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Prove $ax^2+bx+c=0$ has no rational roots if $a,b,c$ are odd If $a,b,c$ are odd, how can we prove that $ax^2+bx+c=0$ has no rational roots?
I was unable to proceed beyond this: Roots are $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
and rational numbers are of the form $\frac pq$.
| $x= \dfrac{-b+ \sqrt{b^2-4ac}}{2a} $ or $\dfrac{-b- \sqrt{b^2-4ac}}{2a}$
If $x=\dfrac{p}{q} \implies b^2-4ac=k^2$, and $k$ is odd. ($odd-even=odd$).
Considering $a,b,c$ odd.
$k^2 \equiv 1 \mod 8$
$b^2 \equiv 1 \mod 8$
$4ac \equiv 4 \mod 8$
$b^2-4ac=-3 \mod 8 \implies 5 \mod 8$, a contradiction.(Either of $a$ or $c$ has to be even), since $k^2 \equiv 1 \mod 8$
$\therefore$ You don't get rational roots when all are odd.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/339605",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 11,
"answer_id": 7
} |
Stirling's Approximation and Ramanujan's Proof of Bertrand's Postulate Perhaps, the most astounding step in Ramanujan's proof of Betrand's postulate is his application of Stirling's approximation.
He starts with the following inequality:
$\log\Gamma(x) - 2\log\Gamma(\frac{1}{2}x + \frac{1}{2}) \le \log[x]! - 2\log[\frac{1}{2}x]! \le \log\Gamma(x+1) - 2\log\Gamma(\frac{1}{2}x + \frac{1}{2})$
Then, applying Stirling's approximation, Ramanujan gets to:
$\log[x]! - 2\log[\frac{1}{2}x]! < \frac{3}{4}x$ if $x > 0$
and
$\log[x]! - 2\log[\frac{1}{2}x]! > \frac{2}{3}x$ if $x > 300$
I would be very interested in understanding how Stirling's approximation gets us to these two conclusions.
As I understand it, Ramanujan is refering to Stirling's Approximation for the Gamma function which as I understand to be this (from Wikipedia):
$\Gamma(z) = \sqrt{\frac{2\pi}{z}}(\frac{z}{e})^{z}(1 + O(\frac{1}{z}))$
If someone could provide the details, I would greatly appreciate it! :-)
| To make the bounds explicit we can start with the earlier form of Stirling's Approximation
$$
\log \Gamma(z) = \left(z-\frac{1}{2}\right)\log z - z + \frac{1}{2}\log (2\pi)
+ \sum_{n=1}^{k-1} \frac{B_{2n}}{2n(2n-1)z^{2n-1}}+R_k(z)
$$
If we take $k-1$ terms in the sum then $|R_k|$ is bounded by the $k$th term (see the answers to this question and the citations). In particular take $k=1$ then $|R_1|\le\frac{1}{12z}$.
Then
$$
\begin{align}
\log \Gamma(z)-2\log\Gamma\left(\frac{z+1}{2}\right)
&= z\log2-z\log\frac{z+1}{z}-\frac{1}{2}\log\frac{2\pi z}{e^2}+R_1(z)-2R_1\left(\frac{z+1}{2}\right)\\
&=z\log2+\Delta_1(z) \\
&=\frac{2}{3}z + (Az + \Delta_1(z))
\end{align}
$$
where $A=\log 2-2/3=0.02648\cdots$. Now we need to find the minimum $z$ that guarantees $Az+\Delta_1(z)>0$. Using $|R_1(z)-2R_1((z+1)/2)|< \frac{5}{12z}$ numerically we find $z>126$ is sufficient.
For the other side
$$
\begin{align}
\log \Gamma(z+1)-2\log\Gamma\left(\frac{z+1}{2}\right)
&= z\log2+\frac{1}{2}\log\frac{z+1}{2\pi}+R_1(z+1)-2R_1\left(\frac{z+1}{2}\right)\\
&=z\log2+\Delta_2(z) \\
&=\frac{3}{4}z - (Bz - \Delta_2(z))
\end{align}
$$
where $B=3/4-\log 2=0.05685\cdots$. Using $|R_1(z+1)-2R_1((z+1)/2)|\le \frac{5}{12(z+1)}$ we find that $Bz-\Delta_2(z)>0$ for all $z>0$.
| {
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"url": "https://math.stackexchange.com/questions/342520",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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} |
The minimum value of $a^2+b^2+c^2+\frac1{a^2}+\frac1{b^2}+\frac1{c^2}?$ I came across the following problem :
Let $a,b,c$ are non-zero real numbers .Then the minimum value of $a^2+b^2+c^2+\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}?$ This is a multiple choice question and the options are $0,6,3^2,6^2.$
I do not know how to progress with the problem. Can someone point me in the right direction? Thanks in advance for your time.
| Did you try $AM-GM$?
$\dfrac{(a^2+b^2+c^2+\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2})}{6} \ge 1$
$a^2+b^2+c^2+\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2} \ge 6$
OR
$\sum (a_i-\dfrac{1}{a_i})^2 \ge 0$
$\sum (a_i)^2+\dfrac{1}{(a_i)^2} \ge 2 \cdot i$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/345379",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 3
} |
Finding determinant of a simple matrix Can someone please explain how to compute the determinant of $J_n - I_n$ where $j_n$ it a matrix of ones?
E.g. for $n=5$ we get the following matrix
$$\left(\begin{array}{ccccc}
0 & 1 & 1 &1 &1 \\
1 & 0 & 1 &1 &1 \\
1 & 1 & 0 &1 &1 \\
1 & 1 & 1 &0 &1 \\
1 & 1 & 1 &1 &0 \end{array} \right)$$
By computing it for the first few $n$'s it seems to be $(-1)^{n+1}(n-1)$ but I couldn't prove it.
| Let us put
$$
D =\left|\begin{array}{ccccc}
0 & 1 & 1 &1 &1 \\
1 & 0 & 1 &1 &1 \\
1 & 1 & 0 &1 &1 \\
1 & 1 & 1 &0 &1 \\
1 & 1 & 1 &1 &0
\end{array} \right|
$$
Performing the operation $L_1 \gets \sum_{k=1} L_k$ and putting $m=n-1$, we obtain
$$
D =\left|\begin{array}{ccccc}
m & m & m &m &m \\
1 & 0 & 1 &1 &1 \\
1 & 1 & 0 &1 &1 \\
1 & 1 & 1 &0 &1 \\
1 & 1 & 1 &1 &0
\end{array} \right|=
m\left|\begin{array}{ccccc}
1 & 1 & 1 &1 &1 \\
1 & 0 & 1 &1 &1 \\
1 & 1 & 0 &1 &1 \\
1 & 1 & 1 &0 &1 \\
1 & 1 & 1 &1 &0
\end{array} \right|
$$
Performing $L_j \gets L_j-L_1$ for all $j\gt 1$, we obtain
$$
D =
m\left|\begin{array}{ccccc}
1 & 1 & 1 &1 &1 \\
0 & -1 & 0 &0 &0 \\
0 & 0 & -1 &0 &0 \\
0 & 0 & 0 &-1 &0 \\
0 & 0 & 0 &0 &-1 \\
\end{array} \right|
=m(-1)^m
$$
as you expected.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/346489",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Max and Min of $f(x,y)$ Let $f(x,y)=x(y \log y-y)-y \log x$. Find $\smash{\displaystyle\max_{\frac{1}{2} \leq x \leq 2}}(\smash{\displaystyle\min_{\frac{1}{2} \leq y \leq 1} f(x,y)})$.
| Note that $$\smash{\displaystyle\max_{\frac{1}{2} \leq x \leq 2}}(\smash{\displaystyle\min_{\frac{1}{2} \leq y \leq 1} f(x,y)}) \leq \smash{\displaystyle\max_{\frac{1}{2} \leq x \leq 2}}f(x,\frac{1}{2})=\smash{\displaystyle\max_{\frac{1}{2} \leq x \leq 2}}(-\frac{x\log 2+x+\log x}{2})$$
Since $-\frac{x\log 2+x+\log x}{2}$ is a decreasing function of $x$, we have
$$\smash{\displaystyle\max_{\frac{1}{2} \leq x \leq 2}}(-\frac{x\log 2+x+\log x}{2})=-\frac{\frac{1}{2}\log 2+\frac{1}{2}+\log \frac{1}{2}}{2}=\frac{\log 2-1}{4}$$
with equality when $x=\frac{1}{2}$. We claim that equality holds throughout and thus
$$\smash{\displaystyle\max_{\frac{1}{2} \leq x \leq 2}}(\smash{\displaystyle\min_{\frac{1}{2} \leq y \leq 1} f(x,y)})=\frac{\log 2-1}{4}$$
It now suffices to show that
$$\smash{\displaystyle\min_{\frac{1}{2} \leq y \leq 1} f(\frac{1}{2},y)}=f(\frac{1}{2}, \frac{1}{2})$$
We have
$$f(\frac{1}{2}, y)=\frac{1}{2}y\log y-\frac{1}{2}y+y\log 2$$
$$\frac{d f(\frac{1}{2}, y)}{dy}=\frac{1}{2}\log y+\log 2>\frac{1}{2}\log \frac{1}{2}+\log 2=\frac{1}{2}\log 2>0$$
Therefore $f(\frac{1}{2}, y)$ is a non-decreasing function of $y$, and we are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/347386",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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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.
| Or, use $$\frac{x^2}{1+x^4}=\frac{1}{2} \frac{2x^2}{1+x^4}=\frac{1}{2}\left(\frac{x^2-1}{1+x^4}+\frac{x^2+1}{1+x^4}\right)$$ and this idea.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/349424",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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If $x \neq 0,y \neq 0,$ then $x^2+xy+y^2$ is ..... I came across the following problem that says:
If $x \neq 0,y \neq 0,$ then $x^2+xy+y^2$ is
1.Always positive
2.Always negative
3.zero
4.Sometimes positive and sometimes negative.
