Q
stringlengths 70
13.7k
| A
stringlengths 28
13.2k
| meta
dict |
|---|---|---|
Solve the recurrence $T(n) = 2T(n-1) + n$ Solve the recurrence $T(n) = 2T(n-1) + n$
where $T(1) = 1$ and $n\ge 2$.
The final answer is $2^{n+1}-n-2$
Can anyone arrive at the solution?
|
\begin{align}
T(n) & = 2 T(n-1) + n = 2(2T(n-2) + n-1) + n = 4T(n-2) + 2(n-1) + n\\
& = 8 T(n-3) + 4(n-2) + 2(n-1) + n = 2^k T(n-k) + \sum_{j=0}^{k-1} 2^j (n-j)\\
& = 2^{n-1} T(1) + \sum_{j=0}^{n-2}2^j (n-j) = 2^{n-1} + \sum_{j=0}^{n-2}2^j (n-j)
\end{align}
\begin{align}
\sum_{j=0}^{n-2}2^j (n-j) & = n \sum_{j=0}^{n-2}2^j - \sum_{j=0}^{n-2} j2^j = n(2^{n-1}-1) - \dfrac{n \cdot 2^n - 3 \cdot 2^n + 4}2\\
& = n(2^{n-1}-1) - (n \cdot 2^{n-1} -3 \cdot 2^{n-1} + 2) = 3 \cdot 2^{n-1} -n - 2
\end{align}
Hence,
$$T(n) = 2^{n-1} + 3 \cdot 2^{n-1} -n - 2 = 2^{n+1} - n - 2$$
EDIT (Adding details)
First note that $\displaystyle \sum_{j=0}^{n-2}2^j$ is sum of a geometric progression and can be summed as shown below.$$\sum_{j=0}^{k} x^j = \dfrac{x^{k+1} -1}{x-1}$$
$\displaystyle \sum_{j=0}^{n-2} j2^j$ is a sum of the form $\displaystyle \sum_{j=0}^{k} jx^j$
$$\sum_{j=0}^{k} jx^j = x \sum_{j=0}^{k} jx^{j-1} = x \dfrac{d}{dx} \left( \sum_{j=0}^k x^j\right) = x \dfrac{d}{dx} \left( \dfrac{x^{k+1} - 1}{x-1}\right) = x \left( \dfrac{kx^{k+1} - (k+1) x^k +1}{(x-1)^2} \right)$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/239974",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 9,
"answer_id": 5
}
|
Show that $\int_0^\infty \sin\left(x^2\right)dx$ converges, but that $\int_0^\infty \sqrt{\sin^2\left(x^2\right)}dx$ does not. Show that $\int_0^\infty \sin\left(x^2\right)dx$ converges, but that $\int_0^\infty \sqrt{\sin^2\left(x^2\right)}dx$ does not.
The first part I think I proved using triangles, but I could not prove the second part.
|
Your second integral is
$$
\int_0^\infty|\sin x^2|\,dx.
$$
Note that $\sin t\geq1/2$ if $t\in[\frac\pi6+2k\pi,\frac{5\pi}6+2k\pi]$. So $\sin x^2\geq1/2$ if $x\in[\sqrt{\frac\pi6+2k\pi},\sqrt{\frac{5\pi}6+2k\pi}]$, $k\in\mathbb N$.
So
$$
\begin{eqnarray}
\int_0^\infty|\sin x^2|\,dx&\geq&\sum_{k=0}^\infty\int_{\sqrt{\frac\pi6+2k\pi}}^{\sqrt{\frac{5\pi}6+2k\pi}}|\sin x^2|\,dx\\
&\geq&\frac12\,\sum_{k=0}^\infty\left(\sqrt{\frac{5\pi}6+2k\pi}-\sqrt{\frac{\pi}6+2k\pi}\right)\\
&\geq&\frac12\,\sum_{k=0}^\infty\frac{\frac{4\pi}6}{\sqrt{\frac{5\pi}6+2k\pi}+\sqrt{\frac{\pi}6+2k\pi}}\\
&\geq&\frac12\,\sum_{k=0}^\infty\frac{\frac{4\pi}6}{2\sqrt{\frac{5\pi}6+2k\pi}}\\
&=&\frac\pi8\,\sum_{k=0}^\infty\frac{1}{\sqrt{\frac{5\pi}6+2k\pi}}=\infty
\end{eqnarray}
$$
(the last series clearly diverges as its terms grow as $k^{-1/2}$).
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/245296",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
}
|
Conditional probability of cows the question is that if you are a farmer and own six cows: 3 white, 2 black and one that is black on one side and white on the other. Then if you see two black cows (that is 2 black sides of cows) then what is the probability that one of them is the black and white cow?
Here is my attempted answer:
if $M_1$ and $M_2$ are the events that the first or second cow is the mixed one and $b_1$ and $b_2$ denote that the sides of the first and second cow we see is black. then we are looking for $$P(M_1 \cup M_2 \mid b_1b_2)=\frac{P(M_1b_1b_2)+P(M_2b_1b_2)}{P(b_1b_2)}$$
$$=\frac{2P(M_1)P(b_1\mid M_1)P(b_2\mid M_1b_1)}{P(M_1 \cup M_2)P(b_1b_2\mid M_1 \cup M_2)+[1-P(M_1 \cup M_2)]P(b_1b_2\mid M_1^cM_2^c)}$$
then $P(M_1)$ = $\frac{1}{6}$ as there are $6$ sheep and only $1$ that is mixed. $P(b_1\mid M_1)= \frac{1}{2}$ as it can be one of two sides and $P(b_2\mid M_1b_1)=\frac{2}{5}$ as this is just the probability of choosing a black cow. So the numerator is equal to $\frac{1}{15}$.
$P(M_1 \cup M_2) = P(M_1) + P(M_2)$ as they are disjoint and so is equal to $\frac{1}{3}$. Then $P(b_1b_2\mid M_1 \cup M_2)$ is just the probability that the other cow is black and we see the black side of the mixed sheep and so is $\frac{2}{5} *\frac{1}{2}$. Finally $[1-P(M_1 \cup M_2)]=\frac{2}{3}$ and $P(b_1b_2\mid M_1^cM_2^c)$ is $\frac{1}{10}$ as it is the number of all black pairs over the total number of pairs. so putting it all together. I get $\frac{\frac{1}{15}}{\frac{2}{15}}=\frac{1}{2}$ but the answer is supposedly $\approx .3$
EDIT
The guy who wrote the paper made a mistake in the answers. It should be $\frac{1}{2}$
|
If the question is asking for the probability that either of the two cows is 2-coloured, we have
$$P(\text {1 cow is 2-coloured | both visible sides are black}) = \frac{P(\text {1 cow is 2-coloured and other is black}) \times P(\text {the black side of the 2-coloured cow is seen})}{P(\text{both visible sides are black})}=\frac{\frac{\binom{3}{0}\binom{1}{1}\binom{2}{1}\cdot\frac{1}{2}}{\binom{6}{2}}}{\frac{\binom{3}{0}\binom{1}{1}\binom{2}{1}\cdot\frac{1}{2}}{\binom{6}{2}}+\frac{\binom{3}{0}\binom{1}{0}\binom{2}{2}}{\binom{6}{2}}}=\frac{1}{2}$$
where $$\frac{1}{15}=\frac{\binom{3}{0}\binom{1}{1}\binom{2}{1}\cdot\frac{1}{2}}{\binom{6}{2}}$$ is the probability that the $2$ visible sides are black when one is 2-coloured and the other is black and $$\frac{1}{15}=\frac{\binom{3}{0}\binom{1}{0}\binom{2}{2}}{\binom{6}{2}}$$ is the probability that the $2$ visible sides are black when both cows are black (these exhaust all possibilities for both visible sides being black).
If the question is asking for the probability that a certain (of the two) cow be 2-coloured, the probability is then $\frac{1}{2}\cdot\frac{1}{2}=\frac{1}{4}$
Note that the number of white cows is irrelevant.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/246460",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
}
|
How do we prove that two parametric equations are drawing the same thing? For example, if I have
$$\begin {align}
x(t) &= r\sin t\cos t\\
y(t) &= r\sin^2 t\\
\end {align}$$
and
$$\begin {align}
x(t) &= \frac r 2 \cos t\\
y(t) &= \frac r 2 (\sin t + 1)
\end {align}$$
How do we show that the two parametric equations draw the same line?
|
$\begin{align}
\hline
x(s) &= r\sin s\cos s\\
y(s) &= r\sin^2 s\\
\hline
x(t) &= \frac r2 \cos t\\
y(t) &= \frac r2 (\sin t + 1)\\
\hline
\end {align}$
Let's rewrite the equations to make them more comparable.
$\begin{align}
\hline
x(s) &= \dfrac r2 \sin 2s\\
y(s) &= \dfrac r2(1 - \cos 2s)\\
\hline
x(t) &= \dfrac r2 \cos t\\
y(t) &= \dfrac r2 (1 + \sin t)\\
\hline
\end {align}$
It seems we need to find a relationship between $s$ and $t$ such that
$$\cos t = \sin 2s \quad \text{and} \quad \sin t = -\cos 2s$$
When we let $2s = \dfrac{\pi}{6} = 30^\circ$, we get
$$\cos t=\dfrac 12 \quad \text{and} \quad \sin t=-\dfrac{\sqrt 3}{2}$$
Which corresponds to the angle $t = -\dfrac{\pi}{3} = -60^\circ$. So it seems that
$$t = 2s - \dfrac{\pi}{2} = 2s - 90^\circ.$$
We need to check that.
\begin{align}
x(t)&= \dfrac r2 \cos t\\
&= \dfrac r2 \cos( 2s - \dfrac{\pi}{2})\\
&= \dfrac r2
(
\cos 2s \, \cos \dfrac{\pi}{2}
+ \sin 2s \, \sin \dfrac{\pi}{2}
) \\
&= \dfrac r2 \sin 2s \\
&= r \sin s \cos s \\
&= x(s)\\
y(t)&= \dfrac r2 (1 + \sin t)\\
&= \dfrac r2 (1 + \sin( 2s - \dfrac{\pi}{2}))\\
&= \dfrac r2
(1 +
\sin 2s \, \cos \dfrac{\pi}{2}
- \cos 2s \, \sin \dfrac{\pi}{2}
) \\
&= \dfrac r2(1-\cos 2s)\\
&= r \sin^2 s\\
&= y(s)
\end{align}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/247752",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
}
|
Finding the minimum value of a function in an ellipse So I have this problem for my homework:
Consider the ellipse: $\dfrac {x^2}{a^2} + \dfrac{y^2}{b^2}=1$ where $0<b<a$. For every point $(x, y)$ on the ellipse find the the perpendicular line to the ellipse so that the point $(x, y)$ is on that line. This line cuts the ellipse in another point $(x', y')$. Prove that the distance between these two points is $$D(x, y)=\dfrac {2\left(\dfrac {x^2}{a^4} + \dfrac{y^2}{b^4}\right)^{3/2}}{\dfrac {x^2}{a^6} + \dfrac{y^2}{b^6}}\ .$$
So I've already done that; but after that it says:
Use Lagrange Multipliers to minimize the function $D(x, y)$.
But the equations get complicated and messy and I don't know if I have to consider the line or just the ellipse. Logically the minimum is at $(0, b)$. Can somebody help me?
|
I think it is not necessary to use Lagrange Multipliers. we can replace $\dfrac{x^2}{a^2}$ with $1- \dfrac{y^2}{b^2}$, then we have:
$D(y)=2\dfrac{(\dfrac{1}{a^2}(1- \dfrac{y^2}{b^2})+ \dfrac{y^2}{b^4})^\frac{3}{2}}{\dfrac{1}{a^4}(1- \dfrac{y^2}{b^2})+ \dfrac{y^2}{b^4}}=2\dfrac{(C_1+C_2y^2)^\frac{3}{2}}{C_3+C_4y^2}$, and $C_1=\dfrac{1}{a^2},C_2=\dfrac{1}{b^2}(\dfrac{1}{b^2}-\dfrac{1}{a^2})=\dfrac{a^2-b^2}{a^2b^4}$,$C_3=\dfrac{1}{a^4},C_4=\dfrac{1}{b^2}(\dfrac{1}{b^4}-\dfrac{1}{a^4})=\dfrac{a^4-b^4}{a^4b^6}$, let $z=y^2$,we have a very simple fomula:
$D(z)=2\dfrac{(C_1+C_2z)^\frac{3}{2}}{C_3+C_4z}$
$D'(z)=2*(\dfrac{3}{2}C_2\dfrac{(C_1+C_2z)^\frac{1}{2}}{C_3+C_4z}-C_4\dfrac{(C_1+C_2z)^\frac{3}{2}}{(C_3+C_4z)^2})$
$=2*\dfrac{(C_1+C_2z)^\frac{1}{2}}{C_3+C_4z}*(\dfrac{3C_2}{2}-C_4*\dfrac{C_1+C_2z}{C_3+C_4z})$,
since $\dfrac{(C_1+C_2z)^\frac{1}{2}}{C_3+C_4z}>0$, when $D'(z)=0$, we have:
$\dfrac{3C_2}{2}-C_4*\dfrac{C_1+C_2z}{C_3+C_4z}=0$, then we put all staff in, we get:
$D'(z)=F(z)*Q(z), F(z)>0$,$Q(z)=b^6-2a^2b^4+(a^4-b^4)z$ , if $D'(z)=0$,then $Q(z)=0$,
$z=\dfrac{b^4(2a^2-b^2)}{a^4-b^4}$, now here is a trick:
$z=y^2 \leq b^2$ then $\dfrac{b^4(2a^2-b^2)}{a^4-b^4} \leq b^2$,ie.$a^2 \geq 2b^2$ it means only under this condition, $D'(z)=0$ can be satisfied.
when $a^2 \geq 2b^2$, we have:
$Q(z)=b^4(b^2-z)+a^2(a^2z-2b^4), $when $z=b^2, Q(z) \geq 0$ ie $D'(z)\geq 0$;,when $ z=0, Q(z)=b^4(b^2-2a^2) < 0$,ie $D'(z)<0$ , which means D(z) has min. put $z=\dfrac{b^4(2a^2-b^2)}{a^4-b^4}$,we get:
$ D_{min}(z)=\dfrac{\sqrt{27}a^2b^2}{(a^2+b^2)^\frac{3}{2}} \leq 2b $
when $a^2 < 2b^2$, we have:
$Q(z)=a^2z(a^2-2b^2)+b^2(2a^2-b^2)(z-b^2)<0$, which mean when $z=b^2$, $D(z)$ has its min which is $2b$.
BTW, this is a Japanese Temple Geometry problem which discussed about 300 years ago.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/249812",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
}
|
Why does $\frac{a}{\frac{b}{x}} = x \times \frac{a}{b}$? As much as it embarasses me to say it, but I always had a hard time understanding the following equality:
$$
\frac{a}{\frac{b}{x}} = x \times \frac{a}{b}
$$
I always thought that the left-hand side of the above equation was equivalent to
$$
\frac{a}{\frac{b}{x}} = \frac{a}{b} \div \frac{x}{1} = \frac{a}{b} \times \frac{1}{x}
$$
What am I doing wrong, here?
|
The problem is that $\frac{a}{\frac{b}{x}}=\frac{\frac{a}{b}}{\frac{1}{x}}=x\frac{a}{b}$ because $\frac{1}{\frac{1}{x}}=x$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/251317",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 8,
"answer_id": 2
}
|
How to show two field extensions are equal. How do I go about showing that $\mathbb{Q}(\sqrt{3}+\sqrt{5})=\mathbb{Q}(\sqrt{3},\sqrt{5})$?
Since $\mathbb{Q}(\sqrt{3}+\sqrt{5}) \subset\mathbb{Q}(\sqrt{3},\sqrt{5})$, we need to show the reverse inclusion. Does it suffice to prove that $\sqrt{2} \in \mathbb{Q}(\sqrt{3}+\sqrt{5})$?
Thank you
|
We have
$$\frac{(\sqrt{3} + \sqrt{5})^3 - 14(\sqrt{3} + \sqrt{5})}{4} = \frac{18\sqrt{3} + 14\sqrt{5} - 14\sqrt{3} -14 \sqrt{5}}{4} = \frac{4\sqrt{3}}{4} = \sqrt{3}$$
So $\sqrt{3} \in \mathbb{Q}(\sqrt{3} + \sqrt{5})$, and hence so is $\sqrt{5}$.
The general trick in these situations is to write one of the irrational numbers as a polynomial in their sum, with coefficients from $\mathbb{Q}$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/251741",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
}
|
Closed form for $\sum_{k=0}^{n} \binom{n}{k}\frac{(-1)^k}{(k+1)^2}$ How can I calculate the following sum involving binomial terms:
$$\sum_{k=0}^{n} \binom{n}{k}\frac{(-1)^k}{(k+1)^2}$$
Where the value of n can get very big (thus calculating the binomial coefficients is not feasible).
Is there a closed form for this summation?
|
Suppose we seek to compute
$$S_n = \sum_{k=0}^n {n\choose k} \frac{(-1)^k}{(k+1)^2}.$$
With this in mind we introduce the function
$$f(z) = n! (-1)^n \frac{1}{(z+1)^2} \prod_{q=0}^n \frac{1}{z-q}.$$
We then obtain for $0\le k\le n$
$$\mathrm{Res}_{z=k} f(z) =
(-1)^n \frac{n!}{(k+1)^2} \prod_{q=0}^{k-1} \frac{1}{k-q}
\prod_{q=k+1}^n \frac{1}{k-q}
\\ = (-1)^n \frac{n!}{(k+1)^2}
\frac{1}{k!} \frac{(-1)^{n-k}}{(n-k)!}
= {n\choose k} \frac{(-1)^k}{(k+1)^2}.$$
This means that
$$S_n = \sum_{k=0}^n \mathrm{Res}_{z=k} f(z)$$
and since residues sum to zero we have
$$S_n + \mathrm{Res}_{z=-1} f(z) + \mathrm{Res}_{z=\infty} f(z) = 0.$$
We can compute the residue at infinity by inspection (it is zero) or
more formally through
$$\mathrm{Res}_{z=\infty}
n! (-1)^n \frac{1}{(z+1)^2} \prod_{q=0}^n \frac{1}{z-q}
\\ = - n! (-1)^n \mathrm{Res}_{z=0} \frac{1}{z^2}
\frac{1}{(1/z+1)^2} \prod_{q=0}^n \frac{1}{1/z-q}
\\ = - n! (-1)^n \mathrm{Res}_{z=0}
\frac{1}{(z+1)^2} \prod_{q=0}^n \frac{z}{1-qz}
\\ = - n! (-1)^n \mathrm{Res}_{z=0}
\frac{z^{n+1}}{(z+1)^2} \prod_{q=0}^n \frac{1}{1-qz} = 0.$$
We get for the residue at $z=-1$ that
$$\mathrm{Res}_{z=-1} f(z) =
n! (-1)^n \left. \left(\prod_{q=0}^n \frac{1}{z-q}\right)'\right|_{z=-1}
\\ = - n! (-1)^n \left.
\left(\prod_{q=0}^n \frac{1}{z-q}\right)
\sum_{q=0}^n \frac{1}{z-q} \right|_{z=-1}
\\ = - n! (-1)^n \frac{(-1)^{n+1}}{(n+1)!} \left(-H_{n+1}\right)
= -\frac{H_{n+1}}{n+1}.$$
We thus have
$$S_n -\frac{H_{n+1}}{n+1} = 0$$
or
$$\bbox[5px,border:2px solid #00A000]{
\frac{H_{n+1}}{n+1}.}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/254416",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 9,
"answer_id": 6
}
|
If a,b,c are positive and (a,b)=(b,c)=1 and 1/a + 1/b + 1/c = integer then b=1 and a=c=1 or 2 If you work it out you see that if
$a=b=c=1$, then $(1,1)=(1,1)=1$ and $1/1 + 1/1 + 1/1 = 3 =$ integer
and
$b=1, a=c=2$, then $(2,1)=(1,2)=1$ and $1/2 + 1/1 + 1/2 = 2=$ integer.
But please help me prove this
|
So you have proven that the given cases does indeed work, and you need help to show that these are the only cases. So let's try to find another case. Let's see what the fraction sum turns out to be:
$$
\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \frac{bc + ac + ab}{abc}
$$
Since $b$ have no factors in common with neither $a$ nor $c$, the last fraction cannot be shortened by any factor of $b$, so the denominator will always be a multiple of $b$, and thus the fraction cannot be an integer unless $b=1$.
If $b = 1$ (which we now know it must be for the conditions to hold), then $\frac{1}{a} + \frac{1}{c}$ must also be an integer, and thus none of $a$ and $c$ can be greater than $2$, and they have to be the same number. So either they are both $1$ or they are both $2$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/254929",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Property of Fejer kernel Let
$$
F_n(x) = \frac{1}{n} \left( \frac{ \sin(\frac{1}{2} n x ) } { \sin(\frac{1}{2} x ) } \right)^2
$$
be the n-th Fejer-Kernel. Then
$$
\forall \epsilon > 0, r < \pi : \exists N \in \mathbb{N} : \forall n \ge N : \int_{[-\pi,\pi] \setminus [-r, r]} F_n(t) \, \mathrm{d} t < \epsilon
$$
The proof of this goes like this
\begin{align*}
\int_{[-\pi,\pi] \setminus [-r, r]} F_n(t) \, \mathrm{d} t
& = \int_{[-\pi,\pi] \setminus [-r, r]} \frac{1}{n} \left( \frac{ \sin(\frac{1}{2} n x ) } { \sin(\frac{1}{2} x ) } \right)^2 \, \mathrm{d} t \\
& \le \frac{1}{n} \int_{[-\pi,\pi] \setminus [-r, r]} \frac{1}{\sin^2(\frac{1}{2}t)} \, \mathrm{d} t \\
& \le \frac{1}{n} \int_{[-\pi,\pi]} \frac{1}{\sin^2(\frac{1}{2}r)} \, \mathrm{d} t \qquad (*) \\
& = \frac{2\pi}{n} \frac{1}{\sin^2(\frac{1}{2}r)} \\
& < \epsilon \quad \textrm{for certain } ~ n \ge N.
\end{align*}
The step i marked with (*) is totally unclear to me, why could there replaced $t$ by $r$, because it need not to be the case that $\sin^2(0.5*r) < \sin^2(0.5*t)$ as i think?
|
To see the estimate (*), first note that for $0<r<x<\pi/2$ we have
$$0<\sin r<\sin x<1$$
Hence for $x$ with $0<r<|x|<\pi$
$$0<\sin^2 \frac{r}{2}<\sin^2 \frac{x}{2}<1$$
that is
$$0<\frac{1}{\sin^2 \frac{x}{2}}<\frac{1}{\sin^2 \frac{r}{2}}$$
and we get
$$ \int_{[-\pi,\pi] \setminus [-r, r]} \frac{1}{\sin^2\frac{x}{2}} dx
\le \int_{[-\pi,\pi]\setminus[-r,r]} \frac{1}{\sin^2\frac{1}{2}r} dx \leq
\int_{[-\pi,\pi]} \frac{1}{\sin^2\frac{1}{2}r} dt = \frac{2\pi}{\sin^2\frac{1}{2}r}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/255002",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
}
|
How to correct this diagonalization argument? I want to diagonalize (with a orthogonal change of coordinates) the quadratic form $F(x,y)=x^2-\frac{n-2}{\sqrt{n-1}}xy-y^2$. I already know that the system $(*)$ bellow
\begin{eqnarray}
x=&\frac{(\sqrt{n-1}+1)u+(\sqrt{n-1}-1)v}{\sqrt{2n}}\\
y=&\frac{(-\sqrt{n-1}+1)u+(\sqrt{n-1}+1)v}{\sqrt{2n}},
\end{eqnarray}
does the work but, I'm trying to find this answer by myself. What I did was the following.
The symmetric matrix associated to $F$ is
$$
A=\begin{pmatrix}1 & \frac{2-n}{2\,\sqrt{n-1}}\cr \frac{2-n}{2\,\sqrt{n-1}} & -1\end{pmatrix}
$$
with eigenvalues $\pm\frac{n}{2\sqrt{n-1}}$ and corresponding eigenvectors
\begin{align}
v_1=&(1,\frac{\left( n-2\right) \,\sqrt{n-1}}{\sqrt{n-1}\,n-2\,n+2})\\
v_2=&(1,-\frac{\left( n-2\right) \,\sqrt{n-1}}{\sqrt{n-1}\,n+2\,n-2}).
\end{align}
As many textbooks teach, we define
$$
P=
\begin{pmatrix}1 & 1\cr \frac{\left( n-2\right) \,\sqrt{n-1}}{\sqrt{n-1}\,n-2\,n+2} & -\frac{\left( n-2\right) \,\sqrt{n-1}}{\sqrt{n-1}\,n+2\,n-2}\end{pmatrix}.
$$
The inverse of $P$ is
$$
P^{-1}=
\begin{pmatrix}\frac{\sqrt{n-1}\,\left( {n}^{2}-4\,n+4\right) }{2\,\sqrt{n-1}\,{n}^{2}+4\,{n}^{2}-4\,n} & \frac{n-2}{2\,n}\cr \frac{\sqrt{n-1}\,\left( {n}^{2}-4\,n+4\right) }{2\,\sqrt{n-1}\,{n}^{2}-4\,{n}^{2}+4\,n} & \frac{2-n}{2\,n}\end{pmatrix},
$$
so we have
$$P^{-1}\cdot A\cdot P=\begin{pmatrix}-\frac{n}{2\sqrt{n-1}} & 0\cr 0 & \frac{n}{2\sqrt{n-1}}\end{pmatrix}.$$
That is, the matrix $P$ gives a change of coordinates that diagonalizes $A$ but it is far from being orthogonal.
So, how could I find the change of coordinates $(*)$?
|
It's not far from being orthogonal; you just failed to normalize the columns. Being the eigenvectors of a symmetric matrix, they're already orthogonal; if you normalize them, $P$ will be orthogonal and $P^{-1}AP$ will still be diagonal.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/255460",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
How to find the general solution of $(1+x^2)y''+2xy'-2y=0$. How to express by means of elementary functions?
Find the general solution of $$(1+x^2)y''+2xy'-2y=0$$
in terms of power series in $x$. Can you express this solution by means
of elementary functions?
I know that $y= \displaystyle\sum_{n=0}^{ \infty } a_nx^n$ and $y'= \displaystyle\sum_{n=1}^{ \infty } a_nnx^{n-1}$ and $y''=\displaystyle\sum_{n=2}^{ \infty } a_nn(n-1)x^{n-2}$.
