $ in the basis $B$. And since $\begin{pmatrix}0\\ 0\\ 1\\\end{pmatrix} = 0b_1 + 1b_2 + 1b_3$, it does seem right.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4206926",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Line through two given skew lines and origin Question
Find the direction cosines of the line through the origin which intersects each of the lines
$\displaystyle \frac{x-1}{2} = \frac{y-2}{3} = \frac{z-3}{4}$ and $\displaystyle \frac{x+2}{4} = \frac{y-3}{3} = \frac{z-4}{2}$.
My Attempt
Projection of a vector connecting the two lines on the perpendicular from the required line to the given two lines is zero. Hence
\begin{equation}
\begin{vmatrix}
1&2&3\\
2&3&4\\
a&b&c\\
\end{vmatrix} = 0
\end{equation}
\begin{equation}
\begin{vmatrix}
-2&3&4\\
4&3&2\\
a&b&c\\
\end{vmatrix} = 0
\end{equation}
This gives me two equations
\begin{equation}
a-2b+c=0
\end{equation}
\begin{equation}
3a-10b+9c=0
\end{equation}
I get two different ratios for the direction cosines (a,b,c) as
\begin{equation}
(5,3,1),(2,3,1)
\end{equation}
Both these direction cosines give me a line intersecting only one given line.
Is my approach wrong or does this mean that there is no line that intersects both given lines and also goes through the origin?
Highly appreciate your help on this.Thanks in advance.
| Your process and equations are correct, but I can't figure out how you arrived at your answers from them. Here's what I did:
$$a-2b+c=0 {\tag 1}$$
$$3a-10b+9c=0 {\tag 2}$$
Let $\frac ac=k$, $\frac bc=l$, then, from $(1)$:
$$k-2l+1=0$$
and from $(2)$:
$$3k-10l+9=0$$
Solving these gives $k=2$ and $l=\frac 32$.
Hence, $a=2c$ and $b=\frac {3c}{2}$. Thus, direction ratio are:
$$\left(2c,\frac {3c}{2}, c \right)\equiv \left(2, \frac 32, 1 \right) \equiv (4,3,2)$$
Thus we get only one direction ratio, $(4,3,2)$. Hence equation of the required line is:
$$\frac x4=\frac y3=\frac z2$$
Addendum: Since our original condition is satisfied for both intersecting and parallel lines, it is necessary to check whether this line intersects or is parallel to either of our given lines. As explained in the comments, it turns out that this line is parallel to one of the lines, and hence no line exists which intersects both.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4207398",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to evaluate $\iiint_{\mathbb{R}^3} \exp \left(-x^2 -y^2-z^2 \right) \mathrm{d}x \mathrm{d}y \mathrm{d}z$ How to evaluate
$$
\iiint_{\mathbb{R}^3} \exp \left( -x^2 -y^2 -z^2 \right) \mathrm{d}x \mathrm{d}y \mathrm{d}z,
$$
My Attempt:
If we use the transformation to spherical polar coordinates $R, \theta, \phi$, then we have
$$
\begin{align}
x &= R \sin \theta \cos \phi, \\
y &= R \sin \theta \sin \phi, \\
z &= R \cos \theta,
\end{align}
$$
where $0 \leq r < +\infty$, $0 \leq \theta \leq \pi$, and $0 \leq \phi < 2 \pi$.
And thus the Jacobian determinant
\begin{align}
&\frac{ \partial (x, y, z) }{ \partial (R, \theta, \phi ) } = \left\lvert \begin{matrix} \frac{\partial x}{\partial R } & \frac{\partial x }{ \partial \theta } & \frac{ \partial x }{ \partial \phi} \\ \frac{\partial y}{\partial R } & \frac{\partial y }{ \partial \theta } & \frac{ \partial y }{ \partial \phi} \\ \frac{\partial z }{\partial R } & \frac{ \partial z }{ \partial \theta } & \frac{ \partial z }{ \partial \phi} \end{matrix} \right\rvert \\
&= \left\lvert \begin{matrix} \sin \theta \cos \phi & R \cos \theta \cos \phi & - R \sin \theta \sin \phi \\ \sin \theta \sin \phi & R \cos \theta \sin \phi & R \sin \theta \cos \phi \\ \cos \theta & -R \sin \theta & 0 \end{matrix} \right\rvert \\
&= R^2 \sin \theta \left\lvert \begin{matrix} \sin \theta \cos \phi & \cos \theta \cos \phi & - \sin \phi \\ \sin \theta \sin \phi & \cos \theta \sin \phi & \cos \phi \\ \cos \theta & - \sin \theta & 0 \end{matrix} \right\rvert \\
&= R^2 \sin \theta \left( -\sin \phi \left\lvert \begin{matrix} \sin \theta \sin \phi & \cos \theta \sin \phi \\ \cos \theta & -\sin \theta \end{matrix} \right\rvert - \cos \phi \left\lvert \begin{matrix} \sin \theta \cos \phi & \cos \theta \cos \phi \\ \cos \theta & -\sin \theta \end{matrix} \right\rvert \right) \\
&= R^2 \sin \theta \left( -\sin^2 \phi \left\lvert \begin{matrix} \sin \theta & \cos \theta \\ \cos \theta & -\sin \theta \end{matrix} \right\rvert - \cos^2 \phi \left\lvert \begin{matrix} \sin \theta & \cos \theta \\ \cos \theta & -\sin \theta \end{matrix} \right\rvert \right) \\
&= R^2 \sin \theta \left( -\sin^2 \phi (-1) - \cos^2 \phi (-1) \right) \\
&= R^2 \sin \theta.
\end{align}
And, therefore we have
\begin{align}
\iiint_V &\exp \left( -x^2 -y^2 -z^2 \right) \, dx\, dy\, dz = \int_0^{2\pi} \int_0^\pi \int_0^\infty \exp\left( -R^2 \right)R^2 \sin \theta \, dR \, d\theta \, d\phi \\
&= 4 \pi \int_0^\infty R^2 \exp \left( - R^2 \right) \, dR \\
&= 4 \pi \left( - \frac{ R \exp \left( -R^2 \right) }{2 } \right)_{R=0}^{R=\infty} + 2 \pi \int_0^\infty \exp \left( -R^2 \right) \, dR \\
&= 2 \pi \int_0^\infty \exp \left( -R^2 \right) \, dR \\
&=
\end{align}
Is what I have done so far correct and clear enough? If so, then how to proceed from here? How to evaluate
$$
\int_0^\infty \exp \left( -R^2 \right) \, dR ?
$$
| The last integral is Gaussian integral
$$\int_0^{+\infty}e^{-x^2}dx = \dfrac{\sqrt{\pi}}{2}$$
There are many ways to evaluate this integral, and here is one way https://en.wikipedia.org/wiki/Gaussian_integral
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4207545",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
} |
Minimum of $\sqrt{a^2+b^2}+\sqrt{(a-1)^2+b^2}+\sqrt{a^2+(b-1)^2}+\sqrt{(a-1)^2+(b-1)^2}.$
Let $a$ and $b$ be real numbers. Find the minimum of $$\sqrt{a^2+b^2}+\sqrt{(a-1)^2+b^2}+\sqrt{a^2+(b-1)^2}+\sqrt{(a-1)^2+(b-1)^2}.$$
The problem is from an inequality book. While I was doing the problem myself, I got the following results:
From Minkowski's inequality, we have
$\sqrt{a^2+b^2}+\sqrt{(a-1)^2+b^2}+\sqrt{a^2+(b-1)^2}+\sqrt{(a-1)^2+(b-1)^2}\\ \geq \sqrt{(a+a-1+a+a-1)^2+(b+b-1+b+b-1)^2}\\ =\sqrt{(4a-2)^2+(4b-2)^2}\\ \geq \sqrt{8(2a-1)(2b-1)} \ \ \ \ \ \ \text{[From AM-GM]}$
Equality holds when $4a-2=4b-2\implies a=b$. And the minimum $\sqrt{8(2a-1)(2b-1)}=0$ when $a=b=\frac 12$.
Here, everything seemed right to me. But when I plugged in $a=b=\frac 12$ in the original expression, I didn't get the minimum $0$ rather I got $2\sqrt 2$.
When I checked the solution in the book, a geometric solution was given and the minimum was indeed $2\sqrt 2$. So, I couldn't find where my mistake is. It will be helpful for me if someone can find that. Other algebraic solutions are also welcome.
I'm not good at inequalities, so pardon silly mistakes. Thanks.
| It seems to me that you misapply Minkowski's inequality.
More accurately:
\begin{align*}
&\sqrt{a^2+b^2}+\sqrt{(a-1)^2+b^2}+\sqrt{a^2+(b-1)^2}+\sqrt{(a-1)^2+(b-1)^2}\\
&\geq\sqrt{2a^2+2(a-1)^2+2b^2+2(b-1)^2}\\
&=2\sqrt{\left(a-\frac{1}{2}\right)^2+\left(b-\frac{1}{2}\right)^2+\frac{1}{2}}.\end{align*}
Then it's just like yours.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4215799",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Calculate the measure of segment AB in triangle rectangle ABC For reference:
Given the triangle $ABC$, straight at $B$. The perpendicular bisector of $AC$ intersects at $P$
with the angle bisector of the outer angle $B$, then $AF \parallel BP$ ($F\in BC$) is drawn.
If $FC$ = $a$, calculate BP(x). (Answer: $\frac{a\sqrt2}{2})$
My progress:
Point P is on the circumcircle of ABC because the angle $\measuredangle ABP = 135^o$
$ Where~ AC = 2R\\
\triangle CBP \rightarrow
PC^2 = BP^2 + BC^2 - \sqrt2BCBP\\
but~PC = PA = R\sqrt{2} \text{(since P is in the bisector AC)}\\
2R^2 = BP^2 +BC^2 - \sqrt{2}BCBP\\
0= BP^2 - \sqrt{2}BCBP + BC^2 - 2R^2\\
0= (BP - \frac1{ \sqrt{2}}BC)^2 + \frac{BC^2}2 - 2R^2\\
0= (BP - \frac1{ \sqrt{2}}BC)^2 + \frac{BC^2-4R^2}2\\
0= (BP - \frac1{ \sqrt{2}}BC)^2 - \frac{AB^2}2\\
$
I can't find the relationship between BC, AB and a...
If anyone finds another way to solve by geometry I would be grateful
| An uglier brute-force method by some trigonometric identities:
Let the green marked $\angle BCA$ be $\theta$. Then the corresponding angle at centre $\angle BEA = 2\theta$, and $\angle PEB = 90^\circ - 2\theta$.
By the laws of cosine,
$$\begin{align*}
BP^2 &= EB^2 + EP^2 - 2 EB\cdot EP \cos \angle PEB\\
&= 2R^2 - 2 R^2 \cos (90^\circ - 2\theta)\\
&= 2R^2 - 2R^2\sin 2\theta\\
&= 2R^2\left(\sin^2\theta -2\sin\theta\cos\theta +\cos^2\theta\right)\\
&= \frac12 (2R)^2\left(\cos\theta-\sin\theta\right)^2\\
&= \frac12 \left[(BF+a)-AB\right]^2\\
&= \frac{a^2}{2} & (AB=BF)\\
x = BP &= \frac{a}{\sqrt2}
\end{align*}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4217432",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 2
} |
How to simplify this equation? I'm trying to simplify equation
$$n = \Big\lceil\sqrt{2x +\frac14} - \frac12 \Big\rceil$$
into
$$n = \Big\lfloor\sqrt{2x} + \frac12 \Big\rfloor$$
where $x$ is an integer. Indeed, both equations seem to output the same result when testing for some values. Thus far, I've done:
\begin{align*}
&n = \Big\lceil\sqrt{2x +\frac14} - \frac12 \Big\rceil \\
\implies &\sqrt{2x +\frac14} - \frac12 \leq n < \sqrt{2x +\frac14} + \frac12
\end{align*}
The left side easily reduces as follows:
\begin{align*}
&\sqrt{2x +\frac14} - \frac12 \leq n \\
\implies &\sqrt{2x} - \frac12 < n
\end{align*}
but I'm having trouble with the right side which I would wish to reduce as follows:
\begin{align*}
&n < \sqrt{2x +\frac14} + \frac12 \\
\implies &\texttt{???} \\
\implies &n \leq \sqrt{2x} + \frac12
\end{align*}
so that I'm able to obtain my end result.
\begin{align*}
&\sqrt{2x} - \frac12 < n \leq \sqrt{2x} + \frac12\\
\implies &n = \Big\lfloor\sqrt{2x} + \frac12 \Big\rfloor
\end{align*}
| The first gives $$n-1<\sqrt{2x+\frac{1}{4}}-\frac{1}{2}\leq n$$ or
$$\left(n-\frac{1}{2}\right)^2<2x+\frac{1}{4}\leq\left(n+\frac{1}{2}\right)^2$$ or
$$\frac{n^2-n}{2}<x\leq\frac{n^2+n}{2}$$ and by the same way the second gives:
$$n\leq\sqrt{2x}+\frac{1}{2}<n+1$$ or $$\frac{n^2-n}{2}+\frac{1}{8}\leq x<\frac{n^2+n}{2}+\frac{1}{8}$$ and since $x$ is an integer number, we obtain $$\frac{n^2-n}{2}<x\leq\frac{n^2+n}{2}$$ again.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4218328",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Indefinite integration gives two different answers The question is :
\begin{array}{l}
\text { If } \int \frac{x+1}{\sqrt{2 x-1}} d x=\mathrm{f}(\mathrm{x}) \sqrt{2 x-1}+\mathrm{C}, \text { where } \mathrm{C} \text { is }\\
\text { constant of integration, then } \mathrm{f}(\mathrm{x}) \text { is equal to : }
\end{array}
Here is how we reach the answer :
\begin{array}{l}
\sqrt{2 x-1}=t \Rightarrow 2 x-1=t^{2} \Rightarrow 2 d x=2 t \\
. d t \\
\int \frac{x+1}{\sqrt{2 x-1}} d x=\int \frac{\frac{t^{2}+1}{2}+1}{t} t d t=\int \frac{t^{2}+3}{2} d t \\
=\frac{1}{2}\left(\frac{t^{3}}{3}+3 t\right)=\frac{t}{6}\left(t^{2}+9\right)+c \\
=\sqrt{2 x-1}\left(\frac{2 x-1+9}{6}\right)+c \\
=\sqrt{2 x-1}\left(\frac{x+4}{3}\right)+c \\
\Rightarrow f(x)=\frac{x+4}{3}
\end{array}
Here is how I solved it .
\begin{aligned}
& \int \frac{x+1}{\sqrt{2 x-1}} \\
\text { Let } 2 x-1 &=t \\
& \frac{d t}{d x}=2 \\
& x+1=\frac{t+3}{2} \\
&=\frac{1}{2} \int \frac{t+3}{2 \sqrt{t}} d t \\
&=\frac{1}{4} \int\left(\sqrt{t}+\frac{3}{\sqrt{t}}\right) d t \\
&=\frac{1}{4} \times \frac{2}{3} t^{3 / 2}+\frac{1}{4} \times \frac{2 \times 3}{1} \times t^{1 / 2} \\
=& \frac{1}{2}\left(\frac{1}{3} t^{3 / 2}+3 t^{1 / 2}\right) \\
&=t^{1 / 2}\left(\frac{1}{6} t^{3}+\frac{3}{2}\right) \\
&=\sqrt{2 x-1}\left(\frac{1}{6}(2 x-1)^{3}+\frac{3}{2}\right)
\end{aligned}
PS- It took me hour to write this . Pls tell if i solved it wrong . But Please answer.
| $ \frac{1}{2}\left(\frac{1}{3} t^{3 / 2}+3 t^{1 / 2}\right) \\
=t^{1 / 2}\left(\frac{1}{6} t^{3}+\frac{3}{2}\right) $
This bit is wrong.