I have to determine which of the aforementioned options is right.
Now since $x \neq 0,y \neq 0$, so $ x^2+xy+y^2=(x-y)^2+3xy > 0$,if $x,y$ are of same sign. But if $x,y$ are of different sign,I am not sure about the conclusion.
Can someone point me in the right direction? Thanks in advance for your time.
| I am posting 2 ways of solving this:
$(1)$ If $x,y$ belong to positive real numbers only:
We know that $x^2 + y^2 \geq 2xy$.
Hence we can say that $x^2 + y^2 > xy$
Hence even if $xy$ is negative $x^2 + y^2$,which is positive, is always greater than $xy$ making the sum $x^2 + y^2 + xy$ always positive.
$(2)$ If $x,y$ belong to real numbers:
$$x^2 + y^2 + xy$$
$$ = x^2 + 2(x)(\dfrac{y}{2}) + \dfrac{y^2}{4} + \dfrac{3y^2}{4}$$
$$={(x+\dfrac{y}{2})}^2 + \dfrac{3y^2}{4}$$which is always positive.
Hope the answer is clear now !!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/349940",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Simplify $\sum_{i=0}^n (i+1)\binom ni$ Simplifying this expression$$1\cdot\binom{n}{0}+ 2\cdot\binom{n}{1}+3\cdot\binom{n}{2}+ \cdots+(n+1)\cdot\binom{n}{n}= ?$$
$$\text{Hint: } \binom{n}{k}= \frac{n}{k}\cdot\binom{n-1}{k-1} $$
| \begin{align}\sum_{k=0}^n(k+1) \binom{n}{k}&=\sum_{k=0}^nk \binom{n}{k}+\sum_{k=0}^n \binom{n}{k}=n\sum_{k=1}^n \binom{n-1}{k-1}+2^n\\&=n\sum_{k=0}^{n-1} \binom{n-1}{k}+2^n=n2^{n-1}+2^n
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/351289",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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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.
| Since $a^2, b^2, c^2$ are in arithmetic progression, $a^2+c^2=2b^2$.
If $a$ is even, then so is $c$, so $4 \mid a^2+c^2=2b^2$, so $b$ is also even, giving a contradiction.
Thus $a$ is odd. Similarly $c$ is odd, so $b$ is also odd.
$$(a-b)(a+b)=a^2-b^2=b^2-c^2=(b-c)(b+c)$$
$$\frac{a-b}{2}\frac{a+b}{2}=\frac{b-c}{2}\frac{b+c}{2}$$
By factoring lemma, there exists integers $w, x, y, z$ such that $\frac{a-b}{2}=wx, \frac{a+b}{2}=yz, \frac{b-c}{2}=wy, \frac{b+c}{2}=xz$. Now
$$a=wx+yz, b=yz-wx=wy+xz, c=xz-wy$$
$y(z-w)=x(z+w)$. Again by factoring lemma, there exists integers $d, e, f, g$ such that $y=de, z-w=fg, x=df, z+w=eg$. Now
$$z=\frac{eg+fg}{2}=g\frac{e+f}{2}, w=\frac{eg-fg}{2}=g\frac{e-f}{2}$$
$$a=wx+yz=dfg\frac{e-f}{2}+deg\frac{e+f}{2}=\frac{dg}{2}(e^2+2ef-f^2)$$
$$b=wy+xz=deg\frac{e-f}{2}+dfg\frac{e+f}{2}=\frac{dg}{2}(e^2+f^2)$$
$$c=xz-wy=dfg\frac{e+f}{2}-deg\frac{e-f}{2}=\frac{dg}{2}(-e^2+2ef+f^2)$$
If $dg$ is divisible by $4$ or an odd prime $p$, then $a, b, c$ will not be relatively prime. Thus $dg=\pm 1$ or $\pm 2$.
If $dg=\pm 2$, then $a=\pm (e^2+2ef-f^2), b=\pm (e^2+f^2), c=\pm (-e^2+2ef+f^2)$. Clearly $\gcd(e, f) \mid \gcd(a, b, c)=1$, so $e, f$ are relatively prime. If $e, f$ have the same parity, then $a, b, c$ are all even, a contradiction, so $e, f$ have different parities.
If $dg= \pm 1$, then $a=\pm \frac{e^2+2ef-f^2}{2}, b=\pm \frac{e^2+f^2}{2}, c=\pm \frac{-e^2+2ef+f^2}{2}$. Thus $e, f$ must have the same parity. Put $e+f=2p, e-f=2q$, then $e=p+q, f=p-q$, and
$$a=\pm \frac{e^2+2ef-f^2}{2}=\pm \frac{(p+q)^2+2(p+q)(p-q)-(p-q)^2}{2}=\pm (p^2+2pq-q^2)$$
$$b=\pm \frac{e^2+f^2}{2}=\pm \frac{(p+q)^2+(p-q)^2}{2}=\pm (p^2+q^2)$$
$$c=\pm \frac{-e^2+2ef+f^2}{2}=\pm \frac{-(p+q)^2+2(p+q)(p-q)+(p-q)^2}{2}=\mp(-p^2+2pq+q^2)$$
Again, $p, q$ must be relatively prime and be of different parities.
Finally, combining the 2 cases, $a, b, c$ can be written as
$$a=\epsilon_a (p^2+2pq-q^2), b=\epsilon_b (p^2+q^2), c=\epsilon_c (-p^2+2pq+q^2)$$
where each $\epsilon_a, \epsilon_b, \epsilon_c$ are $1$ or $-1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/351757",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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A question in Complex Analysis $\int_0^{2\pi}\log(1-2r\cos x +r^2)\,dx$ My problem is to integrate this expression:
$$\int_0^{2\pi}\log(1-2r\cos x +r^2)dx.$$
where $r$ is any constant in $[0,1]$.
I know the answer is zero. Can you explain you idea to me or just prove that?
Maybe you will use the "Cauchy integral theorem ".
| Copying from my blog.
Let $$I(a) = \displaystyle \int_0^{\pi} \ln \left(1-2a \cos(x) + a^2\right) dx$$ Some preliminary results on $I(a)$. Note that we have $$I(a) = \underbrace{\displaystyle \int_0^{\pi} \ln \left(1+2a \cos(x) + a^2\right) dx}_{\spadesuit} = \overbrace{\dfrac12 \displaystyle \int_0^{2\pi} \ln \left(1-2a \cos(x) + a^2\right) dx}^{\clubsuit}$$ $(\spadesuit)$ can be seen by replacing $x \mapsto \pi-x$ and $(\clubsuit)$ can be obtained by splitting the integral from $0$ to $\pi$ and $\pi$ to $2 \pi$ and replacing $x$ by $\pi+x$ in the second integral.
Now let us move on to our computation of $I(a)$.
\begin{align}
I(a^2) & = \int_0^{\pi} \ln \left(1-2a^2 \cos(x) + a^4\right) dx = \dfrac12 \int_0^{2\pi} \ln \left(1-2a^2 \cos(x) + a^4\right) dx\\
& = \dfrac12 \int_0^{2\pi} \ln \left((1+a^2)^2-2a^2(1+ \cos(x))\right) dx = \dfrac12 \int_0^{2\pi} \ln \left((1+a^2)^2-4a^2 \cos^2(x/2)\right) dx\\
& = \dfrac12 \int_0^{2\pi} \ln \left(1+a^2-2a \cos(x/2)\right) dx + \dfrac12 \int_0^{2\pi} \ln \left(1+a^2+2a \cos(x/2)\right) dx
\end{align}
Now replace $x/2=t$ in both integrals above to get
\begin{align}
I(a^2) & = \int_0^{\pi} \ln \left(1+a^2-2a \cos(t)\right) dt + \int_0^{\pi} \ln \left(1+a^2+2a \cos(t)\right) dt = 2I(a)
\end{align}
Now for $a \in [0,1)$, this gives us that $I(a) = 0$. This is because we have $I(0) = 0$ and $$I(a) = \dfrac{I(a^{2^n})}{2^n}$$ Now let $n \to \infty$ and use continuity to conclude that $I(a) = 0$ for $a \in [0,1)$. Now lets get back to our original problem. Consider $a>1$. We have
\begin{align*}
I(1/a) & = \int_0^{\pi} \ln \left(1-\dfrac2{a} \cos(x) + \dfrac1{a^2}\right)dx\\
& = \int_0^{\pi} \ln(1-2a \cos(x) + a^2) dx - 2\int_0^{\pi} \ln(a)dx\\
& = I(a) - 2 \pi \ln(a)\\
& = 0 \tag{Since $1/a < 1$, we have $I(1/a) = 0$}
\end{align*}
Hence, we get that
$$I(a) = \begin{cases} 2 \pi \ln(a) & a \geq 1 \\ 0 & a \in [0,1] \end{cases}$$
| {
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"url": "https://math.stackexchange.com/questions/352046",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Linear Programming: Modifying Coefficients of the Objective Function Consider a final tableau with entries:
\begin{array} {|c|c|}\hline BV & x_1 & x_2 & x_3 & x_4 & x_5 & x_6 & x_7 & RHS \\ \hline x_3 & 0 & -\frac{1}{2} & 1 & 1 & 2 & 0 & -1 & 4 \\ \ x_1 & 1 & \frac{1}{2} & 0 & 2 & -1 & 0 & -2 & 2 \\ x_6 & 0 & 2 & 0 & -1 & -\frac{1}{2} & 1 & 3 & 1 \\ \hline \end{array}
and the objective function is $x_1 + 3x_2 + 2x_3 + x_4 +x_5 + 3x_6 + 2x_7$
Is there an easy way to change the cost coefficient value of $x_6$ of 3 without affecting optimality? Similarly, is there a way to change $x_3$ of 2 by (-1/2) and shift some other things around so that everything remains optimal?