As far as I can tell I am to simply plug these summations in the original equation stated above. Also, I shifted the following to get $x^n$. Giving:
$$\sum_{n=0}^{ \infty } a_{n+2}(n+2)(n+1)x^{n}+\sum_{n=2}^{ \infty } a_nn(n-1)x^{n}+\sum_{n=1}^{ \infty }2 a_nnx^{n}-\sum_{n=0}^{ \infty } 2a_nx^n=0$$
Now, I combine the equation into a single summation.
$$\sum_{n=2}^{ \infty }\bigg[a_{n+2}(n+2)(n+1)+a_nn(n-1)+2a_nn-2a_n\bigg]x^n=0$$
Doing this, I am left over with $2(a_2-a_0)=0$ and $6a_3x=0$. I can also form $a_{n+2}=a_n\frac{1-n}{n+1}$ Following this I plugged values in for the third equation, but this is where things start to get difficult. I'm honestly not sure how to continue. The question itself confuses me specifically "in terms of power series in $x$. Can you express this solution by means of elementary functions?".
|
Let $y=\sum\limits_{n=0}^\infty a_nx^n$ ,
Then $y'=\sum\limits_{n=0}^\infty na_nx^{n-1}$
$y''=\sum\limits_{n=0}^\infty n(n-1)a_nx^{n-2}$
$\therefore(1+x^2)\sum\limits_{n=0}^\infty n(n-1)a_nx^{n-2}+2x\sum\limits_{n=0}^\infty na_nx^{n-1}-2\sum\limits_{n=0}^\infty a_nx^n=0$
$\sum\limits_{n=0}^\infty n(n-1)a_nx^{n-2}+\sum\limits_{n=0}^\infty n(n-1)a_nx^n+\sum\limits_{n=0}^\infty2na_nx^n-\sum\limits_{n=0}^\infty2a_nx^n=0$
$\sum\limits_{n=2}^\infty n(n-1)a_nx^{n-2}+\sum\limits_{n=0}^\infty(n^2+n-2)a_nx^n=0$
$\sum\limits_{n=2}^\infty n(n-1)a_nx^{n-2}+\sum\limits_{n=0}^\infty(n+2)(n-1)a_nx^n=0$
$\sum\limits_{n=2}^\infty n(n-1)a_nx^{n-2}+\sum\limits_{n=2}^\infty n(n-3)a_{n-2}x^{n-2}=0$
$\sum\limits_{n=2}^\infty(n(n-1)a_n+n(n-3)a_{n-2})x^{n-2}=0$
$\therefore n(n-1)a_n+n(n-3)a_{n-2}=0$
$a_n=-\dfrac{(n-3)a_{n-2}}{n-1}$
$\therefore\begin{cases}a_1=a_1\\a_{2n+3}=0~\forall n\in\mathbb{Z}^*\\a_0=a_0\\a_{2n}=\dfrac{(-1)^n((-1)1\times3\times......(2n-3))a_0}{1\times3\times5\times......(2n-1)}\forall n\in\mathbb{N}\end{cases}$
$\begin{cases}a_1=a_1\\a_{2n+3}=0~\forall n\in\mathbb{Z}^*\\a_0=a_0\\a_{2n}=\dfrac{(-1)^{n+1}a_0}{2n-1}\forall n\in\mathbb{N}\end{cases}$
$\begin{cases}a_1=a_1\\a_{2n+3}=0~\forall n\in\mathbb{Z}^*\\a_{2n}=\dfrac{(-1)^{n+1}a_0}{2n-1}\forall n\in\mathbb{Z}^*\end{cases}$
$\therefore y=C_1x+C_2\sum\limits_{n=0}^\infty\dfrac{(-1)^{n+1}x^{2n}}{2n-1}=C_1x+C_2\biggl(1+\sum\limits_{n=1}^\infty\dfrac{(-1)^{n+1}x^{2n}}{2n-1}\biggr)=C_1x+C_2\biggl(1+\sum\limits_{n=0}^\infty\dfrac{(-1)^nx^{2n+2}}{2n+1}\biggr)=C_1x+C_2(1+x\tan^{-1}x)$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/255974",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
}
|
Convergence of Sequence with factorial I want to show that
$$
a_n = \frac{3^n}{n!}
$$
converges to zero. I tried Stirlings formulae, by it the fraction becomes
$$
\frac{3^n}{\sqrt{2\pi n} (n^n/e^n)}
$$
which equals
$$
\frac{1}{\sqrt{2\pi n}} \left( \frac{3e}{n} \right)^n
$$
from this can I conclude that it goes to zero because $\frac{3e}{n}$ and $\frac{1}{\sqrt{2\pi n}}$ approaching zero?
|
Alternatively to Stirling:
We have
\begin{align}
a_n =\frac{3^n}{n!}
\end{align}
Now let $n>3$, then
\begin{align}
0\leq a_n &=\frac{3^n}{n!} = \frac{3\cdot3 \cdot 3}{1\cdot 2\cdot 3}\cdot \frac{3^{n-3}}{4\cdot 5 \cdot ...\cdot n}=\frac{3\cdot3 \cdot 3}{1\cdot 2\cdot 3}\cdot \frac{3\cdot 3 \cdot ... \cdot 3}{4\cdot 5 \cdot ...\cdot n}\\
&\leq \frac{9}{2}\cdot \frac{3\cdot 3 \cdot ... \cdot 3}{4\cdot 4 \cdot ...\cdot 4} = \frac{9}{2} \cdot \Bigr(\frac{3}{4}\Bigl)^n\rightarrow0 \text{ as } n\rightarrow \infty
\end{align}
So we have $a_n\rightarrow 0$ as $n\rightarrow \infty$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/260772",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
}
|
Integration Question on $\int \frac{1}{x^2+10x+21} \, dx$ How would i do the following indefinite integration $$\int \frac{1}{x^2+10x+21} \, dx$$
so far I've turned the bottom polynomial into $(x+7)(x+3)$ not too sure where to go from here
|
Hint: Partial Fractions:
$$\frac{1}{(x+7)(x+3)}=\frac{A}{x+7}+\frac{B}{x+3}$$
Now find $A,B$ and use the linearity of the integral:
$$\int \frac{1}{(x+7)(x+3)} dx=\int \frac{A}{x+7}dx+\int \frac{B}{x+3}dx$$
This should be simple now.
EDIT: Evaluating $A,B$:
$$\frac{1}{(x+7)(x+3)}=\frac{A}{x+7}+\frac{B}{x+3}\iff 1=A(x+3)+B(x+7)$$
Setting $x=-3$ and $x=-7$ give $A,B=?$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/263680",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Synthetic division via the greedy strategy I was looking at the expanded synthetic division within Wikipedia. I was stumped by how to come up with and perform the 'compactified' version of synthetic division.
Does anyone know how to do it?
|
I came up with three shorthand methods from noticing certain patterns, but the one I describe below is the simplest:
*
*Write the coefficients of the dividend on a bar
$\begin{array}{cc}
\begin{array}{|rrrrrrrr}
a & b & c & d & e & f & g & h \\
\hline
\end{array}
\end{array}$
*
*Negate the coefficients of the divisor. Write in every coefficient of the divisor but the first (leading coefficient) one on the left.
$\begin{array}{cc}
\begin{array}{rrrr} i &j & k & l \\ \end{array}
&
\begin{array}{|rrrrrrrr}
a & b & c & d & e & f & g & h \\
\hline
\end{array}
\end{array}$
*
*From the number of coefficients placed on the left side, count the number of dividend coefficients above the bar and then place a vertical bar on the row below. This vertical bar marks the separation between the quotient and the remainder.
$\begin{array}{cc}
\begin{array}{rrrr} i &j & k & l \\ \\ \end{array}
&
\begin{array}{|rrrr|rrrr}
a & b & c & d & e & f & g & h \\
\hline
& & & & & & & \\
\end{array}
\end{array}$
*
*Drop the first coefficient of the dividend below the bar.
$\begin{array}{cc}
\begin{array}{rrrr} i &j & k & l \\ \\ \end{array}
&
\begin{array}{|rrrr|rrrr}
a & b & c & d & e & f & g & h \\
\hline
m & & & & & & & \\
\end{array}
\end{array}$
*
*Multiply the last dropped/summed number to each negated coefficients on the left (starting with the left most); skip this step if the summed number is zero. Place each product on top of the subsequent columns.
$\begin{array}{cc}
\begin{array}{rrrr} \\ i &j & k & l \\ \\ \end{array}
&
\begin{array}{|rrrr|rrrr}
& mi & mj & mk & ml & & & \\
a & b & c & d & e & f & g & h \\
\hline
m & & & & & & & \\
\end{array}
\end{array}$
*
*Perform an column-wise addition on the next column.
$\begin{array}{cc}
\begin{array}{rrrr} \\ i &j & k & l \\ \\ \end{array}
&
\begin{array}{|rrrr|rrrr}
& mi & mj & mk & ml & & & \\
a & b & c & d & e & f & g & h \\
\hline
m & n & & & & & & \\
\end{array}
\end{array}$
*
*Repeat the previous two steps. Stop when you performed the previous two steps on the number just before the vertical bar.
$\begin{array}{cc}
\begin{array}{rrrr} \\ \\ i &j & k & l \\ \\ \end{array}
&
\begin{array}{|rrrr|rrrr}
& & ni & nj & nk & & & \\
& mi & mj & mk & ml & nl & & \\
a & b & c & d & e & f & g & h \\
\hline
m & n & o & & & & & \\
\end{array}
\end{array}$
$\begin{array}{cc}
\begin{array}{rrrr} \\ \\ \\ i &j & k & l \\ \\ \end{array}
&
\begin{array}{|rrrr|rrrr}
& & & oi & oj & & & \\
& & ni & nj & nk & ok & & \\
& mi & mj & mk & ml & nl & ol & \\
a & b & c & d & e & f & g & h \\
\hline
m & n & o & p & & & & \\
\end{array}
\end{array}$
$\begin{array}{cc}
\begin{array}{rrrr} \\ \\ \\ \\ i &j & k & l \\ \\ \end{array}
&
\begin{array}{|rrrr|rrrr}
& & & & pi & & & \\
& & & oi & oj & pj & & \\
& & ni & nj & nk & ok & pk & \\
& mi & mj & mk & ml & nl & ol & pl \\
a & b & c & d & e & f & g & h \\
\hline
m & n & o & p & q & & & \\
\end{array}
\end{array}$
*
*Perform the remaining column-wise additions on the subsequent columns (getting the remainder).
$\begin{array}{cc}
\begin{array}{rrrr} \\ \\ \\ \\ i &j & k & l \\ \\ \end{array}
&
\begin{array}{|rrrr|rrrr}
& & & & pi & & & \\
& & & oi & oj & pj & & \\
& & ni & nj & nk & ok & pk & \\
& mi & mj & mk & ml & nl & ol & pl \\
a & b & c & d & e & f & g & h \\
\hline
m & n & o & p & q & r & s & t \\
\end{array}
\end{array}$
*
*The results below the horizontal bar would be interpreted with increasing degree from right to left beginning with degree zero for both the remainder and the result.
EDIT: I went ahead and edited the wiki.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/264739",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
}
|
How is this a property of Pascal's triangle? For all non-negative integers $k$ and $n$,
$$
\dbinom{k}{k} + \dbinom{k+1}{k} + \dbinom{k+2}{k} + \ldots + \dbinom{n}{k} = \dbinom{n +1}{k+1}
$$
How is this a property of Pascal's triangle? I do not see at all how it relates. I'm looking for an explanation, as opposed to some calculation with algebraic manipulations.
|
You're asking, for example, for the sum of all the indicated cells of Pascal's triangle
$$ \begin{matrix} \cdot
\\ \cdot & \cdot
\\ \cdot & \cdot & \cdot
\\ \cdot & \cdot & \cdot & \bullet &
\\ \cdot & \cdot & \cdot & \bullet & \cdot
\\ \cdot & \cdot & \cdot & \bullet & \cdot & \cdot
\\ \cdot & \cdot & \cdot & \bullet & \cdot & \cdot & \cdot
\\ \cdot & \cdot & \cdot & \bullet & \cdot & \cdot & \cdot & \cdot
\\ \cdot & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot
\end{matrix} $$
which is the same thing as the sum of the cells
$$ \begin{matrix} \cdot
\\ \cdot & \cdot
\\ \cdot & \cdot & \cdot
\\ \cdot & \cdot & \cdot & \bullet & \circ
\\ \cdot & \cdot & \cdot & \bullet & \cdot
\\ \cdot & \cdot & \cdot & \bullet & \cdot & \cdot
\\ \cdot & \cdot & \cdot & \bullet & \cdot & \cdot & \cdot
\\ \cdot & \cdot & \cdot & \bullet & \cdot & \cdot & \cdot & \cdot
\\ \cdot & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot
\end{matrix} $$
because I've added in zero. Now, what do you know about the sum of adjacent cells of Pascal's triangle?
Of course, this method is still a calculation with algebraic manipulations, just organized in picture form.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/265146",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
}
|
Prove that $\cot^2{(\pi/7)} + \cot^2{(2\pi/7)} + \cot^2{(3\pi/7)} = 5$ Prove that $\cot^2{(\pi/7)} + \cot^2{(2\pi/7)} + \cot^2{(3\pi/7)} = 5$ .
I am sure this is derived from using roots of unity and Euler's complex number function, but I am very uncomfortable in these areas so some help would be great. It is evident that $(a + b + c)^2 - 2(ab + ac + bc) = a^2 + b^2 + c^2$ .
So, using a polynomial of degree 3 and the coefficients on the $x^2$ and $x$ terms will get where we need to be.
|
Using this,
the roots of $\displaystyle z^3+z^2-3z-1=0\qquad (1)$ are $\displaystyle 2\cos\frac{2\pi}7, 2\cos\frac{4\pi}7, 2\cos\frac{6\pi}7$
If $\displaystyle\cot^2\frac{r\pi}7=u, \cos\frac{2r\pi}7=\frac{1-\tan^2\frac{r\pi}7}{1+\tan^2\frac{r\pi}7}=\frac{\cot^2\frac{r\pi}7-1}{\cot^2\frac{r\pi}7+1}=\frac{u-1}{u+1}$
$\displaystyle\implies\frac{2(u-1)}{u+1}=2\cos\frac{2r\pi}7($ where $r=1,2,3)$ will satisfy $(1)$
$\displaystyle\implies \left(\frac{2(u-1)}{u+1}\right)^3+\left(\frac{2(u-1)}{u+1}\right)^2-3\left(\frac{2(u-1)}{u+1}\right)-1=0$
On simplification, $\displaystyle 7 u^3-35 u^2+21 u-1=0$ whose roots are $\displaystyle\cot^2\frac{r\pi}7($ where $r=1,2,3)$
Now, use Vieta's formulas, to find $\displaystyle \sum \cot^2\frac{r\pi}7=\frac{35}7$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/265229",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 0
}
|
Permutation and Combination with divisibility? How many five digit positive integers that are divisible by 3 can be formed using the digits 0, 1, 2, 3, 4 and 5, without any of the digits getting repeating?
my explanation:
total number of permutations with 0, 1, 2, 3, 4 and 5
to have 5 digits = 6.5.4.3.2=720 (includes numbers with 0 at the beginning)
now let us find the numbers with zero at the beginning = 5.4.3.2=120
so total no of digits = 720 - 120 = 600
but this is not the answer? the correct answer is 216.
can somebody correct me whats wrong with my approach and suggest a better solution.
|
You only substracted the numbers with 0 at the peginning however: the divisibilty rule for three says that for a number to be a multiple of 3 the sum of its digits needs to be a multiple of three. So really it does not matter what order you put the numbers in but the sum of the numbers themselves. There are in total 6 combinations of numbers you can pick. which are
$1+2+3+4+5 = 15$
$0+2+3+4+5 = 14$
$0+1+3+4+5 = 13$
$0+1+2+4+5 = 12$
$0+1+2+3+5 = 11$
$0+1+2+3+4 = 10$
Therefore only the combinations that use 1,2,3,4,5 or 0,1,2,3,4 are divisible by 3. There are 5! of the first one = 120. However we cant pick combinations which start with zero so there are 5!-4!of the second one= 96. Add them to get 216.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/265383",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
}
|
simple congruence system problem. This is excercise 1 in George E. Andrews number theory. page 51. How am i supposed to solve this? Thanks very much in advance. I used the cancellation law to get isolate the x. But I don’t know what to do next.
$5x\equiv 4 \pmod 3$
$7x\equiv 6 \pmod 5$
$9x\equiv 8 \pmod 7$
|
$$5 x \equiv 1 \pmod 3 \implies x \equiv 2 \pmod 3$$
$$7 x \equiv 1 \pmod 5 \implies x \equiv 3 \pmod 5$$
$$9 x \equiv 1 \pmod 7 \implies x \equiv 4 \pmod 7$$
$$x \equiv 2 \pmod 3 \text{ and }x \equiv 3 \pmod 5 \implies x \equiv a \pmod{15}$$
Since $x \equiv 2 \pmod 3$, we have $a \in \{2,5,8,11,14\}$. Since $x$ is also $3 \pmod 5$, we get that $$x \equiv 8 \pmod {15}$$
$$x \equiv 8 \pmod {15} \text{ and }x \equiv 4 \pmod 7 \implies x \equiv b \pmod{105}$$
Since $x \equiv 8 \pmod 15$, we have $b \in \{8,23,38,53,68,83,98\}$. Since $x$ is also $4 \pmod 7$, we get that $$x \equiv 53 \pmod {105}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/266658",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
}
|
How to compute the determinant of a tridiagonal matrix with constant diagonals? How to show that the determinant of the following $(n\times n)$ matrix
$$\begin{pmatrix}
5 & 2 & 0 & 0 & 0 & \cdots & 0 \\
2 & 5 & 2 & 0 & 0 & \cdots & 0 \\
0 & 2 & 5 & 2 & 0 & \cdots & 0 \\
\vdots & \vdots& \vdots& \vdots & \vdots & \vdots & \vdots \\
0 & \cdots & \cdots & 0 & 2 & 5 & 2 \\
0 & \cdots & \cdots & \cdots & \cdots & 2 & 5
\end{pmatrix}$$
is equal to $\frac13(4^{n+1}-1)$?
More generally:
How does one compute the determinant of the following tridiagonal matrix (where the three diagonals are constant)?
$$M_n(a,b,c) = \begin{pmatrix}
a & b & 0 & 0 & 0 & \cdots & 0 \\
c & a & b & 0 & 0 & \cdots & 0 \\
0 & c & a & b & 0 & \cdots & 0 \\
\vdots & \vdots& \vdots& \vdots & \vdots& \vdots & \vdots \\
0 & \cdots & \cdots & 0 & c & a & b \\
0 & \cdots & \cdots & \cdots & \cdots & c & a
\end{pmatrix}$$
Here $a,b,c$ can be taken to be real numbers, or complex numbers.
In other words, the matrix $M_n(a,b,c) = (m_{ij})_{1 \le i,j \le n}$ is such that
$$m_{ij} = \begin{cases}
a & i = j, \\
b & i = j - 1, \\
c & i = j + 1, \\
0 & \text{otherwise.}
\end{cases}$$
There does not seem to be an easy pattern to use induction: the matrix is not a diagonal block matrix of the type $M = \bigl(\begin{smallmatrix} A & C \\ 0 & B \end{smallmatrix}\bigr)$ (where we could use $\det(M) = \det(A) \det(B)$ for the induction step), and there are no lines or columns with only one nonzero entry, so Laplace expansion gets complicated quickly.
Is there a general pattern that one could use? Or is the answer only known on a case-by-case basis? It's possible to compute the determinant by hand for small $n$:
$$\begin{align}
\det(M_1(a,b,c)) & = \begin{vmatrix} a \end{vmatrix} = a \\
\det(M_2(a,b,c)) & = \begin{vmatrix} a & b \\ c & a \end{vmatrix} = a^2 - bc \\
\det(M_3(a,b,c)) & = \begin{vmatrix} a & b & 0 \\ c & a & b \\ 0 & c & a \end{vmatrix} = a^3 - 2abc
\end{align}$$
But there is no readily apparent pattern and the computation becomes very difficult when $n$ gets large.
|
The determinant can also be verified using block determinants.
Define $A^{(k)}$ to be the $k\times k$ matrix of the form$$A^{(k)}:=
\begin{bmatrix}
5 & 2 & 0 &0&0&\cdots & 0\\
2 & 5 & 2 & 0&0&\cdots & 0\\
0 & 2 &5&2&0 & \cdots & 0\\
\vdots & \vdots& \vdots& \vdots& \vdots& \vdots& \vdots\\
0 & \cdots& & 0& 2 &5& 2\\
0 & \cdots & & \cdots&\cdots& 2 &5
\end{bmatrix}.
$$
This can be expressed as
$$A^{(k)}=\begin{bmatrix}A^{(k-1)}&u\\u^T&5\end{bmatrix},$$
where $u=[0,0,0,\ldots,2]^T$ is a $k\times1$ vector.
It follows form the block determinant form that
$$
\det\left(A^{(k)}\right)=\det\left(A^{(k-1)}\right)\det\left(5-u^T{\left(A^{(k-1)}\right)}^{-1}u\right),
$$
which can be simplified dramatically by the structure of $u$ as
$$
\det\left(A^{(k)}\right)=\det\left(A^{(k-1)}\right)\left(5-4{\left(A^{(k-1)}\right)}^{-1}_{n,n}\right).
$$
Furthermore,
$$\left(A^{(k-1)}\right)^{-1}_{n,n}=\frac{(-1)^{n+n}\det\left(A^{(k-2)}\right)}{\det\left(A^{(k-1)}\right)}=\frac{\det\left(A^{(k-2)}\right)}{\det\left(A^{(k-1)}\right)}.$$
The recurrence relation for the determinant can then be written as, upon simplifying,
$$\det\left(A^{(k)}\right)=5\det\left(A^{(k-1)}\right)-4\det\left(A^{(k-2)}\right),$$
from which the required form of the determinant can be verified similarly to the previous answers.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/266998",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "24",
"answer_count": 5,
"answer_id": 1
}
|
$\frac{n}{2} < 1+\frac{1}{2} + \frac{1}{3}+ \ldots +\frac{1}{2^n-1}How to show that for $n\geqslant 2$
$$\frac{n}{2} < 1+\frac{1}{2} + \frac{1}{3}+ \ldots +\frac{1}{2^n-1}<n$$
|
You can group the sum as follows:
$$\sum_{k=1}^{2^n}{\frac{1}{k}}> 1 + \frac{1}{2} + \left(\frac{1}{4} + \frac{1}{4}\right) + \left(\frac{1}{8} + \frac{1}{8}+ \frac{1}{8}+ \frac{1}{8}\right)...$$
So that:
$$\sum_{k=1}^{2^n-1}{\frac{1}{k}} > 1 + \frac{n}{2}-\frac{1}{2^n}>\frac{n}{2}$$
If instead you group them by the lower power of two you easily get the other limit.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/267191",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
Why is $ab+bc+ac = 0$ in some situation? This is originally a Computer Science question, but I ran a equation that is too hard to solve. Here goes.
So the problem is quite simple, given positive integers $a$, $b$, $c$, and calculate $\sqrt{a^2+b^2+c^2}$
But here's the problem, if I calculate any of the squares, I will trigger an integer overflow, where the computer have not enough allocated space to save the result.
So the problem gives me an additional condition: the difference between the minimum of the three and the bigger other two is perfect squares.
That said, if $c$ is the smallest of the three. $b-c$ and $a-c$ will both be perfect squares. And the answer itself is an integer, which means $a^2+b^2+c^2$ is also a perfect square.
So, here's what I got so far.
assume $c$ is the smallest of the three.
let $m=\sqrt{a-c}$, and $n=\sqrt{b-c}$, so $2m^2n^2=2ab-2ac-2bc+2c^2$
and $a^2+b^2+c^2=(a+b-c)^2-2ab+2ac+2bc$
Organize the two, and I am able to get:
$(a+b-c)^2-2(a-c)(b-c)=a^2+b^2+c^2$
And the answer to this problem simply output: $a+b-c$, so $2(a-c)(b-c)$ must be zero under the given condition.
Here's some sample to test it out.
2, 2, 1 - 3
3, 6, 2 - 7
4, 12, 3 - 13
And the following is wrong, since $\sqrt{a^2+b^2+c^2}$ is not an integer:
2, 5, 1 - 6, which is wrong, should be sqrt(30)
7, 6, 6 - 7, which is wrong, should be 11, here's what I think
- it violate the clues set by the problem. The problem state
- the smallest ONE, and the bigger TWO.
P.S. I tagged it under elementary number theory, if you think it's better off to be somewhere else, comment below. Need more information? Comment below. Thanks!
|
In your question you say:
So the problem gives me an additional condition: the difference between the minimum of the three and the bigger other two is perfect squares.
That said, if c is the smallest of the two. $b−c$ and $a−c$ will both be perfect squares. And the answer itself is an integer, which means $a^2+b^2+c^2$ is also a perfect square.
The second paragraph contradicts the first-the first only guarantees that one of $a-c$ and $b-c$ is a square. If the numbers are as you say, you will still get overflow of $(a+b-c)^2$, which is greater than $a^2$
Your derivation is incorrect. $2m^2n^2=2(a-c)^2(b-c)^2=2(a^2-2ac+c^2)(b^2-2bc+c^2)$ which has a term $2a^2b^2$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/269170",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
}
|
$ \frac{1}{3a^2+1}+\frac{1}{3b^2+1}+\frac{1}{3c^2+1}+\frac{1}{3d^2+1} \geq \frac{16}{7}$ Let $a,b,c,d >0$ and $a+b+c+d=2$. Prove this:
$$ \frac{1}{3a^2+1}+\frac{1}{3b^2+1}+\frac{1}{3c^2+1}+\frac{1}{3d^2+1} \geq \frac{16}{7}$$
|
Let $a=\frac{x}{2}$, $b=\frac{y}{2}$, $c=\frac{z}{2}$ and $d=\frac{t}{2}$.
Hence, $x+y+z+t=4$ and we need to prove that $\sum\limits_{cyc}\frac{1}{3x^2+4}\geq\frac{4}{7}$.
Let $f(x)=\frac{1}{3x^2+4}$.
Hence, $f''(x)=\frac{6(9x^2-4)}{(3x^2+4)^2}>0$ for all $1\leq x\leq4$.
Thus, by Vasc's RCF Theorem it's enough to prove our inequality
for $z=y=x$ and $t=4-3x$, where $1\leq x\leq\frac{4}{3}$.