$t^{1/2} \times t^{3} \neq t^{3/2}.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4220621",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
} |
$\sin\frac{A}{2}+\sin\frac{B}{2}+\sin\frac{C}{2}\geq 2(\sin\frac{A}{2}\sin\frac{B}{2}+\sin\frac{B}{2}\sin\frac{C}{2}+\sin\frac{A}{2}\sin\frac{C}{2})$ Let $A,B,C$ be the three angles of a triangle. Prove that:
$\sin\dfrac{A}{2}+\sin\dfrac{B}{2}+\sin\dfrac{C}{2}\geq 2(\sin\dfrac{A}{2}\sin\dfrac{B}{2}+\sin\dfrac{B}{2}\sin\dfrac{C}{2}+\sin\dfrac{A}{2}\sin\dfrac{C}{2})$
Here all I did:
$p=\dfrac{a+b+c}{2}$
$\sin \dfrac{A}{2} = \sqrt { \dfrac {(p-b)(p-c)}{bc}} $
$\Rightarrow \sin \dfrac{A}{2} \sin \dfrac{B}{2} = \dfrac{(p-c)}{c} \sqrt { \dfrac {(p-a)(p-b)}{ba}}$
$\Rightarrow 2 \sin \dfrac{A}{2} \sin \dfrac{B}{2} \le\dfrac{(p-c)}{c} \dfrac {2p-a-b}{\sqrt{ba}}$
$\Rightarrow 2 \sin \dfrac{A}{2} \sin \dfrac{B}{2} \le\dfrac{(p-c)}{c} \dfrac {c}{\sqrt{ba}}$
$\Rightarrow 2 \sin \dfrac{A}{2} \sin \dfrac{B}{2}\le \dfrac{(p-c)}{\sqrt{ba}}$
So we need to prove that : $\sum \dfrac{(p-c)}{\sqrt{ba}} \le \sum\sqrt { \dfrac {(p-b)(p-c)}{bc}} $. But I still have no idea (I'm not sure that's true either). I hope to get help from everyone. Thanks a lot
| Let $a=\frac{\frac{1}{y}+\frac{1}{z}}{2},$ $b=\frac{\frac{1}{x}+\frac{1}{z}}{2}$ amd $c=\frac{\frac{1}{x}+\frac{1}{y}}{2},$ where $x$, $y$ and $z$ are positives,
Thus, we need to prove that:
$$\sum_{cyc}z\sqrt{x+y}\geq2\sum_{cyc}\frac{xy}{\sqrt{x+y}}$$ or
$$\sum_{cyc}\frac{y(z-x)-x(y-z)}{\sqrt{x+y}}\geq0$$ or
$$\sum_{cyc}(x-y)\left(\frac{z}{\sqrt{y+z}}-\frac{z}{\sqrt{x+z}}\right)\geq0$$ or
$$\sum_{cyc}(x-y)^2z\sqrt{x+y}\geq0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4221326",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Find the maximum of the value $F=x^3y+y^3z+z^3x$ let $x,y,z$ be real number.if $x+y+z=3$,show that
$$x^3y+y^3z+z^3x\le \dfrac{9(63+5\sqrt{105})}{32}$$
and the inequality $=$,then $x=?,y=?,z=?$
I can solve if add $x,y,z\ge 0$,also see: Calculate the maximum value of $x^3y + y^3z + z^3x$ where $x + y + z = 4$ and $x, y, z \ge 0$.
But for real $x,y,z$ I can't solve it
| A SOS (Sum of Squares) solution with computer:
Denote $K = \frac{9(63+5\sqrt{105})}{32}$.
We have
$$
K\left(\frac{x + y + z}{3}\right)^4 - (x^3y + y^3z + z^3x)
= \alpha_1 f_1^2
+ \alpha_2 f_2^2
+ \alpha_3 f_3^2
$$
where
\begin{align*}
\alpha_1 &= \frac{567 - 16K}{13778100} > 0, \\[5pt]
\alpha_2 &= \frac{472K - 5607}{292287082500} > 0,\\[5pt]
\alpha_3 &= \frac{998K - 29538}{1491260625} > 0, \\[8pt]
f_1 &= \left( 10\,{x}^{2}+20\,xy+20\,xz-14\,{y}^{2}+44\,yz-14\,{z}^{2}
\right) K \\
&\qquad -405\,xy+459\,{y}^{2}-999\,yz+459\,{z}^{2},\\[8pt]
f_2 &= \left( 944\,xy-112\,xz-592\,{y}^{2}+1440\,yz-480\,{z}^{2} \right) K \\
&\qquad -
22239\,xy+8442\,xz+21987\,{y}^{2}-40635\,yz+17325\,{z}^{2},\\[8pt]
f_3 &= \left( 16\,{y}^{2}-48\,yz \right) K+45\,xz-261\,{y}^{2}+918\,yz-405\,
{z}^{2}.
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4222935",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
$\lim\limits_{n\rightarrow\infty} \frac{n}{\log n}(n^{\frac{1}{n}}-1)$ How to find $\lim\limits_{n\rightarrow\infty} \frac{n}{\log n}(n^{\frac{1}{n}}-1)$? Here is my attempt.
Put $f(x)=\frac{x}{\log x}(x^{\frac{1}{x}}-1)$ for $x>1$. Then
\begin{aligned}
\lim\limits_{n\rightarrow\infty} \frac{n}{\log n}(n^{\frac{1}{n}}-1) &= \lim\limits_{x\rightarrow\infty} f(x) \\
&= \lim\limits_{x\rightarrow\infty}\frac{x^{\frac{1}{x}}-1}{\frac{\log x}{x}} \\
&= \lim\limits_{x\rightarrow\infty} \frac{x^{\frac{1}{x}-2}(1-\log x)}{\frac{x^2+\log x}{x^4}} \\
&= \lim x^{\frac{1}{x}}\lim x^2\lim\frac{1-\log x}{x^2+\log x} \\
&= 1 \cdot (+\infty)\cdot (0)\mbox{.}
\end{aligned}
This method does not work.
| The expression in your limit is $\frac{n^{1/n}-1}{\frac{\log n}{n}}$
Notice that $\lim_{n\rightarrow\infty}\frac{\log n}{n}=0$,
and $n^{1/n}=\exp(\tfrac{1}{n}\log(n))$.
Setting $h_n=\frac{\log n}{n}$ yields
$$\frac{n^{1/n}-1}{(\log n)/n}=\frac{\exp(\tfrac{\log n}{n})-1}{\frac{\log n}{n}}=\frac{e^{h_n}-1}{h_n}\xrightarrow{n\rightarrow\infty}e^0=1$$
By definition of derivative of the exponential at $0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4225441",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Characterizing integer solutions to $a^2 \mid \bigl((b^2)^2 + (b^2+1)^2\bigr)$ I’m looking to characterize all integers $a$ and $b$ satisfying $$a^2 \mid \bigl((b^2)^2 + (b^2+1)^2\bigr), \qquad a > b \ge 1.$$
Brute force searches have so far turned up the two solutions $(a,b) = (13,7)$ and $(a,b)=(9601,5441)$. I’m guessing there are an infinite number, but I’d like to reduce the numbers I have to check. Obviously, $a$ must be odd and the sum of two squares, so I’m already taking that into account.
Any thoughts about what the next solution might be, or how to attack the characterization of any/all solutions?
If $b$ is odd, then there exists an integer $c$ such that $a=b+2c$. Substituting, we find $a \parallel (32c^4+8c^2+1) = \bigl((4c^2)^2+(4c^2+1)^2\bigr)$, and then $a^2$ divides a related quotient. Based on that, I feel like there might be a Vieta-jumping solution… but I still can’t quite find it.
| This problem in a way is related to equation $x^2+(x+1)^2=y^2$ which has infinite solution.There is a proof in this site.I will try to show this relation by an examples:
$119^2+120^2=13^4$
$119=7\times 17$
$(7\times17)^2+(7\times17+1)=13^4$$\space\space\space\space\space\space(1)$
and we see that:
$7^4+(7^2+1)^2=29\times 13$
Relation (1) is symmetric for 7 and 17 so relation $b^4+(b^2+1)^2= km^2$ must also be true for 17, we check this:
$17^4+(17^2+1)^2=86021=509\times 13^2$
that is if $b$ is a primes multiple of a primitive solution, that prime is also a solution. $x$ and $x+1$ make following sets:
$x\in\{3, 119, 4059, 137903\cdot\cdot\cdot\}$
$(x+1)\in\{4, 120, 4060, 137904\cdot\cdot\cdot\}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4229887",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
How to find the maximum value of $\frac{2\sin\theta\cos\theta}{(1+\cos\theta)(1+\sin\theta)}$? Let $\theta\in\left(0,\frac{\pi}{2}\right)$, then find the maximum value of $\dfrac{2\sin\theta\cos\theta}{(1+\cos\theta)(1+\sin\theta)}$
I found the derivative, which is equal to $\dfrac{(\cos\theta−\sin\theta)(\sin\theta+\cos\theta+1)^2}{(1+\sin\theta+\cos\theta+\sin\theta\cos\theta)^2}$
But there were complicated calculations involved. Is there a simpler way to find the maximum of this function?
| Change the variable:
$t=\tan{\frac{\theta}{2}}; t\in (0,1)$
$\tan{\frac{\theta}{2}}$ is one to one in this interval.
$\sin{\theta}=\frac{2t}{1+t^2}$
$\cos{\theta}=\frac{1-t^2}{1+t^2}$=>
$$f(\theta)=g(t)=2\frac{\frac{2t}{1+t^2}\frac{1-t^2}{1+t^2}}{\left(1+\frac{2t}{1+t^2}\right)\left(1+\frac{1-t^2}{1+t^2}\right)}=2t\frac{1-t}{1+t}$$
Now is easy to find the maximum!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4230143",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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Compute $\frac 17$ in $\Bbb{Z}_3$
Compute $\frac 17$ in $\Bbb{Z}_3.$
We will have to solve $7x\equiv 1\pmod p,~~p=3.$
*
*We get $x\equiv 1\pmod 3.$
*Then $x\equiv 1+3a_1\pmod 9,$ so $7(1+3a_1)\equiv 1 \pmod 9$ basically lifting the exponent of $p=3,$ we get $1+3a_1\equiv 4\pmod 9\implies a_1\equiv 1\pmod 3.$
*So let $$x\equiv 1+3\cdot 1+3^2\cdot a_2 \pmod 2\implies 7(4+3^2\cdot a_2)\equiv 1\pmod {27}\implies 4+3^2\cdot a_2\equiv 4\pmod {27}\implies a_2\equiv 0 \pmod 3.$$
*So let $$ x\equiv 1+3\cdot 1+3^2\cdot 0+ 3^3\cdot a_3 \pmod {81}\implies 7(4+3^2\cdot 0+3^3\cdot a_3)\equiv 1\pmod {81}\implies 4+3^3\cdot a_3\equiv 58\pmod {81}\implies a_2\equiv 2 \pmod 3.$$
*So let $$ x\equiv 1+3\cdot 1+3^2\cdot 0+ 3^3\cdot 2+3^4\cdot a_4 \pmod {243}\implies 7(4+3^2\cdot 0+3^3\cdot 2+3^4\cdot a_4)\equiv 1\pmod {243}\implies 1+3+54+3^4\cdot a_4\equiv 139\pmod {243}\implies a_4\equiv 1 \pmod 3.$$
*So let $$ x\equiv 1+3\cdot 1+3^2\cdot 0+ 3^3\cdot 2+3^4\cdot 1+ 3^5\cdot a_5 \pmod {729}\implies 7(4+3^2\cdot 0+3^3\cdot 2+3^4\cdot 1+3^5\cdot a_5)\equiv 1\pmod {729}\implies 1+3+54+81\equiv 625\pmod {243}\implies a_5\equiv 2 \pmod 3.$$
I haven't worked out but I think $a_6$ is $0.$
So the sequence we are getting is $(a_0,a_1,a_2,a_3,a_4,\dots)=(1,1,0,2,1,2,\dots).$
But I am not sure if it's correct, since it's not being periodic. Any help?
| lets use it 3-adic expansion here.
*
*The theorem says: A rational number with p-adic absolute value 1 has a purely periodic p-adic expansion if and only if it lies in the real interval [-1, 0). and 1/7 is periodic.
*we will use first ${-\frac{1}{7}}$ and then negate it.
*we know that ${\overline {n_0n_1...n_{k-1}} = \frac {n_0n_1...n_{k-1}}{1-p^k}}$
*The least ${k \geq 1}$ making ${3^k \equiv 1 (mod 7)}$ is k=6. so we can take
*${3^6=729-1 =728\equiv 0 (mod7)}$, now ${728=104 \cdot 7}$ hence our equation becomes
*${-\frac{1}{7}}$=${- \frac{1\cdot104}{7\cdot104}}$ = ${- \frac{104}{3^6-1}}$ = ${-\frac{10212_3}{3^6-1}}$ = ${\frac{021201}{1-3^6}}$
*${-\frac{1}{7}=}$ ${\overline{021201...}}$, now negate to get answer, as we know
*if ${x=c_dp^d+c_{d+1}p^{d+1}+...+ c_ip^i+... }$ here ${c_d \neq 0}$ then
*${-x = (p-c_d)p^d+ (p-1-c_{d+1})p^{d+1}+...+(p-1-c_i)p^i+...}$
*${\frac {1}{7}= {\overline {11021}}}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4230782",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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Transform ODE $(u-x)u_x + u + x = 0$ to polar coordinates According to Peter Olver in his book “Applications of Lie Groups to Differential Equations”, p. 104, Example 2.32, the ODE:
$$(u-x)u_x + u + x = 0 \tag{1}$$
is transformed in polar coordinates with $x = r\cos \theta$ and $u = r\sin\theta$ to:
$$\frac{dr}{d\theta} = r.$$
How do I transform $u_x=\partial_x(r\sin\theta)$ to the new coordinates? I tried the following, but the equation I got is very far from $\frac{dr}{d\theta} = r$. I rewrote $\partial_x$ in terms of $\partial_r$ and $\partial_\theta$ using the chain rule. Supposing $f = f(r,\theta)$ is a function of $r$ and $\theta$, we should have:
$$\begin{aligned} \partial_x f(r,\theta) &= (\partial_x r)\partial_r f + (\partial_x \theta)\partial_\theta f\\
&=\cos\theta\partial_r f + \left(\partial_x \arctan\frac ux\right) \partial_\theta f\\
&=\cos\theta\partial_r f + \left(\frac {1}{1+\theta^2}2\theta\frac{xu_x - u}{x^2}\right) \partial_\theta f.\end{aligned}$$
Thus:
$$\begin{aligned} u_x &=\cos\theta\partial_r (r\sin\theta) + \left(\frac {1}{1+\theta^2}2\theta\frac{xu_x - u}{x^2}\right) \partial_\theta (r\sin\theta)\\
&=\cos\theta\sin\theta + \left(\frac {2\theta}{1+\theta^2}\frac{xu_x - u}{(r\cos\theta)^2}\right) (-r\cos\theta + r_\theta \sin\theta)\\
&=\cos\theta\sin\theta - \left(\frac {2\theta}{1+\theta^2}\frac{r(u_x\cos\theta - \sin\theta)}{r\cos\theta}\right) + \left(\frac {2\theta}{1+\theta^2}\frac{\cos\theta u_x - \sin\theta}{r\cos^2\theta}\right) r_\theta \sin\theta.\end{aligned}$$
Rearranging:
$$\begin{aligned} u_x\left[1 + \frac{2\theta}{1+\theta^2}\left(1+\frac{r_\theta}{r}\tan\theta \right)\right] &= \cos\theta\sin\theta + \frac {2\theta}{1+\theta^2}\left(\tan\theta - \frac{r_\theta}{r}\tan^2\theta\right).\end{aligned}$$
So:
$$ u_x = \frac{\cos\theta\sin\theta + \frac {2\theta}{1+\theta^2}\left(\tan\theta - \frac{r_\theta}{r}\tan^2\theta\right)}{1 + \frac{2\theta}{1+\theta^2}\left(1+\frac{r_\theta}{r}\tan\theta \right)}$$
Then, rewriting (1) as $u_x = \frac{x+u}{x-u}$, I obtained:
$$\frac{\cos\theta\sin\theta + \frac {2\theta}{1+\theta^2}\left(\tan\theta - \frac{r_\theta}{r}\tan^2\theta\right)}{1 + \frac{2\theta}{1+\theta^2}\left(1+\frac{r_\theta}{r}\tan\theta \right)}=\frac{\cos\theta+\sin\theta}{\cos\theta-\sin\theta}.$$
| You get by the chain rule, where it is applicable
$$
\frac{du}{dx}=\frac{\frac{du}{dθ}}{\frac{dx}{dθ}}
=\frac{r'(θ)\sinθ+r(θ)\cosθ}{r'(θ)\cosθ-r(θ)\sinθ}
$$
which on the other hand is equal due to the given equation to
$$
=\frac{\sinθ+\cosθ}{\cosθ-\sinθ}
$$
Now solve this by cross-multiplying
$$
(r'(θ)\sinθ+r(θ)\cosθ)(\cosθ-\sinθ)=(r'(θ)\cosθ-r(θ)\sinθ)(\sinθ+\cosθ)
\\~\\
\implies -r'(θ)+r(θ)=0
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Exponent of Matrix A when A is not diagonalizable I want to compute the Matrix $A^{19}$ where
$$A = \begin{pmatrix}
6&1&0\\0&6&1\\0&0&6\end{pmatrix}$$
Since the only eigenvalue of the matrix is 6, it's not diagonalizable. How do you proceed to calculate the exponent matrix of A then?