| After having calculating all the missing $C^\pi_{NBV}$ of the given tableau, we should have following optimal tableau:
\begin{array} {|c|c|}\hline BV & z & x_1 & x_2 & x_3 & x_4 & x_5 & x_6 & x_7 & RHS \\ \hline z & 1 & 0 & \frac{5}{2} & 0 & 0 & \frac{1}{2} & 0 & 3 & 13 \\ \hline x_3 & 0 & 0 & -\frac{1}{2} & 1 & 1 & 2 & 0 & -1 & 4 \\ \ x_1 & 0 & 1 & \frac{1}{2} & 0 & 2 & -1 & 0 & -2 & 2 \\ x_6 & 0 & 0 & 2 & 0 & -1 & -\frac{1}{2} & 1 & 3 & 1 \\ \hline \end{array}
We can change the basic variable $x_6$ cost coefficient by looking at the following formula for calculating the reduced costs of the non-basic variables:
$$C^\pi_{x_{NBV}}=\!\!\!\!\!\!\!\!\!\!\underbrace{C^T_{BV}}_{\text{Change of coeff BV}}\!\!\!\!\!\!\!\!B^{-1}A_{x_{NBV}}-\!\!\!\!\!\!\!\!\!\underbrace{C_{x_{NBV}}}_{\text{Change of coeff NBV}}$$
Where the change of the basic variable coefficients change $C^T_{BV}$, and the change of non-basic variables coefficients change $C_{NBV}$.
Thus, if we don't want to change the basis when we mess around with the coefficient of $x_6$, we need to need to measure the acceptable range of coefficient values $x_6$ is allowed to have that prevents a pivot from the above tableau. Thus, this is measured as such:
$$C^\pi_{x_2}=\begin{bmatrix}2&1&\Delta\end{bmatrix}\begin{bmatrix}-\frac{1}{2}\\\frac{1}{2}\\2\end{bmatrix}-3=2\Delta-\frac{7}{2}\ge0\Longrightarrow \Delta\ge\frac{7}{4}$$
$$C^\pi_{x_4}=\begin{bmatrix}2&1&\Delta\end{bmatrix}\begin{bmatrix}1\\2\\-1\end{bmatrix}-1=3-\Delta\ge0\Longrightarrow \Delta\le3$$
$$C^\pi_{x_5}=\begin{bmatrix}2&1&\Delta\end{bmatrix}\begin{bmatrix}2\\-1\\-\frac{1}{2}\end{bmatrix}-1=2-\frac{1}{2}\Delta\ge0\Longrightarrow \Delta\le4$$
$$C^\pi_{x_7}=\begin{bmatrix}2&1&\Delta\end{bmatrix}\begin{bmatrix}-1\\-2\\3\end{bmatrix}-2=3\Delta-6\ge0\Longrightarrow \Delta\ge2$$
Thus, the allowable range to change the coefficient of $x_6$ in the objective function is between the range $2\le\Delta\le3$. If we were to change the coefficient of $x_6$ in the objective function to say $\frac{3}{2}$, it will still preserve the current optimality of the tableau.
If we want to find the acceptable range of changing the non-basic variable $x_3$, we'll do the following:
$$C^\pi_{x_j}=\begin{bmatrix}2&1&3\end{bmatrix}B^{-1}A_j-\Delta\ge0$$
For each non-basic variable in the tableau like what has been demonstrated above for the change of basic variables.
| {
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"timestamp": "2023-03-29T00:00:00",
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Linear algebra mapping question Does there exist a matrix $A$ such that
$$
A\left( \begin{array}{ccc}
1 \\
0 \\
0 \\
0 \\
\end{array} \right) = \left( \begin{array}{ccc}
1 \\
0 \\
0 \\
0 \\
\end{array} \right)$$
$$
A\left( \begin{array}{ccc}
0 \\
1 \\
0 \\
0 \\
\end{array} \right) = \left( \begin{array}{ccc}
0 \\
1 \\
0 \\
0 \\
\end{array} \right)$$
$$
A\left( \begin{array}{ccc}
0 \\
0 \\
1 \\
0 \\
\end{array} \right) = \left( \begin{array}{ccc}
0 \\
1 \\
0 \\
0 \\
\end{array} \right)$$
$$
A\left( \begin{array}{ccc}
0 \\
0 \\
0 \\
1 \\
\end{array} \right) = \left( \begin{array}{ccc}
0 \\
0 \\
1 \\
0 \\
\end{array} \right)$$
if so, what?
| If $A$ exists, $A$ satisfies
$$A\begin{pmatrix}
1&0&0&0\\
0&1&0&0\\
0&0&1&0\\
0&0&0&1\end{pmatrix}=
\begin{pmatrix}
1&0&0&0\\
0&1&1&0\\
0&0&0&1\\
0&0&0&0\end{pmatrix}$$
So
$$A=
\begin{pmatrix}
1&0&0&0\\
0&1&1&0\\
0&0&0&1\\
0&0&0&0
\end{pmatrix}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/352708",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Finding all $\alpha$ such that a matrix is positive definite I have
$A = $
$
\left[\begin{array}{rrr}
2 & \alpha & -1 \\
\alpha & 2 & 1 \\
-1 & 1 & 4
\end{array}\right]
$
and I want to find all $\alpha$ such that $A$ is positive definite.
I tried
$ x^tAx = $
$
\left[\begin{array}{r}
x & y & z
\end{array}\right]
$
$
\left[\begin{array}{rrr}
2 & \alpha & -1 \\
\alpha & 2 & 1 \\
-1 & 1 & 4
\end{array}\right]
$
$
\left[\begin{array}{r}
x \\
y \\
z
\end{array}\right]
$
$=$
$
\left[\begin{array}{r}
2x + \alpha y - z & \alpha x + 2y + z & -x + y + 4z
\end{array}\right]
$
$
\left[\begin{array}{r}
x \\
y \\
z
\end{array}\right]
$
$= 2x^2 + \alpha xy - xz + \alpha xy + 2y^2 + yz - xz + yz + 4z^2$
$= 2 \alpha xy + 2x^2 + 2y^2 - 2xz + 2yz + 4z^2$
and I wanted to solve the inequality $2 \alpha xy + 2x^2 + 2y^2 - 2xz + 2yz + 4z^2 > 0$ for $\alpha$, but I wasn't sure what to do next.
Am I doing this correctly?
| The matrix is positive definite if and only if their eigenvalues are positive, so we calculate the characteristic polynomial $\chi_A(x)=\det(xI-A)$ and we solve for $x$ and we find:
$$\lambda_1=\frac{1}{2}\sqrt{\alpha^2+4\alpha+12}-\frac{1}{2}\alpha+3,\quad\lambda_2=3-\frac{1}{2}\sqrt{\alpha^2+4\alpha+12}- \frac{1}{2}\alpha,\quad \lambda_3=\alpha+2$$
hence we should find the $\alpha$ for which $\lambda_i>0, i=1,2,3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/353827",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
} |
An integral inequality related to Taylor expansion Problem. Let $f:[a,b]\to\mathbb{R}$ be a function such that $ f\in C^3([a,b])$ and $f(a)=f(b)$. Prove that $$ \left|\int\limits_{a}^{\frac{a+b}{2}}f(x)dx-\int\limits_{\frac{a+b}{2}}^{b}f(x)dx\right|\leq\frac{(b-a)^4}{192}\max_{x\in [a,b]}|f'''(x)|.$$
Any idea are welcome.
| Idea : Making use of successive Integration by parts ,
$\int P(x)f^{(3)}(x)\,dx = P(x)f^{(2)}(x)-P^{(1)}(x)f^{(1)}(x)+P^{(2)}(x)f(x)-\int P^{(3)}(x)f(x)\,dx$
Consider the two third degree monic polynomials, $P_1(x)$ and $P_2(x)$.