Id est, we need to prove that
$$\frac{3}{3x^2+4}+\frac{1}{3(4-3x)^2+4}\geq\frac{4}{7},$$
which is $(x-1)^2(3x+2)(4-3x)\geq0$.
Done!
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/269223",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
}
|
Algebra question involving fractions How would I perform the indicated operation.
$$\frac{t+2}{t^2+5t+6}+\frac{t-1}{t^2+7t+12}-\frac{2}{t+4}.$$
I simplified it to
$$
\frac{t+2}{(t+3)(t+2)} + \frac{t-1}{(t+4)(t+3)}-\frac{2}{t+4}.
$$
Then I did the lowest common denominator but I still have problems.
According to my book the final answer $\;\dfrac{-3}{(t+3)(t+4)},\;$ but I cannot seem to get it.
|
You did quite well, but note that there's a term $(t+2)$that cancels in your first "reduced"/simplified fraction: $$\frac{t+2}{(t+3)(t+2)} = \frac{1}{t+3}\tag{$t\ne -2$}$$
From the start:
$$\frac{(t+2)}{(t^2+5t+6)}+\frac{(t-1)}{(t^2+7t+12)}-\frac{2}{(t+4)}$$
Factoring denominators gives us:
$$= \frac{(t+2)}{(t+3)(t+2)} + \frac{(t-1)}{(t+3)(t+4)} - \frac{2}{(t+4)} $$
Canceling the term $(t+2)$ from numerator and denomintor in the first fraction:
$$ = \frac{1}{(t+3)} + \frac{(t-1)}{(t+3)(t+4)} - \frac{2}{(t+4)} \tag{$t \ne -2$}$$
The rest is finding the common denomitor $(t+3)(t+4)$ and simplifying:
$$ = \frac{(t+4)+(t-1)-2(t+3)}{(t+3)(t+4)} = \frac{2t - 2t + 4 - 1 - 6}{(t+3)(t+4)} =\frac{-3}{(t+3)(t+4)} $$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/274912",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
}
|
If x,y,z are positive reals, then the minimum value of $x^2+8y^2+27z^2$ where $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1$ is what If $x,y, z$ are positive reals, then the minimum value of $x^2+8y^2+27z^2$ where $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1$ is what?
$108$ , $216$ , $405$ , $1048$
|
x:y:z=3:2:1 then x=6,y=6/3 and z=6/2 ('.' 1^3=1, 2^3=8, 3^3=27 and 1+2+3=6)
substitute in the equation
6^2+8(3^2)+27(2^2)=216
u can exchange the value of x y z but he has asked for minimum value for expression.so taking these value exactly for x y and z would be appropriate.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/275153",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 5
}
|
How to solve $|x-5|=|2x+6|-1$? $|x-5|=|2x+6|-1$.
The answer is $0$ or $-12$, but how would I solve it by algebraically solving it as opposed to sketching a graph?
$|x-5|=|2x+6|-1\\
(|x-5|)^2=(|2x+6|-1)^2\\
...\\
9x^4+204x^3+1188x^2+720x=0?$
|
You have $|x-5|=|2x+6|-1$.
1)If $x\geq5$, $|x-5|=x-5$ and, $|2x+6|=2x+6$, so you have
$x-5=2x+6-1$
$x-5=2x+5$
$x=-10$ (but is not valid)
2)If $-3/2\leq x<5$, $|x-5|=-(x-5)$ and $|2x+6|=2x+6$, then
$-(x-5)=2x+6-1$
$-x+5=2x+5$
$-x=2x$
$x=0$
3)If $x<-3/2$, $|x-5|=-(x-5)$ and $|2x+6|=-(2x+6)$, then
$-(x-5)=-(2x+6)-1$
$-x+5=-2x-6-1$
$-x+5=-2x-7$
$-x+5=-2x-7$
$x=-12$
So, the answer for $x$ is $x=0$ or $x=-12$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/275928",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 4
}
|
What will be the one's digit of the remainder in: $\left|5555^{2222} + 2222^{5555}\right|\div 7=?$ What will be the ones digit of the remainder in: $$\frac{\left|5555^{2222} + 2222^{5555}\right|} {7}$$
|
This is not a solution. I was trying to find out the possible values of $n$ such that $${\underbrace{22\cdots22}_{n\text{ digits }}}^{\underbrace{55\cdots55}_{n\text{ digits }}}+{\underbrace{55\cdots55}_{n\text{ digits }}}^{\underbrace{22\cdots22}_{n\text{ digits }}}$$ is divisible by $7$.
Let $$\underbrace{11\cdots11}_{n\text{ digits }} =\frac{10^n-1}9=M$$
Clearly $M$ is odd.
Now, $$(2M)^{5M}+(5M)^{2M}\equiv (2M)^{5M}+(-2M)^{2M}\pmod 7\equiv (2M)^{5M}+(2M)^{2M}=(2M)^{2M}\{(2M)^{3M}+1\}$$ which will be divisible by $7$
if (i) $7\mid \{(2M)^{3M}+1\}\iff 7\mid \{M^{3M}+1\}$ as $2^3\equiv1$
$\iff 7\mid \{\left(\frac{10^n-1}9\right)^{3M}+1\}\iff 7\mid \{(10^n-1)^{3M}+1\}$ as $9^3=3^6\equiv1\pmod 7$
So, we need $(10^n-1)^{3M}\equiv-1\pmod 7\implies (10^n-1)^3\equiv-1$ as $3M\equiv3\pmod {\phi(7)}$ as $M$ is odd.
Taking Discrete Logarithm wrt a primitive root $3\mod 7,$
$3\cdot ind_3(10^n-1)\equiv 3\pmod 6=6c+3$ for some integer $c$
So, $$ ind_3(10^n-1)=2c+1$$
So, $10^n-1\equiv 3^{2c+1}\pmod 7\equiv 3,6,5\iff 10^n\equiv4,6,7\pmod7$
$10^n\not\equiv7\pmod 7$ as $(10,7)=1$
$10^1\equiv3,10^2\equiv3^3\equiv2,10^3\equiv2\cdot10\equiv6,10^4\equiv2^2=4,10^5\equiv2\cdot6\equiv5,10^6\equiv 6^2\equiv1\pmod 7$
So, $n\equiv3,4\pmod 6$
or if (ii)$7\mid (2M)^{2M}\iff 7\mid M^{2M}\implies 7\mid M\implies 10^n\equiv1\pmod 7\iff 6\mid n$
So, if $n\equiv0,3,4\pmod 6,$ $${\underbrace{22\cdots22}_{n\text{ digits }}}^{\underbrace{55\cdots55}_{n\text{ digits }}}+{\underbrace{55\cdots55}_{n\text{ digits }}}^{\underbrace{22\cdots22}_{n\text{ digits }}}$$ is divisible by $7$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/279333",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 0
}
|
Radical questions algebra Hello everyone how would I simplify the following radicals.
$$3 \sqrt{2a^3b^5} \sqrt{32ab^{2}}$$
I got $$3 \sqrt{64a^4b^7}$$ I know $64$ square root is $8$ and $a^4$ square root is $a^2$
My second question is how would I simplify the following
$$\frac{3}{4}\sqrt{3t^3}$$
I know $ \sqrt[\large4]{3t^3}$ is equal to $(3t)^{\frac{3}{4}}$ so would I multiply by $\frac{1}{4}$
My final question is how would I simplify the following
$$ \sqrt[\large 6]{x^6y^4}$$
Whomever helps Fernando with these questions shall receive his eternal gratitude.
|
$3 \sqrt{2a^3b^5} \sqrt{32ab^{2}}$
$$3 \sqrt{2a^3b^5} \sqrt{32ab^{2}} = 3 \sqrt{64a^4b^7} = 3\sqrt{64a^4}\sqrt{b^7} = 3\cdot 8 a^2 \sqrt{b^7} = 24a^2 \sqrt{b^7} = 24a^2b^{\large\frac72}\quad\quad\quad\tag{1}$$
"$ \;\sqrt[\large4]{3t^3}\;$ is equal to $\;(3t)^{\frac{3}{4}}\;$"
Not quite:
$$\sqrt[\large 4]{3t^3} \;=\; (3\,t^3)^{\large \frac 14} \;= \;3^{\large \frac{1}{4}}\,t^{\large\frac{3}{4}}\tag{2}$$
EDIT: to address comment/question below
If your original expression (to simplify) was $\;\dfrac{3}{\sqrt[\large 4]{3t^3}}\;$ then using the simplification above, we have $$\;\dfrac{3}{\sqrt[\large 4]{3t^3}},\;= \;\frac{3}{3^{\large \frac{1}{4}}\,t^{\large\frac{3}{4}}}\;=\;\frac{3^{\large\frac{4}{4}}\cdot 3^{-\large\frac{1}{4}}}{t^{\large\frac{3}{4}}} \;=\; \frac{3^{\large\frac{3}{4}}}{t^{\large\frac{3}{4}}}\;=\;\frac{(3^3)^{\large\frac{1}{4}}}{(t^3)^{\large\frac{1}{4}}} \;=\;\left(\frac{27}{t^3}\right)^{\large\frac{1}{4}} \;= \; \sqrt[\large 4]{\frac{27}{t^3}}$$
My final question is how would I simplify the following: $\quad \sqrt[\large 6]{x^6y^4}\;$?
$$\sqrt[\large 6]{x^6y^4}\; = \;\left(x^6y^4\right)^{\large \frac{1}{6}}\; =\; x^{\large \frac{6}{6}}y^{\large \frac{4}{6}} \;=\; xy^{\large \frac{2}{3}}\; = \;x\sqrt[\large 3]{y^2}\tag{3}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/280475",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
}
|
How to solve for $x$ in $x^3+8^x-9=0$ How to find real numbers $x$ that are solutions to
$$x^3+8^x-9=0$$
Please help me.
|
Note that
\begin{align}
x^3+8^x-9=0\iff
&
\begin{cases}
x^3-1=2^3-(2^{x})^3\\
x^3-8=1^3-(2^{x})^3\\
\end{cases}
\\
\iff
&
\begin{cases}
(x-1)(x^2+x+1)=[2-(2^{x})][2^2+2(2^{x}) + (2^{x})^2] \\
(x-2)(x^2+2x+4)=[1-(2^{x})][1+(2^{x})+(2^{x})^2] \\
\end{cases}
\\
\end{align}
Then is easy to see that $x=1 \implies x^3+8^x-9=0 $.
We have that $f(x)=x^3+8^x-9\implies f^\prime(x)=3x^2+e^x\cdot\log_e(8)>0$ and then $f(x)$ have only one root. That is, $f(x)=0\implies x=1$.
Then $x=1$ the unique root.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/282227",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
}
|
$\frac{(b+c-a)^2}{(b+c)^2+a^2}+ \frac{(c+a-b)^2}{(c+a)^2+b^2}+ \frac{(a+b-c)^2}{(a+b)^2+c^2} \ge \frac{3}{5}$ Let $a,b,c$ be positive numbers. Prove that $$\dfrac{(b+c-a)^2}{(b+c)^2+a^2}+ \frac{(c+a-b)^2}{(c+a)^2+b^2}+ \frac{(a+b-c)^2}{(a+b)^2+c^2} \ge \frac{3}{5}$$
|
A full expanding gives
$$\sum_{cyc}(3a^6+a^5b+a^5c-a^4b^2-a^4c^2+2a^3b^3+3a^4bc-6a^3b^2c-6a^3c^2b+4a^2b^2c^2)\geq0,$$
which is true by Schur and Muirhead:
$$\sum_{cyc}(3a^6+a^5b+a^5c-a^4b^2-a^4c^2+2a^3b^3+3a^4bc-6a^3b^2c-6a^3c^2b+4a^2b^2c^2)=$$
$$=\sum_{cyc}(a^6-a^3b^3)+2\sum_{cyc}(a^6-a^2b^2c^2)+\sum_{cyc}(a^5b+a^5c-a^4b^2-a^4c^2)+$$
$$+3\sum_{cyc}(a^3b^3-a^3b^2c-a^3c^2b+a^2b^2c^2)+3\sum_{cyc}(a^4bc-a^3b^2c-a^3c^2b+a^2b^2c^2)\geq0.$$
Done!
By the way, the last inequality is true for all reals $a$, $b$ and $c$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/282477",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
}
|
Converting to polar form
Write each of the given numbers in the polar form $re^{i\theta}$.
a.) $\frac{1-i}{3}$
b.) $-8\pi (1+\sqrt 3 i)$
For a, I got:
r = $\frac{\sqrt 2}{3}$ and $e^{i7\pi /2}$ since $\frac{\frac{1}{3}}{\frac{\sqrt 2}{3}} = \frac{\sqrt 2}{2}$ and $\frac{-\frac{1}{3}}{\frac{\sqrt 2}{3}} = -\frac{\sqrt 2}{2}$, but the answer for this is $\frac{\sqrt 2}{3}e^{-i\frac{\pi}{4}}$, how did they get that?
For b, I got:
$-16\pi e^{i\frac{\pi}{3}}$ but the answer for this is $16\pi e^{-i\frac{2\pi}{3}}$.
|
Following two points need to be used here:
(1) The principal value of $\arctan \frac yx$ lies in $[-\frac\pi2,\frac\pi2]$
(2)From this, we need to identify the principal value of the argument of a complex number.
Let $$\frac{1-i}3=re^{i\theta}=r(\cos\theta+ i\sin\theta)$$ where $r>0$
Equating the real & the imaginary parts,
$r\cos\theta=\frac13$ and $r\sin\theta=-\frac13 $
Squaring and adding we get, $r^2=(\frac13)^2+(-\frac13)^2=\frac29$
So, $r=\frac{\sqrt2}3$ and $\cos\theta=\frac{\frac13}{\frac{\sqrt2}3}=\frac1{\sqrt2}$ and $\sin\theta=\frac{-\frac13}{\frac{\sqrt2}3}=-\frac1{\sqrt2}$
so $\tan\theta=-1$ and $\theta$ lies in the 4th quadrant hence $\theta=-\frac\pi4$
Similarly, if we put $-8\pi(1+\sqrt3i)=r(\cos\theta+ i\sin\theta)$ where $r>0$
Applying the same method, $r^2=(-8\pi)^2+(-8\pi\sqrt3)^2=(8\pi)^24\implies r=16\pi$
So, $\cos\theta=\frac{-8\pi}{16\pi}=-\frac12$ and $\sin\theta=\frac{-8\pi\sqrt3}{16\pi}=-\frac{\sqrt3}2$
so, $\tan\theta=\sqrt3$ and $\theta$ lies in the 3th quadrant hence $\theta=-\pi+\frac\pi3=-\frac{2\pi}3$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/282540",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Hyperbola is a pair of straight lines? I'm confused by this question:
If $f(x) = 2x^2 - 6y^2+xy+2x-17y-12=0$ is to represent a pair of
straight lines, one of which has equation $x+2y+3=0$, what must be the
equation of the other line? Verify that $f(x)=0$ does, indeed,
represent a pair of straight lines.
Given the general form of a conic section $Ax^2+By^2+Cxy+Dx+Ey+F=0$ we know that if $C^2 > 4AB$ as here, it's a hyperbola. Therefore I don't get how the equation can represent 2 straight lines. Any clues?
|
Divide the function $f(x, y)=2x^2-6y^2+xy+2x-17y-12$ by given factor $(x+2y+3)$ to get the second linear factor as $(2x-3y-4)$
Hence, the equation of second straight line: $2x-3y-4=0$
For general case, a quadratic equation of two variables $x$ & $y$ is given as
$$ax^2+2hxy+by^2+2gx+2fy+c=0$$
The above equation will represent a pair of straight lines iff the discriminant ($\Delta$) is satisfied as follows
$$\Delta=abc+2fgh-af^2-bg^2-ch^2=0$$
As per your question, we have
$a=2$, $h=\frac{1}{2}$, $b=-6$, $g=1$, $f=\frac{-17}{2}$ & $c=-12$
$$\Delta=(2)(-6)(-12)+2\left(\frac{-17}{2}\right)(1)\left(\frac{1}{2}\right)-(2)\left(\frac{-17}{2}\right)^2-(-6)(1)^2-(-12)\left(\frac{1}{2}\right)^2=0$$
Hence, the given equation:$2x^2-6y^2+xy+2x-17y-12=0$ represents a pair of straight lines.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/283591",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
}
|
Derive the Quadratic Equation Find the Quadratic Equation whose roots are $2+\sqrt3$ and $2-\sqrt3$.
Some basics:
*
*The general form of a Quadratic Equation is $ax^2+bx+c=0$
*In Quadratic Equation, $ax^2+bx+c=0$, if $\alpha$ and $\beta$ are the roots of the given Quadratic Equation, Then,
$$\alpha+\beta=\frac{-b}{a}, \alpha\beta=\frac{c}{a}$$
I am here confused that how we can derive a Quadratic Equations from the given roots
|
This appears to be a standard question from the R.D. Sharma textbook used in Indian schools. A common trick I use for this type of sums is
$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$
Comparing with the given roots, we get
$$x = \frac{4 \pm \sqrt{4 - 4c}}{2}$$
In this step we find the value of $a$ to be $1$, $b$ to be $-4$.
To find $c$, simply multiply the roots and divide by $a$ to get c.
$$c = \frac{(2 + \sqrt 3)(2 - \sqrt 3)}{1}$$
$$ c = \frac{4 - 3}{1}$$
$$c = 1$$
We know that the standard form of a quadratic equation is
$$ax^2 + bx + c = 0$$
Hence substituting $a$, $b$ and $c$ into the standard form yields
$$x^2 - 4x + 1 = 0$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/284338",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
}
|
The last digit of $2^{2006}$ My $13$ year old son was asked this question in a maths challenge. He correctly guessed $4$ on the assumption that the answer was likely to be the last digit of $2^6$. However is there a better explanation I can give him?
|
The last digit of $2^{2006}$ is $2^{2006} (\text{mod }10)$. So let's look at the behavior of powers of $2$ mod $10$.
$2^2 = 4$,
$2^3 = 8$ (nothing interesting so far),
$2^4 = 16 = 6$ (remember we are working mod $10$, we only keep the last digit).
$2^5 = 32 = 2$, Now THIS is interesting. This suggests we divide $2006$ by $5$.
So let's do that: $2006 = (401)(5) + 1$
going back to working mod $10$, we have:
$2^{2006} = 2^{(5)(401)+1} = (2^5)^{401}(2)= 2^{402}$ so we have knocked down the size of our exponent a great deal. Repeating this procedure again leads to:
$2^{402} = 2^{(5)(80) + 2} = (2^5)^{80}(2^2) = 2^{82}$ (still mod $10$).
Another go:
$2^{82} = 2^{(5)(16)+2} = (2^5)^{16}(2^2) = 2^{18}$ (mod $10$),
One last time:
$2^{18} = 2^{(5)(3)+3} = (2^5)^3(2^3) = 2^6$ We can stop now, it's clear that the last digit we are looking for is $4$ (the last digit of $64 = 2^6$). But if we wished (and we were for some reason unwilling to physically compute $2^6$) we could continue for one last step:
$2^6 = (2^5)(2^1) = (2)(2) = 4$ (mod $10$).
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/284902",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "61",
"answer_count": 8,
"answer_id": 3
}
|
Prove $\int_0^1 \frac{t^2-1}{(t^2+1)\log t}dt = 2\log\left( \frac{2\Gamma \left( \frac{5}{4}\right)}{\Gamma\left( \frac{3}{4}\right)}\right)$ I am trying to prove that
$$\int_0^1 \frac{t^2-1}{(t^2+1)\log t}dt = 2\log\left( \frac{2\Gamma \left( \frac{5}{4}\right)}{\Gamma\left( \frac{3}{4}\right)}\right)$$
I know how to deal with integrals involving cyclotomic polynomials and nested logarithms but I have no idea with this one.
|
There have been similar integrals such as this with a factor of $1/\log{t}$ in the integrand. The way I have attacked these is to use the substitution $t=e^{-x}$; here, this produces
$$-\int_0^{\infty} dx \: \frac{e^{-x}}{x} \frac{1-e^{-2 x}}{1+e^{-2 x}} $$
$$ = -\int_0^{\infty} dx \: \frac{1}{x} (e^{-x} - e^{-3 x}) \sum_{k=0}^{\infty} (-1)^k e^{-2 k x} $$
Now, we reverse the order of the sum and integral. This is justified by Abel's theorem, although I will leave the details for another time or another person to fill in:
$$ = -\sum_{k=0}^{\infty} (-1)^k \int_0^{\infty} dx \: \frac{1}{x} (e^{-(2 k+1) x} - e^{-(2 k+3) x}) $$
The integrals have a simple closed form, and the result is now a sum:
$$ = \sum_{k=0}^{\infty} (-1)^{k+1} \log{ \left [ \frac{(2 k+1)}{(2 k+3)} \right ] } $$
You can form a log of a Wallis-type product from this sum:
$$ = \log{\left [ \prod_{k=0}^{\infty} \frac{4 k+3}{4 k+1} \frac{4 k+3}{4 k+5} \right ]} $$
Partial products of the above product may be evaluated:
$$ \prod_{k=0}^{n} \frac{4 k+3}{4 k+1} \frac{4 k+3}{4 k+5} = \frac{9 \Gamma{\left ( \frac{5}{4} \right )} \Gamma{\left ( \frac{9}{4} \right )} \Gamma{\left ( n + \frac{7}{4} \right )}^2}{5 \Gamma{\left ( \frac{7}{4} \right )}^2 \Gamma{\left (n+ \frac{5}{4} \right )} \Gamma{\left ( n + \frac{9}{4} \right )}}$$
You may show that the above expression converges as $n \rightarrow \infty$. A little manipulation produces the stated result.
EDIT
@MikeSpivey points my attention to an equation out of Whittaker & Watson that applies here:
$$ \prod_{k=0}^{\infty} \frac{4 k+3}{4 k+1} \frac{4 k+3}{4 k+5} = \prod_{k=0}^{\infty} \frac{(k+\frac{3}{4})^2}{(k+\frac{1}{4})(k+\frac{5}{4})} = \frac{\Gamma{\left ( \frac{1}{4} \right )} \Gamma{\left ( \frac{5}{4} \right )}}{\Gamma{\left ( \frac{3}{4} \right )^2}} = \frac{2^2 \Gamma{\left ( \frac{5}{4} \right )^2}}{\Gamma{\left ( \frac{3}{4} \right )^2}} $$
The result immediately follows.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/285130",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "33",
"answer_count": 3,
"answer_id": 2
}
|
limit of the sum $\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{2n} $ Prove that : $\displaystyle \lim_{n\to \infty} \frac{1}{n+1}+\frac{1}{n+2}+\frac{1}{n+3}+\cdots+\frac{1}{2n}=\ln 2$
the only thing I could think of is that it can be written like this :
$$ \lim_{n\to \infty} \sum_{k=1}^n \frac{1}{k+n} =\lim_{n\to \infty} \frac{1}{n} \sum_{k=1}^n \frac{1}{\frac{k}{n}+1}=\int_0^1 \frac{1}{x+1} \ \mathrm{d}x=\ln 2$$
is my answer right ? and are there any other method ?(I'm sure there are)
|
Set $$a_n = \frac{1}{n+1}+\frac{1}{n+2}+\frac{1}{n+3}+\cdots+\frac{1}{2n}.$$
We can use the following uniform bound
$$a_n = \sum_{k=1}^n \frac{1}{n+k} \leq\sum_{k=1}^n\frac{1}{n+1} = \frac{n}{n+1} < 1$$
Since each $a_n$ is positive, we have that $0\leq a_n < 1$ for all $n$. For monotonicity we use the calculation
$$\begin{align*}a_{n}-a_{n+1} &= \sum_{k=1}^n\frac{1}{n+k} - \sum_{k=1}^{n+1}\frac{1}{(n+1) + k}\\
&= \frac{1}{n+1} - \frac{1}{2n+1} - \frac{1}{2n+2}\\
&= \frac{(2n+1)(2n+2)- (n+1)(2n+2) - (n+1)(2n+1)}{(n+1)(2n+1)(2n+2)}\\
&= -\frac{(n+1)}{(n+1)(2n+1)(2n+2)}\\
&= -\frac{1}{(2n+1)(2n+2)} < 0\\
\end{align*}$$
We thus have that $\{a_n\}$ is bounded and the statement above shows that $a_n < a_{n+1}$ for any $n$ so that $\{a_n\}$ is also monotonic. Hence the sequence converges.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/285308",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 7,
"answer_id": 5
}
|
How to transform $2^{n-2}\frac{(2n-5)(2n-7)...(3)(1)}{(n-1)(n-2)...(3)(2)(1)}$ into $\frac{1}{n-1}\binom{2n-4}{n-2}$? Just an algebraic step within the well known solution for the number of triangulations of a convex polygon!
|
$$\begin{align*}
\frac{1}{n-1}\binom{2n-4}{n-2}&=\frac{(2n-4)!}{(n-1)!(n-2)!}\\\\
&=\frac{\Big((2n-4)(2n-6)\dots(2)\Big)\Big((2n-5)(2n-7)\dots(3)(1)\Big)}{(n-1)!(n-2)!}\\\\
&=\frac{2^{n-2}(n-2)!\Big((2n-5)(2n-7)\dots(3)(1)\Big)}{(n-1)!(n-2)!}\\\\
&=2^{n-2}\frac{(2n-5)(2n-7)\dots(3)(1)}{(n-1)!}
\end{align*}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/286884",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
}
|
Integration by parts $ \int \sqrt{x-x^2} dx $ I have to integrate $$ \int \sqrt{x-x^2} dx $$
The answer on my textbook is $ \frac 14 \left( \arcsin(\sqrt x)-(1-2x)\sqrt{x-x^2} \right) $ but I want to solve this by myself so can you please give me a clue ? :) Thank you.
|
For the real values of $\sqrt{x-x^2},$ we need $x-x^2\ge 0\implies 0\le x\le 1$
If we put $x=\sin^2t$ where $0\le t\le \frac\pi2, 1-x=\cos^2t\ge 0\implies \sqrt{1-x}=+\cos t$ and $dx=2\sin t\cos tdt$
So, $$\int\sqrt{x-x^2}dx=\int\sqrt{x(1-x)}dx$$
$$=\int\sin t\cos t2\sin t\cos tdt$$
$$=\frac12\int (\sin2t)^2dt \text{, as }\sin2y=2\sin y\cos y$$
$$=\frac14\int (1-\cos 4t)dt \text{, as } \cos2y=1-2\sin^2y$$
$$=\frac14\left(t-\frac{\sin4t}4\right)+c$$
Now, $x=\sin^2t\implies \cos2t=1-2\sin^2t=1-2x$
$\implies \sin2t=+\sqrt{1-(1-2x)^2}=2\sqrt{x-x^2}$ as $\sin2t\ge 0$ as $0\le 2t\le \pi$
So, $\sin4t=2\sin2t\cos2t=2\cdot 2\sqrt{x-x^2}(1-2x)$
So, $$\int\sqrt{x-x^2}dx=\frac14\left(\arcsin \sqrt x-4(1-2x)\sqrt{x-x^2}\right)+c$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/287238",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
}
|
Show that $\frac{1}{x^2+x+1}=\frac{2}{\sqrt3}\sum_{n=0}^{\infty}\sin\bigg(\frac{2\pi(n+1)}{3}\bigg)x^n$ Need to show that:
$$\frac{1}{x^2+x+1}=\frac{2}{\sqrt3}\sum_{n=0}^{\infty}\sin\bigg(\frac{2\pi(n+1)}{3}\bigg)x^n$$
There is a hint given that $x^3-1=(x-1)(x^2+x+1)$ but I don't seem to get how I could use it. If I try to integrate the expression on the left I get:
$$\frac{2}{\sqrt3}\arctan\bigg(\frac{2x+1}{\sqrt3}\bigg)$$
Which has some resemblance to the expression I am trying to come up with.