| Let$$B=\begin{bmatrix}0&1&0\\0&0&1\\0&0&0\end{bmatrix}.$$Then $A=6\operatorname{Id}_3+B$. Besides, $6\operatorname{Id}_3$ and $B$ commute, and therefore\begin{align}A^{19}&=(6\operatorname{Id}_3+B)^{19}\\&=6^{19}\operatorname{Id}_3+19\times6^{18}B+\binom{19}26^{17}B^2\\&=\begin{bmatrix}6^{19}&19\times 6^{18}&\binom{19}26^{17}\\0&6^{19}&19\times6^{18}\\0&0&6^{19}\end{bmatrix},\end{align}since $B^n=0$ when $n>2$.
| {
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What is the measurement of the $CT$ segment in the square below? For reference The side of square $ABCD$ measures $4$. Calculate TC. (answer: 4)
My progress:
Por propriedade BTE é $90 ^ \circ$
G is the barycenter of the $\triangle ADC$
We have a 2:4 ratio in the $\triangle DEA$ so $\measuredangle EAD = \frac{53^\circ}{2}$
| Let's say $\angle ABF=\theta$. It's not hard to see that
$$BT=4\cos\theta$$
$$\angle TBC=90-\theta$$
By Law of Cosines on $\angle TBC$ in $\Delta TBC$, we have
$$TC^2=BT^2+BC^2-2\cdot BT\cdot BC\cos\angle TBC$$
$$TC^2=16\cos^2\theta+16-32\cos\theta\sin\theta$$
$$TC^2=16(1+\cos^2\theta-2\cos\theta\sin\theta)$$
Since $\sin\theta=\frac{1}{\sqrt{5}}$ and $\cos\theta=\frac{2}{\sqrt{5}}$, we have
$$TC^2=16(1+\frac{4}{5}-\frac{4}{5})$$
$$TC^2=16$$
$$\boxed{TC=4}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4233845",
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"source": "stackexchange",
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"answer_count": 5,
"answer_id": 2
} |
Find the total number of ordered pairs $(m,n)$ such that $m^2n=20^{20}$ where $m,n$ are positive integers. Find the total number of ordered pairs $(m,n)$ such that $m^2n=20^{20}$ where $m,n$ are positive integers.
My Approach: Simplifying $$m^2n=20^{20}=(2^2 \times 5)^{20}=2^{40} \times 5^{20}$$
Hence one way is $n=5^{20}, m=2^{20}$ or vice-versa or $m=2^{10},n=10^{20}$ and $m=5^{10},n=4^{20}$. These are the only ways I could think of initially but there are several combinations possible for values of $m,n$. If anyone can help spot a pattern or what are the possible values of $m,n$, I 'll try again to proceed further. I believe that there are many integers possible but I am not able to apply the correct logic.
Thank You.
| Let $m=2^a5^b$ and $n=2^c5^d$.
So,
\begin{align*}
m^2n&=2^{2a}5^{2b}2^c5^d\\
&=2^{2a+c}5^{2b+d}=2^{40}5^{20}
\end{align*}
which implies
$$2a+c=40\\2b+d=20$$
Since $2a+c=40$, $c$ has to even. So, we can put $c=2,4,6\dots$ and find the corresponding values of $a$. We can do the same for $b$ and $d$.
Note that $40\leq c\leq 0$. So, there are $21$ solutions for the first equation, and similarly $11$ solutions for the second equation. Since each such $(a,c)$ and each $(b,d)$ gives you a solution, the total number of solutions is $21\times 11=231$.
This completes the answer.
| {
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"source": "stackexchange",
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How to find the limit of $\lim_{x \to 0} \frac{x \tan^{-1} x}{1-\cos x}$ How to find the limit of the following function?
$$\lim_{x \to 0} \frac{x \tan^{-1} x}{1-\cos x}$$
What I tried is as follows.
$$\tan^{-1} x = y \implies \tan y = x $$
$$\frac{x \tan^{-1} x}{1-\cos x}=\frac{\tan y \cdot y}{1-\cos(\tan y) }$$
But it didn't work.
Please Help me.
| Surprised no one explicitly used L'Hopital's rule.
$\displaystyle \frac{d}{dx} \tan^{-1}(x) = \frac{1}{1 + x^2}.$
Derivative of the numerator is therefore
$\displaystyle \tan^{-1}(x) + \frac{x}{1 + x^2}.$
Derivative of the denominator is $\sin(x)$
Since both derivatives evaluate as $(0)$ at $(x = 0)$, L'Hopital's rule must be re-applied.
Next derivative of numerator is
$\displaystyle \frac{1}{1 + x^2} + \frac{1 + x^2 - 2x^2}{\left(1 + x^2\right)^2}.$
At $(x=0)$, this evaluates to $(2)$.
Next derivative of denominator is $\cos(x)$.
At $(x = 0)$, this evaluates to $(1)$.
Therefore, the limit is $\displaystyle \frac{2}{1} = 2.$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 5
} |
$\sum_{cyc}^{}\frac{a(a+b+c)}{9b^2+(a+b+c)^2} \geq \frac{1}{2}$ I was attempting this question: https://www.youtube.com/watch?v=600X-ZGNBbk
Which is the following:
$$\sum_{cyc (a,b,c)}{}\frac{a}{b^2+1} \geq \frac{3}{2}$$
Where $a+b+c = 3$ and $a,b,c > 0 $
The video itself used a solution where they added $3$ and then subtracted $a+b+c$ to rewrite the inequality into something more manageable. Though I understand this solution, it seems arbitrary to me - thinking to add 3 and then subtract the condition is somewhat out of the blue.
What I did was to first try to homogenize the inequality to get rid of the condition. This meant writing $1$ as $\frac{(a+b+c)^2}{9}$ and multiplying the numerator by $\frac{a+b+c}{3}$:
$$\sum_{cyc}^{}\frac{\frac{a(a+b+c)}{3}}{b^2+\frac{(a+b+c)^2}{9}} \geq \frac{3}{2}$$
$$\leftrightarrow $$
$$\sum_{cyc}^{}\frac{a(a+b+c)}{9b^2+(a+b+c)^2} \geq \frac{1}{2}$$
I'm fairly certain this inequality is true without the condition of $a+b+c=3$, but I can't see a way to prove it. Multiplying out seems like a lot of trouble; is there an easier way (ideally with Olympiad mathematics methods)?
| To prove your inequality (not the original inequality), I would introduce the condition $a + b + c = 1$. Note that
\begin{align*}
\frac{1}{9b^2+1} &\geq 1 - \frac{3}{2}b\\
\iff 0 &\leq \frac{3}{2}b\left(1 - 3b\right)^2
\end{align*}
(this is the Tangent Line Trick). Hence, we have
\begin{align*}
\sum_{\text{cyc}} \frac{a(a+b+c)}{9b^2+(a+b+c)^2} &= \sum_{\text{cyc}} \frac{a}{9b^2 + 1}\\
&\geq \sum_{\text{cyc}} a\left(1 - \frac{3}{2}b\right)\\
&= 1 - \frac{3}{2}(ab+bc+ca)\\
&\geq 1 - \frac{1}{2}(a+b+c)^2 && \text{by AM-GM}\\
&= \frac{1}{2}
\end{align*}
and so we're done.
| {
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Orbits of $z_{n+1} = z_n ^2 - 1$ Consider the sequence $z_{n+1} = z_n ^2 - 1$ defined for an arbitrary complex number $z_0$. I am trying to determine all $z_0$ such that the sequence eventually becomes periodic.
Here is my progress so far:
If $|z_0|> \frac{1+\sqrt{5}}{2}$ the sequence is absolutely increasing since
$$|z_{n+1}|=|z_n^2 - 1|\geq|z_n|^2-1 > |z_n| ,$$
which is true since $ |z_n| > \frac{1+\sqrt{5}}{2}$ holds inductively.
If $z_0=\frac{1+\sqrt{5}}{2}$ the sequence would be the constant sequence of $\frac{1+\sqrt{5}}{2}$ and hence periodic.
Similar to the first case if $|z_0|<\frac{1}{2}$ or more precisely if $|z_0|$ less than the root of $\alpha^3 +2\alpha -1 = 0,$ the sequence $\{z_{2i} \}$ becomes strictly decreasing since
$$|z_{2i}| = |z_{2i-1}^2 - 1 |= |z_{2i-2}^4 - 2z_{2i-2}^2|< |z_{2i-2}| \Leftrightarrow |z_{2i-2}^3 - 2z_{2i-2}| < 1.$$
Which holds true if
$$ |z_{2i-2}^3 - 2z_{2i-2}| \leq |z_{2i-2}^3|+| 2z_{2i-2}| <1.$$
And as a result $\{z_i\}$ cannot be periodic.
| Let us find the alternative solutions.
Firstly, let us consider the case of the exact periodicity.
Assuming the periods $1,2,3,\dots,$ easily to get
*
*$z=z^2-1,\quad z_{11}=\phi=\dfrac{1+\sqrt5}2,\quad z_{12}=-\dfrac1\phi=\dfrac{1-\sqrt5}2;$
*$z=(z^2-1)^2-1,\quad z(z+1)(z^2-z-1)=0,$
$\qquad z_{21}=z_{11},\quad z_{22}=z_{12},\quad z_3=0,\quad z_4=1;$
*
*$z=((z^2-1)^2-1)^2-1,\quad (z^2-z-1)(z^6+z^5-2z^4-z^3+z^2+1)=0$
$\qquad z_{31}=z_{11},\quad z_{32}=z_{12},$
$\quad z_{33,34}\approx -1.42203\pm 0.114188i,\quad z_{35,36}\approx-0.0871062\pm0.655455i,\quad z_{37,38}=1.00914\pm0.324759i;$
*
*$\dots$
Also, the roots of the equations
$$z^2-1=z_{mn},$$
$$((z^2-1)^2-1)=z_{mn},\dots$$
leads to the required sequences.
| {
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Proving a function is increasing in $n$. I am trying to prove that the function $\frac{n}{2n+1}$, defined for $n \in \mathbb{N}$, decreases in $\mathbb{N}$. I attempted it by induction, but I'm not convinced that I fully need induction. Why can I not prove that for an arbitrary $n$, $f(n) \leq f(n+1)$ and deduce that, because $n$ was arbitrary, this holds for all $n$? The only thing left out would be the base case, but. I'm not fully sure why I need it here.
Regardless, here is my attempt at the induction:
Let $f: \mathbb{N} \to \mathbb{R}$ be defined by $f(n) = \frac{n}{2n+1}$. We prove by induction on $n$ that $f$ is increasing in $n$. If $n = 1$, we notice that
\begin{align*}
f(1) = \frac{1}{3} \leq \frac{2}{5} = f(2).
\end{align*}
Suppose inductively that we have $f(n) \leq f(n+1)$ for some $n \geq 1$. So we have
$\frac{n}{2n+1} \leq \frac{n+1}{2n+3}$. First, we have
\begin{align*}
\frac{n+1}{2n+3} \leq \frac{n+3}{2n+3}.
\end{align*}
Furthermore, $2n + 5 \geq 2n + 3$, so $\frac{1}{2n + 5} \leq \frac{1}{2n+3}$, so $\frac{n+3}{2n + 3} \leq \frac{n+3}{2n+5}$. Therefore, it follows that
\begin{align*}
\frac{n+1}{2n+3} \leq \frac{n+3}{2n+3} \leq \frac{n+3}{2n + 5} = \frac{(n+2) + 1}{2(n+2) + 1},
\end{align*}
so $f(n+1) \leq f(n+2)$, which closes the induction
| A proof by induction would express that if $f$ is increasing at $n$, then $f$ is increasing at $n+1$, or
$$\frac n{2n+1}<\frac{n+1}{2n+3}\implies\frac{n+1}{2n+3}<\frac{n+2}{2n+5},$$
which can be rewritten as
$$-\frac1{(2n+1)(2n+3)}<0\implies-\frac1{(2n+3)(2n+5)}<0.$$
This proposition is true, but a little nonsense as both members are tautologies.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4243093",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
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How many negative eigenvalues can a $3 \times 3$ symmetric matrix have? Suppose that I have a symmetric matrix of the form
\begin{equation}
M= \begin{pmatrix}
1 & B & C \\
B & 1 & E \\
C & E & 1\\
\end{pmatrix},
\end{equation}
where $B, C, E$, all belong to $[-1, 1]$. What is the most number of negative eigenvalues that this matrix can have? The trace of the matrix is $3$, so it cannot have more than $2$ negative eigenvalues.
| Here's a strange application of the formula for the roots of a cubic equation. For a degree-3 polynomial in depressed form
$$
p(x) = x^3 + px +q,
$$
there is a formula for the roots, given by
$$
r_k = 2\sqrt{-\frac{p}{3}}\cos\left(
\frac{1}{3}\arccos\left(
\frac{3q}{2p}\sqrt{-\frac{3}{p}}
\right)
-\frac{2\pi k}{3}
\right),~~~~~k=0,1,2.
$$
We'll apply this to the characteristic polynomial of the matrix, given by
$$p(x) = (x - 1)^3 + (-b^2-c^2-e^2)(x - 1) - 2 e b c.$$
Then, the expression for the roots is
$$
r_k = 1 + 2\sqrt{\frac{b^2+e^2+c^2}{3}}\cos\left(
\frac{1}{3}\arccos\left(
{\frac{3\sqrt{3}ebc}{(b^2+e^2+c^2)^{3/2}}}
\right)
-\frac{2\pi k}{3}
\right),~~~~~k=0,1,2.
$$
The smallest this expression can be is when the $\cos$ evaluates to $-1$, and since the off-diagonal elements of the matrix are restricted to be between $-1$ and 1, the largest that the factor multiplying the $\cos$ can be is $2$, i.e.,
$$
2\sqrt{\frac{b^2+e^2+c^2}{3}}\leq2\sqrt{\frac{1^2+1^2+1^2}{3}}=2.
$$
Thus, the smallest a root can be is $-1$. Similarly the largest root is at most 3, and so, as mentioned in another answer, this means that there can be at most one negative eigenvalue (since the trace is 3).