Now, we compute the difference of the definite integrals:
$\int\limits_{\frac{a+b}{2}}^b P_2(x)f^{(3)}(x)\,dx-\int\limits_a^{\frac{a+b}{2}} P_1(x)f^{(3)}(x)\,dx$
$=[P_2(b)f^{(2)}(b)+P_1(a)f^{(2)}(a)-P_2(\frac{a+b}{2})f^{(2)}(\frac{a+b}{2})-P_1(\frac{a+b}{2})f^{(2)}(\frac{a+b}{2})] - [P_2'(b)f^{(1)}(b)+P_1'(a)f^{(1)}(a)-P_2'(\frac{a+b}{2})f^{(1)}(\frac{a+b}{2})-P_1'(\frac{a+b}{2})f^{(1)}(\frac{a+b}{2})] + [P_2''(b)f(b)+P_1''(a)f(a)-P_2''(\frac{a+b}{2})f(\frac{a+b}{2})-P_1''(\frac{a+b}{2})f(\frac{a+b}{2})] - [\int\limits_{\frac{a+b}{2}}^b P_2'''(x)f(x)\,dx - \int\limits_a^{\frac{a+b}{2}} P_1'''(x)f(x)\,dx]$
We want our polynomials to satisfy the relations:
$P_2(b)=0,P_1(a)=0,P_2(\frac{a+b}{2})+P_1(\frac{a+b}{2})=0$ ... (i)
$P_2'(b)=0,P_1'(a)=0,P_2'(\frac{a+b}{2})+P_1'(\frac{a+b}{2})=0$ ... (ii)
$P_2''(b)+P_1''(a)=0,P_2''(\frac{a+b}{2})+P_1''(\frac{a+b}{2})=0$ ... (iii)
(In (iii) we made use of $f(a)=f(b)$)
From, (ii) and (iii), it follows that $P_1(x)=(x-a)^2(x-c_1)$ and $P_2(x)=(x-b)^2(x-c_2)$, plugging in the other conditions we get $c_1=\frac{a+3b}{4}$ and $c_2=\frac{b+3a}{4}$
Therefore,
$\int\limits_a^{\frac{a+b}{2}} P_1(x)f^{(3)}(x)\,dx-\int\limits_{\frac{a+b}{2}}^b P_2(x)f^{(3)}(x)\,dx$
$=6\int\limits_{\frac{a+b}{2}}^b f(x)\,dx-6\int\limits_a^{\frac{a+b}{2}}f(x)\,dx$
Making use of the fact $P_1(x)\le0$ in the interval $[a,\frac{a+b}{2}]$ (Since, $\frac{a+3b}{4} \ge \frac{a+b}{2}$)
and $P_2(x)\ge 0$ in the interval $[\frac{a+b}{2},b]$ (Since, $\frac{b+3a}{4} \le \frac{a+b}{2}$)
and, $\int\limits_a^{\frac{a+b}{2}}(x-a)^2(x-\frac{a+3b}{4})=-\frac{1}{64}(b-a)^4$
$\int\limits_{\frac{a+b}{2}}^b (x-b)^2(x-\frac{b+3a}{4})=\frac{1}{64}(b-a)^4$
Therefore, $\left|\int\limits_{\frac{a+b}{2}}^b f(x)\,dx-\int\limits_a^{\frac{a+b}{2}}f(x)\,dx\right| \le \frac{2}{64\times 6}(b-a)^4\max_{x\in [a,b]}|f'''(x)|=\frac{1}{192}(b-a)^4\max_{x\in [a,b]}|f'''(x)|$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 1
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Solving $x^2+19=y^5$ I was given several exercises and there is a particular one, I am not able to solve.
Let it be given that $Pic(\mathbb{Z}[\sqrt{−19}])$ is a finite group of order $3$. Use this to find all integral solutions of the equation $x^2 + 19 = (x + \sqrt{−19})(x-\sqrt{−19})= y^5$.
I have absolutely no idea what to do. Could anybody help me.
Thanks in advance!
| Let $x$ and $y$ be integers which satisfy $x^2 + 19 = y^5$. Then $(x + \sqrt{-19})(x - \sqrt{-19}) = (y)^5$.
If you can find that $(x + \sqrt{-19})$ and $(x - \sqrt{-19})$ are coprime ideals, (if $19 \mid x$ would hold, then $19 \mid y$ and $19 = y^5 - x^2$ where the RHS is divisible by $19^2$ - contradiction) then $(x + \sqrt{-19})$ is the fifth power of some ideal $I$; since the class group would have order $3$ (it doesn't), $I$ must be principal, and $x + \sqrt{-19}$ a perfect fifth power.
Equating $x + \sqrt{-19}=(a + \frac{1+\sqrt{-19}}{2}b)^5$ gives you $$x = a^5 +\frac{5}{2}a^4b - 45a^3b^2 - 70a^2b^3 + \frac{155}{2}ab^4 + \frac{101}{2}b^5$$ for $a,b \in \mathbb{Z}$ such that $$5a^4b + 10a^3b -40a^2 b^3 - 45ab^4 + 11b^5 = 2;$$ the lower equation implies that $b = \pm 1$ or $b = \pm 2$. We calculate
if $b=1$, then $5a^4 + 10a^3 - 40a^2 -45a + 9 = 0$ with no integral solutions;
if $b=-1$, then $5a^4 -10a^3 - 40a^2 + 45a + 13 = 0$ with no integral solutions;
$b = 2$ gives solutions $a = 5$ and $a = -7$ and therefore $a+\frac{1+\sqrt{-19}}{2}b = \pm 6 + \sqrt{-19}$;
$b = -2$ again gives none.
So you must have $x + \sqrt{-19} = (\pm6 + \sqrt{-19})^5 = \pm 22434 + \sqrt{-19}$;
so $x = \pm 22434$ and $y = (22434^2 + 19)^{(1/5)} = 55$ is the only integral solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/354069",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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} |
If $z$ is a complex number of unit modulus and argument theta If $z$ is a complex number such that $|z|=1$ and $\text{arg} z=\theta$, then what is $$\text{arg}\frac{1 + z}{1+ \overline{z}}?$$
| $z$ can be written as $\cos\theta+i\sin\theta\implies \bar z=\cos\theta-i\sin\theta$
So, $$\frac{1+z}{1+\bar z}=\frac{1+\cos\theta+i\sin\theta}{1+\cos\theta-i\sin\theta}$$
$$=\frac{2\cos^2\frac \theta2+2i\cos\frac\theta2\sin\frac\theta2}{2\cos^2\frac \theta2-2i\cos\frac\theta2\sin\frac\theta2}$$
$$=\frac{\cos\frac\theta2+i\sin\frac\theta2}{\cos\frac\theta2-i\sin\frac\theta2}$$
$$=\frac{\left(\cos\frac\theta2+i\sin\frac\theta2\right)^2}{\left(\cos\frac\theta2-i\sin\frac\theta2\right)\left(\cos\frac\theta2+i\sin\frac\theta2\right)}$$ assuming $\cos\frac\theta2\ne 0$ i.e., $\frac\theta2\ne (2n+1)\frac\pi2$ i.e., $\theta\ne(2n+1)\pi$ where $n$ is any integer
as $\theta=(2n+1)\pi, 1+\cos\theta\pm i\sin\theta=0$
$$\frac{1+z}{1+\bar z}=\cos\theta+i\sin\theta\text{ using de Moivre's formula} $$
$$\implies \frac{1+z}{1+\bar z}=z $$
$$\implies arg\left(\frac{1+z}{1+\bar z}\right)= arg(z) $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/354922",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
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.
| There is a mechnical procedure, as follows.
Any polynomial function of $r = \sqrt 2 + \sqrt 3$ must have the form $a + b\sqrt 2 + c\sqrt 3 + d\sqrt 6$ for rational $a,b,c,d$. Consider the set of numbers of that form as a vector space $V$ over the rationals. It has dimension 4.
Now calculate $r^0, r^1, r^2, r^3, r^4$. These are five elements of the vector space $V$, and since $V$ has dimension only 4, they cannot be linearly independent. Therefore there must exist rationals $a_0,\ldots, a_4$ such that $a_4r^4 + a_3r^3 + a_2r^2 + a_1r^1 + a_0r^0 = 0$. These can be found by well-known mechanical methods for changing the basis of a vector space. Then our polynomial is $a_4x^4 + a_3x^3 + a_2x^2 + a_1x^1 + a_0$.
(There are a couple of fine points I skipped here: $a_4$ might be zero; $r^3$ might not be independent of $r^0, r^1, $ and $r^2$. None of this is hard to deal with.)
Here is an example.