I know the expansion for $\arctan(x)$ function, but it does not involve sines. So can I arrive at this expression explicitly of should make some useful observations allowing me to incorporate sines into solution?
|
$$let : C_x = \sum_{n=0}^{\infty} \cos \left( \frac{2\pi(n+1)}{3} \right )x^n$$
$$and : S_x = \sum_{n=0}^{\infty} \sin \left( \frac{2\pi(n+1)}{3} \right )x^n$$
$$C_x + iS_x = \large{\sum_{n=0}^{\infty} e^{\frac{2\pi i(n+1)}{3}} x^n = e^{i\frac{2\pi}{3}} \sum_{n=0}^{\infty} (xe^{\frac{2\pi i}{3}} )^n}$$
$$= \frac{e^{i\frac{2\pi}{3}}}{1 - xe^{i\frac{2\pi}{3}}} \Rightarrow S_x = \Im \left( \frac{e^{i\frac{2\pi}{3}}}{1 - xe^{i\frac{2\pi}{3}}} \right ) \overset{\mathbf{do \ it \ by \ yourself}}{=} = \frac{\sqrt{3}}{2(x^2 + x + 1)}$$
$$\Rightarrow \frac{2}{\sqrt{3}}S_x = \frac{1}{x^2+x+1}$$
$$note : \Im = Im $$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/287594",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
Describe all solutions of $Ax=0$. Describe all solutions of $Ax=0$ in parametric vector form, where $A$ is row equivalent to the given matrix.
$$ \begin{bmatrix}
1 & -2 & 3 & -6 & 5 & 0\\
0 & 0 & 0 & 1 & 4&-6\\
0&0&0&0&0&1\\
0&0&0&0&0&0
\end{bmatrix}$$
I know that I should get this into row reduced echelon form, but I'm having trouble doing so. I attempted it below.
$$ \begin{bmatrix}
1 & -2 & 3 & -6 & 5 & 0\\
0 & 0 & 0 & 1 & 4&0\\
0&0&0&0&0&1\\
0&0&0&0&0&0
\end{bmatrix}$$
$$ \begin{bmatrix}
1 & -2 & 3 & 0 & 29 & 0\\
0 & 0 & 0 & 1 & 4&0\\
0&0&0&0&0&1\\
0&0&0&0&0&0
\end{bmatrix}$$
I'm not quite sure where to go from here, also I don't know how I would describe all solutions of $Ax=0$.
|
Think about what $Ax=0$ means for your $A$:
$$\begin{bmatrix}
1 & -2 & 3 & -6 & 5 & 0\\
0 & 0 & 0 & 1 & 4&-6\\
0&0&0&0&0&1\\
0&0&0&0&0&0
\end{bmatrix}
\begin{bmatrix}x_1\\x_2\\x_3\\x_4\\x_5\\x_6\end{bmatrix}=\begin{bmatrix} 0\\0\\0\\0\end{bmatrix},$$
that is,
\begin{align}
x_1-2x_2+3x_3-6x_4+5x_5+0x_6&=0,\\
x_4+4x_5-6x_6&=0,\\
x_6&=0.
\end{align}
Clearly the last equation says $x_6=0$ and then the second equation says $x_4=-4x_5$ while the first equation tells you $x_1=2x_2-3x_3+6x_4-5x_5 \implies x_1=2x_2-3x_3-29x_5$.
Putting this in vector form, you can say
$$
x=\begin{bmatrix}x_1\\x_2\\x_3\\x_4\\x_5\\x_6\end{bmatrix}
=\begin{bmatrix}2x_2-3x_3-29x_5\\ x_2\\x_3\\-4x_5\\x_5\\0\end{bmatrix},$$
where $x_2,x_3,x_5$ are viewed as parameters (sometimes renamed $a$, $b$, $c$).
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/288199",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
}
|
Bell Numbers: How to put EGF $e^{e^x-1}$ into a series? I'm working on exponential generating functions, especially on the EGF for the Bell numbers $B_n$.
I found on the internet the EGF $f(x)=e^{e^x-1}$ for Bell numbers. Now I tried to use this EGF to compute $B_3$ (should be 15). I know that I have to put the EGF into a series and have a look at the coefficients. Using $e^{f(x)}=1+f(x)+\frac{f(x)^2}{2!}+\frac{f(x)^3}{3!}+\ldots$ I get
\begin{eqnarray*}
e^{e^x-1}&=&1+(e^x-1)+\frac{(e^x-1)^2}{2!}+\frac{(e^x-1)^3}{3!}\\
&=&1+e^x-1+\frac{e^{x^2}-2e^x+1}{2!}+\frac{e^{x^3}-3e^{x^2}+3e^x-1}{3!}\\
&=&e^x+\frac{e^{x^2}}{2!}-e^x+\frac{1}{2!}+\frac{e^{x^3}}{3!}-\frac{e^{x^2}}{2!}+\frac{e^{x}}{2!}-\frac{1}{3!}\\
&=&\frac{e^{x}}{2!}+\frac{e^{x^3}}{3!}-\frac{1}{2!}+\frac{1}{3!}\\
&=&\frac{1}{2!}e^x+\frac{1}{3!}e^{x^3}-\frac{1}{3}\\
&=&\frac{1}{2!}\left( 1+x^2+\frac{x^4}{2!}+\frac{x^8}{3!} \right)+\frac{1}{3!}\left( 1+x^3+\frac{x^6}{2!}+\frac{x^9}{3!}\right)\\
&=&\frac{1}{2}+\frac{x^2}{2}+\frac{x^4}{4}+\frac{x^8}{24}+\frac{1}{6}+\frac{x^3}{3!}+\frac{x^6}{12}+\frac{x^9}{36}\\
&=&\frac{1}{2}+\frac{x^2}{2}+\frac{x^3}{6}+\ldots
\end{eqnarray*}
I think I can stop here, because the coefficient in front of $\frac{x^3}{3!}$ is not $15$.
Perhaps someone can help me out and give a hint to find my mistake?
|
\begin{eqnarray*}
e^{e^x-1}&=&1+(e^x-1)+\frac{(e^x-1)^2}{2!}+\frac{(e^x-1)^3}{3!}\\
&=&1+e^x-1+\frac{e^{2x}-2e^x+1}{2!}+\frac{e^{3x}-3e^{2x}+3e^x-1}{3!}\\
&=&e^x+\frac{e^{2x}}{2!}-e^x+\frac{1}{2!}+\frac{e^{3x}}{3!}-\frac{e^{2x}}{2!}+\frac{e^{x}}{2!}-\frac{1}{3!}\\
&=&\frac{e^{x}}{2!}+\frac{e^{3x}}{3!}-\frac{1}{2!}+\frac{1}{3!}\\
&=&\frac{1}{2!}e^x+\frac{1}{3!}e^{3x}-\frac{1}{3}\\
&=&\frac{1}{2!}\left( 1+x+\frac{x^2}{2!}+\frac{x^3}{3!} \right)+\frac{1}{3!}\left( 1+3x+\frac{9x^2}{2!}+\frac{27x^3}{3!}\right)-\frac{1}{3}\\
&=&\frac{1}{2}+\frac{x}{2}+\frac{x^2}{4}+\frac{x^3}{24}+\frac{1}{6}+x+\frac{3x^2}{4}+\frac{9x^3}{12}-\frac{1}{3}\\
&=&\frac{2}{3}+\frac{3x}{2}+\frac{5x^2}{4}+\frac{9x^3}{12}+\ldots
\end{eqnarray*}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/289097",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
}
|
Prove that $1^3 + 2^3 + ... + n^3 = (1+ 2 + ... + n)^2$ This is what I've been able to do:
Base case: $n = 1$
$L.H.S: 1^3 = 1$
$R.H.S: (1)^2 = 1$
Therefore it's true for $n = 1$.
I.H.: Assume that, for some $k \in \Bbb N$, $1^3 + 2^3 + ... + k^3 = (1 + 2 +...+ k)^2$.
Want to show that $1^3 + 2^3 + ... + (k+1)^3 = (1 + 2 +...+ (k+1))^2$
$1^3 + 2^3 + ... + (k+1)^3$
$ = 1^3 + 2^3 + ... + k^3 + (k+1)^3$
$ = (1+2+...+k)^2 + (k+1)^3$ by I.H.
Annnnd I'm stuck. Not sure how to proceed from here on.
|
IMHO, this fact is a coincidence; a better approach is to prove the closed-form formula for both. As we know
$$ 1 + 2 + \cdots + k = \frac{k(k+1)}{2} $$
the corresponding claim to prove is
$$ 1^3 + 2^3 + \cdots + k^3 = \frac{k^2(k+1)^2}{4} $$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/294213",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 3,
"answer_id": 2
}
|
Find the spanning set of the range of the linear transformation $T(x)=Ax$. Let
$$
A=
\begin{bmatrix}
-4 & -4 & 12 & 0 \\
-4 & -4 & 12 & 0 \\
4 & -2 & 0 &-6 \\
1 &-4 &7 &-5 \\
\end{bmatrix}
$$
Find the spanning set of the range of the linear transformation $T(x)=Ax$.
I have found the row reduced echelon form of A.
$$
RREF(A)=
\begin{bmatrix}
1 & 0 & -1 & -1 \\
0 & 1 & -2 & 1 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
\end{bmatrix}
$$
I don't know what to do with it after.
|
The range of $T$ is the column space of $A$. So the columns of $A$ already form a spanning set. If you want to find a linearly independent spanning set, you should find a column echelon form of $A$ instead of a row echelon form. I found that $\{(-4-4,4,1)^T,\,(0,0,-6,-5)^T\}$ is an answer, but depending on the column operations you perform, you may get a different answer.
Edit: for a starter,
$$
\begin{bmatrix}
-4 & -4 & 12 & 0 \\
-4 & -4 & 12 & 0 \\
4 & -2 & 0 &-6 \\
1 &-4 &7 &-5 \\
\end{bmatrix}
\stackrel{C_2-C_1,\, C_3+3C_1}{\longrightarrow}
\begin{bmatrix}
-4 & 0 & 0 & 0 \\
-4 & 0 & 0 & 0 \\
4 & -6 & 12 &-6 \\
1 &-5 &10 &-5 \\
\end{bmatrix}
\,\longrightarrow\cdots
\begin{bmatrix}
-4 & 0 & 0 & 0 \\
-4 & 0 & 0 & 0 \\
4 & -6 & 0 & 0 \\
1 &-5 & 0 & 0 \\
\end{bmatrix}
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/294344",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
}
|
Problems regarding $\{x_n \}$ defined by $x_1=1$; $x_n$ is the smallest distinct natural number such that $x_1+...+x_n$ is divisible by $n$. Let me denote a sequence of distinct natural numbers by $x_n$ whose terms are determined as follows:
$x_1$ is $1$ and $x_2$ is the smallest distinct natural number $n$ such that $x_1+x_2$ is divisible by $2$, which is $3$. Similarly $x_n$ is the smallest distinct natural number such that $x_1+...+x_n$ is divisible by $n$. Thus $x_1=1$, $x_2=3$, $x_3=2$.
Now the problem is to show that
(i) $x_n$ is surjective
(ii) $x_{(x_n)}$ is $n$, for example $x_2=3$ and $x_3=2$
|
As other posters have noticed, this is closely related to the golden ratio $\varphi = \frac{1+\sqrt{5}}{2}$.
@ Ross Millikan your answer is a great place to start.
$\frac{1}{\varphi}+\frac{1}{\varphi ^2}=1$, so we can partition the natural numbers greater than 1 into numbers of the forms $\lceil {n \varphi} \rceil, \lceil {n \varphi ^2} \rceil$. (Add 1 to the beatty sequences)
Define $a_1=1, a_{\lceil {n \varphi} \rceil}=\lceil {n \varphi ^2} \rceil, a_{\lceil {n \varphi ^2} \rceil}=\lceil {n \varphi} \rceil$. Clearly $a_{a_n}=n$.
That's where the next crucial observation that $\sum\limits_{i=1}^{n}{a_i}=n \lceil{\frac{n}{\varphi}} \rceil$ comes in.
We proceed by induction on $n$ to show that $\sum\limits_{i=1}^{n}{a_i}=n \lceil{\frac{n}{\varphi}} \rceil$ and that $a_n$ is the smallest integer satisfying the divisiblity relation $n \mid \sum\limits_{i=1}^{n}{a_i}$.
When $n=1$, we have $1=1 \lceil{\frac{1}{\varphi}} \rceil$, which is true.
When $n=2$, we have $1+3=2 \lceil{\frac{2}{\varphi}} \rceil$, which is true.
Suppose it holds for $n=k \geq 2$. Then $\sum\limits_{i=1}^{k}{a_i}=k \lceil{\frac{k}{\varphi}} \rceil$. Consider 2 cases:
Case 1: $k+1=\lceil{m \varphi} \rceil$ for some $m \in \mathbb{Z}^+$.
Then $a_{k+1}=\lceil{m \varphi ^2} \rceil=m+\lceil{m \varphi} \rceil=m+k+1$.
Now $k+1=m \varphi +1- \{m \varphi \}$ so $m-1<m-\frac{\{m \varphi \}}{\varphi}=\frac{k}{\varphi}<m<m+\frac{1-\{m \varphi \}}{\varphi}=\frac{k+1}{\varphi}<m+1$. Thus
$\lceil{\frac{k}{\varphi}} \rceil=m, \lceil{\frac{k+1}{\varphi}} \rceil=m+1$.
$\sum\limits_{i=1}^{k+1}{a_i}=k \lceil{\frac{k}{\varphi}} \rceil+a_{k+1}=km+(m+k+1)=(k+1)(m+1)=(k+1) \lceil{\frac{k+1}{\varphi}} \rceil$.
Minimality: Note that $a_{k+1}=m+k+1<2(k+1)$, and because we have a partition, $a_{k+1} \not =a_i \, \forall i$, so it suffices to show that $a_i=m$ for some $i$ with $1 \leq i \leq k$, or equivalently, $a_m \leq k$. If $m=\lceil{b \varphi ^2 } \rceil$ for some $b \in \mathbb{Z}^+$, we are done since $a_m<m \leq k$. Otherwise $m=\lceil{b \varphi} \rceil$ for some $b \in \mathbb{Z}^+$, so $a_m=\lceil{b \varphi ^2 } \rceil \leq \lceil{\lceil b \varphi \rceil \varphi } \rceil=\lceil{m \varphi } \rceil=k+1$. Since $k+1=\lceil{m \varphi} \rceil \not =\lceil{b \varphi ^2 } \rceil = a_m$ (Partition), we now have the desired inequality $a_m \leq k$.
Case 2: $k+1=\lceil{m \varphi ^2} \rceil=m+\lceil{m \varphi} \rceil$ for some $m \in \mathbb{Z}^+$.
Then $a_{k+1}=\lceil{m \varphi} \rceil=k+1-m$.
Now $k+1=m \varphi ^2 +1-\{m \varphi ^2 \}$.
As we have a partition, $k+1$ cannot be written in the form $\lceil{i \varphi} \rceil$.
Thus $\exists c \in \mathbb{Z}^+$ such that $\lceil{c \varphi} \rceil<k+1<\lceil{(c+1) \varphi} \rceil$. (Since $k>1$.)
This gives $\lceil{c \varphi} \rceil=k, \lceil{(c+1) \varphi} \rceil=k+2$.
Thus: $k=c \varphi +1-\{c \varphi \}, k+2=(c+1) \varphi +1-\{(c+1) \varphi \}$, so
\begin{align}
c<c+\frac{1-\{c \varphi \}}{\varphi}=\frac{k}{\varphi}=m \varphi -\frac{\{m \varphi ^2 \}}{\varphi}<m \varphi & <m \varphi +\frac{1-\{m \varphi ^2 \}}{\varphi} \\
& =\frac{k+1}{\varphi} \\
& =c+1-\frac{\{(c+1) \varphi \}}{\varphi} \\
& <c+1
\end{align}
We get $\lceil \frac{k}{\varphi} \rceil =\lceil m \varphi \rceil =\lceil \frac{k+1}{\varphi} \rceil =c+1$, so $k+1=m+c+1, a_{k+1}=c+1$.
$\sum\limits_{i=1}^{k+1}{a_i}=k \lceil{\frac{k}{\varphi}} \rceil +a_{k+1}=k(c+1)+(c+1)=(k+1)(c+1)=(k+1) \lceil{\frac{k+1}{\varphi}} \rceil$
Minimality: Also note that $a_{k+1}=k+1-m<k+1$, so $a_{k+1}$ is the smallest integer satisfying the divisibility relation, and because we have a partition, $a_{k+1} \not =a_i \, \forall i$.
We are thus done by induction. Now, since $x_n$ is unique and $a_n$ satisfies the given conditions, $x_n=a_n \, \forall n \in \mathbb{Z}^+$ and we are done. Both i) $x_n$ is surjective and ii) $x_{x_n}=n$ follow trivially.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/295042",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "42",
"answer_count": 6,
"answer_id": 3
}
|
More on the generalized integral Refer to my previous topic: Definite integral, quotient of logarithm and polynomial: $I(\lambda)=\int_0^{\infty}\frac{\ln ^2x}{x^2+\lambda x+\lambda ^2}\text{d}x$
I think we get this :
$$\frac{\sin \theta}{1-2\cos \theta x+x^2}=\sum_{k=1}^{\infty}\sin (k\theta )x^{k-1}$$
Then $$\int_0^1\frac{\ln ^2x}{1-2\cos \theta x+x^2}\text{d}x=\frac{2}{\sin \theta}\sum_{k=1}^{\infty}\frac{\sin (k\theta)}{k^3}=\frac{2}{\sin \theta}\left(\frac{\pi ^2\theta}{6}-\frac{\pi \theta ^2}{4}+\frac{\theta ^3}{12}\right)$$
Moreover $$\begin{align}
& I\left( \lambda ,\theta \right)=\int_{0}^{\infty }{\frac{{{\ln }^{2}}x}{{{x}^{2}}-2\lambda x\cos \theta +{{\lambda }^{2}}}\text{d}x}=\int_{0}^{\infty }{\frac{{{\ln }^{2}}x}{{{\lambda }^{2}}{{x}^{2}}-2\lambda x\cos \theta +1}\text{d}x} \\
& \ \ \ \ \ \ \ \ \ \ \ =\frac{1}{\lambda }\int_{0}^{\infty }{\frac{{{\left( \ln x-\ln \lambda \right)}^{2}}}{{{x}^{2}}-2\cos \theta x+1}\text{d}x}=\frac{{{\ln }^{2}}\lambda }{\lambda }\int_{0}^{\infty }{\frac{1}{{{x}^{2}}-2\cos \theta x+1}\text{d}x}+\frac{1}{\lambda }\int_{0}^{\infty }{\frac{{{\ln }^{2}}x}{{{x}^{2}}-2\cos \theta x+1}\text{d}x} \\
& \ \ \ \ \ \ \ \ \ \ \ =\frac{{{\ln }^{2}}\lambda }{\lambda }\cdot \frac{\pi -\theta }{\sin \theta }+\frac{4}{\lambda \sin \theta }\left( \frac{{{\pi }^{2}}\theta }{6}-\frac{\pi {{\theta }^{2}}}{4}+\frac{{{\theta }^{3}}}{12} \right) \\
\end{align}$$
Can anybody verify my result? Or perhaps show more method? :)
|
Some food for thought that hopefully one of us will pursue in more detail soon.
The function $(1-2 x \cos{\theta} + x^2)^{-1}$ is a generating function for the Chebyshev polynomials of the second kind $U_n(\cos{\theta})$:
$$(1-2 x \cos{\theta} + x^2)^{-1} = \sum_{n=0}^{\infty} U_n(\cos{\theta}) x^n$$
So your integral is
$$\int_0^1 dx \: \frac{\ln ^2x}{1-2\cos \theta x+x^2} = \sum_{n=0}^{\infty} U_n(\cos{\theta}) \int_0^1 dx \: x^n \, \log^2{x} $$
Now, I found this very interesting: It turns out that
$$\int_0^1 dx \: x^n \, \log^2{x} = \frac{2}{(n+1)^3} $$
The integral becomes
$$\int_0^1 dx \: \frac{\ln ^2x}{1-2\cos \theta x+x^2} = 2\sum_{n=0}^{\infty} \frac{U_n(\cos{\theta})}{(n+1)^3}$$
EDIT
I will derive the result for the integral above. Integrate by parts, as I indicated:
$$\begin{align}\int_0^1 dx \: x^n \, \log^2{x} &= \underbrace{[x^{n+1} \log{x} (\log{x} - 1)]_0^1}_{0} - \int_0^1 dx \: x (\log{x} - 1) \frac{d}{dx} [x^n \log{x}]\\ &= -\int_0^1 dx \: x^n (\log{x} - 1) (1+n \log{x})\\ &= -n \int_0^1 dx \: x^n \, \log^2{x} + (n-1) \int_0^1 dx \: x^n \, \log{x} + \int_0^1 dx \: x^n \\\end{align}$$
so that
$$\begin{align} (n+1)\int_0^1 dx \: x^n \, \log^2{x} &= \frac{1}{n+1} - \frac{n-1}{(n+1)^2} \\ &= \frac{2}{(n+1)^2} \\ \end{align}$$
The result follows from this.
EDIT$_2$
Well, in my hubris, I did not catch the defining relation of $U_n$:
$$U_n(\cos{\theta}) = \frac{\sin{[(n+1) \theta]}}{\sin{\theta}}$$
This reproduces the sum you evaluated above.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/295377",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
Simplifying $\frac{1}{x} + \frac{5+x}{x+1} - \frac{7x^2 + 3}{(x+2)^2}$ I'm having trouble simplifying this expression:
$$\frac{1}{x} + \frac{5+x}{(x+1)} - \frac{7x^2 + 3}{(x+2)^2}$$
Would you first do the addition or subtraction?
What's the steps to solve this?
The final answer is
$$\frac{-6x^4 + 3x^3 + 26x^2 + 25x + 4}{x^4 + 5x^3 + 8x^2 + 4x}.$$
Thanks.
|
Or, if you set $I=\frac{1}x+\frac{x+5}{x+1}-\frac{7x^2+3}{(x+2)^2}$, then by multiplying $I$ by $x(x+1)(x+2)^2$, you get:
$$x(x+1)(x+2)^2\times I=x(x+1)(x+2)^2\left(\frac{1}x+\frac{x+5}{x+1}-\frac{7x^2+3}{(x+2)^2}\right)\\=(x+1)(x+2)^2+(x+5)x(x+2)^2-(7x^2+3)x(x+1)\\=3x^3+26x^2+25x+4-6x^4$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/297013",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
Solving non-linear congruence $x^2+2x+2\equiv{0}\mod(5)$, $7x\equiv{3}\mod(11)$
My attempt:
$x^2+2x+2\equiv{0}\mod(5)$
$(x+1)^2\equiv-1\mod(5)$, we have $x+1\equiv-1\mod(5)$
since $5$ and $11$ are coprime. We have a solution in $\mathbb{Z}_{11}$
With $[3]$ represent $3$, $[13]$ works for $7x\equiv3\mod11$
so $[3]$ is the solution in $\mathbb{Z}_{55}$
General solution is $x=-2+k55$.
But the answer is wrong..
|
$\rm mod\ 5\!:\ 0 \equiv x^2+2x+2 \equiv x^2-3x+2 \equiv (x-1)(x-2)\:\Rightarrow\: x\equiv \color{#C00}1,\,\color{#0A0}2.$
$\rm mod\ 11\!:\ 7x\equiv 3\equiv 14\:\Rightarrow\:x\equiv 2.\ $ Solving these congruences yields
*
*$\rm\ \ x\equiv 2\ mod\ 11,\ x\equiv \color{#0A0}2\ mod\ 5\,\Rightarrow \: x\equiv 2\ mod\ 55.\ $
*$\rm\ \ x\equiv 2\ mod\ 11,\ x\equiv\color{#C00} 1\ mod\ 5\!:\, 1 \equiv x\equiv 2+11n\equiv 2+n\Rightarrow n\equiv -1,\,$ thus
$\rm\qquad\qquad\qquad\qquad\qquad\qquad\qquad\ \ x = 2 + 11(-1 + 5k)\equiv -9\equiv 46\ mod\ 55.$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/300940",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
Permutation question. $qpq^{-1}$, $q,p,r,s \in S_{8}$. Let $p,r,s,q \in S_{8}$ be the permutation given by the following products of cycles:
$$p=(1,4,3,8,2)(1,2)(1,5)$$
$$q=(1,2,3)(4,5,6,8)$$
$$r=(1,2,3,8,7,4,3)(5,6)$$
$$s=(1,3,4)(2,3,5,7)(1,8,4,6)$$
Compute $qpq^{-1}$ and $r^{-2}sr^{2}.$
thanks for your help.