From this formula, we can see that if all the off-diagonals are equal to $-1$, then the $k=2$ root is $r_2=-1$, with the other two roots both being $2$.
| {
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"url": "https://math.stackexchange.com/questions/4243791",
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"source": "stackexchange",
"question_score": "3",
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Finding the equation of a straight line joining the origin and the point of intersection of two straight lines Question:
Find the equation of the straight line joining the origin and the point of intersection of the straight lines $\frac{x}{a}+\frac{y}{b}=1...(i)$ & $\frac{x}{b}+\frac{y}{a}=1...(ii)$ (Answer: $x-y=0$)
My attempt:
Rewriting (i) and (ii) as (iii) and (iv) in slope-intercept form respectively:-
$y=\frac{-b}{a}x+b...(iii)$
$y=\frac{-a}{b}x+a...(iv)$
Now, from (iii) and (iv),
$$\frac{-b}{a}x+b=\frac{-a}{b}x+a$$
$$\implies x(\frac{a}{b}-\frac{b}{a})=a-b$$
$$\implies x=\frac{ab(a-b)}{a^2-b^2}...(v)$$
Rewriting (i) & (ii) as (vi) & (vii) respectively,
$x=-\frac{a}{b}y+a...(vi)$
$x=-\frac{b}{a}y+b...(vii)$
Now, from (vi) & (vii),
$$-\frac{a}{b}y+a=-\frac{b}{a}y+b$$
$$\implies y(\frac{b}{a}-\frac{a}{b})=b-a$$
$$\implies y=\frac{ab(b-a)}{b^2-a^2}...(viii)$$
$$\implies y=\frac{ab(a-b)}{a^2-b^2}...(viii)$$
Now, the straight line formed by $(0,0)$ & $(\frac{ab(a-b)}{a^2-b^2}, \frac{ab(a-b)}{a^2-b^2}$ if $(x,y)$ is a point on that line,
$$\frac{y-0}{x-0}=\frac{\frac{ab(a-b)}{a^2-b^2}-0}{\frac{ab(a-b)}{a^2-b^2}-0}$$
$$\implies x=y$$
$$\implies x-y=0$$
So, I was able to get at the correct answer, but is my process valid? In (v) & (viii), for example, I divided by $(a^2-b^2)$, and I assumed that $a^2-b^2\neq0$. Was this assumption of mine correct? Is there an alternative way to do this problem that is quicker to do and that doesn't include dividing by ($a^2-b^2$)?
| Yes you work is correct but it will be good to state in the beginning that the lines intersect only if $a \ne b$. For $a = b$, both equations in fact represent the same line.
You can also simplify your work a bit. Say, the intersection point is $(x_1, y_1)$. Then you have,
$\frac{x_1}{a}+\frac{y_1}{b}=1...(i)$
$\frac{x_1}{b}+\frac{y_1}{a}=1...(ii)$
From $(i)$ and $(ii)$, $\frac{x_1}{a}+\frac{y_1}{b} = \frac{x_1}{b}+\frac{y_1}{a}$
$x_1 (\frac{1}{a} - \frac{1}{b}) = y_1 (\frac{1}{a} - \frac{1}{b})$
If $a \ne b$, you have $~ y_1 = x_1$
So the equation of line passing through the origin and intersection of the given lines is,
$(y-0) = \frac{y_1 - 0}{x_1 - 0} ~ (x-0) ~ $ or $ ~ y = x$
| {
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"question_score": "1",
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What is the mistake in my reasoning in calculating limit where $x \to -\infty$ I had this question in an exam today:
$$
\lim \limits_{x \, \to \,-\infty} \; \frac{\sqrt{x^2+4x}-2x}{2x}
$$
My question is: I have tried to solve it and I also found two other methods to solve it, one in my textbook and one from somebody else. All three methods give different answers, and I can't see the issue with either of the methods which are not mine. I need help figuring out what the invalid operation(s) in each of the wrong methods are, and which 2 (or 3) methods are wrong.
I didn't solve it in the exam, but afterwards tried this:
$$
\lim \limits_{x \, \to \,-\infty} \; \frac{\sqrt{x^2+4x}-2x}{2x} \\
= \; \lim \limits_{x \, \to \,-\infty} \; \Bigg(\frac{\sqrt{x^2+4x}}{2x}\Bigg) - 1 \\
= \; \frac{1}{2}\lim \limits_{x \, \to \,-\infty} \; \Bigg(\frac{\sqrt{x^2+4x}}{x}\Bigg) - 1 \\
= \; \frac{1}{2}\lim \limits_{x \, \to \,-\infty} \; \Big(\sqrt{x^2+4x}\cdot \frac{1}{x}\Big) - 1 \\
= \; \frac{1}{2} \; \Big(\lim \limits_{x \, \to \,-\infty}\sqrt{x^2+4x} \, \cdot \, \lim \limits_{x \, \to \,-\infty}\frac{1}{x}\Big) - 1 \\
= \; \frac{1}{2} \; \Bigg(\sqrt{\lim \limits_{x \, \to \,-\infty}x^2+4x}\,\cdot \, \lim \limits_{x \, \to \,-\infty}\frac{1}{x}\Bigg) - 1 \\
= \; \frac{1}{2} \; \Bigg(\sqrt{\lim \limits_{x \, \to \,-\infty}x^2+4x}\,\cdot \, 0\Bigg) - 1 \\
= \; -1 \\
$$
I understand my mistake: I multiply an infinite amount with 0 and say it is equal to 0 in my last step.
However, our study guide has that exact example with the following solution:
$$
\lim \limits_{x \, \to \,-\infty} \; \frac{\sqrt{x^2+4x}-2x}{2x} \\
= \; \lim \limits_{x \, \to \,-\infty} \; \frac{\sqrt{x^2(1+\frac{4}{x})}-2x}{2x} \\
= \; \lim \limits_{x \, \to \,-\infty} \; \frac{|x|\sqrt{(1+\frac{4}{x})}-2x}{2x} \\
= \; \lim \limits_{x \, \to \,-\infty} \; \frac{-x\sqrt{(1+\frac{4}{x})}-2x}{2x} \; ; \; (x < 0) \\
= \; \lim \limits_{x \, \to \,-\infty} \; \frac{-x(\sqrt{(1+\frac{4}{x})}+2)}{2x} \\
= \; \lim \limits_{x \, \to \,-\infty} \; \frac{-(\sqrt{(1+\frac{4}{x})}+2)}{2} \\
= \; \frac{-\Bigg(\sqrt{\lim \limits_{x \, \to \,-\infty} \; (1+\frac{4}{x})}+2\Bigg)}{2} \\
= \; \frac{-(\sqrt{1+0}+2)}{2} \\
= \; \frac{-(\sqrt{1}+2)}{2} \\
= \; \frac{-(1+2)}{2} \\
= \; -\frac{3}{2}
$$
To complicate matters even more: someone else solved the problem like this and got a third answer:
$$
\lim \limits_{x \, \to \,-\infty} \; \frac{\sqrt{x^2+4x}-2x}{2x} \\
= \; \lim \limits_{x \, \to \,-\infty} \; \frac{\sqrt{x^2+4x}}{2x} - 1 \\
= \; \lim \limits_{x \, \to \,-\infty} \; \sqrt{\frac{x^2+4x}{4x^2}} - 1 \\
= \; \lim \limits_{x \, \to \,-\infty} \; \sqrt{\frac{1}{4} + \frac{1}{x}} - 1 \\
= \; \; \sqrt{\lim \limits_{x \, \to \,-\infty} \; \Big(\frac{1}{4} + \frac{1}{x}\Big)} - 1 \\
= \; \sqrt{\lim \limits_{x \, \to \,-\infty} \; \frac{1}{4} + \lim \limits_{x \, \to \,-\infty} \; \frac{1}{x}} - 1 \\
= \; \sqrt{\frac{1}{4} + 0} - 1 \\
= \; \sqrt{\frac{1}{4}} - 1 \\
= \; \frac{1}{2} - 1 \\
= \; -\frac{1}{2}
$$
| As noticed the second method is the correct one, to check the result we can make the limit simpler at positive $\infty$ by $u=-x\to \infty$, then
$$\lim \limits_{x \, \to \,-\infty} \; \frac{\sqrt{x^2+4x}-2x}{2x} =\lim \limits_{u \, \to \,\infty} \; \frac{\sqrt{u^2-4u}+2u}{-2u}=\lim \limits_{u \, \to \,\infty} \; \frac{\sqrt{1-\frac4 u}+2}{-2}=-\frac 3 2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4253310",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Binomial Expansion for n < 1 Write Down the First 3 terms of $\sqrt[3]{1 + \frac{x}{8}}$
My First step : After Evaluation, I got $$\Bigl(\frac{1}{8}x + 1\Bigl)^\frac{1}{3}$$
but, $${{\frac{1}{3}} \choose {r} } = \text{undefined}$$
How am I supposed to do this or am I missing something crucial ?
| $$\sqrt[3]{1 + \frac{x}{8}}=1+\bigg(\frac{\frac{1}{3}}{1!}\bigg)\bigg(\frac{x}{8}\bigg)^1+\bigg(\frac{\frac{1}{3}\times(\frac{1}{3}-1)}{2!}\bigg)\bigg(\frac{x}{8}\bigg)^2+\cdots=1+\frac{x}{24}-\frac{x^2}{576}+\cdots$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4254436",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
how to find formula for the summation series I need a formula for the summation of the series $$
n+2(n−1)+2^2(n−2)+2^3(n−3)\cdots $$
Is there any way to simplify it further? I am guessing it will be of form
$$
\sum_{j=0}^{n} (n-j)a^j
$$
For the above summation series i have:
$
n + 2(n-1) + 4(n-2) + 8(n-3)
$
$
2^k(n-k) + \sum_{j=0}^{k-1} (n-j)a^j
$
n-k is zero so
$
2^k + n(2^ -1) + \sum_{j=0}^{n-1} ja^j
$
I am just not able to get it into geometric series form.
| Your sum is:
$\begin{align*}
S_n
&= \sum_{0 \le k \le n} 2^k (n - k) \\
&= n \sum_{0 \le k \le n} 2^k - \sum_{0 \le k \le n} k \cdot 2^k
\end{align*}$
First one is just a geometric sum:
$\begin{align*}
\sum_{0 \le k \le n} 2^k
&= \frac{2^{n + 1} - 1}{2 - 1} \\
&= 2^{n + 1} - 1
\end{align*}$
Second one can be computed by a trick. Start with:
$\begin{align*}
\sum_{0 \le k \le n} z^k
&= \frac{1 - z^{n + 1}}{1 - z} \\
z \frac{\mathrm{d}}{\mathrm{d} z} \sum_{0 \le k \le n} z^k
&= z \frac{\mathrm{d}}{\mathrm{d} z} \frac{1 - z^{n + 1}}{1 - z} \\
\sum_{0 \le k \le n} k z^k
&=
\end{align*}$
Evaluate the derivative on the right hand side at $z = 2$ to get $(2 n - 2) 2^n + 2$. Pull both together:
$\begin{align*}
\sum_{0 \le k \le n} 2^k (n - k)
&= n \cdot (2^{n + 1} - 1) - (2 n - 2) \cdot 2^n - 2 \\
&= 2^{n + 1} - n - 2
\end{align*}$
| {
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"url": "https://math.stackexchange.com/questions/4255020",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Finding the area of a region in an isosceles triangle that contains squares of area $256$ and $49$ There was a quiz posted on F*cebook by someone. Here's the problem.
And here's my attempt:
First, as you can see there, I drew a line from one of the corner of the pink square to the one of the corner of the green square. The degree has been symbolized as alpha ($\alpha$).
$$\begin{align}
\alpha &= \tan^{-1}\left(\frac{16}{23}\right)\\
\end{align}$$
That means the degree between the line (almost diagonally to the green square) and the side line of the green square is $90-\alpha$. Let's say the side green square is $x$
We know $\cot(x)=\tan(90^{0} - x)$, so
$$\begin{align}
\tan(90^{0} - \alpha) = \cot(\alpha) &= \frac{16}{x}\\
x &= 16 \tan(\alpha)\\
&= 16 \tan\left(\tan^{-1}\left(\frac{16}{23}\right)\right)\\
&= \frac{16^2}{23} = \frac{256}{23}
\end{align}$$
Now, I assume (I can't tell reason, it's just my guesswork) that $x$ I've found earlier has the same length as the blue side one. I also assume those two blue are right triangles.
Finally, the last calculation is:
$$\begin{align}
& \left(\frac{1}{2}\cdot \frac{256}{23}\cdot 16\right) + \left(\frac{1}{2}\cdot 7 \left( \frac{256}{23} + 9\right)\right)\\
&= 159.5
\end{align}$$
The poster gave me cry emoji and didn't say anything, I conclude my answer is incorrect. Where's the mistake?
| Let $\Delta ABC$ be our triangle, $AB=BC$, $PQRL$ be a square,
where $P\in AB$, $Q\in BC$, $R\in QC$, $L\in AR$, $PQ=16.$
Also, let $MRNK$ be a square, $M\in LR$, $N\in RC$ and $K\in AC$.
Let $AL=x$.
Thus, since $\Delta ALP\sim\Delta PQB,$ we obtain $$\frac{BQ}{16}=\frac{16}{x}$$ or $$BQ=\frac{256}{x}.$$
Also, since $\Delta KNC\sim\Delta AMK,$ we obtain:
$$\frac{NC}{7}=\frac{7}{x+9}$$ or
$$NC=\frac{49}{x+7}.$$
Thus, $$BC=\frac{256}{x}+23+\frac{49}{x+9}$$ and from $\Delta ARB$ we obtain:
$$(x+16)^2+\left(16+\frac{256}{x}\right)^2=\left(\frac{256}{x}+23+\frac{49}{x+9}\right)^2$$ or
$$(x+16)^2+\left(16+\frac{256}{x}\right)^2=\left(\frac{256}{x}+16+\frac{7(x+16)}{x+9}\right)^2$$ or
$$1+\frac{256}{x^2}=\left(\frac{16}{x}+\frac{7}{x+9}\right)^2$$ or
$$(x-12)(x^2+30x+168)=0,$$ which gives $x=12.$
Can you end it now?
I got $169.5.$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that there are don't exist integers $a,b,c$ such that for every integer $x$ the number $A=(x+a)(x+b)(x+c)-x^3-1$ is divisible by $9$. Problem: Prove that there are don't exist integers $a,b,c$ such that for every integer $x$ the number $$A=(x+a)(x+b)(x+c)-x^3-1$$ is divisible by $9$.
We can consider $0\le a,b,c<9$ as we only care $\pmod 9.$
So $(x+a)(x+b)(x+c)\equiv 1,2,0\mod 9.$
Note that $$\frac{\prod_{x=0}^8 (x+a)}{(a+2)(a+5)(a+8)}\cdot \frac{\prod_{x=0}^8(x+b)}{(b+2)(b+5)(b+8)}\cdot \frac{\prod_{x=0}^8(x+c)}{(c+2)(c+5)(c+8)}= [1\cdot 2\cdot 4\cdot 5\cdot 7\cdot 8]^3\equiv -1^3\equiv -1 $$
I am stuck. Solutions are appreciated.
| I'd start by proving some algebraic relations between $a,b,c$ in $\mathbb{Z}/9\mathbb{Z}.$ For instance, by plugging in $x=0$ we have
$$abc\equiv 1 \pmod 9$$
so each of $a,b,c$ is a unit: congruent to either $1,2,4,5,7$ or $8$. It follows that the square of each of $a,b,c$ is either $1,4$ or $7$ (mod $9$).
Expanding $A$ and using the above relation we have
$$x^2(a+b+c) + x(ab+ac+bc) \equiv 0 \pmod 9$$
and plugging in $x=1$ and $x=-1$ and summing we get
$$a+b+c\equiv 0 \pmod 9$$
so that
\begin{align*}
x(ab + ac + bc) &\equiv 0 \pmod 9\\
x ((a+b+c)^2 - a^2-b^2-c^2) &\equiv 0 \pmod 9\\
x(a^2+b^2+c^2) &\equiv 0 \pmod 9.
\end{align*}
Plug in $x=1$ to get $a^2+b^2+c^2\equiv 0 \pmod 9$.