Calculate powers of $r = \sqrt2 + \sqrt3$, and tabulate them:
$$\begin{array}{crrrr}
% & 1 & \sqrt2 & \sqrt3 &\sqrt 6\\
%\hline
r^0 = & 1 &&&\\
r^1 = & & \sqrt2 & + \sqrt3 & \\
r^2 = & 5 & && + 2\sqrt6\\
r^3 = & &11\sqrt2 &+ 9\sqrt3 \\
r^4 = & 49 &&& + 20\sqrt 6
\end{array}$$
Now we want to find rational $a,b,c,d$ such that $r^4 = ar^3 + br^2 + cr^1 + dr^0$. Such rationals must exist. (Unless $r^0\ldots r^3$ are not independent, in which case we are looking for a polynomial of lower degree, and we can use the same method with even less effort.) The relations in the table above impose relations on $a,b,c,d$ that we can read off from the table, one relation for each column:
$$
\begin{array}{rrrrl}
& 5b & & + d &=49\\
11a&& + c &&= 0\\
9a&&+c&&=0\\
&2b&&& = 20
\end{array}
$$
We can solve the equations mechanically (they are particularly simple in this case; you can just read off the answer) and find that $a=0, b=10, c=0, d=-1$. So we have calculated, entirely mechanically, that $r^4 = 10r^2-1$, which means that $r$ is a zero of the polynomial $$x^4-10x^2+1.$$
(I wrote this up in detail on my blog a few years back, and just happened to use $\sqrt 2 + \sqrt 3$ as an example.)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/359054",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
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Is there a formula for $\sum_{n=1}^{k} \frac1{n^3}$? I am searching for the value of $$\sum_{n=k+1}^{\infty} \frac1{n^3} \stackrel{?}{=} \sum_{n = 1}^{\infty} \frac1{n^3} - \sum_{n=1}^{k} \frac1{n^3} = \zeta(3) - \sum_{n=1}^{k} \frac1{n^3}$$
For which I think I need to know the value of $$\sum_{n=1}^{k} \frac1{n^3}$$
Does anyone know of a formula (and a reference if it is complicated)?
| We have
$$S_n = \sum_{k=n+1}^{\infty} \dfrac1{k^3} = \int_{n^+}^{\infty} \dfrac{d \lfloor t \rfloor}{t^3} = \left. \dfrac{\lfloor t \rfloor}{t^3} \right \vert_{t=n^+}^{\infty} + 3 \int_{n^+}^{\infty} \dfrac{\lfloor t\rfloor dt}{t^4} = -\dfrac1{n^2}+3 \int_{n^{+}}^{\infty} \dfrac{t-\{t\}}{t^4} dt$$
Hence, we get that
$$S_n = -\dfrac1{n^2} + 3 \int_{n^+}^{\infty} \dfrac{dt}{t^3} - 3 \int_{n^+}^{\infty} \dfrac{\{t\}}{t^4}dt = \dfrac1{2n^2} - 3 \underbrace{\int_{n^+}^{\infty} \dfrac{\{t\}}{t^4}dt}_{\mathcal{O}(1/n^3)}$$
You can get better approximations by repeating the above procedure, and this is called as Euler-Maclaurin Summation. To get a higher order approximation, we need to get a good approximation of $\displaystyle \int_{n^+}^{\infty} \dfrac{\{t\}}{t^4}dt$. This is done as follows.
$$\int_{n^+}^{\infty} \dfrac{\{t\}}{t^4}dt = \int_{n^+}^{\infty} \dfrac{1/2}{t^4}dt+ \int_{n^+}^{\infty} \dfrac{\{t\}-1/2}{t^4}dt = \dfrac1{6n^3}+\int_{n^+}^{\infty} \dfrac{B_1(\{t\})dt}{t^4}\\ = \dfrac1{6n^3} + \overbrace{\left.\dfrac{B_2(\{t\})}{2t^4} \right \vert_{t=n^+}^{\infty}}^0 + 4 \underbrace{\int_{n^+}^{\infty} \dfrac{B_2(\{t\})}{t^5} dt}_{\mathcal{O}(1/n^4)}$$
where $B_n(x)$ are Bernoulli polynomials of order $n$.
Hence, a better asymptotic for $S_n$ is
$$S_n = \dfrac1{2n^2} - \dfrac1{2n^3} + \mathcal{O}(1/n^4)$$
Crank this repeatedly to get higher order estimates.
The Euler MacLaurin for infinite sums, where the integrand is well-behaved (by which I mean all the derivatives of the function vanish as $x \to \infty$) is
$$\sum_{k=n}^{\infty} f(k) = \int_{n}^{\infty} f(x)dx + \dfrac{f(n)}2 - \sum_{k=1}^{\infty} \dfrac{B_{2k}}{(2k)!} f^{(2k-1)'}(n)$$ where $B_{2k}$ are the Bernoulli numbers.
In your case, $f(x) = 1/x^3$ and $f^{(2k-1)'}(x) = -\dfrac{(2k+1)!}{2 \cdot x^{2k+2}}$.
Hence, we get that
$$\sum_{k=n}^{\infty} \dfrac1{k^3} = \dfrac1{2n^2} + \dfrac1{2n^3} + \sum_{k=1}^{\infty} \dfrac{(2k+1)B_{2k}}{2 \cdot n^{2k+2}}$$
$$\sum_{k=n+1}^{\infty} \dfrac1{k^3} = \dfrac1{2n^2} - \dfrac1{2n^3} + \sum_{k=1}^{\infty} \dfrac{(2k+1)B_{2k}}{2 \cdot n^{2k+2}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/361386",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
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Show that $ \frac{\tan x}{1+\sec x}+\frac{1+\sec x}{\tan x}= 2 \csc x$ Verify the following identity:
$$ \frac{\tan x}{1+\sec x}+\frac{1+\sec x}{\tan x}= 2 \csc x$$
| Method $1:$
As
$\displaystyle\tan^2x=\sec^2x-1=(\sec x-1)(\sec x+1),\frac{\tan x}{\sec x-1}=\frac{\sec x+1}{\tan x}$
$$\implies\frac{\tan x}{1+\sec x}+\frac{1+\sec x}{\tan x}=\frac{\sec x-1}{\tan x}+\frac{1+\sec x}{\tan x}=\frac{2\sec x}{\tan x}=\frac{\dfrac2{\cos x}}{\dfrac {\sin x}{\cos x}}=\frac2{\sin x}$$
Method $2:$
$$\frac{\tan x}{1+\sec x}=\frac{\frac {\sin x}{\cos x}}{1+\frac1{\cos x}}=\frac{\sin x}{1+\cos x}=\frac{\sin x(1-\cos x)}{(1+\cos x)(1-\cos x)}=\frac{1-\cos x}{\sin x}\text{ as } \sin^2x=1-\cos^2x$$
and
$$\frac{1+\sec x}{\tan x}=\frac{1+\frac1{\cos x}}{\frac{\sin x}{\cos x}}=\frac{1+\cos x}{\sin x}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/361883",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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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.
| By Riemann sum we have
$$\frac{1}{n^3}\sum_{k=1}^nk(k+1)=\frac{1}{n}\sum_{k=1}^n\frac{k}{n}(\frac{k+1}{n})\sim_\infty \frac{1}{n}\sum_{k=1}^n\left(\frac{k}{n}\right)^2\to\int_0^1x^2dx=\frac{1}{3}$$
| {
"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": 1
} |
Evaluating the integral $ \int_{-\infty}^{\infty} \frac{\cos \left(x-\frac{1}{x} \right)}{1+x^{2}} \ dx$ I'm curious about the proper way to evaluate $$ \int_{-\infty}^{\infty} \frac{\cos \left(x-\frac{1}{x} \right)}{1+x^{2}} \, dx = \text{Re} \int_{-\infty}^{\infty} \frac{e^{i(x- \frac{1}{x})}}{1+x^{2}} \, dx$$ using contour integration.
If I let $f(z) = \frac{e^{i(z- \frac{1}{z})}}{1+z^{2}}$, there is an essential singularity at the origin.
So if I integrate around a closed semicircle in the upper half-plane, the contour goes right through the singularity.
Can you indent a contour around an essential singularity?
| Let $x-1/x = t$. Now note that $t^2 = x^2 + \dfrac1{x^2} -2 \implies \left(x+\dfrac1x \right)^2 = t^2+4$. We then get that
$$\left(1 + \dfrac1{x^2} \right)dx = dt \implies \dfrac{dx}{1+x^2} = \dfrac{x^2}{(1+x^2)^2} dt = \dfrac{dt}{\left(x+1/x \right)^2} = \dfrac{dt}{t^2+4}$$
Hence, the integral becomes
$$\int_{-\infty}^{\infty} \dfrac{\cos(x-1/x)}{x^2+1} dx = 2 \int_{0}^{\infty} \dfrac{\cos(x-1/x)}{x^2+1} dx = 2\int_{-\infty}^{\infty} \dfrac{\cos(t)}{t^2+4} dt = \dfrac{\pi}{e^2}$$
where the last integral can be obtained from the post below.
Calculating the integral $\int_{0}^{\infty} \frac{\cos x}{1+x^2}\mathrm{d}x$ without using complex analysis
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/365663",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
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Determine all $2\times2$ matrices $A$ such that $A^2=0$. This is from Lang's introduction to Linear Algebra page no 61.
Determine all $2\times 2$ matrices $A$ such that $A^2 = 0$.
Let $A=\begin{pmatrix}
a & b \\
c & d
\end{pmatrix}$
$A^2=\begin{pmatrix}
a^2+bc & ab+bd \\
ac+cd & d^2+cb
\end{pmatrix}$
Equating all the four terms of the matrix to zero and solving,
$a^2+bc=0$ and $d^2+bc=0$ gives $a=\pm d$
Solving $ab+bd=0$ and $ac+cd=0$ gives $a=-d$ OR $b=c$, which gives $a=d=0$
My question: I know that if $A$ is an $n\times n$ matrix with all the diagonal elements and all the elements below it equal to zero then $A^n=0$, so in this question
\begin{equation}
\begin{pmatrix}
0 & x \\
0 & 0
\end{pmatrix}
\end{equation} qualifies as $A$, but it does not agree with $b=c$ as in above solution, so where did I go wrong?