I want to write the following permutations like :
$p=\begin{pmatrix}
1 & 2&3&4&5&6&7&8\\4&1&8&3&?&?&?&2\end{pmatrix}$
$q=\begin{pmatrix}
1 & 2&3&4&5&6&7&8\\2&3&1&5&6&8&7&4\end{pmatrix}$
$r=\begin{pmatrix}
1 & 2&3&4&5&6&7&8\\ 2&3&8&3&6&5&4&7\end{pmatrix}$
$s=\begin{pmatrix}
1 & 2&3&4&5&6&7&8\\ 3&?&4&1&?&?&?&?\end{pmatrix}$
can you help me plese to fill the $?$ mark. Is there another method to compute $qpq^{-1}$?
thanks:)
|
If you write all right then
$$
p= \begin{pmatrix}
1 & 2&3&4&5&6&7&8\\4&1&8&3&5&6&7&2
\end{pmatrix}
\begin{pmatrix}
1 & 2&3&4&5&6&7&8\\2&1&3&4&5&6&7&8
\end{pmatrix}
\begin{pmatrix}
1 & 2&3&4&5&6&7&8\\5&2&3&4&1&6&7&8
\end{pmatrix}
$$
and so on (sorry, I write badly the formula). So you have first to calculate $p,q,r,s$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/307512",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
}
|
Absolute value inequality - Please guide further
Prove that if the numbers $x$, $y$ are of one sign, then $\left|\frac{x+y}{2}-\sqrt{xy}\right|+\left|\frac{x+y}{2}+\sqrt{xy}\right|=|x|+|y|$.
Expanding the LHS,
$$\left|\frac{x+y}{2}-\sqrt{xy}\right|= \left|\frac{x+y -2\sqrt{xy}}{2}\right| = \frac{(\sqrt{x}-\sqrt{y})^2}{2}$$
and similarly
$$\left|\frac{x+y}{2}+\sqrt{xy}\right| = \left|\frac{ (\sqrt{x}+\sqrt{y})^2}{2}\right|$$
|
Hint: It may be clearer if you write $x=(-1)^t2a^2$, $y=(-1)^t2b^2$ for some $a,b>0$ and $t\in\{0,1\}$.
Mouse over the gray box for more details if you get stuck.
You now want to show that $$|(-1)^t(a^2+b^2)-2ab|+|(-1)^t(a^2+b^2)+2ab|=2a^2+2b^2.$$ Shuffling the minus signs, we get $$|a^2+b^2+(-1)^{t+1}2ab|+|a^2+b^2+(-1)^t2ab|=2a^2+2b^2.$$ Whether or not $t=0$ or $t=1$, one of the two terms on the left is equal to $$|a^2+b^2-2ab|=|(a-b)^2|=(a-b)^2=a^2+b^2-2ab,$$ and the other term is equal to $$|a^2+b^2+2ab|=|(a+b)^2|=(a+b)^2=a^2+b^2+2ab.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/307836",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
Verify that: $2\cot{x}/\tan{2}x = \csc^2x-2$ Verify the following:
$$\frac{2\cot{x}}{\tan{2}x} = \csc^2x-2\;.$$
|
Using the same technique as presented here (where it explains why we want to do things this way) we put $z = e^{ix}$ to obtain
$$ \cot x = i \frac{z + \frac{1}{z}}{z - \frac{1}{z}} =
i \frac{z^2+1}{z^2-1}$$
and
$$ \tan 2x = \frac{1}{i} \frac{z^2 - \frac{1}{z^2}}{z^2 + \frac{1}{z^2}} =
\frac{1}{i} \frac{z^4 - 1}{z^4 + 1}.$$
and
$$ \csc^2 x - 2 = \left( \frac{2i}{z - \frac{1}{z} }\right)^2 -2 =
-4 \left( \frac{1}{z - \frac{1}{z}}{} \right)^2 -2 =
-4 \left(\frac{z}{z^2-1} \right)^2 - 2 \\
= \frac{-4z^2}{(z^2-1)^2} -2
= -2 \frac{z^4+1}{(z^2-1)^2}$$
But now $$ \frac{2\cot x}{\tan 2x} =
\frac{ 2i \frac{z^2+1}{z^2-1}}{\frac{1}{i} \frac{z^4 - 1}{z^4 + 1}} =
-2 \frac{(z^2+1)(z^4+1)}{(z^4-1)(z^2-1)} =
-2 \frac{z^4+1}{(z^2-1)(z^2-1)}$$
and we are done.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/308160",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
}
|
integrating $\oint_C \frac{3z^3 + 2}{(z-1)(z^2 + 9)}dz$ on $|z|=4$ I am doing $\oint_C \dfrac{3z^3 + 2}{(z-1)(z^2 + 9)}dz$ on $|z|=4$ and I find that there are poles within the contour at $z = 1$ and at $z = 3i$, both simple poles. I find that the integral $I = 2\pi i \,\text{Res}(1) + 2 \pi i \,\text{Res}(3i)$, or $I = \pi i + 2 \pi i \,\,\,\text{Res}(3i)$, with only that second residue left to find. Is this correct, or should I include $\text{Res}(-3i)$ as well?
|
Let $\,C_1,\,C_2,\,C_3\,$ be little circles (say, of radius $\,0.1\,$) around each of the poles $\,1,\,-3i,\,3i\,$ resp., of the function, so by Cauchy's Integral Theorem:
$$\int\limits_{|z|=4}\frac{3z^3+2}{(z-1)(z^2+9)}dz=\int\limits_{C_1}\frac{\frac{3z^3+2}{(z^2+9)}}{z-1}dz+\int\limits_{C_2}\frac{\frac{3z^3+2}{(z-1)(z-3i)}}{z+3i}dz+\int\limits_{C_3}\frac{\frac{3z^3+2}{(z-1)(z+3i)}}{z-3i}dz=$$
$$=2\pi i\left(\left.\frac{3z^3+2}{(z^2+9)}\right|_{z=1}+\left.\frac{3z^3+2}{(z-1)(z-3i)}\right|_{z=-3i}+\left.\frac{3z^3+2}{(z-1)(z+3i)}\right|_{z=3i}\right)=$$
$$=2\pi i\left(\frac{1}{2}+\frac{2+81i}{-18+6i}+\frac{2-81i}{-18-6i}\right)=$$
$$=2\pi i\left(\frac{1}{2}-\frac{1}{6}\left(-5\right)\right)=\frac{8\pi i}{3}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/308721",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
}
|
Limit of a sequence with indeterminate form Let $\displaystyle u_n =\frac{n}{2}-\sum_{k=1}^n\frac{n^2}{(n+k)^2}$. The question is: Find the limit of the sequence $(u_n)$.
The problem is if we write $\displaystyle u_n=n\left(\frac{1}{2}-\frac{1}{n}\sum_{k=1}^n\frac{1}{(1+\frac{k}{n})^2}\right)$ and we use the fact that the limit of Riemann sum $\displaystyle \frac{1}{n}\sum_{k=1}^n\frac{1}{(1+\frac{k}{n})^2}$ is $\displaystyle \int_0^1 \frac{dx}{(1+x)^2}=\frac{1}{2}$ we find the indeterminate form $\infty\times 0$. How can we avoid this problem? Thanks for help.
|
$$\dfrac1{(1+k/n)^2} = \left( 1 + \dfrac{k}n \right)^{-2} = 1 + \dfrac{(-2)}{1!} \dfrac{k}n + \dfrac{(-2)(-3)}{2!} \dfrac{k^2}{n^2} + \dfrac{(-2)(-3)(-4)}{3!} \dfrac{k^3}{n^3} + \cdots $$
Now recall that
$$\sum_{k=1}^n \left(\dfrac{k}n \right)^m = \dfrac{n}{m+1} + \dfrac12 + \mathcal{O}(1/n)$$
Hence,
\begin{align}
\sum_{k=1}^n \dfrac1{(1+k/n)^2} & = n + \dfrac{(-2)}{1!} \left(\dfrac{n}2 + \dfrac12 + \mathcal{O}(1/n)\right)\\
& + \dfrac{(-2)(-3)}{2!} \left(\dfrac{n}3 + \dfrac12 + \mathcal{O}(1/n)\right)\\
& + \dfrac{(-2)(-3)(-4)}{3!} \left(\dfrac{n}4 + \dfrac12 + \mathcal{O}(1/n)\right)\\
& + \dfrac{(-2)(-3)(-4)(-5)}{4!} \left(\dfrac{n}5 + \dfrac12 + \mathcal{O}(1/n)\right)\\
& + \cdots
\end{align}
The leading order term is $n-n+n-n+n-n + \cdots$, which when regularized gives us $\dfrac{n}2$. The next term is
$$\dfrac12 \left(-2+3-4+5-6 \pm \cdots\right)$$ which when regularized gives us $$\dfrac{-1+\eta(-1)}2 = \dfrac{-1+\dfrac14}2 = - \dfrac38$$
Hence,
$$\sum_{k=1}^n \left(1+\dfrac{k}n\right)^{-2} = \dfrac{n}2 - \dfrac38 + \mathcal{O}(1/n)$$
Hence, the limit is $\dfrac38$.
The argument, though it appears to be non-rigorous, can be made rigorous with careful regularization.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/309939",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 2
}
|
If $a,b \in \mathbb R$ satisfy $a^2+2ab+2b^2=7,$ then find the largest possible value of $|a-b|$ I came across the following problem that says:
If $a,b \in \mathbb R$ satisfy $a^2+2ab+2b^2=7,$ then the largest possible value of $|a-b|$ is which of the following?
$(1)\sqrt 7$, $(2)\sqrt{7/2}$ , $(3)\sqrt {35}$ $(4)7$
My Attempt: We notice, $(a-b)^2=7-(4ab+b^2)$ and hence $|a-b|=\sqrt {7-(4ab+b^2)}$ and so $|a-b|$ will be maximum whenever $(4ab+b^2)$ will be minimum. But now I am not sure how to progress further hereon.Can someone point me in the right direction? Thanks in advance for your time.
|
Here's a trick that doesn't involve calculus:
You have seen that
$$(a - b)^2 + 4ab + b^2 = 7$$
Now, adding this equation to S times the original equation, you get:
$$(a - b)^2 + Sa^2 + (4 + 2S)ab + (1 + 2S)b^2 = 7 + 7S$$
You would like to turn the last three terms into a perfect square, ie:
$$Sa^2 + (4 + 2S)ab + (1 + 2S)b^2 = (\sqrt{S}a + \sqrt{1 + 2S}b)^2$$
So, matching the middle terms, we get:
$$4 + 2S = 2\sqrt{S}\sqrt{1 + 2S}$$
$$(2 + S)^2 = S(1 + 2S)$$
$$4 + 3S- S^2 = 0$$
Now, $S = 4$ and $S = -1$ are solutions, but we want the positive one. Plugging in $S = 4$, we go back to the second line:
$$(a - b)^2 + 4a^2 + 12ab + 9b^2 = 7 + 7 \cdot 4$$
$$(a - b)^2 + (2a + 3b)^2 = 35$$
Since squares are positive, the largest possible value of $(a - b)^2$ is clearly $35$, and it is achieved exactly when $2a + 3b = 0$.
You might think this is a convoluted and needlessly tricky way of solving this problem. That's probably true, but I think the technique of creating terms which are perfect squares is handy in proving inequalities, because of the fact that perfect squares are nonnegative.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/310348",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 3
}
|
Find $\delta$ such that $0<|x-3|<\delta \Rightarrow |\frac{1}{x} - \frac{1}{3}| < 10^{-4}$ I was working on a calculus problem:
Find $\delta$ such that $0<|x-3|<\delta \Rightarrow |\frac{1}{x} - \frac{1}{3}| < 10^{-4}$
I did some algebra on the consequent obtaining:
$$
\frac{|x-3|}{|3x|} < 10^{-4}
$$
Then I noted that if $0< |x-3| < 1$ for $\delta = 1$. We have:
$$
\frac{|x-3|}{|3x|} < \frac{1}{6}
$$
So it's sufficient to have $\delta = 6\cdot 10^{-4} $,
$$
\frac{|x-3|}{|3x|} < \frac{|x-3|}{6} < \frac{1}{6} \cdot 6\cdot 10^{-4}
$$
Basically this follows because we have a lower bound on the denominator. Is this reasoning good, or is there something I'm not accounting for?
|
Note that
$0<|x-3|<\delta\iff x\in(3-\delta, 3+\delta)\iff 3-\delta < x < 3+\delta$.
Therefore, you can write $x = 3 + \rho$, such that $|\rho| < \delta$. So you have that
$\bigg\lvert \frac{1}{x}-\frac{1}{3}\bigg\lvert = \bigg\lvert \frac{1}{3 + \rho}-\frac{1}{3}\bigg\lvert = \bigg\lvert \frac{3- 3-\rho}{3(3+\rho)}\bigg\lvert=\bigg\lvert\frac{\rho}{3(3+\rho)}\bigg\lvert$.
If you take $\delta$ such that $\delta\leq10^{-4}$, then $|\rho| <\delta\Rightarrow \bigg\lvert \frac{1}{x}-\frac{1}{3}\bigg\lvert =\bigg\lvert\frac{\rho}{3(3+\rho)}\bigg\lvert < \frac{\delta}{3\cdot2}<\frac{10^{-4}}{6}<10^{-4}$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/311484",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
}
|
Prove that $\sum_{j = 0}^{n} (-\frac{1}{2})^j = \frac{2^{n+1} + (-1)^n}{3 \times 2^n}$ whenever $n$ is a nonnegative integer. I'm having a really hard time with the algebra in this proof. I'm supposed to use mathematical induction (which is simple enough), but I just don't see how to make the algebra work.
$\sum_{j = 0}^{k} (-\frac{1}{2})^k + (-\frac{1}{2})^{k + 1} = \frac{2^{k+1} + (-1)^k}{3 \times 2^k}+(-\frac{1}{2})^{k + 1}$, by adding $(-\frac{1}{2})^{k + 1}$ to both sides.
I want to show that the right side is equal to:
$\frac{2^{k+1+1} + (-1)^{k+1}}{3 \times 2^{k+1}}$
Thank you!
|
You have $$\frac{2^{k+1}+(-1)^k}{3\cdot 2^k}+\left(-\frac{1}{2}\right)^{k+1}.$$Getting a common denominator and combining, we have $$\frac{2^{k+2}+2(-1)^k+3\cdot(-1)^{k+1}}{3\cdot 2^{k+1}}.$$Now we can factor out $(-1)^k$ in the part of the numerator which has it and we get $$\frac{2^{k+2}+(2-3)(-1)^k}{3\cdot 2^{k+1}}=\frac{2^{k+2}+(-1)^{k+1}}{3\cdot 2^{k+1}}.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/311709",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 0
}
|
Solving Equations with Factorials I am attempting to solve an equation ${{n-2} \choose {2}} + {{n-3} \choose {2}} + {{n-4} \choose {2}} = 136$. With the formula for a combination being $\frac{n!}{r!(n - r)!}$, I simplified the given equation to:
$(n-2)! + (n-3)! + (n-4)! = 272n - 1088$
However, I am not sure how I would solve for $\\n$ in the simplified version. I have looked into Sterling's Approximation, hoping to find some way to solve this there, but I believe that is going about it incorrectly.
Any help that can be offered in solving this equation is greatly appreciated.
|
Note that
$$\dbinom{n-2}2 = \dfrac{(n-2)(n-3)}{2}$$
$$\dbinom{n-3}2 = \dfrac{(n-3)(n-4)}{2}$$
$$\dbinom{n-4}2 = \dfrac{(n-4)(n-5)}{2}$$
Hence,
$$\dbinom{n-2}2 + \dbinom{n-3}2 + \dbinom{n-4}2 = \dfrac{(n-2)(n-3)}{2} + \dfrac{(n-3)(n-4)}{2} + \dfrac{(n-4)(n-5)}{2}$$
Now solve the quadratic to get the answer.
EDIT
$$\dbinom{m}2 = \dfrac{m!}{2! (m-2)!} = \dfrac{m(m-1) \cdot (m-2)!}{2! \cdot (m-2)!} = \dfrac{m(m-1)}2$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/312249",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
}
|
Factorization of cyclic polynomial
Factorize $$a(b^2-c^2)+b(c^2-a^2)+c(a^2-b^2)$$
Since this is a cyclic polynomial, factors are also cyclic
$$f(a) = a(b^2-c^2)+b(c^2-a^2)+c(a^2-b^2)$$
$$f(b) = b(b^2-c^2)+b(c^2-b^2)+c(b^2-b^2) = 0 \Rightarrow a-b$$
is a factor of the given expression. Therefore, other factors are $(b-c)$ and $(c-a)$. The given expression may have a coefficient a constant factor which is nonzero. Let it be $m$.
$$\therefore a(b^2-c^2)+b(c^2-a^2)+c(a^2-b^2) = m(a-b)(b-c)(c-a)$$
Please guide further on how to find this coefficient.
|
You can put arbitrary values in $f(a,b,c)$ and get the value of $m$
Let's put $a=1, b=2, c=0$ in $$ a(b^2-c^2)+b(c^2-a^2)+c(a^2-b^2) = m(a-b)(b-c)(c-a)$$
we get, $m = 1$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/314105",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
}
|
Integral $\int\limits_0^\infty \prod\limits_{k=0}^\infty\frac{1+\frac{x^2}{(b+1+k)^2}}{1+\frac{x^2}{(a+k)^2}} \ dx$ Does anybody know how to prove this identity?
$$\int_0^\infty \prod_{k=0}^\infty\frac{1+\frac{x^2}{(b+1+k)^2}}{1+\frac{x^2}{(a+k)^2}} \ dx=\frac{\sqrt{\pi}}{2}\frac{\Gamma \left(a+\frac{1}{2}\right)\Gamma(b+1)\Gamma \left(b-a+\frac{1}{2}\right)}{\Gamma(a)\Gamma \left(b+\frac{1}{2}\right)\Gamma(b-a+1)}$$
I found this on Wikipedia.
|
While this is fairly advanced we can at least bring it into a manageable form which allows specific values to be calculated, assuming $a$ and $b$ are positive integers. Start by evaluating the inner product, which has two parts.
First, $$f_1(x) = \prod_{k=0}^\infty \frac{1}{1+ \frac{x^2}{(a+k)^2}} =
\prod_{k=1}^{a-1}\left( 1+ \frac{x^2}{k^2} \right)
\prod_{k=1}^\infty \frac{1}{1+ \frac{x^2}{k^2}} =
\prod_{k=1}^{a-1} \frac{(k+ix)(k-ix)}{k^2} \frac{\pi x}{\sinh \pi x} \\
= \frac{1}{\Gamma(a)^2} \prod_{k=1}^{a-1} (k+ix)(k-ix) \frac{\pi x}{\sinh \pi x} =
\frac{1}{\Gamma(a)^2} \frac{\Gamma(a+ix)\Gamma(a-ix)}{\Gamma(1+ix)\Gamma(1-ix)}
\frac{\pi x}{\sinh \pi x} \\
= \frac{\Gamma(a+ix)\Gamma(a-ix)}{\Gamma(a)^2},$$
where we have used Euler's reflection formula in the last step.
Next, by the same reasoning,
$$f_2(x) = \prod_{k=0}^\infty \left( 1+ \frac{x^2}{(b+1+k)^2}\right) =
\frac{\Gamma(b+1)^2}{\Gamma(b+1+ix)\Gamma(b+1-ix)},$$
so the integral becomes
$$ I(a, b) = \frac{\Gamma(b+1)^2}{\Gamma(a)^2}
\int_0^\infty \frac{\Gamma(a+ix)\Gamma(a-ix)}{\Gamma(b+1+ix)\Gamma(b+1-ix)} dx \\
= \frac{1}{2} \frac{\Gamma(b+1)^2}{\Gamma(a)^2}
\int_{-\infty}^\infty \frac{\Gamma(a+ix)\Gamma(a-ix)}{\Gamma(b+1+ix)\Gamma(b+1-ix)} dx.$$
Now suppose that $b>a.$ We have
$$ \frac{\Gamma(a+ix)}{\Gamma(b+1+ix)} = \prod_{k=a}^b \frac{1}{k+ix}
\quad \text{and} \quad
\frac{\Gamma(a-ix)}{\Gamma(b+1-ix)} = \prod_{k=a}^b \frac{1}{k-ix}.$$
We evaluate the inner integral $J(a,b)$ with the Cauchy Residue Theorem using a contour consisting in the limit of the x axis and a half circle in the upper half plane. The poles are simple and located at $x=im$ with $a\le m\le b$ and we get
$$ J(a,b) = 2\pi i \sum_{m=a}^b
\operatorname{Res}
\left(\prod_{k=a}^b \frac{1}{k+ix} \prod_{k=a}^b \frac{1}{k-ix}; x=im \right) \\=
2\pi i \sum_{m=a}^b \operatorname{Res}
\left(\prod_{k=a}^b \frac{1}{k^2+x^2}; x=im \right).$$
Using the fact that
$$ I(a,b) = \frac{1}{2} \frac{\Gamma(b+1)^2}{\Gamma(a)^2} J(a,b)$$
we may now calculate specific values. Suppose that $b=a+1$ and obtain
$$ I(a, a+1) = \frac{\pi}{2} \frac{a(a+1)}{2a+1}.$$
For $b=a+2$ we get
$$ I(a, a+2) = \frac{3\pi}{4} \frac{a(a+1)(a+2)}{(2a+1)(2a+3)}.$$
Setting $b = a+3$ we find
$$ I(a, a+3) = \frac{5\pi}{4} \frac{a(a+1)(a+2)(a+3)}{(2a+1)(2a+3)(2a+5)}.$$
Setting $b = a+4$ we find
$$ I(a, a+4) = \frac{35\pi}{16} \frac{a(a+1)(a+2)(a+3)(a+4)}{(2a+1)(2a+3)(2a+5)(2a+7)}.$$
The pattern here is readily apparent, with the coefficient at the front of $I(a,a+n)$ being $$\frac{1}{2^n}{2n-1\choose n}.$$ These formulas including the factorials in the binomial coefficient may of course be rewritten in terms of $\Gamma$ functions. Edit. Note that the product in the original source becomes finite in this case which might make it more manageable. This latter representation should of course simplify to the same result as the long computation and indeed by the looks of it this transformation should be a trivial computation.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/314856",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 3,
"answer_id": 2
}
|
Find of $\int e^x \cos(2x) dx$ I did the following.
Using the LIATE rule:
$$\begin{align*}
u &=& \cos(2x)\\
u\prime &=& -2 \sin(2x)\\
v &=& e^x\\
v\prime&=&e^x
\end{align*}$$
We get:
$$\int e^x \cos(2x)dx = e^x \cos(2x) +2 \int e^x \sin(2x)dx $$
Now we do the second part.
$$\begin{align*}
u &=& \sin(2x)\\
u\prime &=& 2 \cos(2x)\\
v &=& e^x\\
v\prime&=&e^x
\end{align*}$$
We get:
$$\int e^x \sin(2x)dx = e^x \sin(2x) -2 \int e^x \cos(2x)dx$$
Putting it together we get:
$\begin{align*}
\int e^x \cos(2x)dx &=& e^x \cos(2x) +2 \int e^x \sin(2x)dx \\
&=&e^x \cos(2x) + 2[e^x \sin(2x) -2 \int e^x \cos(2x)dx]\\
&=&e^x \cos(2x) + 2e^x \sin(2x) -4 \int e^x \cos(2x)dx
\end{align*}$
$\begin{align*}
\int e^x \cos(2x)dx &=&e^x \cos(2x) + 2e^x \sin(2x) -4 \int e^x \cos(2x)dx\\
5\int e^x \cos(2x)dx &=&e^x(\cos(2x) + 2 \sin(2x))\\
\int e^x \cos(2x)dx &=&\frac{e^x( \cos(2x) + 2 \sin(2x))}{5}
\end{align*}$
I am not sure if this is right, if it is right, is there a better way of doing this.
|
Hint: Take the derivative of both sides with respect to $x$ to see if you did everything right.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/316639",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
}
|
Please, help me to find where is a mistake in the solutions of the equation. I have this equation and I will be very thankful to anyone who can provide me any help with the one discrepancy in my solution and the solution from the self-learning website:
$$
\frac{1+\tan(x) + \tan^2(x) + ... + \tan^n(x) + ...}{1-\tan(x) + \tan^2(x) - ... +(-1)^n\tan^n(x) ...}= 1+\sin(2x)
$$
My solution
I solved it in not too graceful way. First I found the roots for this equation:
$$
\require{cancel}
\begin{align}
&\boxed{\frac{1+\tan(x)}{1-\tan(x)}= 1+\sin(2x)} \\ \\
&\frac{(1+\tan(x))\cancel{(1-\tan(x))}}{\cancel{1-\tan(x)}}= (1+\sin(2x))(1-\tan(x))\\ \\
&\cancel{1}+\tan(x)=\cancel{1} - \tan(x) + \sin(2x) - \sin(2x)\tan(x) \\
&\frac{\sin(2x)}{\tan(x)} - \frac{\sin(2x)\cancel{\tan(x)}}{\cancel{\tan(x)}} -\frac{2\cancel{\tan(x)}}{\cancel{\tan(x)}} = 0\\ \\
&\bbox[Beige]{\boxed{\sin(2\alpha)=2\sin(\alpha)\cos(\alpha)}} \\ \\
&\frac{\cancel{2}\cancel{\sin(x)}\cos(x)\cos(x)}{\cancel{\sin(x)}} - \cancel{2}\sin(x)\cos(x) - \cancel{2} = 0 \\ \\
&\cos^2(x) - \sin(x)\cos(x) -1 = 0\\ \\
&\bbox[Beige]{\boxed{\sin^2(\alpha) + \cos^2(\alpha) = 1}} \\ \\
&\cancel{\cos^2(x)} - \sin(x)\cos(x) - \sin^2(x) - \cancel{\cos^2(x)} = 0\\ \\
&\sin^2(x) + \sin(x)\cos(x) = 0 \\ \\
&\sin(x)(\sin(x) + \cos(x)) = 0 \\ \\
&\sin(x) = 0 \implies \boxed{x_1=\pi k ,\space k \in \mathbb{Z}} \\ \\
&\sin(x) + \cos(x) = 0 \implies \tan(x) = -1 \implies \boxed{x_2=\frac{\pi}{4}\left(4m - 1\right) ,\space m \in \mathbb{Z}}
\end{align}
$$
Also I checked that there were no parasite and missing roots, cause I've divided the equation by $\tan(x)$ and multiplied it by $1-\tan(x)$. There were no such ones cause $\tan(x)$ cannot assume $1$, when $\sin(x)=0$ or $\tan(x) = -1$, and the missing roots with $\tan(x) = 0$ are a subset of $x_1$.