Well, the only combination of $\{1,4,7\}$ that sums to zero mod 9 is (WLOG) $a^2\equiv 7, b^2\equiv 1,c^2\equiv 1.$ But then $a\equiv \pm 4$ and $b,c\equiv \pm 1$ and no combination of signs will give you $abc\equiv 1.$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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What is the value of $ \angle x $ shown below? For reference: AB=CF
My progress:
This is a typical Peruvian exercise where the solutions are provided by auxiliary lines. I tried one way but I didn't find the solution
| Construct equilateral triangle $ABE$. Because $\angle AEB=2 \angle BCA$, $E$ is the circumcenter of $ABC$. Thus, $AB=FC=CE=BE=AE$. $\angle EAF=60^\circ-24^\circ=36^\circ, \angle AEC=180^\circ-72^\circ=108^\circ, \angle BEC=108^\circ-60^\circ=48^\circ, $ $\angle ECB=90^\circ-24^\circ=66^\circ, \angle ACE=36^\circ, \angle CFE=72^\circ, \angle FEA=36^\circ \implies AF=FE$ Therefore $\triangle ABF \cong \triangle BFE$ (SSS)Finally, $2\angle AFB=360^\circ-108^\circ, \angle AFB=126^\circ=x+30^\circ, x=96^\circ$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4257751",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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What is the relationship between the length of the circumradius and the inradius in $ \triangle ABC $? For reference: In a right angle triangle ABC, an interior bisector BD is traced, where I is or incenter, $\measuredangle B = 90 ^ o$ and $3BI = 4ID$. Find the relationship between the circumraio and inraio lenght of $\triangle ABC$. (Answer:3)
My progress:
I made the drawing
Inradius = r
Circumradius = R
$r=\frac{a+c-b}{2}=\frac{ab}{a+b+c}\\
R = \frac{b}{2}\\
\frac{R}{r} = \frac{b}{a+c-b}$
| Using the bisector’s property:
$$
\begin{align}
\frac{AB}{BC}&=\frac{AD}{DC}\\
\\
\frac{BC}{DC}&=\frac{BI}{ID}
\end{align}
$$
And $AD+DC=AC$ & $\frac{BI}{ID}=\frac{3}{4}$ can you get $\frac{3}{4}\left(AB+BC\right)=AC$?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4263318",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
How do I prove that $\sum_{cyc}\left(\dfrac{1}{x^2-xy+y^2}\right)+15\ge6(\sqrt{xy}+\sqrt{yz}+\sqrt{zx})$ given $x,y,z > 0$ and $x+y+z=3$? I tried to apply AM-GM inequality:
$$ \dfrac{1}{x^2-xy+y^2} + (x^2-xy+y^2) \ge 2 \implies \dfrac{1}{x^2-xy+y^2} \ge 2 - (x^2-xy+y^2) $$
Then,
$$ \sum_{cyc}\left(\dfrac{1}{x^2-xy+y^2}\right)+15\ge 21-2(x^2+y^2+z^2)+xy+yz+zx $$
I am left to prove
$$ 21-2(x^2+y^2+z^2)+xy+yz+zx \ge 6\left(\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\right) $$
which is equivalent to
$$ 6\left(\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\right)+2(x^2+y^2+z^2)-(xy+xz+zx) \le 21. $$
Before I attempted the problem, I noticed that the equality case for the inequality to prove is at $x=y=z=1$. While these values would satisfy the inequality above, I found out from WolframAlpha that the max of LHS is greater than $21$ (approximately $22.4545$).
Thanks in advance for the help!
| We need to prove that:
$$\sum_{cyc}\frac{1}{x^4-x^2y^2+y^4}+15\geq6(xy+xz+yz),$$ where $x$, $y$ and $z$ are positives such that $x^2+y^2+z^2=3$.
Now, let $x+y+z=3u$, $xy+xz+yz=3v^2$ and $xyz=w^3$.
Thus, $3u^2-2v^2=1$ and by C-S we obtain:
$$\sum_{cyc}\frac{1}{x^4-x^2y^2+y^4}=\sum_{cyc}\frac{z^2}{x^4z^2-x^2y^2z^2+y^4z^2}\geq\frac{(x+y+z)^2}{\sum\limits_{cyc}(x^4y^2+x^4z^2-x^2y^2z^2)}=$$
$$=\frac{9u^2}{(x^2+y^2+z^2)(x^2y^2+x^2z^2+y^2z^2)-6x^2y^2z^2}=\frac{3u^2}{3(3u^2-2v^2)(3v^4-2uw^3)-2w^6}$$ and it's enough to prove that $f(w^3)\geq0,$ where
$$f(w^3)=\frac{u^2}{3(3u^2-2v^2)(3v^4-2uw^3)-2w^6}+\frac{5}{(3u^2-2v^2)^2}-\frac{6v^2}{(3u^2-2v^2)^3}.$$
But, $$f'(w^3)=-\frac{u^2(-3(3u^2-2v^2)2u-4w^3)}{(3(3u^2-2v^2)(3v^4-2uw^3)-2w^6)^2}>0,$$ which says that it's enough to prove $f(w^3)\geq0$ or $$\frac{(x+y+z)^2}{\sum\limits_{cyc}(x^4y^2+x^4z^2-x^2y^2z^2)}+\frac{135}{(x^2+y^2+z^2)^2}\geq\frac{162(xy+xz+yz)}{(x^2+y^2+z^2)^3}$$ for the minimal value of $w^3$, which by reasoning like here says us, that it's enough to check two cases:
*
*$y=z=1$, which gives $$(x-1)^2(x^6+6x^5+291x^4-48x^3-270x^2-120x+248)\geq0,$$ which is true by AM-GM;
*$z\rightarrow0^+$ and $y=1$, which gives:
$$x^6+2x^5+138x^4-158x^3+138x+2x+1\geq0,$$ which is true by AM-GM again.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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Prove that a certain Group is Abelian Let $G$ be a group with following two properties:
*
*for all $a, b \in G$ we have $(ab)^2=(ba)^2$
*every element $a \in G$ of order $2$, ie $a^2=e_G$, is already the neutral element $a=e_G$
question: is this group already Abelian?
what I tried: we can derive a nice identity of commuators $[a,b]:=aba^{-1}b^{-1}$:
$$ aba^{-1}b^{-1} = b^{-1}a^{-1}ababa^{-1}b^{-1}= b^{-1}a^{-1}(ab)^2a^{-1}b^{-1}= b^{-1}a^{-1}(ba)^2a^{-1}b^{-1} = b^{-1}a^{-1}ba $$
Therefore $[a,b]=[b^{-1},a^{-1}]$ and we have to show that $[a,b]=e_G$.
| As $\left(ab\right)^2 = \left(ba\right)^2$, then $e_G = b^{-1}a^{-1}b^{-1}a^{-1}baba$, for all $a,b \in G$.
Now
\begin{align}
\left[a,b\right]^2 &= \left(aba^{-1}b^{-1}\right)^2 \\
&= aba^{-1}b^{-1}aba^{-1}b^{-1} \\
&= ab\left(b^{-1}a^{-1}b^{-1}a^{-1}baba\right)a^{-1}b^{-1}ab\left(b^{-1}a^{-1}b^{-1}a^{-1}baba\right)a^{-1}b^{-1} \\
&= \left(b^{-1}a^{-1}ba\right)\left(b^{-1}a^{-1}ba\right) \\
&= \left(b^{-1}a^{-1}ba\right)^2,
\end{align}
and
$$
\left[a,b\right]^2 = \left(\left(ab\right)\left(a^{-1}b^{-1}\right) \right)^2 = \left(a^{-1}b^{-1}ab\right)^2.
$$
Then
\begin{align}
\left[a,b\right]^4 &= \left(a^{-1}b^{-1}ab\right)^2 \left(b^{-1}a^{-1}ba\right)^2 \\
& = \left(a^{-1}b^{-1}ab\right)\left(a^{-1}b^{-1}ab\right)\left(b^{-1}a^{-1}ba\right)\left(b^{-1}a^{-1}ba\right)\\
& = e_G.
\end{align}
By the second property $\left[a,b\right]^2 = e_G$, therefore $\left[a,b\right] = e_G$, then G is Abelian.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find all monic irreducible quadratics in $\mathbb{Z}_3[x]$ The general form for any monic quadratic in $\mathbb{Z}_3[x]$ is: $x^2+Ax+C$
We get the following list of possibilities:
$x^2$
$x^2+1$
$x^2+2$
$x^2+x$
$x^2+x+1$
$x^2+x+2$
$x^2+2x$
$x^2+2x+1$
$x^2+2x+2$
A polynomial is said to be irreducible if it's degree is greater than or equal to 1 and if factored s.t. $f(x)=g(x)h(x)$ then one of the factors must be a constant polynomial. Also the factors must also be an element in the polynomialring.
Here's what I've done so far:
$x^2+2x$ is reducible since it can be factored of the form $x(x+2)$. Similar argument can be made for $x^2+x$. Then is $x^2$ also reducible? We can argue that $x^2=x*x$ and as such it is reducible, is this correct?
$x^2+x+1=(x+2)(x+2)=x^2+4x+4=x^2+x+1,\hspace{0.2cm} \text{ in } \hspace{0.2cm} \mathbb{Z}_3[x]$
$x^2+2x+1=(x+1)(x+1)=x^2+2x+1$
$x^2+2=(x+1)(x+2)=x^2+3x+2=x^2+2$
The ones I've found to be irreducible are:
$x^2+x+2$
$x^2+2x+2$
$x^2+1$
For the one with only a constant factor we can argue that one of the polynomials is simply the constant polynomial of $h(x)=1$.
For the other two I have simply not been able to find any such factorization, how can I argue for this in a better sense?
A statement my professor said was that if the polynomial has a degree less than or equal to three and there are no linear factors, it is irreducible. Then shouldn't $x^2+2$ and $x^2$ be irreducible?
I'm specifically looking for how I should argue for the irreducible ones, I hope I've atleast made an attempt on this exercise and found the correct answer. But I need help with the argumentative part!
| $x^2+2$ is irreducible in for instance $\mathbb Z[x]$, but $\mathbb Z[x]/3\mathbb Z$ is a different domain. As you find in your answers $x^2+2\equiv(x+1)(x+2)\bmod 3$, and that settles reducibility. This factorization means while $x^2+2$ does not equal $0$ for any integer $x$, it can equal some other multiple of $3$ and then it will be $\equiv0\bmod3$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find the minimum natural number $n$, such that the equation $\lfloor \frac{10^n}{x}\rfloor=1989$ has integer solution $x$
Find the minimum natural number $n$, such that the equation $\lfloor \frac{10^n}{x}\rfloor=1989$ has integer solution $x$.
My work-
$\frac{10^n}{x}-1<\lfloor \frac{10^n}{x}\rfloor≤\frac{10^n}{x}\Rightarrow\frac{10^n}{x}-1<1989≤\frac{10^n}{x}\Rightarrow\frac{10^n}{1990}<x≤\frac{10^n}{1989}$
I am unable to proceed beyond this. Any help or other method is appreciated.
| Rewrite the equation as
$$\frac{10^n}{x} = 1989+\epsilon$$
where $0 < \epsilon < 1$ and $x$ is a positive integer.
Then $\dfrac{10^n}{1989 + \epsilon} = x$ and
$\dfrac{10^n}{1989}$ must be slightly larger than an integer.
The first few digits of $\dfrac{1}{1989}$ are $.00050276520864...$
$$\lfloor\dfrac{10^4}{5}\rfloor = 2000$$
$$\lfloor\dfrac{10^5}{50}\rfloor = 2000$$
$$\lfloor\dfrac{10^6}{502}\rfloor = 1992$$
$$\lfloor\dfrac{10^7}{5027}\rfloor = 1989$$
| {
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How to prove:$\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}+a+b+c\ge\sum_{cyc}{\sqrt{2(a^2+b^2)}}$ Problem: Let $a,b,c>0. $ Prove that: $$\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}+a+b+c\ge\sqrt{2(a^2+b^2)}+\sqrt{2(b^2+c^2)}+\sqrt{2(c^2+a^2)}$$
I have seen problem before, and I tried to prove: $$2(a+b+c)\ge\sqrt{2(a^2+b^2)}+\sqrt{2(b^2+c^2)}+\sqrt{2(c^2+a^2)}(!)$$
since by C-S inequality: $\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\ge a+b+c$. But (!) is not true.
Anyone can help me give a hint to solve this nice problem?. Thanks!
| Another way:
$$\sum_{cyc}\left(\frac{a^2}{b}+a-\sqrt{2(a^2+b^2)}\right)=\sum_{cyc}\left(\frac{a^2}{b}-2a+b-\left(\sqrt{2(a^2+b^2)}-a-b\right)\right)=$$
$$=\sum_{cyc}(a-b)^2\left(\frac{1}{b}-\frac{1}{\sqrt{2(a^2+b^2)}+a+b}\right)\geq0.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Finding all positive integers $n$ such that $\left \lfloor{\frac{2^n}{n}}\right \rfloor$ is a power of $2$. I am trying to figure out a question I have, I don't really know where to start. The question would be
Find all positive integers $n$ such that $\displaystyle \left \lfloor{\frac{2^n}{n}}\right \rfloor$ is a power of $2$.
I don't have many ideas on how to tackle this problem, so I don't really have much to give.
One idea was to find the largest $k$ such that $2^{n-k} \geq n$ and deriving some properties from that.
| Given an integer $n$, by long division there are uniquely determined integers $q_n$ and $r_n$ with $0\leq r_n<n$ such that
$$2^n=q_n\cdot n+r_n.$$
Then $\lfloor\tfrac{2^n}{n}\rfloor=q_n$.
First, if $r_n=0$ then $q_n$ divides $2^n$ and so $q_n$ is a power of $2$. In this case $n$ also divides $2^n$, so $n$ is also a power of $2$. Conversely, if $n$ is a power of $2$, say $n=2^k$, then clearly $$\lfloor\frac{2^n}{n}\rfloor=2^{2^k-k}$$ is a power of $2$.
Next, if $r_n>0$ and $q_n$ is a power of $2$, say $q_n=2^m$ for some nonnegative integer $m$, then
$$2^n=2^m\cdot n+r_n.$$
Clearly $2^m\leq2^n$ and so $2^m$ also divides $r_n$. Then from $r_n<n$ it follows that $2^m<n$ and hence
$$2^n=2^m\cdot n+r_n<n^2+n.$$
Of course the left hand side grows faster than the right hand side, so this is only possible for small values of $n$; already for $n=5$ we see that $2^5>5^2+5$, so $n\leq4$. The values $n=1,2,4$ are powers of $2$ and so we just need to check for $n=3$ that
$$\lfloor\frac{2^3}{3}\rfloor=\lfloor\frac{8}{3}\rfloor=2,$$
is also a power of $2$.
| {
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"url": "https://math.stackexchange.com/questions/4279172",
"timestamp": "2023-03-29T00:00:00",
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$1-\frac{1}{5\cdot 3^2}-\frac{1}{7\cdot 3^3}+\frac{1}{11\cdot 3^5}+\frac{1}{13\cdot 3^6}--++\cdots.$ I want to evaluate the series
$$1-\frac{1}{5\cdot 3^2}-\frac{1}{7\cdot 3^3}+\frac{1}{11\cdot 3^5}+\frac{1}{13\cdot 3^6}--++\cdots.$$
I can rewrite this as
$$1+\sum_{n\geq 1} (-3)^{-3n}\left(\frac{3}{6n-1}+\frac{1}{6n+1}\right)$$
The answer should be $\frac{\ln7}{2}$. Although I can't see explicitly, since I get a log function, maybe it can be solved by differentiating some test function $f(x)$ and then substituting an appropriate number.
| Consider the series,
$$x-\frac{x^5}{5\cdot 3^2}-\frac{x^7}{7\cdot 3^3}+\frac{x^{11}}{11\cdot 3^5}+\frac{x^{13}}{13\cdot 3^6}--++\cdots.$$
Differentiate this and get
$$1-\frac{x^4}{ 3^2}-\frac{x^6}{ 3^3}+\frac{x^{10}}{ 3^5}+\frac{x^{12}}{3^6}--++\cdots.$$
This is a geometric series, and sums to
$$3\frac{9-x^4}{27+x^6}$$
Integrating,
$$\int 3\frac{9-x^4}{27+x^6}dx=\frac{1}{2}(\ln(x^2+3x+3)-
\ln(x^2-3x+3))+C$$
Note that $C=0$ by letting $x=0$
Therefore
$$x-\frac{x^5}{5\cdot 3^2}-\frac{x^7}{7\cdot 3^3}+\frac{x^{11}}{11\cdot 3^5}+\frac{x^{13}}{13\cdot 3^6}--++\cdots.$$
$$=\frac{1}{2}(\ln(x^2+3x+3)-
\ln(x^2-3x+3))$$
Setting $x=1$ we get $\frac{1}{2}\ln 7$.
| {
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"url": "https://math.stackexchange.com/questions/4279594",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 1
} |
help with $\lim_{n\to\infty}\left(\frac{1}{n^{2}}+\frac{2}{n^{2}}+\cdots+\frac{n-1}{n^{2}}\right)$ Is my solution right?
\begin{align}
\lim_{n\to\infty}\left(\frac{1}{n^{2}}+\frac{2}{n^{2}}+\cdots+\frac{n-1}{n^{2}}\right) &=\lim_{n\to\infty}\left(\frac{1}{n^{2}}\left(1+2+3+\cdots+n-1\right)\right)\\
&=\lim_{n\to\infty}\left(\frac{1}{n^{2}}\left(\frac{(n-1)(1+n-1)}{2}\right)\right)\\
&=\lim_{n\to\infty}\left(\frac{1}{n^{2}}\left(\frac{n(n-1)}{2}\right)\right)\\
&=\lim_{n\to\infty}\left(\frac{n-1}{2n}\right)\\
&=\lim_{n\to\infty}\left(\frac{1}{2}\right)\\
&=\boxed{\frac{1}{2}}
\end{align}
| Correct solution. Another way is to use Riemann sums:
\begin{align*}
\sum_{k=1}^{n-1}\frac k{n^2}
&=\frac 1n\sum_{k=1}^n\frac kn-\frac1n
\end{align*}
and\begin{align*}
\lim_n\left(\frac 1n\sum_{k=1}^n\frac kn\right)=\int_0^1x\;dx=\frac12.