And if I go with $b=c$, then $A^2$ does not equal $0$.
| Well you have your 4 equations:
$$\cases{a^2+bc=0 \\ ab+bd=0 \\ ac+cd=0 \\ d^2+bc=0}$$
And you know that if $A^2=0$ then $det(A^2)=det(A)^2=0$ and thus $det(A)=0$. So you have a fifth equation:
$$ad-bc=0 \Rightarrow ad=bc$$
Now replace this in your system and factorize it:
$$\cases{a^2+ad=a(a+d)=0 \\ ab+bd=b(a+d)=0 \\ ac+cd=c(a+d)=0 \\ d^2+ad=d(a+d)=0 \\ ad=bc}$$
*
*If $a+d\neq0$ then $a=b=c=d=0$
*If $a+d=0$ then all these equations are equivalent and you're left with two equations (you already know you'll probably have 2 degrees of liberty):
$$\cases{a+d=0 \\ ad=bc}\Rightarrow\cases{d=-a \\ -a^2=bc}$$
*
*If $b=0$ then $a=d=0$ and you're just left with $c\in\Bbb{R}$
*If $b\neq0$ then $\cases{d=-a \\ c=-\frac{a^2}{b}}$
Now this means that you have two possible matrices s.t. $A^2=0$:
$A=\left(\begin{array}{cc} a && b \\ -\frac{a^2}{b} && -a \end{array}\right),(a,b)\in\Bbb{R}\times\Bbb{R}^*$ or $A=\left(\begin{array}{cc} 0 && 0 \\ c && 0 \end{array}\right),c\in\Bbb{R}$
You also have their transposes $\left((A^T)^2=A^TA^T=(AA)^T=(A^2)^T=0^T=0\right)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/366506",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"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.
| HINT
$$\dfrac{\sqrt{x^4+1}}{\sqrt[3]{x^6+1}} = \dfrac{x^2 \sqrt{1+1/x^4}}{x^2 \sqrt[3]{1+1/x^6}} = \dfrac{\sqrt{1+1/x^4}}{\sqrt[3]{1+1/x^6}}$$
| {
"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": 2
} |
Derivation of the general forms of partial fractions I'm learning about partial fractions, and I've been told of 3 types or "forms" that they can take
(1) If the denominator of the fraction has linear factors:
$${5 \over {(x - 2)(x + 3)}} \equiv {A \over {x - 2}} + {B \over {x + 3}}$$
(2) If the denominator of the fraction has quadratic factors that don't factorise:
$${{2x + 3} \over {(x - 1)({x^2} + 4)}} \equiv {A \over {x - 1}} + {{Bx + C} \over {{x^2} + 4}}$$
(3) If the denominator has a factor that repeats:
$${{5x + 3} \over {(x - 2){{(x + 3)}^2}}} \equiv {A \over {x - 2}} + {B \over {x + 3}} + {C \over {{{(x + 3)}^2}}}$$
I'd appreciate it if someone could explain to me how these equivalence relationships are derived, essentially how do they take the forms that they do? I sort of understand the first one, but the other two I don't.
Thank you
| It comes from two facts for polynomials $f(x)$ and $g(x)$:
*
*We can find $a(x)$ and $b(x)$ such that $\gcd(f(x),g(x))=a(x)f(x)+b(x)g(x)$.
*We can write $f(x)=q(x)g(x)+r(x)$, where $\deg(r)<\deg(g)$.
For your second example, we can use fact 1 to write $1=a(x)(x-1)+b(x)(x^2+4)$. Multiplying by $2x+3$ yields:
$$2x+3=[(2x+3)a(x)](x-1)+[(2x+3)b(x)](x^2+4)$$
If $\deg((2x+3)a(x))\ge \deg(x^2+4)=2$, then we can use fact 2 to write $(2x+3)a(x)=q(x)(x^2+4)+r(x)$, where $\deg(r)<2$. Then:
$$2x+3=r(x)(x-1)+[q(x)(x-1)+(2x+3)b(x)](x^2+4)$$
Comparing degrees, we see that:
$$2x+3=r(x)(x-1)+s(x)(x^2+4)$$
where $\deg(r)<2$, and $\deg(s)<1$. Then:
$$\frac{2x+3}{(x-1)(x^2+4)}=\frac{r(x)(x-1)+s(x)(x^2+4)}{(x-1)(x^2+4)}=\frac{r(x)}{x^2+4}+\frac{s(x)}{x-1}=\frac{Ax+B}{x^2+4}+\frac{C}{x-1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/368665",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 8,
"answer_id": 2
} |
Proving the inequality $\ln(\cos x)\ge \frac{-x^2}{\cos^2(x)}$ for $x\in [0,\frac{\pi}{2}]$ I want to prove the following inequality: $\ln(\cos x)\ge \dfrac{-x^2}{\cos^2(x)}$
Please I'm stuck with this problem, maybe considering the equivalent inequality $\ln(\sec x)\ge (x \sec x)^2$ would help, but I'm not sure.
I want to prove it by the use of the Mean Value theorem.
| $$
\ln (1+x) = x - \frac{x^2}2 + \frac{x^3}3 - \cdots.
$$
$$
\ln\cos x = \ln\left(1 -\frac{x^2}{2} + \frac{x^4}{24}-\cdots\right) \ge\ln\left(1 -\frac{x^2}{2}\right)
$$
$$
=\frac{-x^2}{2} -\frac{x^4}{8} + \cdots \ge -\frac{x^2}{1.5} \ge-\frac{x^2}{\cos^2 x}\text{ for $x$ close enough to $0$.}
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
How to find $\gcd(a^{2^m}+1,a^{2^n}+1)$ when $m \neq n$? How to prove the following equality?
For $m\neq n$,
$\gcd(a^{2^m}+1,a^{2^n}+1) = 1 $ if $a$ is an even number
$\gcd(a^{2^m}+1,a^{2^n}+1) = 2 $ if $a$ is an odd number
Thanks in advance.
| Let $p$ divides $a^{2^n}+1\implies a^{2^n}\equiv-1\pmod p$
If $m>n,$ taking $2^{m-n}$th power, $(a^{2^n})^{2^{m-n}}\equiv(-1)^{2^{m-n}}\pmod p\implies a^{2^m}\equiv1\pmod p$
So, $p$ divides $ a^{2^m}-1$
If $p$ divides $ a^{2^m}+1,p$ will divide $(a^{2^m}+1)-(a^{2^m}-1)=2$
So, if $a$ is even, $a^{2^n}+1$ will be odd $\implies p=1\implies $ gcd$(a^{2^m}+1,a^{2^n}+1)=1$
Similarly, for odd $a$
Alternatively, WLOG let $m>n,a^{2^n}=b$(say) and $m=n+r$ where integer $r>0$
$\implies a^{2^m}+1=a^{2^{n+r}}+1=(a^{2^n})^{2^r}+1=b^{2^r}+1$
Now, $b^2-1=(b+1)(b-1)$
$(b^2)^2-1=(b^2+1)(b^2-1)=(b^2-1)(b+1)(b-1)$
$\cdots$
$b^{2^r}-1=(b^{2^{r-1}}+1)(b^{2^{r-1}}-1)\cdots(b^4+1)(b^2+1)(b+1)$
Now, $gcd(b^{2^r}+1,b^{2^r}-1)=gcd(b^{2^r}+1-(b^{2^r}-1),b^{2^r}-1)$
$=gcd(2,b^{2^r}-1)= \begin{cases} 2 &\mbox{if } b \mbox{ is odd} \\
1 & \mbox{if } b \mbox{ is even }\end{cases} $
As $(b+1)$ divides $(b^{2^r}-1),gcd(b^{2^r}+1, b+1)$ will divide $gcd(b^{2^r}+1,b^{2^r}-1)$
Observe that if $b$ is odd, $gcd(b^{2^r}+1,b^{2^r}-1)=2$ and
$b^{2^r}+1, b+1$ are both even , hence $ gcd(b^{2^r}+1, b+1)=2\implies gcd(a^{2^m}+1,a^{2^n}+1)=2$
Similarly, for even $b$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/373675",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Find the points of discontinuity: $f(x) = (x^4+x^3+2x^2)/\tan^{-1}(x)$ if $x\ne0$ and $f(0)=10$ Here is the question:
Find the points of discontinuity:
$$f(x) =
\begin{cases}
\frac{x^4+x^3+2x^2}{\tan^{-1}x} & \text{if} \ x\ne0 \\ \\
10 & \text{if} \ x=0 \\
\end{cases}.$$
$\frac{x^4+x^3+2x^2}{\tan^{-1}x}$ being the division of a polynomial function and a inverse trigonometric function (both continuous) is continuous at each $x\ne0$ and $10$ being a constant function is also continuous.