Then I checked which of these roots fitted the next equation:
$$
\frac{1+\tan(x) + \tan^2(x)}{1-\tan(x) + \tan^2(x)}= 1+\sin(2x)
$$
And it turned out that only $x_1$ did. Then my logic was: if there were another roots for the next steps of this sequence, e.g. $\frac{1+\tan(x) + \tan^2(x) + \tan^3(x)}{1-\tan(x) + \tan^2(x) - \tan^3(x)}= 1+\sin(2x)$ they would not fit the first step while $x_1$ would fit all steps. So the only solution is $x_1 = \pi k ,\space k \in \mathbb{Z}$
Possibly the wrong solution with my correction. You can see the original one at the www.bymath.com
It is obvious that the fraction numerator and denominator are geometrical sequences (progressions) with the common ratios $\tan(x)$ and $-\tan(x)$ correspondingly. Note, that here $|\tan(x)| < 1$, otherwise the left-hand side expression is meaningless. Therefore it is possible to transform the fraction numerator and denominator by the sum formula for the unboundedly decreasing geometric sequence (progression) $\bbox[Beige]{\boxed{S = \frac{b_1}{1-q}}}$, where $b$ is the first member of a sequence and $q$ in a ratio.
$$
\begin{align}
&\frac{1}{1 - \tan(x)} : \frac{1}{1 - (-\tan(x))} = 1 + \sin(2x) \\
&\frac{1}{1 - \tan(x)} * (1 + \tan(x)) = 1 + \sin(2x) \\
&\bbox[Beige]{\boxed{\sin(\alpha) = \frac{2\tan(\frac{\alpha}{2})}{1+\tan^2(\frac{\alpha}{2})}}} \\
&\frac{1 + \tan(x)}{1 - \tan(x)} - 1 - \frac{2\tan(x)}{1+\tan^2(x)} = 0\\ \\
&\frac{(1 + \tan(x))(1+\tan^2(x)) - (1 - \tan(x))(1+\tan^2(x)) - 2\tan(x)(1 - \tan(x))}{(1 - \tan(x))(1+\tan^2(x))} = 0\\ \\
&\frac{1 + \tan^2 x + \tan x + \tan^3 x -(1 + \tan^2 x - \tan x - \tan^3 x ) -2\tan x + 2\tan^2 x}{(1 - \tan x)(1+\tan^2 x)} =0 \\
\end{align}
$$
So here is the point where this second solution is wrong. The right is this:
$$
\begin{align}
&\frac{\cancel{1} + \cancel{\tan^2 x} + \cancel{\tan x} + \tan^3 x -\cancel{1} - \cancel{\tan^2 x} + \cancel{\tan x} + \tan^3 x -\cancel{2\tan x} + 2\tan^2 x}{(1 - \tan x)(1+\tan^2 x)} =0 \\
&\frac{2\tan^3 x + 2\tan^2 x}{(1 - \tan x)(1+\tan^2 x)} =0 \implies \tan x = 0, \ \tan x = -1 \\
\end{align}
$$
Whereas the author of the site got this: $\frac{4\tan^3 x}{(1 - \tan x)(1+\tan^2 x)} =0 \implies \tan x = 0$ and lost one root $x_2=\frac{\pi}{4}\left(4m - 1\right) ,\space m \in \mathbb{Z}$
ACTUALLY QUESTION
So when I looked to the second solution I was pretty confused with this:
in that solution it is said that unless $|tan(x)| < 1$ the left-hand side of the equation in meaningless. Why, can somebody explain, please? I suppose that this is not true cause:
*
*while the transformation with the sum formula was correct, the solution itself was wrong
*the right solution gives us two roots one of which does not fit all possible steps of the sequences, hence the numerator and denominator are not a unboundedly decreasing geometric sequence (progression).
Is my conclusion correct?
My sincerest appreciation.
|
$$
\require{cancel}
\begin{align}
&\frac{\sin(2x)}{\tan(x)} - \frac{\sin(2x)\cancel{\tan(x)}}{\cancel{\tan(x)}} -\frac{2\cancel{\tan(x)}}{\cancel{\tan(x)}} = 0\\ \\
\end{align}
$$
Here you lost the solution $\tan(x)=0$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/317149",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
}
|
Finding a generating-function using partitions Find a generating function for a , the number of partitions of r into
(a.) Even integers
(b.) Distinct odd integers.
I am at a loss of starting this.
|
Let $p_E(n)$ be the number of partitions of $n$ into even parts, where we set $p_E(0)=1$. For each $k\in\Bbb Z^+$ let $p_k(n)$ be the number of ways to partition $n$ into parts of size $2k$; clearly
$$p_k(n)=\begin{cases}
1,&\text{if }2k\mid n\\
0,&\text{otherwise}\;,
\end{cases}$$
so $$\sum_{n\ge 0}p_k(n)x^n=1+x^{2k}+x^{2(2k)}+x^{3(2k)}+\ldots=\frac1{1-x^{2k}}\;.$$
By the product rule for ordinary generating functions,
$$\begin{align*}
\sum_{n\ge 0}p_E(n)x^n&=\left(1+x^2+x^4+x^6+\ldots\right)\left(1+x^4+x^8+x^{12}+\ldots\right)\dots\\
&=\prod_{k\ge 1}\frac1{1-x^{2k}}\;.
\end{align*}$$
It’s not hard to see that this infinite product makes sense: the factor $\frac1{1-x^{2k}}$ contributes to the $x^n$ term only if $n\le 2k$.
As an example of what’s going on, the $x^8$ terms are $1\cdot1\cdot1\cdot x^{1\cdot8}$, $x^{1\cdot2}\cdot1\cdot x^{1\cdot6}$, $x^{2\cdot4}$, $x^{2\cdot2}\cdot x^{1\cdot4}$, and $x^{4\cdot2}$, corresponding to the partitions $8$, $2+6$, $4+4$, $2+2+4$, and $2+2+2+2$, respectively.
Suppose that we want to count the partitions of $n$ into distinct parts no larger than $4$. For each of the sizes $1,2,3$, and $4$ we can have either $0$ or $1$ part of that size, so the generating function is $(1+x)(1+x^2)(1+x^3)(1+x^4)$: e.g., the term $1\cdot x^2\cdot1\cdot x^4$, for instance, corresponds to the partition $2+4$ of $6$. Thus, if $p_d(n)$ is the number of partitions of $n$ into distinct parts, we must have
$$\sum_{n\ge 0}p_d(n)x^n=\prod_{k\ge 1}\left(1+x^k\right)\;.$$
How should you modify this to get the generating function for the number of partitions of $n$ into distinct odd parts?
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/317695",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
$xy=1 \implies $minimum $x+y=$? If $x,y$ are real positive numbers such that $xy=1$, how can I find the minimum for $x+y$?
|
Note that $y=1/x$ and $$\left(x+\frac{1}{x}\right)^2=\left( x-\frac{1}{x}\right)^2+4\cdot x\cdot \frac{1}{x}\geq4$$ Hence $x+y\geq 2$. Also for $x=y=1$, you have $x+y=2$, hence $2$ indeed the minimum value.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/317831",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 13,
"answer_id": 0
}
|
Evaluating $\frac{d^2y}{dx^2}$ I want to evaluate $\frac{d^2y}{dx^2}$ from the given data: $x=a\cos^3\theta$ and $y=b\sin^3\theta$.
I tried in this way- $\frac{x^2}{a^2}+\frac{y^2}{b^2}=\cos^6\theta+\sin^6\theta=1-3(\frac{xy}{ab})^\frac{2}{3}$. After that I seem to be lost and will appreciate some help in doing this efficiently.
|
$\displaystyle x=a\cos^3\theta \Rightarrow \frac {dx}{d\theta}=-3a\cos^2\theta\sin\theta$
$\displaystyle y=b\sin^3\theta \Rightarrow \frac {dy}{d\theta}=3b\sin^2\theta\cos\theta$
$\displaystyle \Rightarrow\frac {dy}{dx}=\frac{\frac {dy}{d\theta}}{\frac {dx}{d\theta}}=\frac{3b\sin^2\theta\cos\theta}{-3a\cos^2\theta\sin\theta}=-\frac b a \tan\theta$
$\displaystyle \Rightarrow \frac{d^2y}{dx^2}=-\frac b a \sec^2\theta=-\frac b a (\frac a x)^\frac 3 2=-\frac {\sqrt a b} {\sqrt{x^3}}$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/319464",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
}
|
Partial Fraction Decomposition $\frac{2x^2+3x+3}{(x+1)(x^2+1)}$ How would one go about decomposing this fraction?
$$
2x^2+3x+3\over
(x+1)(x^2+1)
$$
Here is what I have so far:
$$
{2x^2+3x+3\over
(x+1)(x^2+1)
} =
{ A\over x+1 } + { Bx+C\over x^2+1} $$
$$
{2x^2+3x+3\over
(x+1)(x^2+1)
} =
{ A(x^2+1) } + (Bx+C)(x+1) $$
I know how to solve for A - it's 1. But how do I get B and C?
|
You should have
$2x^2 + 3x + 3 = A(x^2+1) + (Bx+C)(x+1)$
Now, try suitable values of $x$ to find $A, B, C$. For e.g. $x = -1$ will easily give you $A$. Hint, there is another value which easily gives you $C$ after you know $A$.
In case no suitable $x$ values occur to you, just use a few simple values to get as many equations as the coefficients, and then solve a set of linear equations.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/320260",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
Integration by anti-differentiation I have a question in my assignment, and I highly suspect there is a typo somewhere.
Evaluate the following integral:
$\int \frac{1}{(1-t)^{\frac{3}{2}}} dt = \frac{t}{\sqrt{1-t^2}} + c$
What I did was to perform differentiation and got the following:
$$
\begin{align*}
\frac{d}{dt}\frac{t}{\sqrt{1-t^2}} &= \frac{(0.5)(1-t^2)^{-\frac 1 2}(-2t)(t) - \sqrt{1-t^2}}
{1-t^2} \space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\spadesuit
\\&=\frac{-t(1-t^2)^{-\frac 1 2} - \sqrt{1-t^2}}{1-t^2}
\\&=-\sqrt{1-t^2}
\end{align*}
$$
At the same time, by integrating formally,
$$
\begin{align*}
\int \frac{1}{(1-t)^{\frac{3}{2}}} dt
&= 2\int \frac{1}{2(1-t)^{\frac{3}{2}}} dt
\\&=(1-t)^{-\frac 1 2} + c\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\heartsuit
\end{align*}
$$
Because
$$
\begin{align*}
\frac{d}{dx}(1-t)^{-\frac 1 2}
&= (-0.5)(1-t)^{-\frac 3 2}(-1)
\\&= \frac{1}{2(1-t)^{\frac 3 2}}
\end{align*}
$$
Could somebody please check my working?
UPDATE: Solved. I will preserve this version of the question to highlight all the common mistakes that careless people like me make - which is to utilize the quotient rule wrongly at $\spadesuit$ and forgetting to include the factor of $2$ at $\heartsuit$
|
You made an error in differentiating. It should be
\begin{align*}
\frac{d}{dt}\frac{t}{\sqrt{1-t^2}} &= \frac{\sqrt{1-t^2} - (0.5)(1-t^2)^{-\frac 1 2}(-2t)(t)}
{1-t^2}
\\&=\frac{\sqrt{1-t^2} + t^2(1-t^2)^{-\frac 1 2}}{1-t^2}
\\&=\frac{1 - t^2 + t^2}{(1 - t^2)^{3/2}}
\\&=\frac{1}{(1 - t^2)^{3/2}}.
\end{align*}
So the integral on the left is wrong as well, the $t$ should actually be $t^2$.
Also, you missed the factor of two in your answer for the integral. It should be $2(1 - t)^{-1/2} + c$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/323186",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Trig Integral (Definite) How do I evaluate this definite integral?
$$\int_{0}^{\frac{\pi}{12}}{\sin^4x \, \cos^4x\, \operatorname{d}\!x}$$
I know this is a trig. function.
|
From here, we have
$$\int_0^{\pi/2} \sin^{2n}(x) dx = \dfrac{2n-1}{2n} \cdot \dfrac{2n-3}{2n-2} \cdots \dfrac34 \cdot \dfrac12 \cdot\dfrac{\pi}2$$
Hence,
$$\int_0^{\pi/2} \sin^4(x) \cos^4(x) dx = \dfrac1{16} \int_0^{\pi/2} \sin^4(2x) dx = \dfrac1{32} \int_0^{\pi} \sin^4(t) dt = \dfrac2{32} \int_0^{\pi/2} \sin^4(t) dt$$
Hence, the answer is
$$\dfrac1{16} \cdot \dfrac34 \cdot \dfrac12 \cdot \dfrac{\pi}2 = \dfrac{3 \pi}{256}$$
If the integral is from $0$ to $\pi/12$, then, we get that
\begin{align}
\int_0^{\pi/{12}} \sin^4(x) \cos^4(x) dx & = \dfrac1{16} \int_0^{\pi/12} \sin^4(2x) dx = \dfrac1{32} \int_0^{\pi/6} \sin^4(t) dt\\
& = \dfrac1{32} \int_0^{\pi/6} \dfrac{(1-\cos(2t))^2}{4} dt\\
& = \dfrac1{128} \int_0^{\pi/6} (1+\cos^2(2t) - 2 \cos(2t)) dt\\
& = \dfrac1{128} \int_0^{\pi/6} \left(1+\dfrac{1+\cos(4t)}2 - 2 \cos(2t) \right) dt\\
& = \dfrac1{128} \int_0^{\pi/6} \left(\dfrac32+\dfrac{\cos(4t)}2 - 2 \cos(2t) \right) dt\\
& = \dfrac1{128} \left(\dfrac{\pi}4 + \dfrac{\sin(4 \pi/6)}{8}-\sin(2 \pi/6)\right)\\
& = \dfrac1{128} \left(\dfrac{\pi}4 + \dfrac{\sqrt3}{16} - \dfrac{\sqrt3}2\right)\\
& = \dfrac1{128} \left(\dfrac{\pi}4 - \dfrac{7\sqrt3}{16} \right) = \dfrac{4 \pi - 7 \sqrt3}{2048}
\end{align}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/323783",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
Results of the multivariate function $f(x,y) = 3x^2 - 2xy + y^3 $ I was asked to derive 2 experessions for this multivariate function
$$
f(x,y) = 3x^2 -2xy+y^3
$$
The first is $\large{\frac{f(x+h,y) - f(x,y)}{h}}$ and the other is $\large{\frac{f(x,y+k) - f(x,y)}{k}}$
The working is as follows:
$$
\begin{align*}
\frac{f(x+h,y) - f(x,y)}{h} &= \frac{3(x+h)^2 - 2y(x+h) + y^3 -(3x^2-2xy+y^3)}{h}
\\&= \frac{3(x+h)^2 - 3x^2 -2y(x+h) + 2xy}{h}
\\&=\frac{3(x+h+x)(x+h-x) - 2y(h)}{h}
\\&=\frac{3(x+h+x)(x+h-x) - 2y(h)}{h}
\\&= 3(2x+h) -2y
\end{align*}
$$
and
$$
\begin{align*}
\frac{f(x,y+k) - f(x,y)}{k} &= \frac{3x^2 - 2x(y+k) + (y+k)^3-(3x^2-2xy+y^3)}{k}
\\&= \frac{-2x(y+k) +(y+k)^3 + 2xy-y^3}{k}
\\&=\frac{2x(-y-k+y) + (y+k)^3 - y^3}{k}
\\&=\frac{-2xk + (y+k-y)[(y+k)^2 + y(y+k) + y^2]}{k}
\\&=\frac{-2xk + k[(y+k)^2 + y(y+k) + y^2]}{k}
\\&= -2x + (y+k)^2 + y(y+k) + y^2
\end{align*}
$$
Are the calculations correct?
UPDATE
Thanks to the community, I have verified the answer. For completeness, I shall continue to derive the partial derivatives using the above answers.
$$
\begin{align*}
\frac{\partial f}{\partial x}(x,y)=\lim_{h\to0}\frac{f(x+h,y)-f(x,y)}{h}
&= \lim_{h \to 0}3(2x+h) - 2y
\\&= 6x + 2y
\end{align*}
$$
and
$$
\begin{align*}
\frac{\partial f}{\partial y}(x,y)=\lim_{k\to0}\frac{f(x,y+k)-f(x,y)}{k}
&= \lim_{k \to 0} -2x + (y+k)^2 + y(y+k) + y^2
\\&= -2x + y^2 + y^2 + y^2
\\&= -2x + 3y^2
\end{align*}
$$
|
Yes. The calculations are correct.
Note that when $h\to 0$, the first expression is the partial derivative of $f(x,y)$ by $x$; and when $k\to 0$, the second expression is the partial derivative of $f(x,y)$ by $y$.
That is, we define the partial derivative $$\frac{\partial f}{\partial x}(x,y)=\lim_{h\to0}\frac{f(x+h,y)-f(x,y)}{h}.$$
See that after getting rid of $h$ and $k$ you get the same results as a "symbolic differentiation" would have given you (i.e. $\frac d{dx}(3x^2)=6x$ and so on).
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/324769",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
Show determinant of matrix is non-zero I have $a,b,c\in\mathbb{Q}$ not all zero. ($a^2+b^2+c^2\ne 0$), I want to show that the following determinant is then non-zero. I failed to arrive at an appropriate form of the polynomial. Help please.
$$\left|\begin{bmatrix} a & 2c & 2b\\b & a & 2c\\ c & b & a\end{bmatrix}\right| = a^3+2 b^3-6 a b c+4 c^3$$
Second question, what is the easiest way to argue that $\{1,\sqrt[3]{2},(\sqrt[3]{2})^2\}$ is linearly independent in $\mathbb{Q}$?
Motivation:
Prove that $\mathbb{Q}[\sqrt[3]{2}] = \{a+b\sqrt[3]{2}+c(\sqrt[3]{2})^2\;|\;a,b,c\in\mathbb{Q}\}$ forms a field.
Proof: Since $\mathbb{Q}[\sqrt[3]{2}] \subset \mathbb{R}$, we prove $\mathbb{Q}[\sqrt[3]{2}]$ is a subfield of $(\mathbb{R},+,\cdot)$
$\forall (a+b\sqrt[3]{2}+c(\sqrt[3]{2})^2)\in \mathbb{Q}[\sqrt[3]{2}]\backslash\{0\}.$ We want to find $(a+b\sqrt[3]{2}+c(\sqrt[3]{2})^2)$ such that
$(a+b\sqrt[3]{2}+c(\sqrt[3]{2})^2)(d+e\sqrt[3]{2}+f(\sqrt[3]{2})^2) =$
$ (ad+2ec+2bf)+(ae+bd+2cf)\sqrt[3]{2}+(af+cd+be)(\sqrt[3]{2})^2 = 1$
Since $\{1,\sqrt[3]{2},(\sqrt[3]{2})^2\}$ is linearly independent (?) over $\mathbb{Q}$, we show there is unique solution to:
$$\begin{bmatrix} a & 2c & 2b\\b & a & 2c\\ c & b & a\end{bmatrix} \cdot \left[\begin{array}{l l} d\\e\\f \end{array}\right] = \left[\begin{array}{l l} 1\\0\\0 \end{array}\right] $$
Which is equivalent in showing the determinant is non-zero
$$\left|\begin{bmatrix} a & 2c & 2b\\b & a & 2c\\ c & b & a\end{bmatrix}\right| = a^3+2 b^3-6 a b c+4 c^3=(?)$$
By subfield test, 1)2)3)4) is enough to say that $(\mathbb{Q}[\sqrt[3]{2}],+,\cdot)$ is a subfield of $(\mathbb{R},+,\cdot)$ therefore a field.
EDIT: If you have shorter way that prove the proposition without touching my 2 questions, that is even better.
|
Since $a,\ b$ and $c$ are rational, we may clear denominators in $$a^3 + 2b^3 -6abc +4c^3 = 0$$ The above equation is a homogenous equation of degree $3$ so we may cancel common factors. If there exists non-trivial solutions to the equation, we may therefore assume without loss of generality that $a,\ b$ and $c$ are integers with $\gcd(a,\ b,\ c)=1$.
Reducing modulo $2$, we find that $a\equiv 0\pmod 2$. Let $a=2\alpha$. Making the substitution and cancelling common factors, we arrive at
$$4\alpha^3 +b^3 - 6\alpha bc + 2c^3 = 0$$
Reducing mod $2$ again, we get $b\equiv 0\pmod2$. So let $b=2\beta$ to obtain
$$2\alpha^3 + 4\beta^3 - 6\alpha\beta c + c^3 = 0$$
Reducing modulo $2$ one last time gives $c\equiv 0\pmod 2$. This contradicts the fact that $\gcd(a,\ b,\ c)=1$. Therefore there are no non-trivial integer solutions to the above equation. It follows that the determinant is non-zero since $a,\ b$ and $c$ are not all zero.
To show the linear independence of $\left\{1,\ \sqrt[3]{2},\ \left(\sqrt[3]{2}\right)^2\right\}$ in $\mathbb{Q}$, suppose to the contrary that there exists some non-trivial rational linear combination such that
$$r_0 + r_1\sqrt[3]{2} + r_2\left(\sqrt[3]{2}\right)^2 = 0$$
Then clearing denominators, there exists a non-trivial integral linear combination of the above set to $0$. Specifically, there exists an integral polynomial $p(x)$ of degree $2$ such that $\sqrt[3]{2}$ is a root. But the minimal polynomial of $\sqrt[3]{2}$ is $x^3 - 2$. This is a contradiction.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/327884",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 5,
"answer_id": 1
}
|
$x^2-y^2=n$ has solution in $\Bbb Z$ IFF : $n \not \equiv 2 \pmod 4$ suppose $n\in \Bbb Z $ then how to prove this statement:
$x^2-y^2=n$ has solution in $\Bbb Z$ IFF : $n \not \equiv 2 \pmod 4$
|
Observe that $x+y=x-y+2y$
So, $x+y,x-y$ have the parity
If they are odd, $n=x^2-y^2$ will be odd
If they are even, $n=x^2-y^2$ will be divisible by $4$
Conversely, if $n$ is odd $=2m+1$(say)
$(x+y)(x-y)=2m+1$ where $m$ is any integer
So, we can write $x+y=2m+1,x-y=1\implies x=m+1,y=m$
or $x+y=-(2m+1),x-y=-1\implies x=-(m+1),y=-m$
If $4\mid n,n=4r$(say)
So, we can write $x+y=2r,x-y=2\implies x=r+1,y=r-1$
or $x+y=-2r,x-y=-2\implies x=-(r+1),y=-(r-1)$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/328195",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
}
|
Range of $\frac{1}{2\cos x-1}$ How can we find the range of $$f(x) =\frac{1}{2\cos x-1}$$
Since range of $\cos x$ can be given as : $-1 \leq \cos x \leq 1$
therefore we can proceed as :$$\begin{array}{rcl}
-2 \leq & 2\cos x & \leq 2 \\
-2-1 \leq & 2\cos x -1 & \leq 2-1\\
-3 \leq & 2\cos x -1 & \leq 1 \\
\frac{-1}{3} \leq & \frac{1}{2\cos x-1} & \leq 1
\end{array}$$
Is it the range? Please suggest and guide.
|
As $2\cos x-1$ varies through $[-3,0)$, $\frac{1}{2\cos x-1}$ varies through $(-\infty,-\frac13]$.
As $2\cos x-1$ varies through $(0,1]$, $\frac{1}{2\cos x-1}$ varies through $[1,+\infty)$.
Note that division by $0$ is undefined, so we do not consider the case $2\cos x-1=0$.
Hence the range is $(-\infty,-\frac13]\cup [1,+\infty)$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/328493",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
}
|
$y^2=x^3+7$ has an integral soln. trivial issues that are imposing me from understanding a proof from a book. I have some trivial issues that are imposing me from understanding a proof from a book.
(i) if $y^2=8k^3+7$ where $k\in \mathbb{Z}$ $\implies$ $y^2\equiv7 \,[8]$ which is not possible, because $(\frac{7}{8})=-1$. But can someone tell me an easy way to interpret why
$(\frac{7}{8})=-1$?
(ii) if $x$ is odd then I know why $(x-1)^2+3\equiv3 \,[4]$, but why does this imply that exists prime $p$ st $p|(x-1)^2+3\equiv3 \, [4]$ where $p\equiv3 \, [4]$? (according to book, otherwise all its prime factors are 1 modulo 4 $\implies$ $(x-1)^2+3\equiv1 \, [4]$) fair enough, but why does this $\implies$ $(x-1)^2+3\not\equiv3 \, [4]$?
can a number not be congruent 1 [4] and 3 [4] simultaneously?
And does it have to be congruent to at least one of these?
(iii) $p\equiv3 \, [4]$ $\implies$ $(\frac{-1}{p})=-1$ ?
(iv) if $y^2=x^3+7$ has an integral soln. (which it does not) then $x$ has to be odd and $y$ even, why is $y$ necessarily even?
(v) $y^2=x^3+7$ reducing equation modolu $4$ $\implies$ $0\equiv(x+3) \,[4]$ I see what they have done, e.g. $7\equiv3 \, [4]$ so 7 becomes 3.
how can we prove that $0\equiv(x+3) \, [4]$ is even true?
(vi) $x$ odd and $x+2>0$ and $x+2\equiv3 \, [4]$ $\implies$ it has prime factor $p\equiv3 \, [4]$
why?
thanks a bunch
|
Here, I will try to be as explicit as possible to show that $y^2=x^3+7$ has no integer solution.
Case 1: Suppose that $x$ is even, so $x=2k$. It follows that $y^2=8k^3+7$. This implies that $y^2\equiv 7\mod 8$. This is a contradiction because there is no $y$ such that $y^2\equiv 7\mod 8$. You can check this by squaring the numbers $0,1,\ldots,7$, and showing that none of them give a remainder of $7$ when you divide by $8$.
Case 2: Suppose that $x$ is odd. Notice that $x^3+7$ will be even, so that $y$ must also be even (else $y^2=x^3+7$ will be odd). Note that $x^3\equiv x\mod 4$ (check this for $x=1,3$), $y^2\equiv 0\mod 4$ (because $y$ is even), and $7\equiv 3\mod 4$. Thus, reducing our equation mod $4$ gives the equation $0\equiv (x+3)\mod 4$, or $x\equiv 1\mod 4$.
Now write $y^2+1=x^3+8=(x+2)(x^2-2x+4)$. Since $x+2\equiv 3\mod 4$, there must exist some prime $p\equiv 3\mod 4$ dividing $x+2$. This follows from the fact that all odd numbers are congruent to either $1$ or $3\mod 4$, and the product of two numbers congruent to $1\mod 4$ is also congruent to $1\mod 4$. Then for this prime $p$ we can write $y^2+1\equiv 0\mod p$. This is impossible by quadratic reciprocity.