\end{align*}
| {
"language": "en",
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"source": "stackexchange",
"question_score": "5",
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} |
How to show $\displaystyle a_{n}=\frac{4^{2n+1}}{n^{2n}}$ is a descending sequence? I have to show that the sequence
$$a_{n}=\frac{4^{2n+1}}{n^{2n}},\qquad n\geq1$$
is descending.
I thought about proving
$$\frac{a_{n+1}}{a_{n}}<1$$
However, since
$$\frac{a_{n+1}}{a_{n}}=\frac{16n^{2n}}{(n+1)^{2n+2}}$$
I have no clue at all how I could make this.
| We have
\begin{eqnarray}
a_1&=&4^3=64\\
a_2&=&\frac{4^5}{2^4}=64
\end{eqnarray}
and for $n\ge 3$,
\begin{eqnarray}
\frac{a_{n+1}}{a_n}&=&\frac{4^{2n+3}\cdot n^{2n}}{(n+1)^{2n+2}\cdot 4^{2n+1}}\\
&=&\frac{16}{(n+1)^2}\cdot\left(\frac{n}{n+1}\right)^{2n}\\
&<&\frac{16}{(3+1)^2}\cdot 1^{2b}\\
&=&1
\end{eqnarray}
i.e. $a_{n+1}/a_n<1$ for $n\ge 3$. Therefore the sequence is decreasing.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 3
} |
Spherical integration of vector function $\frac{\vec{r}\times(\vec{\mathbb{e}_z}\times\vec{r})}{r^3} \exp^{-2 r}\mathrm{d}{\vec{r}}$ In the context of a problem about the magnetic response of the electron cloud in the hydrogen atom I need the solution of the following volume integral of a vector function:
\begin{align} I &= \frac{1}{4\pi}\int_{\mathbb{R}^3} \frac{\vec{r}\times(\vec{\mathbb{e}_z}\times\vec{r})}{r^3} \exp^{-2 r}\mathrm{d}{\vec{r}} \\ &= \frac{1}{4\pi}\int_{\mathbb{R}^3} \begin{pmatrix}-xz \\ -yz \\ x^2+y^2 \end{pmatrix} \frac{\exp^{-2 r}}{r^3}\mathrm{d}{x}\mathrm{d}{y}\mathrm{d}{z}\end{align}
with $\vec{r}=\begin{pmatrix}x \\ y \\ z \end{pmatrix},\;\vec{\mathbb{e}}_z=\begin{pmatrix}0 \\ 0 \\ 1 \end{pmatrix},\; r=|\vec{r}|=\sqrt{x^2+y^2+z^2}.$
I have tried solving it with Mathematica, but it seems to evaluate for ever. Can anyone help me?
| By symmetry the $x$ and the $y$ component is $= 0$. For the $z$ component we obtain
\begin{align}
I_z & = \frac{1}{4\pi} \int_0^\infty\int_0^\pi \int_0^{2\pi} (r^2-z^2) \frac{\exp^{-2r}}{r} \sin{(\vartheta)} dr d\vartheta d\varphi \\
& = \frac{1}{4\pi} \int_0^\infty r \exp^{-2r} dr\int_0^\pi \sin^3(\vartheta)d\vartheta \int_0^{2\pi} d\varphi \\
& = \frac{1}{4\pi}\frac{1}{4} \frac{4}{3} 2\pi \\
& = \frac{1}{6}
\end{align}
Hence \begin{align}
I & = \begin{pmatrix} 0\\ 0\\ \frac{1}{6}\end{pmatrix}
\end{align}
| {
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$\frac1{12}(\sin^2A+\sin^2B+\sin^2C)\leq \sin^2\frac12B\sin^2\frac12C+\sin^2\frac12C\sin^2\frac12A+\sin^2\frac12A\sin^2\frac12B$ for $A+B+C=180^\circ$
If $A+B+C=180^{\circ}$, prove that
$$\begin{align}
\tfrac{1}{12}(\sin^2 A+\sin^2 B+\sin ^2C)
&\leq
\sin^2\tfrac{1}{2}B\sin^2\tfrac{1}{2}C \\[0.5em]
&+\sin^2\tfrac{1}{2}C\sin^2\tfrac{1}{2}A \\[0.5em]
&+\sin^2\tfrac{1}{2}A\sin^2\tfrac{1}{2}B
\end{align}$$
This is probably not that difficult, but I have a mental block when it comes to inequalities.
I have been trying the identity,
$$\sin^2\frac{1}{2}A=\frac{1}{2}(1-\cos A)$$
and
$$\sin^2 A+\sin^2 B+\sin ^2C=2+2\cos A\cos B\cos C$$ which holds when
$A+B+C=180^{\circ}$.
What is the insight or trick I am missing in this problem ?
| This is only a special case
I managed to prove the inequality when $A,B,C$ are the angles of a triangle. I expect some kind of transformation to ensure that the general case $A+B+C = 180^\circ$ can be accommodated, but I acknowledge that I haven't investigated this.
If we do assume that $A,B,C$ are the angles of a triangle with side lengths $a,b,c$ respectively opposite $A,B,C$, then there are certain well-known formulas that tell us what the sine terms are, when expressed using $a,b,c$ and the quantity $s = \frac{a+b+c}{2}$ (the semi-perimeter).
These are (note : the cyclic counterparts of these equalities hold for $B,C$) :
$$
\sin\frac A2 = \sqrt{\frac{(s-b)(s-c)}{bc}} \\
\sin A = \sqrt{\frac{4s(s-a)(s-b)(s-c)}{b^2c^2}}
$$
They can be found on page 3 of this document that discusses trigonometric inequalities in general.
With this, we proceed to rewrite the inequality using the quantities $s,a,b,c$. Before doing this, we revert to a common inequality notation which is considered useful even for expressing inequalities. Suppose we have a function $f(x,y,z)$ of three arguments. Within the realm of this answer (and more generally as well), $\displaystyle\sum_{cyc} f(p,q,r)$ is defined as the quantity $f(p,q,r) + f(q,r,p)+f(r,p,q)$. It's the "cyclic" sum , if one likes.
With this, we can now write :
$$
\frac{\sin^2 A + \sin^2 B + \sin^2 C}{12} = \sum_{cyc} \frac{s(s-a)(s-b)(s-c)}{3b^2c^2}
$$
and $$
\sin^2 \frac{B}{2} \sin^2\frac A2 + \sin^2 \frac{C}{2} \sin^2\frac A2+\sin^2 \frac{B}{2} \sin^2\frac C2 \\ = \sum_{cyc} \frac{(s-a)^2(s-b)(s-c)}{a^2bc}
$$
which means that we have to prove :
$$
\sum_{cyc} \frac{s(s-a)(s-b)(s-c)}{3b^2c^2}\leq \sum_{cyc} \frac{(s-a)^2(s-b)(s-c)}{a^2bc}
$$
Now, we want to get rid of the common denominator : for this, we multiply by $3a^2b^2c^2$ on both sides to get the equivalent inequality :$$
\sum_{cyc} a^2s(s-a)(s-b)(s-c) \leq \sum_{cyc} 3bc(s-a)^2(s-b)(s-c)
$$
The term $(s-a)(s-b)(s-c)$ is common to each summation : it's a positive quantity, and dividing out by it, we get :
$$
\sum_{cyc} a^2s \leq \sum_{cyc} 3bc(s-a)
$$
We need one more transform now.
At this point, we use the Ravi substitution (named after Ravi Vakil). In short, we set $s-a = z, s-b = y , s-c = x$ . The idea of the substitution is the following : the condition that $a,b,c$ are the sides of a triangle with semi-perimeter $s$, is equivalent to the condition that $x,y,z$ are positive real numbers.
Under the substitution, we have $a=x+y,b=x+z,c=y+z$ , as can be easily checked. From now on, all cyclic inequalities will cycle over $x,y,z$ instead of $a,b,c$. So, we get :
$$
\sum_{cyc} a^2 s = \sum_{cyc} (x+y)^2(x+y+z)
$$
and $$
\sum_{cyc} 3bc(s-a) = \sum_{cyc} 3z(x+z)(y+z)
$$
Thus, we come down to comparing :
$$
\sum_{cyc} (x+y)^2(x+y+z) \leq \sum_{cyc} 3z(x+z)(y+z)
$$
Now, cyclicity often hides a lot of terms that can be cancelled upon viewing for a short period. Let me clarify this by expanding the brackets on both sides :
$$
\sum_{cyc} (x^3+y^3 +3x^2y+3xy^2+x^2z+y^2z+2xyz) \leq \sum_{cyc} (3z^3+3yz^2+3xz^2+3xyz)
$$
Note that $\sum_{cyc} x^3 = \sum_{cyc} y^3 = \sum_{cyc} z^3$. Therefore, that part cancels to leave the equivalent $$
\sum_{cyc} (3x^2y+3xy^2+x^2z+y^2z+2xyz) \leq \sum_{cyc} (z^3+3yz^2+3xz^2+3xyz)
$$
Now notice that there's a $2xyz$ on the left and $3xyz$ on the right. Resolving that:$$
\sum_{cyc} (3x^2y+3xy^2+x^2z+y^2z) \leq \sum_{cyc} (z^3+3yz^2+3xz^2+xyz)
$$
Now observe, that $\sum_{cyc} x^2y = \sum_{cyc} z^2 x$, and observe that $\sum_{cyc} xy^2 = \sum_{cyc} yz^2$ (write this down if it's uncomfortable initially!). Thus, we are reduced to :
$$
\boxed{\sum_{cyc} (x^2z+xz^2) \leq \sum_{cyc}{(z^3+3xyz)}}
$$
This boxed inequality is true and well-known, see the $t=1$ special case here. It's the most typical application of the Schur inequality for three variables.
Having worked through the substitutions and manipulations in reversible fashion, we conclude that the original inequality is true as well. Furthermore, the equality case for the Schur-inequality is attained when $x=y=z$ (or some of them are zero,which is ruled out but can be thought of as a triangle becoming exceedingly obtuse and closing in on a straight line) which would correspond to an equilateral triangle.
I hope someone can finish the whole problem from here.
| {
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Showing $\frac{\pi}{4}= 2 \arctan\frac13+\arctan\frac17$ by using multiplication of complex numbers Imagine three right triangles on top of each other. The legs of the first two triangles have the ratio $1/3$ and the third triangle $1/7$. The total angle of all three angles is $\pi/4$.
I can get that with the arctan, of course, but I don't know how to use the complex numbers. Because the task is:
Show that
$$\frac{\pi}{4}= 2 \arctan\frac13+\arctan\frac17$$
by using multiplication of complex numbers.
| Since
$$
\tan^{-1}x+\tan^{-1}y =\tan^{-1}[\tan(\tan^{-1}x+\tan^{-1}y )]
$$
i.e.
$$
\tan^{-1}x+\tan^{-1}y=\tan^{-1}\left(\frac{x+y}{1-xy}\right)
$$
we have
$$
2\tan \frac{1}{3}=\tan^{-1}\left(\frac{\frac{2}{3}}{1- \frac{1}{9}}\right)=\tan^{-1}\frac{6}{8}=\tan^{-1}\frac{3}{4}
$$
and
$$
2\tan \frac{1}{3}+\tan^{-1}\frac{1}{7}=\tan^{-1}\left(\frac{\frac{3}{4}+\frac{1}{7}}{1- \frac{3}{28}}\right)=\tan^{-1}\frac{\frac{25}{28}}{\frac{25}{28}}=\tan^{-1}(1)=\frac{\pi}{4}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4290217",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 3
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Find inverse of a function $f(x)=x^3+3x^2+4x-1$ If $f(x)=x^3+3x^2+4x-1$, then $f(x)$ is one to one as it is increasing function for $x \in R$. I want to find $x$ such that
$\frac{-1}{4}x-\frac{7}{16}=2f^{-1}(\frac{-x}{2}-1)$
my attempt:
One option is that calculate $f^{-1}$ , inverse of $f$, and then changes $x$ to $\frac{-x}{2}-1$ in $f^{-1}$ and then solve the equation.
I don't know how to find inverse here
Could someone help me with this?
| Knowing that an inverse exists tells us that the equation $\frac{-x}{4} - \frac{7}{16} = 2 f^{-1}\bigl(\frac{-x}{2} - 1\bigr)$ is equivalent to
$$
f\Bigl(\frac{-x}{8} - \frac{7}{32}\Bigr) = \frac{-x}{2} - 1.
$$
Or $f(y) = 4 y - \frac{1}{8}$, where $y = \frac{-x}{8} - \frac{7}{32}$, which can be rewritten as
$$\begin{align}
y^3 + 3 y^2 - 7/8 &= 0 \\
8y^3 + 24y^2 - 7 &= 0.
\end{align}$$
We can avoid using the cubic formula by testing, according to the rational root theorem, the numbers $\pm 1, \pm \frac{1}{2}, \pm \frac{1}{4}, \pm \frac{1}{8}, \pm 7, \pm \frac{7}{2}, \pm \frac{7}{4}, \pm \frac{7}{8}$. This gives $y = 1/2$ as a rational root, and dividing by $y - 1/2$ and using the quadratic formula gives the remaining two roots, $-7/4 \pm \sqrt{21}/4$.
Using the formula $x = -8y - 7/4$ on these three numbers gives the three solutions to our original equation
$$
x = -\frac{23}{4}, \frac{49}{4} \pm 2\sqrt{21}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4290810",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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The probability of sum 5 after n time’s Assuming the die is unbiased, what is the probability of getting sum of 5 after $n$ time (use only one cube at a time).
It's clear to me that more than 5 time is wrong.(Because the lowest number is $1$ so $1+1+1+1+1$ is $\max$)
I tried this way: $\frac{1}{6}+\frac{4}{6}*\frac{1}{6}…$
First time only 5 is ok,in the second time 4 possible numbers are ok(1,2,3,4) then there is one complementary number.
Is this the way solution?
| The number of cases corresponding to the sum $5$ for $n$ throws of a fair dice will be the coefficient of $x^5$ in the following polynomial:
$p(x)=(x^1+x^2+\ldots+x^6)^n$
$=x^n(1+x+\ldots+x^5)^n=x^n\frac{(1-x^6)^n}{(1-x)^n}=x^n(1-x^6)^n(1-x)^{-n}$
$=x^n(1-nx^6+\ldots+(-1)^nx^{6n})\left(1+{n \choose 1}x+{n+1 \choose 2}x^2+{n+2 \choose 3}x^3+{n+3 \choose 4}x^4+{n+4 \choose 5}x^5+\ldots\right)$
$=(x^n-nx^{6+n}+\ldots)\left(1+{n \choose 1}x+{n+1 \choose 2}x^2+{n+2
\choose 3}x^3+{n+3 \choose 4}x^4+{n+4 \choose 5}x^5+\ldots\right)$
As can be seen from above,
*
*When $1 \leq n \leq 5$, the coefficient of $x^5$ in the above polynomial product will be exactly same as the coefficient of $x^{5-n}$ in the RHS polynomial $(1-x)^{-n}$, i.e., it will be ${n+5-n-1 \choose 5-n} = {4 \choose 5-n}$. Hence the requeired probability will be $\frac{{4 \choose 5-n}}{6^n}$. For example, we can list the probabilities in the following table:
$\begin{align}
\quad \quad n \quad & prob \\
\quad \quad 1 \quad & \frac{{4 \choose 4}}{6^1}=\frac{1}{6}\\
\quad \quad 2 \quad & \frac{{4 \choose 3}}{6^2}=\frac{1}{9}\\
\quad \quad 3 \quad & \frac{{4 \choose 2}}{6^3}=\frac{1}{36}\\
\quad \quad 4 \quad & \frac{{4 \choose 1}}{6^4}=\frac{1}{324}\\
\quad \quad 5 \quad & \frac{{4 \choose 0}}{6^5}=\frac{1}{7776}\\
\end{align}$
*
*When $n>5$, the required probability is $0$.