At $x=0,$
$$\lim_{x\to 0}\frac{x^4+x^3+2x^2}{\tan^{-1}x}=?$$
How do I simplify this?
| Applying L'Hospital:
$$\lim_{x\to 0}\frac{x^4+x^3+2x^2}{\tan^{-1}x}=\lim_{x\to 0}\frac{4x^3+3x^2+4x}{\frac{1}{1+x^2}}=0\ne 10$$
Therefore, $f$ is not continuous at $0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/374970",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Prove this inequality $a \sqrt{1-b^2}+b\sqrt{1-a^2}\le1 $ Prove that for $a,b\in [-1,1]$:
$$a\sqrt{1-b^2}+b\sqrt{1-a^2}\leq 1$$
| By Cauchy-Schwarz
$$(a\sqrt{1-b^2}+b\sqrt{1-a^2})^2 \leq (a^2+b^2)(1-b^2+1-a^2)=2(a^2+b^2)-(a^2+b^2)^2$$
$$=1-(1-a^2-b^2)^2\leq 1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/375260",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
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} |
How to find the solutions $x$ of $ 2\sin{11^{\circ}}\sin{71^{\circ}}\sin{(x^{\circ}+30^{\circ})}=\sin{2013^{\circ}}\sin{210^{\circ}}$ Let
$$2\sin{11^{\circ}}\sin{71^{\circ}}\sin{(x^{\circ}+30^{\circ})}=\sin{2013^{\circ}}\sin{210^{\circ}}$$
where $90^{\circ}<x<180^{\circ}$.
My idea: $$2\sin{11^{\circ}}\sin{71^{\circ}}\sin{(x^{\circ}+30^{\circ})}=\sin{33^{\circ}}\sin{30^{\circ}}$$
and
$$\Big[\cos{(71^{\circ}-11^{\circ})}-\cos{(71^{\circ}+11^{\circ})}\Big]\sin{(x^{\circ}+30^{\circ})}=\dfrac{1}{2}\sin{33^{\circ}}$$
and
$$(1-2\sin{8^{\circ}})\sin{(x^{\circ}+30^{\circ})}=\sin{33^{\circ}}$$
I used Wolfram|Alpha to find $x=19$ or $101$.
My question: we can't use Wolfram|Alpha. How to find this $x$? Thank you.
| $$2\sin 11^\circ*\sin 71^\circ*\sin (x+30)^\circ=\sin 2013^\circ*\sin 210^\circ$$
$$2\sin 71^\circ*\sin 11^\circ*\sin (x+30)^\circ=\sin 213^\circ*-\sin 30^\circ$$
$$2\sin 71^\circ*\sin 11^\circ*\sin (x+30)^\circ=-\sin 30^\circ*(\sin 213^\circ-\sin 71^\circ+\sin 71^\circ)$$
$$2\sin 71^\circ*\sin 11^\circ*\sin (x+30)^\circ=-\sin 30^\circ*(2\cos 142^\circ*\sin 71^\circ+2\cos 60^\circ*\sin 71^\circ)$$
$$2\sin 71^\circ*\sin 11^\circ*\sin (x+30)^\circ=-\sin 30^\circ*2sin 71^\circ*(\cos 142^\circ+\cos 60^\circ)$$
$$2\sin 71^\circ*\sin 11^\circ*\sin (x+30)^\circ=-\sin 71^\circ*(\cos 142^\circ+\cos 60^\circ)$$
$$2\sin 11^\circ*\sin (x+30)^\circ=(\cos 142^\circ+cos 60^\circ)$$
$$-2\cos 101^\circ*\sin (x+30)^\circ=-2\cos 101^\circ*cos 41^\circ$$
$$\sin (x+30)^\circ=\cos 41^\circ$$
$$\sin (x+30)^\circ=\sin 49^\circ$$
For $$(x+30)^\circ=49^\circ$$ or $$x^\circ=19^\circ$$
Also $$(x+30)^\circ=131^\circ$$ or $$x^\circ=101^\circ$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/377057",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Is there a way to compute $\lim\limits_{x\to\pi/3}\frac{\sqrt{3+2\cos x}-2}{\ln(1+\sin3x)}$ without using L'hopital? I can compute
$$\lim_{x\to\pi/3}\frac{\sqrt{3+2\cos x}-2}{\ln(1+\sin3x)}$$ using L'hopital and the limit equals $\frac{\sqrt{3}}{12}$, but is there another way to compute this limit without using L'hopital, please do it for me if there is.
Many thanks.
| Let's replace $x$ by $x+\pi/3$ and let $x \to 0$.
I will be as simple-minded as I can.
We use $\sin(\pi/3) = \sqrt{3}/2$ and $\cos(\pi/3) = 1/2$.
$\cos(x+\pi/3) = \cos(x) \cos(\pi/3) - \sin(x)\sin(\pi/3)
= \cos(x)/2 - \sin(x)\sqrt{3}/2
$.
As $x \to 0$,
$\cos(x+\pi/3) \to 1/2$.
$\sin(3(x+\pi/3))
= \sin(3x+3\pi/3)
= \sin(3x+\pi)
= \sin(3x)\cos(\pi) + \cos(3x)\sin(\pi)
= -\sin(3x)
\to 0
$.
So, initially, this goes to $0/0$.
Let's use $\sin(x) \approx x$ and
$\cos(x) \approx 1-x^2/2$ as $x \to 0$.
$\cos(x+\pi/3) \approx
(1-x^2/2)/2 - x\sqrt{3}/2
\approx 1/2- x\sqrt{3}/2
$
and
$\sin(3(x+\pi/3))
\approx
-3x$.
Then $\sqrt{3+2\cos (x+\pi/3}-2 \approx
\sqrt{3+2(1/2- x\sqrt{3}/2)}
=\sqrt{4-x\sqrt{3}}
=2\sqrt{1-x\sqrt{3}/4}
\approx 2(1-x\sqrt{3}/8)
=2-x\sqrt{3}/4
$
so the numerator is about
$-x\sqrt{3}/4$ as $x \to 0$.
For the denominator,
we have $\ln(1+x) \approx x$ as $x \to 0$,
so
$\ln(1+\sin(3(x+\pi/3))
\approx \ln(1-3x)
\approx -3x$
so the ratio is
$\frac{-x\sqrt{3}/4}{-3x}
=\frac{\sqrt{3}}{12}
$.
This can be made rigorous if we can use
$\sin(x) = x +o(x^2)$
or
$\sin(x) \approx x +O(x^3)$
as $x \to 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/377811",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 2
} |
finding Moment generating function and CDF with pmf The random variable X has the pmf f(-1)=1/4, f(0)=1/8, f(1)=1/4, f(2)=3/8
a) How would you draw the c.d.f with points (-2,F(-2)), (-1,F(-1)), (0,F(0)), (1,F(1)), (2, F(2)), (3,F(3))
b)Write the MGF of this distribution
c) Range of and p.m.f of X^(2)
For the cdf i tried to come up with an equation that satisfies the values but couldn't I'm guessing the cdf is the antiderivative of such equation?
for the pmf should include f(-1)=1/4 ? i thought pmf f(x)>0
| (1) I do not think you are expected to give single "equation."
We have $F(-2)=\Pr(X\le -2)=0$.
Similarly, $F(-1)=\Pr(X\le -1)=\frac{1}{4}$.
Also, $F(0)=\Pr(X\le 0)=\frac{1}{4}+\frac{1}{8}=\frac{3}{8}$.
Also, $F(1)=\Pr(X\le 1)=\frac{5}{8}$.
Also, $F(2)=1$. And $F(3)=1$.
To understand what we did, remember we are calculating the cumulative distribution function.
(2) For the moment generating function, note that we want $E(e^{Xt})$. Note that $X=-1$ with probability $\frac{1}{4}$, $X=0$ with probability $\frac{1}{8}$, $X=1$ with probability $\frac{1}{4}$, and $X=2$ with probability $\frac{3}{8}$. So $e^{Xt}$ takes values $e^{-t}$, $1$, $e^t$, and $e^{2t}$ with the probabilities just listed. The mgf is therefore
$$\frac{1}{4}e^{-t}+\frac{1}{8}+\frac{1}{4}e^{t}+\frac{3}{8}e^{2t}.$$
(3) for the mass function of $X^2$, note that $X^2$ takes on values $0$, $1$, and $4$. The event $X^2=0$ happens precisely if $X=0$ happens. The probability of this is $\frac{1}{8}$.
The event $X^2=1$ can happen in $2$ ways, if $X=-1$ and if $X=1$. Thus the probability that $X^2=1$ is $\frac{1}{4}+\frac{1}{4}$.
Finally, $\Pr(X^2=4)=\frac{3}{8}$.
Remark: For the cumulative distribution function $F(x)$ of this kind of discrete distribution, the traditional integral is not of much use. In our case, $F(x)=0$ for all $x\lt -1$. Then at $x=-1$, $F(x)$ jumps to $\frac{1}{4}$. It remains at $\frac{1}{4}$ for all $x$ with $-1\le x\lt 0$. Then at $0$ it jumps to $\frac{3}{8}$, where it remains for $0\le x\lt 1$. At $1$ it jumps to $\frac{5}{8}$, then jumps to $1$ at $x=2$, and stays there for all $x\ge 2$.
| {
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"url": "https://math.stackexchange.com/questions/378922",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find all matrices $A$ of order $2 \times 2$ that satisfy the equation $A^2-5A+6I = O$
Find all matrices $A$ of order $2 \times 2$ that satisfy the equation
$$
A^2-5A+6I = O
$$
My Attempt:
We can separate the $A$ term of the given equality:
$$
\begin{align}
A^2-5A+6I &= O\\
A^2-3A-2A+6I^2 &= O
\end{align}
$$
This implies that $A\in\{3I,2I\} = \left\{\begin{pmatrix}
3 & 0\\
0 & 3
\end{pmatrix},
\begin{pmatrix}
2 & 0\\
0 & 2
\end{pmatrix}\right\}$.