Hence, no integer solutions exist.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/332345",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
}
|
Show that $(n - 1)2^{n+1} + 2 + (n+1)2^{n+1} = n(2^{n+2})+2$ I'm having a really hard time showing this equality is true, I've tried several ways of going about it and I just can't seem to make it work. Help!
$(n - 1)2^{n+1} + 2 + (n+1)2^{n+1} = n(2^{n+2})+2$
Thanks!
|
Just let $A=2^n$. Then $2^{n+1}=2\cdot 2^n=2A$ and $2^{n+2}=2^2\cdot 2^n = 4A$. Then your equality reads:
$$2(n - 1)A + 2 + 2(n+1)A = 4nA+2.$$
Let us show that this is indeed true:
$$2(n - 1)A + 2 + 2(n+1)A = 2A(n-1+n+1)+2=2A(2n)+2=4nA+2,$$
as desired.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/333375",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
}
|
Find $\int_0^\infty \frac{\ln ^2z} {1+z^2}{d}z$ How to find the value of the integral $$\int_{0}^{\infty} \frac{\ln^2z}{1+z^2}{d}z$$ without using contour integration - using usual special functions, e.g. zeta/gamma/beta/etc.
Thank you.
|
\begin{align}J&=\int_0^\infty \frac{\ln^2 x}{1+x^2}dx\\
K&=\int_0^\infty\int_0^\infty \frac{\ln^2(xy)}{(1+x^2)(1+y^2)}dxdy\\
&\overset{u(x)=xy}=\int_0^\infty\int_0^\infty \frac{y\ln^2 u}{(u^2+y^2)(1+y^2)}dudy\\
&=\frac{1}{2}\int_0^\infty \ln^2 u\left[\frac{\ln\left(\frac{1+y^2}{u^2+y^2}\right)}{u^2-1}\right]_{y=0}^{y=\infty}du\\
&=\int_0^\infty \frac{\ln^3 u}{u^2-1}du=\int_0^1 \frac{\ln^3 u}{u^2-1}du+\underbrace{\int_1^\infty \frac{\ln^3 u}{u^2-1}du}_{z=\frac{1}{u}}=2\int_0^1 \frac{\ln^3 u}{u^2-1}du\\
&=2\int_0^1 \frac{\ln^3 u}{u-1}du-2\underbrace{\int_0^1 \frac{u\ln^3 u}{u^2-1}du}_{z=u^2}=2\int_0^1 \frac{\ln^3 u}{u-1}du-\frac{1}{8}\int_0^1 \frac{\ln^3 z}{z-1}dz=\frac{15}{8}\int_0^1 \frac{\ln^3 z}{z-1}dz\\
&=\frac{15}{8}\times 6\zeta(4)=\boxed{\frac{\pi^4}{8}}
\end{align}
On the other hand,
\begin{align}K&=\int_0^\infty\int_0^\infty\frac{\ln^2 x}{(1+x^2)(1+y^2)}dxdy+\int_0^\infty\int_0^\infty\frac{\ln^2 y}{(1+x^2)(1+y^2)}dxdy+\\&
2\underbrace{\int_0^\infty\int_0^\infty\frac{\ln x\ln y}{(1+x^2)(1+y^2)}dxdy}_{=0}\\
&=2J\underbrace{\int_0^\infty \frac{1}{1+x^2}dx}_{=\frac{\pi}{2}}=\boxed{\pi J}
\end{align}
Therefore,
\begin{align}\boxed{J=\frac{\pi^3}{8}}\end{align}
NB:
I assume,
\begin{align}\int_0^1 \frac{\log^3 x}{x-1}dx=6\zeta(4)=\frac{\pi^4}{15}\end{align}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/333672",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 4
}
|
Trying to find a function whose derivative is $\cfrac{4x^3}{x^4+1}$ any ideas? The closet I can get is
$\arctan(x^4) $
the derivative of $\arctan(x^4)$ is $\dfrac{4x^3}{x^8 + 1} ...$
Any tips?
|
Hint: Note that the derivative of $x^4 + 1\;$ is $4x^3$.
If we let $f(x) = x^4 + 1$, then we know $f'(x) = 4x^3$.
Then note that $$\frac{4x^3}{x^4 + 1} = \dfrac{f'(x)}{f(x)}\tag{1}$$
And all integrals of the form $\displaystyle \int \dfrac{f'x}{f(x)}\,dx $ evaluate as $$\int \dfrac{f'x}{f(x)}\,dx = \ln|f(x)| + C \tag{where C is some constant}$$
*
*Just note that $\;\frac{d}{dx}(\ln|f(x)| + C) = \dfrac{f'(x)}{f(x)}$ (by the chain rule).
If you already know how to integrate, you'll find we can integrate by substitution:
Let $u = f(x) =x^4 + 1$, then $f'(x) = du/dx = 4x^3 \implies \; du = 4x^3 \,dx$.
$$
\begin{align} \int \dfrac{4x^3\,dx}{x^4 + 1}\tag{By $(1)$}
& = \int \dfrac{du}{u} = \ln|u| + C \\ \\
& = \ln(x^4 + 1) + C \\ \\
\end{align}
$$
(because $x^4 + 1 \gt 0$, we don't need the absolute value sign: $|x^4 + 1|)$
If you haven't learned integration, note that $\; \dfrac{4x^3}{x^4 + 1}\;$ is precisely the derivative of $\ln(x^4 + 1)$.
You can check for yourself: using the chain rule, we get $$\frac d{dx}(\ln(x^4 + 1)) = \dfrac d{dx}(x^4 + 1)\cdot \dfrac {1}{x^4 + 1} = 4x^3 \cdot \dfrac{1}{x^4 + 1}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/334394",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 0
}
|
Solving recursion with generating function I am trying to solve a recursion with generating function, but somehow I ended up with mess.....
$$y_n=y_{n-1}-2y_{n-2}+4^{n-2}, y_0=2,y_1=1 $$
\begin{eqnarray*}
g(x)&=&y_0+y_1x+\sum_2^{\infty}(y_{n-1}-2y_{n-2}+4^{n-2})x^n\\
&=&2+x+\sum^{\infty}_2y_{n-1}x^n-2\sum_2^{\infty}y_{n-2}x^n+\sum_2^{\infty}4^{n-2}x^n\\
&=&2+x+x\sum_1^{\infty}y_{n-1}x^{n-1}-2x^2\sum_0^{\infty}y_{n-2}x^{n-2}+\frac{1}{4^2}\sum^{\infty}_{2}(4x)^{n}\\
&=&2+x+x(g(x)-1)-2x^2g(x)+\frac{1}{4^2}\left(\frac{1}{1-4x}-1-4x\right)\\
g(x)(1-x+2x^2)&=&2+\frac{1}{4^2}\frac{1}{1-4x}-\frac{1}{4^2}-\frac{x}{4}\\
g(x)&=& \frac{x^2-8x+2}{(1-4x)(1-x+2x^2)}
\end{eqnarray*}
How do I go from here to get $y_n$ as a complete solution, and also I noticed that $(1-x+2x^2)$ has imaginary roots, what does it mean? no solution?
|
I get a slightly different result from you. Let
$$y(x) = \sum_{n=0}^{\infty} y_n x^n$$
Then, summing the recurrence relation from $n=2$ on, I get
$$y(x) - y_0 - y_1 x - x [y(x)-y_0] + 2 x^2 y(x) = \frac{x^2}{1-4 x}$$
Simplifying, using the initial conditions $y_0=2$ and $y_1=1$:
$$(2 x^2-x+1)y(x) = \frac{x^2}{1-4 x} + 2-x = \frac{5 x^2-9 x+2}{1-4 x}$$
Therefore the generating function is
$$y(x) = \frac{5 x^2-9 x+2}{(1-4 x)(2 x^2-x+1)}$$
Your concern about whether the denominator has complex roots is unfounded. The roots of the quadratic in the denominator are based on the characteristic equation for the recurrence that leads to the homogeneous solution. When the roots of this equation are complex, then there is both exponential growth and oscillatory behavior; this is quite normal.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/335289",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
}
|
Prove $\sqrt{3a + b^3} + \sqrt{3b + c^3} + \sqrt{3c + a^3} \ge 6$
If $a,b,c$ are non-negative numbers and $a+b+c=3$, prove that:
$$\sqrt{3a + b^3} + \sqrt{3b + c^3} + \sqrt{3c + a^3} \ge 6.$$
Here's what I've tried:
Using Cauchy-Schawrz I proved that:
$$(3a + b^3)(3 + 1) \ge (3\sqrt{a} + \sqrt{b^3})^2$$
$$\sqrt{(3a + b^3)(4)} \ge 3\sqrt{a} + \sqrt{b^3}$$
$$\sqrt{(3a + b^3)} \ge \frac{3\sqrt{a} + \sqrt{b^3}}{2}$$
Also I get:
$$\sqrt{(3b + c^3)} \ge \frac{3\sqrt{b} + \sqrt{c^3}}{2}$$
$$\sqrt{(3c + a^3)} \ge \frac{3\sqrt{c} + \sqrt{a^3}}{2}$$
If I add add 3 inequalities I get:
$$\sqrt{(3a + b^3)} + \sqrt{(3b + c^3)} + \sqrt{(3c + a^3)} \ge \frac{3\sqrt{a} + \sqrt{b^3}}{2} + \frac{3\sqrt{b} + \sqrt{c^3}}{2} + \frac{3\sqrt{c} + \sqrt{a^3}}{2}$$
Now i need to prove that:
$$\frac{3\sqrt{a} + \sqrt{a^3}}{2} + \frac{3\sqrt{b} + \sqrt{b^3}}{2} + \frac{3\sqrt{c} + \sqrt{c^3}}{2} \ge 6 = 2(a+b+c)$$
It's enough now to prove that:
$$\frac{3\sqrt{a} + \sqrt{a^3}}{2} \ge b+c = 3-a$$
$$\frac{3\sqrt{b} + \sqrt{b^3}}{2} \ge a+c = 3-b$$
$$\frac{3\sqrt{c} + \sqrt{c^3}}{2} \ge b+a = 3-c$$
All three inequalities are of the form:
$$\frac{3\sqrt{x} + \sqrt{x^3}}{2} \ge 3-x$$
$$3\sqrt{x} + \sqrt{x^3} \ge 6-2x$$
$$(3\sqrt{x} + \sqrt{x^3})^2 \ge (6-2x)^2$$
$$9x + x^3 + 6x^2 \ge 36 - 24x + 4x^2$$
$$x^3 + 2x^2 + 33x - 36 \ge 0$$
$$(x-1)(x^2 + 3x + 33) \ge 0$$
Case 1:
$$(x-1) \ge 0 \ \ \ \ \text{ for any }\ x \geq 1$$
$$(x^2 + 3x + 33) \ge 0 \ \ \ \ \text{ for any x in R} $$
Case 2:
$$0 \ge (x-1) \ \ \ \ \text{ for any }\ 1 \geq x$$
$$0 \ge (x^2 + 3x + 33) \ \ \ \ \text{there are no solutions in R} $$
This proves that for $$x \geq 1$$ $$(x-1)(x^2 + 3x + 33) \ge 0$$ is true and so it is
$$\sqrt{3a + b^3} + \sqrt{3b + c^3} + \sqrt{3c + a^3} \ge 6$$, but a, b, c can be every non-negative number. I proved it's true for $$a,b,c \geq 1$$, but i can't for $$a,b,c \geq 0$$
|
By Holder $$\left(\sum_{cyc}\sqrt{3a+b^3}\right)^2\sum_{cyc}\frac{(a+b)^3}{3a+b^3}\geq8(a+b+c)^3=216.$$
Thus, it's enough to prove that:
$$\sum_{cyc}\frac{(a+b)^3}{3a+b^3}\leq6$$ or $$\sum_{cyc}\frac{(a+b)^3}{a(a+b+c)^2+3b^3}\leq2$$ or
$$\sum_{cyc}(2a^8b+5a^7b^2-3a^6b^3+3a^6c^3-a^5b^4+17a^5c^4)+$$
$$+abc\sum_{cyc}(12a^6+20a^5b+11a^5c-10a^4b^2-11a^4c^2-3a^4b^3-19a^3b^2c-32a^3c^2b+9a^2b^2c^2)\geq0$$ or
$$1.5\sum_{cyc}(a^8b+a^4b^5-2a^6b^3)+0.5\sum_{cyc}(a^7b^2+a^3b^6-2a^5b^4)+9abc\sum_{cyc}(a^6-a^4b^2-a^4c^2+a^2b^2c^2)+$$
$$+3abc\sum_{cyc}(a^6-a^3b^3)+2abc\sum_{cyc}(a^5b+a^5c-a^4b^2-a^4c^2)+$$
$$+\sum_{cyc}(0.5a^8b+4.5a^7b^2+2.5a^6c^3+15.5a^5c^4+18a^6b^2c+9a^6c^2b+a^5b^3c-19a^4b^3c^2-32a^4c^3b^2)\geq0,$$
which is true by Schur, Muirhead and AM-GM.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/336367",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 7,
"answer_id": 2
}
|
Analytic Geometry (high school): Why is the sum of the distances from any point of the ellipse to the two foci the major axis? I don't understand where that formula came from. Could someone explain? For example any point $(x,y)$ on the ellipse from the two foci $(-c,0)$ and $(c,0)$ is equal to $2a$ where $2a$ is the distance of the major axis. Where did this idea come from?
|
The formula for an ellipse centred at the origin is $$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \tag{1} $$ where a>b.
Now let us define a constant (note at this point I'm just defining a constant - nothing about foci)
$$f = \sqrt{a^2 - b^2}\tag{2}$$
Let us now pick 2 points on the x axis (f,0) and (-f,0).
The distance from point (f,0) to a point on the ellipse is (by Pythagoras)
$$r_1 = \sqrt{(x-f)^2 + y^2}$$
and the distance from point (-f,0) to the same point on the ellipse is
$$r_2 = \sqrt{(x+f)^2 + y^2}$$
Looking at $r_1$ first
$$r_1 = \sqrt{(x-f)^2 + y^2}\tag{3}$$
From (1)
$$y^2 = (1-\frac{x^2}{a^2})b^2\tag{4}$$
so substituting (4) into (3) we have
$$r_1 = \sqrt{(x-f)^2 + (1-\frac{x^2}{a^2})b^2}$$
Expanding this out we get
$$r_1 = \sqrt{x^2 -2fx + f^2 + b^2 - \frac{x^2b^2}{a^2}}$$
Rearranging
$$r_1 = \sqrt{x^2(1-\frac{b^2}{a^2}) -2fx + f^2 + b^2}$$
$$ r_1 = \sqrt{x^2(\frac{a^2-b^2}{a^2}) -2fx + f^2 + b^2}$$
but from (2) $f^2 = a^2 - b^2$ so
$$ r_1 = \sqrt{x^2(\frac{f^2}{a^2}) -2fx + f^2 + b^2}$$
also from (2) $f^2 + b^2 = a^2 $ so
$$ r_1 = \sqrt{x^2(\frac{f^2}{a^2}) -2fx + a^2}$$
and rearrange under the sq root
$$r_1 = \sqrt{(a - \frac{xf}{a})^2}$$
$$r_1 = a - \frac{xf}{a}$$
Similarly you can show that
$$r_2 = a + \frac{xf}{a}$$
so
$$r_1+r_2 = 2a$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/336622",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 1
}
|
Why $\sum_{k=1}^{\infty} \frac{k}{2^k} = 2$? Can you please explain why
$$
\sum_{k=1}^{\infty} \dfrac{k}{2^k} =
\dfrac{1}{2} +\dfrac{ 2}{4} + \dfrac{3}{8}+ \dfrac{4}{16} +\dfrac{5}{32} + \dots =
2
$$
I know $1 + 2 + 3 + ... + n = \dfrac{n(n+1)}{2}$
|
Since infinite series with nonnegative terms can be rearranged arbitrarily,
$$\sum_{i=1}^\infty \frac{i}{2^i} =
\sum_{i=1}^\infty \sum_{j=1}^i \frac{1}{2^i}
= \sum_{j=1}^\infty \sum_{i=j}^\infty \frac{1}{2^i}
= \sum_{j=1}^\infty \frac{1}{2^{j-1}}
= 2
$$
More graphically,
1/2 + 2/4 + 3/8 + 4/16 + ...
= 1/2 + 1/4 + 1/8 + 1/16 + ... (= 1)
+ 1/4 + 1/8 + 1/16 + ... (= 1/2)
+ 1/8 + 1/16 + ... (= 1/4)
+ 1/16 + ... (= 1/8)
.... ....
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/337937",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "35",
"answer_count": 12,
"answer_id": 2
}
|
Recurrence relation for the number of $n$-digit ternary sequences with no consecutive $1$s or $2$s
Find the recurrence relation for the number of $n$-digit ternary sequences with no consecutive $1$'s or $2$'s.
The solution is
$$
a_n = a_{n-1} + 2a_{n-2} + 2a_{n-3} + 2a_{n-4} + \dots. \tag1
$$
I've thought about this for quite some time and I can't really understand it. I feel like I'm making up my reasoning to fit the solution; rather than understanding why it really is in the above form.
Can someone please explain this combinatorially?
|
Call the number of sequences that interest you that end in 0 $a_n$, if they end in 1 $b_n$, $c_n$ if they end in 2, and $d_n$ for ending in 3. Then clearly:
$$
a_0 = b_0 = c_0 = d_0 = 1
$$
Now think how a sequence that ends in 0 can be built: Add a 0 to any of the others, so $a_{n + 1} = a_n + b_n + c_n + d_n$. To make a sequence ending in 1, anything except a sequence ending in 1 can come before, so it is $b_{n + 1} = a_n + c_n + d_n$. Continuing the same way:
$$
\begin{align*}
a_{n + 1} &= a_n + b_n + c_n + d_n \\
b_{n + 1} &= a_n + c_n + d_n \\
c_{n + 1} &= a_n + b_n + d_n \\
d_{n + 1} &= a_n + b_n + c_n + d_n
\end{align*}
$$
This setup serves for any restrictions you care to set up.
Now define generating functions:
$$
\begin{align*}
A(z) &= \sum_{n \ge 0} a_n z^n \\
B(z) &= \sum_{n \ge 0} b_n z^n \\
C(z) &= \sum_{n \ge 0} c_n z^n \\
D(z) &= \sum_{n \ge 0} d_n z^n
\end{align*}
$$
This gives the system of equations:
$$
\begin{align*}
\frac{A(z) - 1}{z} &= A(z) + B(z) + C(z) + D(z) \\
\frac{B(z) - 1}{z} &= A(z) + C(z) + D(z) \\
\frac{C(z) - 1}{z} &= A(z) + B(z) + D(z) \\
\frac{D(z) - 1}{z} &= A(z) + B(z) + C(z) + D(z)
\end{align*}
$$
By sheer luck, the values we are interested in are just the $a_{n + 1}$. Solving the system of equations we get:
$$
A(z) = \frac{1 + z}{1 - 3 z - 2 z^2}
$$
This can be split into (very ugly) partial fractions, and expand the resulting geometric series:
$$
a_n = - \frac{\sqrt{17} - 1}{\sqrt{17} (\sqrt{17} + 3)}
\cdot \left( - \frac{4}{\sqrt{17} + 3} \right)^n
+ \frac{\sqrt{17} + 1}{\sqrt{17} (\sqrt{17} - 3)}
\cdot \left(\frac{4}{\sqrt{17} - 3} \right)^n
$$
(this would need some more cleanup, sorry for the mess).
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/338417",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
}
|
Use the $\varepsilon$ - $\delta$ definition to prove $\lim_{x\to\,-1}\frac{x}{2x+1}=1$ Use the $\varepsilon$ - $\delta$ definition of limit to prove that $\displaystyle\lim_{x\to\,-1}\frac{x}{2x+1}=1$.
My working:
$\left|\frac{x}{2x+1}-1\right|=\left|\frac{-x-1}{2x+1}\right|=\frac{1}{\left|2x+1\right|}\cdot \left|x+1\right|$
First restrict $x$ to $0<\left|x+1\right|<\frac{1}{4}$ $\Rightarrow$ initial choice of $\delta=\frac{1}{4}$
$\left| 2x+1 \right|$ = $\left|2(x+1)-1\right|$
$\le$ $\left|2(x+1)\right|+\left|-1\right|$ = $2\left|x+1\right|+1$ $> 1$
Thus if $\left|x+1\right|<\frac{1}{4}$ , then
$\left|\frac{x}{2x+1}-1\right|=\frac{1}{\left|2x+1\right|}.\left|x+1\right|$
$<1.\left|x+1\right|$.
Therefore, $\delta = \min\{\frac{1}{4},\varepsilon\}$
$0<\left|x+1\right|<\delta$ $\Rightarrow$ $\left|\frac{x}{2x+1}-1\right| < 1\cdot\left|x+1\right| < 1\cdot\varepsilon = \varepsilon$
Thus, the limit is 1.
|
Hint: Note that
\begin{align}
\left| \frac{x}{2\cdot x +1} - 1 \right| < \epsilon
\Longleftrightarrow
&
1-\epsilon< \frac{x}{2\cdot x +1}< 1+\epsilon
\\
\Longleftrightarrow
&
\left\{
\begin{array}
(2x+1)(1-\epsilon)<x \\
\\
x< (2x+1)(1+\epsilon)
\end{array}
\right.
\end{align}
After algebric manipulations of two last inequalities you get
$$
\left| \frac{x}{2\cdot x +1} - 1 \right| < \epsilon
\Longleftrightarrow
\left\{
\begin{array}
\;x-1<\frac{1+\epsilon}{1+2\epsilon}\\
\\
x-1< \frac{1-\epsilon}{1-2\epsilon}
\end{array}
\quad \mbox{ for } \epsilon <\frac{1}{2}
\right.
$$
This suggests $\delta=\min\{\frac{1-\epsilon}{1-2\epsilon},\frac{1-\epsilon}{1-2\epsilon}\}$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/338933",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
}
|
generating functions / combinatorics Calculate number of solutions of the following equations:
$$ x_1 + x_2 + x_3 + x_4 = 15 $$
where $ 0 \le x_i < i + 4 $
I try to solve it using generating functions/enumerators :
$$ (1+x+x^2+x^3+x^4)(1+x+x^2+x^3+x^4+x^5)(1+x+x^2+x^3+x^4+x^5+x^6)(1+x+x^2+x^3+x^4+x^5+x^6+x^7)$$
and take coefficient near $15$. But I do not know how to quickly calculate it. Maybe there exists any faster way?
|
Use the stars and bars technique to solve for the number of solutions without restrictions, and look at how to count the solutions that violate the restrictions.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/339230",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
}
|
How do I write this matrix in Jordan-Normal Form I have the matrix $A=\begin{pmatrix}2&2&1\\-1&0&1\\4&1&-1\end{pmatrix}$, I want to write it in Jordan-Normal Form. I have $x_1=3,x_2=x_3=-1$ and calculated eigenvectors $v_1=\begin{pmatrix}1\\0\\1\end{pmatrix},v_2=\begin{pmatrix}1\\-4\\5\end{pmatrix},v_3=\begin{pmatrix}0\\0\\0\end{pmatrix}$. But, the matrix $Z=\begin{pmatrix}1&1&0\\0&-4&0\\1&5&0\end{pmatrix}$ is not invertible since $\text{det}(Z)=0$. Does this mean the matrix cannot be written in JNF or do I need to find different eigenvectors?
I have tried to find different eigenvectors, but keep arriving at the same problem, any suggestions?
Thanks
|
You need a generalized eigenvector for the third eigenvalue and it looks like that went wrong somehow, so lets fix it.
We have $\left(A - \lambda_2 I\right)v_3 = v_2$
From this, we get:
$\begin{pmatrix}3&2&1&1\\-1&1&1&-4\\4&1&0&5\end{pmatrix}$
The RREF yields:
$\displaystyle \begin{pmatrix}1&0&-\frac{1}{5}&\frac{9}{5}\\0&1&\frac{4}{5}&-\frac{11}{5}\\0&0&0&0\end{pmatrix}$
This gives us a generalized eigenvector of: $\displaystyle \left(\frac{9}{5}, -\frac{11}{5}, 0 \right)$.
To write the Jordan Normal Form, we form:
$\displaystyle A = S\cdot J\cdot S^{-1} = \begin{pmatrix} 1 & \frac{9}{5} & 1\\
-4 & -\frac{11}{5} & 0 \\ 5 & 0 & 1\end{pmatrix} \cdot \begin{pmatrix} -1 & 1 & 0\\ 0 & -1 & 0\\ 0 & 0 & 3 \end{pmatrix} \cdot \begin{pmatrix} -\frac{11}{80} & -\frac{9}{80} & \frac{11}{80} \\ \frac{1}{4} & -\frac{1}{4} & -\frac{1}{4} \\ \frac{11}{16} & \frac{9}{16} & \frac{5}{16}\end{pmatrix}$.
Notice the structure of the Jordan block. Also, notice what the columns of $S$ and $J$ are made of? Clear?
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/343941",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
}
|
Sum of $\sum_{n=0}^\infty \frac{(x+2)^{n+2}}{3^n} $ Calculate the sum of the next series and for which values of $x$ it converges:
$$\sum_{n=0}^\infty \frac{(x+2)^{n+2}}{3^n}$$
I used D'Alembert and found that the limit is less than 1, so: $-5 < x < 1$ (because the fraction must be less than 1).
and then I assigned the values: $x=-5$ and $x=1$ in the series and got:
for $x=-5$ and $x=1$, it diverges.
then the series converges in the range of $(-5,1)$, $R=3$ and the center point is for $x=2$.