Hence, we have the following required probability:
$\text{Prob}(\text{sum of $5$ in $n$ die throws}) = \left\{\begin{array}{lr}
\frac{{4 \choose 5-n}}{6^n}, & \text{for } & 1\leq n \leq 5\\
0, & \text{ } & \text{otherwise}
\end{array}\right\}$
| {
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Prove that there is such integer $a, b, c, d$ for prime number $p$ which satisfies $p|n^4-1-(n^2+an+b)(n^2+cn+d)$ for all $n \in \mathbb{N}$.
Prove that there is such integer $a, b, c, d$ for prime number $p$ which satisfies $p|\Big(n^4-1-(n^2+an+b)(n^2+cn+d)\Big)$ for all $n \in \mathbb{N}$.
My attempt:
\begin{align}
&n^4-1-(n^2+an+b)(n^2+cn+d) = n^4-1-(n^4+cn^3+dn^2+an^3+acn^2+adn + bn^2+bcn+bd) \\
&= -(a+c)n^3-(ac+b+d)n^2-(ad+bc)n-bd-1 \equiv 0 (\mod p). \\
&\therefore (a+c)n^3+(ac+b+d)n^2+(ad+bc)n+bd \equiv -1 (\mod p).
\end{align}
| Since $n^4 - 1 = (n^2 - 1)(n^2 + 1)$, this suggests using $b = -1$ and $d = 1$. Along with $a = ep$ and $c = gp$ for any integers $e$ and $g$, so those terms are congruent to $0$ modulo $p$, this gives
$$\begin{equation}\begin{aligned}
f(n) & = n^4 - 1 - (n^2 + an + b)(n^2 + cn + d) \\
& = n^4 - 1 - ((n^2 - 1) + epn)((n^2 + 1) + gpn) \\
& = n^4 - 1 - ((n^4 - 1) + gpn(n^2 - 1) + epn(n^2 + 1) + egp^2n^2) \\
& = p(-gn(n^2 - 1) - en(n^2 + 1) - egpn^2)
\end{aligned}\end{equation}\tag{1}\label{eq1A}$$
Somewhat more simply, using congruences, we get
$$\begin{equation}\begin{aligned}
f(n) & \equiv n^4 - 1 - (n^2 + an + b)(n^2 + cn + d) \\
& \equiv n^4 - 1 - ((n^2 - 1) + epn)((n^2 + 1) + gpn) \\
& \equiv n^4 - 1 - (n^2 - 1)(n^2 + 1) \\
& \equiv 0 \pmod{p}
\end{aligned}\end{equation}\tag{2}\label{eq2A}$$
Both equations show $p \mid f(n)$ for all $n \in \mathbb{N}$. Note a basic case of $e = g = 0$ gives $f(n) = 0$.
Update: Similar to $a$ and $c$ being any integers congruent to $0$ modulo $p$, we could also have $b \equiv -1 \pmod{p} \; \to \; b = -1 + hp$, and $d \equiv 1 \pmod{p} \; \to \; d = 1 + ip$, for any integers $h$ and $i$. In addition, the $-1$ and $1$ between $b$ and $d$ could be switched. I didn't include this originally to keep the calculations somewhat shorter & simpler, especially for \eqref{eq1A}, plus the question only asks to show there's at least one solution of a set of integers instead of to determine all possible ones.
| {
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Given $\cos(5\theta)=0$, prove that $\cos(\frac{\pi}{10})\cos(\frac{3\pi}{10}) = \frac{\sqrt{5}}{4}$ Q: (a) By comparing the expressions for $(\cos(\theta) + \sin(\theta)^5$ given by De Moivre's theorem and by the binomial theorem prove that $\cos(5\theta) = 16\cos^5(\theta)-20\cos^3(\theta) + 5\cos(\theta)$
(b) By considering the equation $\cos(5\theta)=0$, prove that $\cos(\frac{\pi}{10})\cos(\frac{3\pi}{10}) = \frac{\sqrt{5}}{4}$
I have completed part (a). I am stuck on part (b) however.
My Workings:
By Factor formulae, $\cos(\frac{\pi}{10})\cos(\frac{3\pi}{10}) = \frac{1}{2} (\cos(\frac{2\pi}{5}) + \cos(\frac{\pi}{5}))$
Considering roots of $z^5 = 1$. The sum of the roots equals 0.
So $\cos(\frac{2\pi}{5}) + \cos(\frac{4\pi}{5}) + \cos(\frac{6\pi}{5}) + \cos(\frac{8\pi}{5}) + \cos(\frac{10\pi}{5})=0$
$\cos(\frac{2\pi}{5}) = \cos(\frac{8\pi}{5})$, $\cos(\frac{4\pi}{5}) = \cos(\frac{6\pi}{5}) = -\cos(\frac{\pi}{5})$
So $2\cos(\frac{2\pi}{5})-2\cos(\frac{\pi}{5}) = -1$
$\cos(\frac{2\pi}{5})-\cos(\frac{\pi}{5}) = -\frac{1}{2}$
Been on stuck on this problem for an hour. Please help me.
| If $\cos(5\theta) = 0$, then $z = e^{i\theta}$, must satisfy $z^5 = \pm i$. Note that the two possibilities correspond to conjugates of each other, which won't change the real part, so let's just assume $z^5 = i$. The roots of $z^5 = i$ are:
$$e^{i\pi/10}, e^{5i\pi/10} \color{red}{( = i)}, e^{9\pi i/10}, e^{13\pi i/10} \color{red}{( = e^{-7\pi i/10})}, e^{17\pi i/10} \color{red}{( = e^{-3\pi i/10})}.$$
So, the roots of the polynomial $\cos(5\theta) = 0$ in terms of $\cos(\theta)$ will be the real parts of the above, specifically:
$$\cos(\pi/10), 0, \cos(9\pi/10),\cos(7\pi/10), \cos(3\pi10),$$
remembering that $\cos$ is even. It's also worth noting that, since $\cos(\pi - x) = -\cos(x)$, we can further simplify to:
$$0, \pm \cos(\pi/10), \pm \cos(3\pi/10).$$
Using Vieta on $\cos(5\theta)/\cos(\theta)$ (to remove the zero root), we can therefore see that the product of the non-zero roots is $\frac{5}{16}$. That is,
$$\frac{5}{16} = (-\cos^2(\pi/10))(-\cos^2(3\pi/10)) = (\cos(\pi/10)\cos(3\pi/10))^2.$$
Obviously $\cos(\pi/10)$ and $\cos(3\pi/10)$ are positive, so
$$\cos(\pi/10)\cos(3\pi/10) = \frac{\sqrt{5}}{4}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4299942",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Why do I get incorrect solutions $ x = 0 $ and $ x = 2 $ for $ x + 1 + \sqrt{4x + 1} = 0 $? Here is my incorrect attempt at solving
$$ x + 1 + \sqrt{4x + 1} = 0. $$
Subtracting $ \sqrt{4x + 1} $ from both sides,
$$ x + 1 = -\sqrt{4x + 1} $$
Squaring both sides,
$$ x^2 + 2x + 1 = 4x + 1 $$
Subtracting $ 4x + 1 $ from both sides,
$$ x^2 - 2x = 0 $$
We have obtained
$$ x(x - 2) = 0 $$
It has two solutions: $ x = 0 $ and $ x = 2 $.
But if we substitute $ x = 0 $ in LHS of the original equation we get
$$ x + 1 + \sqrt{4x + 1} = 1 + \sqrt{1} = 2 $$
If we substitute $ x = 2 $ in LHS of the original equation we get
$$ x + 1 + \sqrt{4x + 1} = 3 + \sqrt{9} = 6 $$
Apparently I have made a mistake in some step that has led to this contradiction. Which step is incorrect in my solution above? It must be the squaring step that changes the minus sign to positive sign.
But squaring both sides is an often used step in many equations. What rules I need to keep in mind while solving such equations so that I do not get incorrect solution after squaring both sides?
| Trick question: there will not be any real-number solutions. If the original equation, $ \ x + 1 + \sqrt{4x + 1} \ = \ 0 \ $ is considered as a curve intersection equation, we would write $ \ \sqrt{4x + 1} \ = \ -x - 1 \ \ . $ The left side of the equation represents a square-root curve which resembles the "upper half" of a "horizontal" parabola $ \ ( y \ \ge \ 0 ) \ $ with its vertex at $ \ \left( -\frac14 \ , \ 0 \right) \ \ , $ since the domain of this function is $ \ x \ \ge \ -\frac14 \ \ . $ The line $ \ y \ = \ -x - 1 \ $ has the $ \ y-$intercept $ \ (0 \ , \ -1) \ $ and the $ \ x-$ intercept $ \ (-1 \ , \ 0) \ \ , $ so it passes "to the left" of the square-root curve's vertex and never enters the first quadrant. Hence, there is no intersection of these curves, which means there is no real-number solution for our equation.
The act of squaring both sides of the equation to obtain $ \ 4x + 1 \ = \ (x + 1)^2 \ \ , $ however, produces a second equation which is consistent with this squared equation: $ \ -\sqrt{4x + 1} \ = \ -x - 1 \ \ . $ Because this new square-root curve is on or "below" the $ \ x-$axis $ \ ( y \ \le \ 0 ) \ \ . $ it can intersect the line $ \ y \ = \ -x - 1 \ $ at the two points that you found:
$$ 4x \ + \ 1 \ \ = \ \ x^2 \ + \ 2x \ + \ 1 \ \ \Rightarrow \ \ x·(x - 2) \ \ = \ \ 0 $$ $$ \Rightarrow \ \ x \ = \ 0 \ \ , \ \ y \ = \ -\sqrt{4·0 + 1} \ = \ -0 - 1 \ = \ -1 \ \ \ ; \ \ \ $$
$$ \Rightarrow \ \ x \ = \ 2 \ \ , \ \ y \ = \ -\sqrt{4·2 + 1} \ = \ -2 - 1 \ = \ -3 \ \ . $$
Because the "negative square-root" curve is not implied by our original equation, we must ignore both of these solutions.
There are no "hard-and-fast rules" for solving non-linear equations, since there are so many possibilities for the functions involved. But it is generally a good idea to consider the properties of the functions you will be working with, and having at least a sketch of the graph for the situation can provide some insight.
[I thought it would be of interest to discuss this problem because it is a rare instance of an equation for which all of its solutions are "spurious".]
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4301240",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Calculate the integral $\int_0^{\infty} x^{11} e^{-x^3}dx$ Calculate the integral $\int_0^{\infty} x^{11} e^{-x^3}dx$.
I wanted to approach this using the reduction formula. Here's what I have tried
$$\int x^2e^{-x^3}dx = -\frac{1}{3}e^{-x^3}$$
Then extending this to $n$
$$I_n = \int_0^{\infty}x^ne^{-x^3}dx $$
Using substitution I can get
$$u = x^{n-2}; du = (n-2)x^{n-3}; dv = x^2e^{-x^3}; v = -\frac{1}{3}e^{-x^3}$$
Plugging this in
$$I_n = \int_0^{\infty}x^ne^{-x^3}dx = \left[-\frac{x^{n-2}}{3}e^{-x^3} \right]_0^{\infty}+\frac{(n-3)}{3} \int_0^{\infty}x^{n-3}e^{-x^3}dx $$
$$\implies I_n = \int_0^{\infty}x^ne^{-x^3}dx =\frac{(n-3)}{3} \int_0^{\infty}x^{n-3}e^{-x^3}dx$$
$$=\frac{n-3}{3}I_{n-3}$$
Then starting from $n=11$ I get
$$\frac{8}{3} \cdot \frac{5}{3} \cdot\frac{2}{3} \int_0^{\infty} x^2e^{-x^3}dx$$
However this does not produce the answer which is $2$. How should I go about the reduction formula for this?
| Your recurrence formula has a little error on the first line in the 2nd term on the right should be $n-2$ not $n-3$ and the result should read:
$$
I_n=\frac{n-2}3 I_{n-3}
$$
Now we get:
$$
I_{11}=\frac93\cdot\frac63\cdot\frac33 I_2=2
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4304028",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
What function does $\sum_{n=0}^\infty \frac{(-1)^n(2n)!}{(1-2n)n!^24^n}\frac{x^n}{y^{2n}}$ correspond to? I have the following Taylor series, and although it looks familiar, I cannot figure out which function it corresponds to! Does anyone recognize this Taylor series?
$$1+\frac{x}{2y^2} - \frac{x^2}{8y^4} + \frac{x^3}{16y^6}+\frac{5x^4}{128y^8}+.... = \sum_{n=0}^\infty \frac{(-1)^n(2n)!}{(1-2n)n!^24^n}\frac{x^n}{y^{2n}}$$
Thank you!
| A good series to know offhand is
$$\frac{1}{\sqrt{1-4t}} = \sum_{n=0}^\infty {2n \choose n}t^n $$
which almost gets us the series we desire with $t = -\frac{x}{4y^2}$
$$\frac{1}{\sqrt{1+\frac{x}{4y^2}}} = \sum_{n=0}^\infty {2n \choose n}\left(-\frac{x}{4y^2}\right)^n $$
except for the offending term $1-2n$ in the denominator. However a small adjustment to the original series with $t = -z^{-2}$
$$\frac{1}{\sqrt{1+\frac{4}{z^2}}} = \frac{|z|}{\sqrt{z^2+4}}= \sum_{n=0}^\infty {2n \choose n}(-1)^nz^{-2n}$$
gives us the series we want via integration
$$\sqrt{z^2+4}\operatorname{sgn}(z) = \sum_{n=0}^\infty {2n \choose n}\frac{z^{1-2n}}{1-2n} \implies \frac{\sqrt{z^2+4}}{|z|} = \sum_{n=0}^\infty {2n \choose n}\frac{z^{-2n}}{1-2n}$$
by substituting $z= \frac{2y}{\sqrt{x}}$
$$\sum_{n=0}^\infty {2n \choose n}\frac{1}{1-2n}\left(-\frac{x}{4y^2}\right)^n = \boxed{\frac{\sqrt{y^2+x}}{|y|}=\sqrt{1+\frac{x}{y^2}}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4304336",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Closed expression for a continued fraction Does anyone know a closed expression for the following continued fraction?
$$G(x) = \cfrac{1}{x+1+\cfrac{1}{x+3+\cfrac{4}{x+5+\cfrac{9}{x+7+\cfrac{16}{x+9+\cdots}}}}}$$
All I know is that $G(0) = \frac{\pi}{4}$ and $G(x)$ converges for $x \geq 0$, while also $G(x) \sim \frac{1}{x} \ (x \to \infty)$.
The equation $G(0) = \frac{\pi}{4}$ follows from a well-known continued fraction expansion for $\arctan$. However, the expansion above is different. It came up in my research, and I can't find it in any continued fraction tables.
| The form of the coefficients of the proposed continued fraction is rather similar to that of the ratio of two hypergeometric functions tabulated here:
\begin{equation}
\frac{\mathbf{F}\left(a,b;c;z\right)}{\mathbf{F}\left(a+1,b+1;c+1;z\right)}={x
_{0}+\cfrac{y_{1}}{x_{1}+\cfrac{y_{2}}{x_{2}+\cfrac{y_{3}}{x_{3}+\cdots}}}}
\end{equation}
where
\begin{align}
x_{n}&=c+n-(a+b+2n+1)z\\
y_{n}&=(a+n)(b+n)z(1-z)
\end{align}
and $\mathbf{F}$ are regularized hypergeometric functions.
In the following, the continued fraction is adapted to this representation by finding a parameter $\alpha$ to fit the above formula:
\begin{align}
G(x) &= \cfrac{1}{x+1+\cfrac{1}{x+3+\cfrac{4}{x+5+\cfrac{9}{x+7+\cfrac{16}{x+9+\cdots}}}}}\\
\frac{\alpha}{G(x)}&=\alpha(x+1)+\cfrac{1^2\alpha^2}{\alpha(x+3)+\cfrac{2^2\alpha^2}{\alpha(x+5)+\cfrac{3^2\alpha^2}{\alpha(x+7)+\cfrac{4^2\alpha^2}{\alpha(x+9)+\cdots}}}}
\end{align}
then we have
\begin{align}
x_n&=\alpha(x+2n+1)\\
y_n&=n^2\alpha^2
\end{align}
From the $y_n$ definition, we must have $a=b=0$ and $z(1-z)=\alpha^2$. Then, from $x_n$, we find $c=1/2+\alpha x$ and $\alpha=1/2-z$. The compatibiliy condition for the parameter $z$ are
\begin{align}
\alpha^2&=z(1-z)\\
\alpha&=\frac12-z
\end{align}
and thus $z=(1-1/\sqrt{2})/2$ and $\alpha=2^{-3/2}$ which imposes $c=(1+x/\sqrt{2})/2$. (The root corresponding to $\left|z\right|<1$ was chosen to avoid possible divergence of the hypergeometric functions). With these parameters,
\begin{align}
\frac{2^{-3/2}}{G(x)}&=\frac{\mathbf{F}\left(0,0;(1+x/\sqrt{2})/2;(1-1/\sqrt{2})/2\right)}{\mathbf{F}\left(1,1;(3+x/\sqrt{2})/2;(1-1/\sqrt{2})/2\right)}\\
&=\frac{(1+x/\sqrt{2})/2}{{}_2F_1\left(1,1;(3+x/\sqrt{2})/2;(1-1/\sqrt{2})/2\right)}
\end{align}
Finally,
\begin{equation}
G(x)=\frac{{}_2F_1\left(1,1;(3+x/\sqrt{2})/2;(1-1/\sqrt{2})/2\right)}{x+\sqrt{2}}
\end{equation}
From $\sin \pi/8=\sqrt{2-\sqrt{2}}/2$, it comes
\begin{equation}
G(x)=\frac{{}_2F_1\left(1,1;(3+x/\sqrt{2})/2;\sin^2\pi/8\right)}{x+\sqrt{2}}
\end{equation}
which seems to be numerically correct.
For $x=0$, we have from here
\begin{align}
{}_2F_1\left(1,1;3/2;s\right)=\frac{\sin^{-1}\sqrt{s}}{\sqrt{1-s}\sqrt{s}}
\end{align}
and thus $G(0)=\pi/4$ as expected. Other special cases can be deduced for $x=p\sqrt{2}$ or $x=p/\sqrt2$ where $p$ is an integer, as sevral explicit expressions for the hypergeometric function ${}_2F_1(1,1;c;s)$ are known. For example
\begin{equation}
G(\sqrt{2})=\frac{1+\sqrt{2}}{2}\ln\left( \frac{8}{3+2\sqrt{2}} \right)
\end{equation}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4304881",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
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Calculating a limit or proving it does not exist I have the following function:
$$\frac{\sqrt[3]{x+3}(\sqrt{x^2+1}-\sqrt{x^2+3x+4})}{\sqrt[6]{x^2+5}}$$
The limit definition we are using is
to $-\infty$ such that $L = \lim_{x\to -\infty}f(x)$ iff
for every $\epsilon > 0$ there exists a $M<0$ such that for every $x<M$ we have that $\lvert f(x)-L\rvert < \epsilon$.
I can see that the function is actually
$$\frac{\sqrt[3]{x+3}}{\sqrt[6]{x^2+5}}(\sqrt{x^2+1}-\sqrt{x^2+3x+4})$$
I found that
$$\lim_{x\to -\infty}\frac{\sqrt[3]{x+3}}{\sqrt[6]{x^2+5}}=1.$$
I can see that
$$(\sqrt{x^2+1}-\sqrt{x^2+3x+4})$$
has no limit but I am unable to disprove it by the definition.
After I will be able to disprove it I guess by arithmetics of limit I will be able to prove that the entire function has no limit.
Any assistance will be welcomed.
| You have\begin{multline}\lim_{x\to-\infty}\frac{\sqrt[3]{x+3}\left(\sqrt{x^2+1}-\sqrt{x^2+3x+4}\right)}{\sqrt[6]{x^2+5}}=\\=\lim_{x\to-\infty}\frac{\sqrt[3]{x+3}\left(-3x-3\right)}{\sqrt[6]{x^2+5}\left(\sqrt{x^2+1}+\sqrt{x^2+3x+4}\right)}=\\=\lim_{x\to-\infty}\frac{\sqrt[3]{x+3}}{\sqrt[6]{x^2+5}}\lim_{x\to-\infty}\frac{-3x-3}{\sqrt{x^2+1}+\sqrt{x^2+3x+4}}=\\=(-1)\times\frac32=-\frac32.\end{multline}We have$$\lim_{x\to-\infty}\frac{x+1}{\sqrt{x^2+1}+\sqrt{x^2+3x+4}}=-\frac12$$because\begin{align}\lim_{x\to-\infty}\frac{x+1}{\sqrt{x^2+1}+\sqrt{x^2+3x+4}}&=\lim_{x\to-\infty}\frac{1+\frac1x}{-\sqrt{1+\frac1{x^2}}-\sqrt{1-\frac3x-\frac4{x^2}}}\\&=\frac1{-1-1}\\&=-\frac12.\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4305414",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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prove that $\int_0^\infty \frac{\sin^2 x-x\sin x}{x^3} \, dx= \frac{1}{2} - \ln 2$
Prove that
$$ \int_{0}^{\infty} \frac{\sin^2 x-x\sin x}{x^3} \, dx = \frac{1}{2} - \ln 2 .$$
Integration by parts gives
\begin{align*}
&\lim_{R\to \infty} \int_{0}^{R} \frac{\sin^2 x-x\sin x}{x^3} \, dx \\
&= \lim_{R\to \infty} \biggl( \int_{0}^{R} \frac{\sin^2x}{x^3} \, dx - \int_{0}^{R} \frac{\sin x}{x^2} \, dx \biggr)\\
&= \lim_{R\to\infty} \biggl( \frac{\sin^2 x}{-2x^2}\Biggr\rvert_{0}^{R} - \int_{0}^{R} \frac{\sin (2x)}{-2x^2} \, dx - \biggl(-\frac{\sin x}{x} \Biggr\rvert_{0}^{R} + \int_{0}^{R} \frac{\cos x}{x} \,dx \biggr) \biggr) \\
&= \lim_{R\to \infty} \biggl(\frac{1}{2} + \int_{0}^{2R} \frac{\sin u}{u^2/2} \, \Bigl(\frac{1}2 \, du\Bigr) - \biggl( 1 + \int_{0}^{R} \frac{\cos x}{x} \, dx \biggr)\biggr) \\
&\hspace{22em}\text{(using the substitution $u\mapsto 2x$)}\\
&= -\frac{1}{2} + \lim_{R\to \infty} \biggl(\int_{0}^{2R} \frac{\sin u}{u^2} \, du - \int_{0}^{R} \frac{\cos x}{x} \,dx \biggr)\\
&= \frac{1}{2} + \lim_{R\to\infty} \biggl(\int_{R}^{2R} \frac{\cos x}{x} \, dx \biggr)
\end{align*}
Thus it suffices to show that $\lim_{R\to\infty} \int_{R}^{2R} \frac{\cos x}{x} \, dx = \ln 2$. The Taylor series expansion of $\cos x$ is given by $\cos x = \sum_{i=0}^{\infty} \frac{(-1)^i x^{2i}}{(2i)!}$.
(If the step below (the one involving the interchanging of an infinite sum and integral) is valid, why exactly is it valid? For instance, does it use uniform convergence?)
The limit equals
$$
\lim_{R\to\infty} \int_{R}^{2R} \biggl( \frac{1}{x} + \sum_{i=1}^{\infty} \frac{(-1)^i x^{2i-1}}{(2i)!} \biggr) \, dx
= \ln 2 + \lim_{R\to\infty} \sum_{i=1}^{\infty} \biggl[\frac{(-1)^ix^{2i}}{(2i)(2i)!}\biggr]_{R}^{2R} . $$
But I don't know how to show $\lim_{R\to\infty} \lim_{R\to\infty} \sum_{i=1}^{\infty} \Bigl[\frac{(-1)^ix^{2i}}{(2i)(2i)!}\Bigr]_{R}^{2R} = -2 \ln 2$.
| Alternatively, we could use the fact that $$\int_{0}^{\infty} t^{2} e^{-xt} \, \, \mathrm dt = \frac{2}{x^{3}}, \quad x >0,$$ and then apply Fubini's theorem.$$\begin{align} \int_{0}^{\infty} \frac{\sin^{2}x - x \sin x}{x^{3}} \, \mathrm dx &= \frac{1}{2}\int_{0}^{\infty} \left(\sin^{2} x - x \sin x \right) \int_{0}^{\infty} t^{2}e^{-xt} \, \mathrm dt \, \mathrm dx \\ &= \frac{1}{2}\int_{0}^{\infty} t^{2} \int_{0}^{\infty} \left(\sin^{2} x - x \sin x \right) e^{-xt} \, \mathrm dx \, \mathrm dt \\ &= \frac{1}{2}\int_{0}^{\infty} t^{2} \left(\frac{1}{2} - \frac{\cos 2x}{2} - x \sin x \right) e^{-xt} \, \mathrm dx \, \mathrm dt \\ &= \frac{1}{2}\int_{0}^{\infty} t^{2} \left(\frac{1}{2t} - \frac{1}{2} \frac{t}{t^{2}+4}+ \frac{\mathrm d}{\mathrm d t}\frac{1}{t^{2}+1} \right) \, \mathrm dt \tag{1}\\ &= \frac{1}{2}\int_{0}^{\infty} t^{2} \left(\frac{1}{2t} - \frac{1}{2} \frac{t}{t^{2}+4}- \frac{2t}{(t^{2}+1)^{2}} \right) \, \mathrm dt \\ &= \frac{1}{2} \int_{0}^{\infty} t^{2}\left(\frac{2}{t(t^{2}+4)} - \frac{2t}{(t^{2}+1)^{2}} \right) \, \mathrm dt \\ &= \frac{1}{2} \int_{0}^{\infty} \left(\frac{2t}{t^{2}+4} - \frac{2t}{t^{2}+1} + \frac{2t}{(t^{2}+1)^{2}} \right) \, \mathrm dt \\ &= \frac{1}{2} \left(\ln \left(\frac{t^{2}+4}{t^{2}+1} \right) - \frac{1}{t^{2}+1} \right)\Bigg|_{0}^{\infty} \\ &= \frac{1}{2} \left(-\ln 4 +1 \right) \\ &= \frac{1}{2} - \ln 2. \end{align} $$
$(1)$ $\int_{0}^{\infty} x \sin(x) e^{-xt} \, \mathrm dx = -\frac{\mathrm d}{\mathrm dt} \int_{0}^{\infty} \sin(x) e^{-xt} \, \mathrm dx$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4305593",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
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What's the measure of the radius of the circle below? For reference: In the figure, calculate $R$. If : $\overset{\LARGE{\frown}}{AB}= 120°$, $CM = 1$ and $MB = 2$
(Answer: $\frac{3}{4}(\sqrt{3}+1$))
My progress:
Draw OP
Th.Chords:
$CM.MP = BM.AM \implies 1.[(R-1)+(R)] = 2.AM\\
\boxed{2AM = 2R-1}: \boxed{R=AM+ \frac{1}{2}}\\
\triangle AOB (isósceles):$\
Draw $AE$
$\implies$
$\triangle EAB(right):
AE^2+(2+AM)^2 = 4R^2\\
AE^2 +4+4AM + AM^2 = 4R^2\\
AE^2 + 4+8R-4 + R^2 - R+\frac{1}{4} = 4R^2\\
4AE^2+16+32R - 16+4R^2-4R+1 = 16R^2\\
4AE^2+28R-12R^2+1 = 0 \implies\\
AE^2 = 12R^2-28R-1\\$
...?
I have not found another relationship with AM
| $\angle AOB = 120^o$
Law of cosines:
$AB^2 = 2R^2 + 2R^2 \cdot \frac 12 = 3R^2 \implies AB = R \sqrt3$
Power point M:
$1 \cdot (2R-1) = 2 \cdot (R \sqrt3 - 2) \iff 2R -1 = 2R\sqrt3 - 4 \iff R = \frac3{2(\sqrt3-1)} = \frac{3(\sqrt3+1)}4$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4310815",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Playing tennis until one of the players wins 3 times (Binomial distribution) Question
Nadal and Federer playing tennis against each other.
Nadal's probability to win one match is $\frac{2}{3}$, independently from the previous results.
The two are playing until one of them wins 3 matches.
What is the probability for Nadal's victory?
My Take
Nadal's probability to win is Nadal's probability to win exactly 3 matches which is:
$$
{5 \choose 3}\left(\frac{2}{3}\right)^{3}\left(\frac{1}{3}\right)^{2}=\frac{80}{243}
$$
The Book's Answer
According to the book I should take into consideration Nadal's chances of winning 3\4\5 times:
$$
{5 \choose 3}\left(\frac{2}{3}\right)^{3}\left(\frac{1}{3}\right)^{2}+{5 \choose 4}\left(\frac{2}{3}\right)^{4}\left(\frac{1}{3}\right)+\left(\frac{2}{3}\right)^{5}=\frac{64}{81}
$$
I can't tell why that's the right answer.
As far as I understand the game should stop after Nadal gets 3 points, so why adding the probability of him winning 4\5 matches?
| $ \displaystyle {5 \choose 3}\left(\frac{2}{3}\right)^{3}\left(\frac{1}{3}\right)^{2} ~ $ considers that there are five matches and Nadal wins exactly three of the five matches.
The book's solution considers that Nadal may win three or more of the five matches which is the correct solution. On your point about matches not continuing if Nadal won the first three matches or three out of four matches, here is an alternate way to think -
$(i)$ Nadal wins straight first three matches and we stop.
$(ii)$ He loses exactly one of the first three matches and wins fourth match. We then stop.
$(iii)$ He loses exactly two of the first four matches and wins fifth match.
So Nadal's chances of winning is:
$ \displaystyle \left(\frac{2}{3}\right)^{3} + {3 \choose 1} \cdot\frac{1}{3} \cdot \left(\frac{2}{3}\right)^{3} + {4 \choose 2} \left(\frac{1}{3}\right)^{2} \left(\frac{2}{3}\right)^{3} = \frac{64}{81}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4312216",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that $\int_{0}^{\infty}\frac{e^{-xt}}{1+t}dt>\frac{1}{2}\ln(1+\frac{2}{x})$ for $x > 0$
Question: Let $x> 0$, show that
$$\int_{0}^{\infty}\frac{e^{-xt}}{1+t}dt>\frac{1}{2}\ln(1+\frac{2}{x}).$$
I tried the integration by parts, but somehow couldn't get the term involving natural logarithm of $x$. I am looking forward to some hint for approaching this problem.
| $$I=\int_{0}^{\infty}\frac{e^{-xt}}{1+t}\,dt=e^x \,\Gamma (0,x)$$
If $x$ is large, we have
$$I=\frac{1}{x}-\frac{1}{x^2}+\frac{2}{x^3}-\frac{6}{x^4}+O\left(\frac{1}{x^5}\right)$$ while
$$J=\frac{1}{2} \log \left(1+\frac{2}{x}\right)=\frac{1}{x}-\frac{1}{x^2}+\frac{4}{3 x^3}-\frac{2}{x^4}+O\left(\frac{1}{x^5}\right)$$
$$I-J=\frac{2}{3 x^3}-\frac{4}{x^4}+O\left(\frac{1}{x^5}\right)$$
The case where $x$ is small seems to be more delicate.
What we know is that
$$I-J=\frac{1}{2} \log \left(\frac{1}{2 x}\right)-\gamma+O\left(x^1\right)$$
What remains to prove is that
$$(I-J)'=e^x \Gamma (0,x)-\frac{x+1}{x(x+2)}$$ is a increasing function which never cancels and tends to $0^-$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4315129",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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