Please let know if there is a mistake and find the sum.
|
for $|(x+2)/3|<1$ it converges to the limit given by multiplication of geometric series limit and polynomial:
$$
\sum_{n=0}^\infty\frac{(x+2)^{n+2}}{3^n}=(x+2)^2\sum_{n=0}^\infty\left(\frac{x+2}{3}\right)^n=(x+2)^2\frac{1}{1-\frac{x+2}{3}} =\frac{3(x+2)^2}{1-x}
$$
for $|(x+2)/3|\geq 1$, the sum is not convergent.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/348079",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
}
|
Integrate rational function $\frac{x^2}{1+x^4}$ Integrate $$\int\frac{x^2dx}{1+x^4}$$
I've factored the denominator to $(x^2-\sqrt{2}x+1)(x^2+\sqrt{2}x+1)$ and got stuck.
|
The next step is to perform a partial fraction decomposition:
$$
\frac{x^2}{(x^2-\sqrt{2}x+1)(x^2+\sqrt{2}x+1)} = \frac{1}{2 \sqrt{2}} \frac{x}{x^2-\sqrt{2}x+1} - \frac{1}{2 \sqrt{2}} \frac{x}{x^2+\sqrt{2}x+1}
$$
And the than use the table anti-derivative for $\int \frac{x}{x^2 + a x+b} \mathrm{d}x$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/349424",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
}
|
Help please on complex polynomials I wanted to know if there's any good approaches to these questions
a)By considering $z^9-1$ as a difference of two cubes, write $1+z+z^2+z^3+z^4+z^5+z^6+z^7+z^8$
as a product of two real factors one of which is a quadratic.
b) Solve $z^9-1=0$ and hence write down the 6 solutions of $z^6+z^3+1=0$
c)By letting $y=z+\frac{1}{z}$ and dividing $z^6+z^3+1=0$ by $z^3$, deduce that:
$cos\frac{2\pi}{9}+cos\frac{4\pi}{9}+cos\frac{8\pi}{9}=0$
Any help would be greatly appreciated.
|
Hint: $a^3-b^3 = (a-b) (a^2+a b+b^2)$. Think about this with $a=z^3$ and $b=1$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/350879",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
Cauchy's Residue Theorem I need to evaluate $\int_C \frac{5z-2}{z(z-1)}dz$ where $C$ is the circle $|z|$=2. I used partial fraction decomposition to get $$\frac{5z-2}{z(z-1)}=\frac{2}{z}+\frac{3}{z-1}$$ I have the answer ($10\pi i$) in my book but I don't fully understand how to get it. I think I have to evaluate in the two domains $0<|z|<1$ and $0<|z-1|<1$, is it because the singularities are at $z=0$ and $z=1$? I ask mainly because of the $|z-1|$ in the second inequality.
|
I don't think you need the Residue Theorem, but just Cauchy's integral formula? Note that $$\dfrac{5z - 2}{z(z-1)} = \frac{3}{z-1} + \frac{2}{z}$$ and so $$\int_{C} \frac{5z-2}{z(z-1)}dz = \int_{C}\frac{3}{z-1} dz + \int_{C}\frac{2}{z} dz = 2\pi i f_1(1) + 2\pi i f_2(0)$$ where $f_1(z) = 3$ and $f_2(z) = 2$, so $2\pi i f_1(1) + 2\pi i f_2(0) = 2\pi i \cdot 3 + 2\pi i \cdot 2 = 10 \pi i$.
However, you can also use the residue theorem. Note that $\frac{3}{z-1}$ is already a Laurent series expanded about $z = 1$ when $0 < |z-1| < 1$ and $\frac{2}{z}$ is one as well expanded about $z = 0$ when $0 < |z| < 1$. Therefore,
$$\dfrac{5z - 2}{z(z-1)} = \frac{3}{z-1} + \frac{2}{z}$$ and so $$\int_{C} \frac{5z-2}{z(z-1)}dz = \int_{C}\frac{3}{z-1} dz + \int_{C}\frac{2}{z} dz = 2\pi i \mbox{Res}\left(\frac{3}{z-1}\right) + 2\pi i\mbox{Res}\left(\frac{2}{z}\right)$$ where
$$\mbox{Res}\left(\frac{3}{z-1}\right) = 3 \quad \mbox{and} \quad \mbox{Res}\left(\frac{2}{z}\right) = 2$$
so you again get $10 \pi i$.
The choice of the domains is because $0 < |z| < 1$ and $0 < |z-1| < 1$ is where the function $(5z-2)/(z(z-1))$ is analytic and so justifies the residues. That is, $2/z$ is already a Laurent series when $0 < |z| < 1$ with $\mbox{Res}(2/z) = 2$ and $3/(z-1)$ is already a Laurent series when $0 < |z-1| < 1$ with $\mbox{Res}(3/(z-1)) = 3$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/351364",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
}
|
arithmetic progression of triangle sides Let $gcd(a,b,c)=1$ such that $a^2, b^2, c^2$ are in arithmetic progression. Show they can be written in the form
$a=-p^2+2pq+q^2$
$b=p^2+q^2$
$c=p^2+2pq-q^2$
for relatively prime integers $p,q$ of different parities.
so if $a^2, b^2, c^2$ are in arithmetic progression, then
$a^2=x, b^2=x+d, c^2=x+2d$, for $x,d\in\mathbb{Z}$
That means $x+x+d=x+d+d \Rightarrow 2x+d=x+2d \Rightarrow x=d$.
So then $a^2=x, b^2=2x, c^2=3x$
Now with primitive pythagorean triples, we have $(2pq, p^2-q^2, p^2+q^2)$. I just don't see how to go from here....i'm sure it's easy, but I'm not seeing something.
|
$a^2, b^2,$ and $c^2$ are in AP if and only if $a^2+c^2=2b^2$.
Looking at $a,b,$ and $c$ modulo $4$, we see that they must all be odd integers.
Then $u=\frac 12(c-a)$ and $w=\frac 12(c+a)$ are also integers.
In fact $a=w-u$ and $c=w+u$.
So $2b^2=a^2+c^2=(w-u)^2+(w+u)^2=2(u^2+w^2)$.
Hence $u^2+w^2 = b^2$.
So there exists coprime $p$ and $q$ such that
If $2pq < q^2-p^2$, then
\begin{align}
u &= 2pq \\
w &= q^2-p^2 \\
b &= q^2+p^2 \\
\hline
a &= q^2-p^2-2pq \\
c &= q^2-p^2+2pq \\
b &= q^2+p^2 \\
\end{align}
If $2pq > q^2-p^2$, then
\begin{align}
u &= q^2-p^2 \\
w &= 2pq \\
b &= q^2+p^2 \\
\hline
a &= 2pq-(q^2-p^2) \\
c &= 2pq+(q^2-p^2) \\
b &= q^2+p^2 \\
\end{align}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/351757",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
Sketch of the ordinate set of $f$ Let $f$ be defined on $[0,1] \times [0,1]$ as follows:
$f(x,y)= \begin{cases} x+y \mbox{ if } x^2 \leq y \leq 2x^2 \\ 0 \mbox{ otherwise} \end{cases}$
I want to make a sketch of the ordinate set of $f$ over $[0,1] \times [0,1]$ and compute the volume of this ordinate set by double integration (Assuming the integral exists)
|
Comment by OP:
If $f$ is nonnegative, the set $S$ of points $(x,y,z)$ in 3-space with
$(x,y)$ in $[0,1]^2$ and $0\le z\le f(x,y)$ is called the ordinate set
of $f$ over $[0,1]$.
My interpretations is as follows. The picture represents a region $R$ bounded by $y=2x^2$, $y=x^2$ (with $0\le x\le 1$) and $y=1$.
$$y=2x^2\text{ (blue)}, y=x^2\text{ (green)}$$
The volume bounded by $R$ and $0\le z\le x+y$ is given by
\begin{eqnarray*}
I &=&\iint_{R}x+y\,dA=\int_{0}^{\sqrt{2}/2}\left(
\int_{x^{2}}^{2x^{2}}x+y\,dy\right) dx+\int_{\sqrt{2}/2}^{1}\left(
\int_{x^{2}}^{1}x+y\,dy\right) dx \\
&=&\int_{0}^{\sqrt{2}/2}x^{3}+\frac{3}{2}x^{4}\,dx+\int_{\sqrt{2}%
/2}^{1}x-x^{3}+\frac{1}{2}-\frac{1}{2}x^{4}\,dx \\
&=&\frac{1}{16}+\frac{3}{80}\sqrt{2}+\frac{37}{80}-\frac{19}{80}\sqrt{2} \\
&=&\frac{21}{40}-\frac{1}{5}\sqrt{2}.
\end{eqnarray*}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/354185",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
Really Confused on a surface area integral can't seem to finish the integral off. Basically the question asks to compute $\int \int_{S} ( x^{2}+y^{2}) dA$ where S is the portion of the sphere $x^{2} + y^{2}+ z^{2}= 4$ and $z \in [1,2]$ we start with a chnage of variables
$x=x $
$y=y$
$ z= 2 \cdot(4-(x^{2} + y^{2}))^{1/2}$
$Det(u,v)= \begin{bmatrix}
i & j& k \\
1 & 0 & \frac {-x}{(4-(x^{2} + y^{2}))^{1/2}} \\
0 & 1 & \frac {-y}{(4-(x^{2} + y^{2}))^{1/2}} \\
\end{bmatrix}=(\frac {x}{(4-(x^{2} + y^{2}))^{1/2}})i + (\frac {y}{(4-(x^{2} + y^{2}))^{1/2}})j + k$
$dA=(\frac {x^{2}+y^{2}}{(4-(x^{2} + y^{2}))} +1)^{1/2}$
$\int \int_{S} (\frac {(x^{2}+y^{2})^{3}}{(4-(x^{2} + y^{2}))}+(x^{2}+y^{2})^{2})^{1/2}$
Projecting when z=1 and z=2 we have $x^{2} + y^{2}= 4-1$ $\to r= 0,(3)^{1/2}$
going to polar we have:
$(\frac {r^{6}}{(4-r^{2})}+r^{4})^{1/2}rdrd\theta=(\frac {4r^{4}}{(4-r^{2})})^{1/2}rdrd\theta$
my problem is $2\int^{2\pi}_{0} \int^{(3)^{1/2}}_{0} (\frac {4r^{4}}{(4-r^{2})})^{1/2}rdrd\theta$ there is no nice way i can think of to integrate this. it can also be written as:
$2\int^{2\pi}_{0} \int^{(3)^{1/2}}_{0} \frac {2r^{2}}{((4-r^{2}))^{1/2}}rdrd\theta$
|
It seems that spherical coordinates are more appropriate for calculating $\iint\limits_{S} ( x^{2}+y^{2}) dA$.
In the last integral
$$
2\int^{2\pi}_{0} \int^{(3)^{1/2}}_{0} \frac {2r^{2}}{((4-r^{2}))^{1/2}}rdrd\theta=
{4\pi}\int^{(3)^{1/2}}_{0} \frac {2r^{2}}{((4-r^{2}))^{1/2}}r\,dr$$
you can make the substitution $r=2\sin{t}.$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/356385",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
}
|
Solve : $ab(a+b)(a-b)=c^2-1$ As we know that $ab(a+b)(a-b)=c^2$ has no integer solution in $Z^+$.However, it seems that $$ab(a+b)(a-b)=c^2-1$$
has infinite positive integer solutions,could you prove it?
Here are some of them:
$(a,b,c)=(3, 1, 5), (5, 1, 11), (7, 3, 29), (7, 5, 29), (8, 2, 31), (8, 7, 29), (9, 8, 35), (13, 3, 79), (15, 8, 139), (15, 11, 131), (17, 7, 169), (20, 6, 209), (20, 14, 239), (21, 5, 209), (27, 8, 379), (28, 2, 209), (29, 16, 521)...$
Thanks in advance!
|
Let us want
$$a(a-b) = c+1 \text{ and }b(a+b) = c-1$$
For this, we need
$$2 = a(a-b) - b(a+b) = a^2 - 2ab -b^2 = (a-b)^2 - 2b^2$$
Hence, all we want is to solve the Pell's equation
$$x^2-2y^2 =2$$
One trial solution to the above is $(2,1)$ and now generate the rest of $x$ and $y$. From this we get $b=y$ and $a = x+y$. This gives us $$c = \dfrac{a^2+b^2}2$$
Hence, to summarize the solution is as follows. Find the solutions to the Pell's equation
$$2x^2-b^2 = 1$$
(We have replaced $x$ by $2x$ and $y$ by $b$. Hence we get this Pell's equation, which is even more easier to obtain infinite solutions for $x$ and $b$).
We then have
\begin{align}
a & = 2x+b\\
c & = 2x^2 + 2xb + b^2
\end{align}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/356748",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 2,
"answer_id": 1
}
|
Computing the integral $\int \sqrt{\frac{x-1}{x+1}}\,\mathrm dx $ How do I compute the next integral:
$$\int \sqrt{\frac{x-1}{x+1}}\,\mathrm dx \;?$$
|
Let $\sqrt{\dfrac{x-1}{x+1}} = t$. We then get that
$$\dfrac{x-1}{x+1} = t^2 \implies x-1 = t^2(x+1) \implies x = \dfrac{1+t^2}{1-t^2} \implies dx = \dfrac{4t}{(1-t^2)^2}dt$$
Hence, we get that
$$\int \sqrt{\dfrac{x-1}{x+1}} dx = \int \dfrac{4t^2}{(1-t^2)^2} dt$$ I trust you can take it from here via the method of partial fractions.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/357306",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
}
|
Constructing a degree 4 rational polynomial satisfying $f(\sqrt{2}+\sqrt{3}) = 0$ Goal: Find $f \in \mathbb{Q}[x]$ such that $f(\sqrt{2}+\sqrt{3}) = 0$.
A direct approach is to look at the following
$$
\begin{align}
(\sqrt{2}+\sqrt{3})^2 &= 5+2\sqrt{6} \\
(\sqrt{2}+\sqrt{3})^4 &= (5+2\sqrt{6})^2 = 49+20\sqrt{6} \\
\end{align}
$$
Putting those together gives
$$
-1 + 10(\sqrt{2}+\sqrt{3})^2 - (\sqrt{2}+\sqrt{3})^4 = 0,
$$
so $f(x) = -1 + 10x^2 - x^4$ satisfies $f(\sqrt{2}+\sqrt{3}) = 0$.
Is there a more mechanical approach? Perhaps not entirely mechanical, but something more abstract.
|
You can guess that the conjugates will be $\pm \sqrt 2 \pm \sqrt 3$, and multiply all the corresponding linear factors together.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/359054",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 5,
"answer_id": 2
}
|
$x^4 + 4x^3 - 2x^2 - 12x + k$ has 4 real roots. Find the condition on k. The question is: $f(x) = x^4 + 4x^3 - 2x^2 - 12x + k$ has 4 real roots. What values can k take? Please drop a hint!
|
Hint:
Lets call vieta for the rescue.
$f(x) = x^4 + 4x^3 - 2x^2 - 12x + k=0$, let $x_1, x_2,x_3$ and $x_4$are the roots.
$\sum x_i=-4$
$\sum x_ix_j=-2$
$\sum x_ix_jx_k=+12$
$x_1x_2x_3x_4=k$
$\sum (x_i)^2=(\sum x_i)^2-2\sum x_ix_j=16-(-4)=12$
$\dfrac{x_1^2+x_2^2+x_3^2+x_4^2}{4} \ge (x_1x_2x_3x_4)^\frac{1}{2}$
$3 \ge k^{\frac{1}{2}} \implies 9 \ge k$. Now check the lower bound using the discriminant formula.
HAIL VIETA!, like I always say.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/359513",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
}
|
How do I evaluate $\lim_{n \rightarrow \infty} \frac{1\cdot2+2\cdot3 +3\cdot4 +4\cdot5+\ldots}{n^3}$ How to find this limit : $$\lim_{n \rightarrow \infty} \frac{1\cdot2+2\cdot3 +3\cdot4 +4\cdot5+\ldots+n(n+1)}{n^3}$$ As, if we look this limit problem viz. $\lim_{x \rightarrow \infty} \frac{1+2+3+\ldots+n}{n^2}$ then we take the sum of numerator which is sum of first n natural numbers and we can write :
$$\lim_{x \rightarrow \infty} \frac{n(n+1)}{2n^2}$$ which gives after simplification :
$$ \frac{1}{2} $$ as other terms contain $\frac{1}{x}$ etc. and becomes zero.
|
$$\sum_{1\le r\le n}r(r+1)=\sum_{1\le r\le n}r^2+\sum_{1\le r\le n}r=\frac{n(n+1)(2n+1)}6+\frac{n(n+1)}2$$
So, $$\lim_{n\to\infty}\frac{\sum_{1\le r\le n}r(r+1)}{n^3}$$
$$=\lim_{n\to\infty}\left(\frac{\frac{n(n+1)(2n+1)}6+\frac{n(n+1)}2}{n^3}\right)$$
$$=\lim_{n\to\infty}\left(\frac{(1+\frac1n)(2+\frac1n)}6+\frac{(\frac1n+\frac1{n^2})}2\right)$$
$$=\frac{1\cdot2+0}6+\frac02=\frac13$$
Alternatively,
$$\frac{n(n+1)(2n+1)}6+\frac{n(n+1)}2=\frac{n(n+1)}6\left(2n+1+3\right)=\frac{n(n+1)(n+2)}3$$
$$\implies\lim_{n\to\infty}\frac{\sum_{1\le r\le n}r(r+1)}{n^3}=\lim_{n\to\infty}\frac{n(n+1)(n+2)}{3n^3}=\frac13\lim_{n\to\infty}\left(1+\frac1n\right)\left(1+\frac2n\right)=\frac13$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/364284",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 2
}
|
Limits at negative infinity Prove that $\lim_{x\to -\infty}$ $\frac{x}{x+2}$ = 1.
Please help me check if this is correct:
Let $\varepsilon > 0$ be given. Choose $M=2-\frac{2}{\varepsilon}$
$$\begin{align}
x<M \Rightarrow x-2 < M-2\\
\frac{1}{x-2} > \frac{1}{M-2}\\
|f(x)-1|=|\frac{x}{x-2}-1|=|\frac{2}{x-2}|\\
<|\frac{2}{M-2}|=\frac{-2}{M-2}=\varepsilon
\end{align}$$
|
Since $x \to -\infty$, we may assume $M < 0$. Then your line $\frac{1}{x-2} > \frac{1}{M-2}$ is a relationship between two negative quantities. Thus $\left|\frac{1}{x-2}\right| < \left|\frac{1}{M-2}\right|$. Can you take it from there?
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/364432",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Computing $\int\frac{7x^{13}+5x^{15}}{(x^7+x^2+1)^3}\,dx$
Compute the indefinite integral
$$
\int\frac{7x^{13}+5x^{15}}{(x^7+x^2+1)^3}\,dx
$$
My Attempt:
$$
\int\frac{7x^{13}+5x^{15}}{x^{21}(x^{-7}+x^{-5}+1)^3}\,dx = \int\frac{7x^{-8}+5x^{-6}}{(x^{-7}+x^{-5}+1)^3}\,dx
$$
Let $t=(x^{-7}+x^{-5}+1)$ such that
$$
\begin{align}
dt&=(-7x^{-8}-5x^{-6})\,dx\\
-dt&=(7x^{-8}+5x^{-6})\,dx
\end{align}
$$
We can change the variables of the integral to get
$$
\begin{align}
-\int \frac{1}{t^3}\,dt &=\frac{1}{2}\cdot\frac{1}{t^2}+C\\
&= \frac{1}{2}.\frac{1}{(x^{-7}+x^{-5}+1)^2}+C\\
&= \frac{x^{14}}{2.(1+x^2+x^7)^2}+C
\end{align}
$$
I'd prefer to compute the integral using methods of differentiation rather than integration, i.e.
$$\frac{7x^{13}+5x^{15}}{(x^7+x^2+1)^3} = \frac{d}{dx} \left(\frac{ax^2+bx+c}{(x^7+x^2+1)^2}\right)
$$
but I could not get the same answer as above.
|
Your first computation is correct.
If $P$ is a polynomial of degree $n$, then
$$
\frac{d}{dx} \Biggl(\frac{P(x)}{(x^7+x^2+1)^2}\Biggr)=\frac{Q(x)}{(x^7+x^2+1)^3}
$$
where
$$
Q(x)=(x^7+x^2+1)P'(x)-2(7\,x^6+2\,x)P(x)
$$
is a polynomial of degree $\le n+6$. To get $x^{15}$ in the numerator you need $n\ge9$ (in fact, in this case $n=14$).
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/365313",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 1,
"answer_id": 0
}
|
Determine the limiting behaviour of $\lim_{x \to \infty}{\frac{\sqrt{x^4+1}}{\sqrt[3]{x^6+1}}}$ Determine the limiting behaviour of $\lim_{x \to \infty}{\dfrac{\sqrt{x^4+1}}{\sqrt[3]{x^6+1}}}$
Used L'Hopitals to get $\;\dfrac{(x^6+1)^{\frac{2}{3}}}{x^2 \sqrt{x^4+1}}$ but not sure what more i can do after that.
|
$$ {\frac{\sqrt{x^4+1}}{\sqrt[3]{x^6+1}}}\sim {\frac{\sqrt{x^4}}{\sqrt[3]{x^6}}}=\frac{x^2}{x^2}=1 $$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/367060",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 3
}
|
How to find $\cos A \cos B - \sin A \sin B$? Given that:
$\tan A=1$
and
$\tan B = \sqrt{3}$
How would you find $\cos A \cos B - \sin A \sin B$?
EDIT: This is what I've tried after reading bhattacharjee's answer:
$$ \tan(A+B) = \tan A+\tan B−\tan A\tan B$$
so,
$\tan(A+B)= {1+\sqrt{3} \over 1-\sqrt{3}}$
from this I get $1 \over \cos(A+B)^2 $ $=1+ \left ( {1+\sqrt{3} \over 1-\sqrt{3}}\right )^2$
=> $ 1 \over \cos^2(A+B) $ $=$ $ 8 \over 4-2 \sqrt{3}$
Is this right, because it seems like a dead end to me? How am I supposed to proceed from here?
|
HINT:
$\cos A \cos B - \sin A \sin B=\cos(A+B)$
and $$\tan(A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}$$
Do you know how to find $\cos \theta$ from $\tan\theta?$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/367477",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
}
|
Finding the general formula $a_n$ for $a_n = \frac{1}{2a_{n-1}} + 2a_{n-2}$ How to calculate the general formula $a_n$ for the following sequence:
$$a_n = \frac{1}{2a_{n-1}} + 2a_{n-2}$$
where $a_1=\frac{1}{2}, a_2=\frac{1}{4}$
|
$$\begin{align}&a_{n} = \frac{1}{2 a_{n-1}} + 2 a_{n-2}\\
\iff & 2 a_{n} a_{n-1} + 1 = 2 ( 2 a_{n-1} a_{n-2} + 1)\\
\implies & 2 a_{n} a_{n-1} + 1 = 2^{n-2} (2 a_2 a_1 + 1 ) = \frac{5}{16} 2^n\\
\implies & a_{n}/a_{n-2} = \frac{\frac{5}{16} 2^n - 1}{\frac{5}{32} 2^n - 1}\\
\implies & a_{n} = \begin{cases}
a_2 \prod_{k=0}^{m-1} (\frac{\frac{5}{16} 2^n - 4^k}{\frac{5}{32} 2^n - 4^k}), & \text{for}\;n = 2m\\
a_1 \prod_{k=0}^{m} (\frac{\frac{5}{16} 2^n - 4^k}{\frac{5}{32} 2^n - 4^k}), &
\text{for}\;n = 2m+1
\end{cases}
\end{align}$$
This give us a pretty ugly list $a_i = ( \frac{1}{2},\frac{1}{4},3,\frac{2}{3},\frac{27}{4},\frac{38}{27},\frac{1053}{76},\frac{3002}{1053},\frac{167427}{6004},\frac{957638}{167427}, \ldots )$ and I cannot see any obvious pattern in it.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/371233",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
}
|
Expanding complex geometric series I'm having trouble with part $ii.$ of the following question:
$i.$ Express the following in terms of N and z:
$$\sum^N_{n=1}2^{-n}z^n$$
Expanding with geometric series:
$$\sum^N_{n=1}2^{-n}z^n \equiv \sum^N_{n=1}\bigg(\frac{z}2\bigg)^n = \bigg(\frac{z}2\bigg) + \bigg(\frac{z}2\bigg)^2 + \bigg(\frac{z}2\bigg)^3 + ... + \bigg(\frac{z}2\bigg)^N$$
$$= \frac{\bigg(\frac{z}2\bigg)\bigg(1-\bigg(\frac{z}2\bigg)^N\bigg)}{1 - \bigg(\frac{z}2\bigg)} = \frac{z\bigg(1-\bigg(\frac{z}2\bigg)^N\bigg)}{2-z}$$
$ii.$ Deduce that:
$$\sum^{10}_{n=1}2^{-n}\sin(\frac1{10}n\pi) = \frac{1025\sin(\frac1{10}\pi)}{2560-2048\cos(\frac1{10}\pi)}$$
Looking instead at the imaginary part of the following series:
$$\sum^{10}_{n=1}2^{-n}e^{i\frac{n\pi}{10}} = \frac12e^{i\frac{\pi}{10}} + \frac14e^{i\frac{2\pi}{10}} + \frac18e^{i\frac{3\pi}{10}} + ... + \frac1{2^{10}}e^{i\pi}$$
$$=\frac{\bigg(e^{i\frac\pi{10}}\bigg)\bigg(1-\bigg(\frac{e^{i\frac\pi{10}}}2\bigg)^{10}\bigg)}{2-e^{i\frac\pi{10}}}
= \frac{\bigg(e^{i\frac\pi{10}}\bigg)\bigg(1-\frac{e^{i\pi}}{2^{10}}\bigg)}{2-e^{i\frac\pi{10}}}$$
Applying $-e^{i\pi} \equiv 1$ and simplifying, then applying $e^{i\theta}\equiv r(\cos\theta+i\sin\theta)$:
$$\implies \sum^{10}_{n=1}2^{-n}e^{i\frac{n\pi}{10}}
= \frac{\bigg(e^{i\frac\pi{10}}\bigg)\bigg(2^{10}+1\bigg)}{2^{10}(2-e^{i\frac\pi{10}})}
= \frac{1025(\cos\frac\pi{10}+i\sin\frac\pi{10})}{2048-1024(\cos\frac\pi{10}+i\sin\frac\pi{10})}$$
My understanding is that, from the final equation, looking at it's imaginary part we would have to take the $i$ term from the numerator and the real term from the denominator, like so:
$$= \frac{1025(\sin\frac\pi{10})}{2048-1024(\cos\frac\pi{10})}$$
But how do I satisfy the denominator to get the required $(2560 - 2048\cos\frac\pi{10})$?
Edit:
On second thought, adding $512-1024\cos\frac\pi{10}$ to the denominator does not balance the equation, which means that this is not the correct sum.
|
HINT:
$$\frac{\cos t+i\sin t}{2-\cos t-i\sin t}$$
$$=\frac{(\cos t+i\sin t)(2-\cos t+i\sin t)}{(2-\cos t)^2+\sin^2 t}$$
$$=\frac{(2\cos t-1+2i\sin t)}{5-4\cos t}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/372610",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.