Are these the only two possible values for $A$, or are there other solutions?If there are other solutions, how can I find them?
| As in general we have
$$
\mathbf{A}^2 - \big( \lambda_1 + \lambda_2 \big) \mathbf{A}
+ \lambda_1 \lambda_2 \mathbf{I} = 0, \tag 1
$$
we see that in this case
$$
\lambda_1 = 2, \quad \lambda_2 = 3. \tag 2
$$
So the general solution is given by
$$
\bbox[16px,border:2px solid #800000]
{ \mathbf{B} \pmatrix{ 3 & 0 \\ 0 & 2} \mathbf{B}^{-1}, } \tag 3
$$
for any matrix such that $\det(\mathbf{B}) \ne 0$.
Note that
$$
\pmatrix{0 & 1 \\ 1 & 0} \pmatrix{3 & 0 \\ 0 & 2} \pmatrix{0 & 1 \\ 1 & 0}
= \pmatrix{ 2 & 0 \\ 0 & 3}. \tag 4
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/379076",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 3
} |
Is $B = A^2 + A - 6I$ invertible when $A^2 + 2A = 3I$? Given:
$$A \in M_{nxn} (\mathbb C), \; A \neq \lambda I, \; A^2 + 2A = 3I$$
Now we define:
$$B = A^2 + A - 6I$$
The question:
Is $B$ inversable?
Now, what I did is this:
$A^2 + 2A = 3I \rightarrow \lambda^2v + 2\lambda v = 3v \rightarrow \lambda_1 = 1, \lambda_2 = -3$
Is what I suggested correct? I know that if so, I just do the same to B and calculate the determinant.
| Note that
$$
\begin{align}
(A-I)B
&=(A-I)(A^2+A-6I)\\
&=(A-2I)(A^2+2A-3I)\\[4pt]
&=0
\end{align}
$$
Thus, unless $A=I$, and therefore, $B=-4I$, $(A-I)B=0$ implies that $B$ is not invertible.
Clarification: Suppose that $B^{-1}$ exists, then
$$
\begin{align}
A-I
&=(A-I)BB^{-1}\\
&=0B^{-1}\\[6pt]
&=0
\end{align}
$$
Thus, if $B^{-1}$ exists, then $A=I$. This is the contapositive of "if $A\ne I$, then $B$ is not invertible".
A bit of explanation
I used the Euclid-Wallis Algorithm to try and write $(A^2+A-6I)x+(A^2+2A-3I)y=I$ to compute an inverse for $A^2+A-6I\bmod A^2+2A-3I$:
$$
\begin{array}{c|c}
&&1&-A+I\\\hline
1&0&1&A-I\\
0&1&-1&-A+2I\\
A^2+A-6I&A^2+2A-3I&-A-3I&0
\end{array}
$$
Unfortunately, this showed that $A+3I$ was the GCD of $A^2+A-6I$ and $A^2+2A-3I$. However, this did show that
$$
(A-I)(A^2+A-6I)-(A-2I)(A^2+2A-3I)=0
$$
which was used in the answer above.
| {
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How to find the number of real roots of the given equation?
The number of real roots of the equation $$2 \cos \left( \frac{x^2+x}{6} \right)=2^x+2^{-x}$$ is
(A) $0$, (B) $1$, (C) $2$, (D) infinitely many.
Trial: $$\begin{align} 2 \cos \left( \frac{x^2+x}{6} \right)&=2^x+2^{-x} \\ \implies \frac{x^2+x}{6}&=\cos ^{-1} \left( \frac{2^x+2^{-x}}{2} \right)\end{align}$$ Then I can't proceed.
| $$\text{So, }(2^x)^2- 2^x\cdot2\cos\left(\frac{x^2+x}6\right)+1=0$$ which is a quadratic equation in $2^x$
As $x$ is real, so would be $2^x$
so,the discriminant must be $\ge 0$
i.e., $$\{2 \cos\left(\frac{x^2+x}6\right)\} ^2-4\cdot1\cdot1\ge0$$
$$\implies4\left(\cos^2\left(\frac{x^2+x}6\right)-1\right)\ge0$$
$$\implies \sin^2\left(\frac{x^2+x}6\right)\le 0$$
$$\implies \sin^2\left(\frac{x^2+x}6\right)= 0\text{ as the square of a real number }\ge 0 $$
So, the discriminant must be $0,\implies \cos\left(\frac{x^2+x}6\right)=\pm1$
$\implies 2^x=\frac{2\cos\left(\frac{x^2+x}6\right)\pm\sqrt{2 \cos\left(\frac{x^2+x}6\right)\}^2-4\cdot1\cdot1}}2=\pm1$
As $x$ is real, $2^x>0$
$\implies 2^x=1\implies x=0$ which also satisfies $ \sin^2\left(\frac{x^2+x}6\right)= 0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/380896",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Showing that $\mathbb Q(\sqrt{17})$ has class number 1 Let $K=\mathbb Q(\sqrt{d})$ with $d=17$. The Minkowski-Bound is $\frac{1}{2}\sqrt{17}<\frac{1}{2}\frac{9}{2}=2.25<3$.
The ideal $(2)$ splits, since $d\equiv 1$ mod $8$. So we get $(2)=(2,\frac{1+\sqrt{d}}{2})(2,\frac{1-\sqrt{d}}{2})$ and $(2,\frac{1\pm\sqrt{d}}{2})$ are two ideals of norm $2$.
Now if we can show that $(2,\frac{1\pm\sqrt{d}}{2})$ are principal ideals, then we know that every ideal class contains a principal ideal, which shows that the class number is $1$.
But how can we show that $(2,\frac{1\pm\sqrt{d}}{2})$ are principal?
| Hint:
$$
\left(\frac{3+\sqrt{17}}2\right)\left(\frac{3-\sqrt{17}}2\right)=\frac{9-17}4=-2.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/382188",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
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} |
Find all real numbers $x$ for which $\frac{8^x+27^x}{12^x+18^x}=\frac76$ Find all real numbers $x$ for which $$\frac{8^x+27^x}{12^x+18^x}=\frac76$$
I have tried to fiddle with it as follows:
$$2^{3x} \cdot 6 +3^{3x} \cdot 6=12^x \cdot 7+18^x \cdot 7$$
$$ 3 \cdot 2^{3x+1}+ 2 \cdot 3^{3x+1}=7 \cdot 6^x(2^x+3^x)$$
Dividing both sides by $6$ gives us
$$2^{3x}+3^{3x}=7 \cdot 6^{x-1}(2^x+3^x)$$
Is this helpful? If so, how should I proceed form here? If not any hints would be greatly appreciated.
| $$
\begin{array}{rcl}
6(8^x + 27^x) &=& 7(12^x + 18^x) \\
6(2^{3x} + 3^{3x}) &=& 7(3^x2^{2x} + 3^{2x}2^x) \\
\end{array}
$$
Substitute $a\!=\!2^x$ and $b\!=\!3^x$ for simplicity:
$$
\begin{array}{rcl}
6(a^3 + b^3) &=& 7(a^2b + ab^2) \\
6(a+b)(a^2 - ab + b^2) &=& 7ab(a+b) \\
6(a^2 - ab + b^2) &=& 7ab \\
6a^2 -13ab + 6b^2 &=& 0 \\
a^2 -\frac{13}{6}ab + b^2 &=& 0
\end{array}
$$
The left hand side can be factorized as $\left(a-\dfrac{3}{2}b\right)\left(a-\dfrac{2}{3}b\right)$.
$$
\left(a-\dfrac{3}{2}b\right)\left(a-\dfrac{2}{3}b\right)=0 \\
\begin{array}{rclcrcl}
2a &=& 3b &\text{or}& 3a &=& 2b \\
2.2^x &=& 3.3^x &\qquad\text{or}\qquad& 3.2^x &=& 2.3^x \\
x &=& -1 &\qquad\text{or}\qquad& x &=& +1
\end{array}
$$
Therefore, $x$ can be either $-1$ or $+1$.
$$ \boxed{x = \mp 1} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/384090",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "47",
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Evaluting the limit $\lim_{x\rightarrow\infty}\frac{1}{\sqrt{x^{2}-4x+1}-x+2}$ I'm attempting to evaluate the limit
$\lim_{x\rightarrow\infty}\frac{1}{\sqrt{x^{2}-4x+1}-x+2}$
I got it reduced to the following
$\lim_{x\rightarrow\infty}\frac{\sqrt{\frac{1}{\left(x-2\right)^{2}}-\frac{3}{\left(x-2\right)^{4}}}+1}{1-\frac{3}{\left(x-2\right)^{2}}-1}$
But putting in $\infty$ I get $\frac{1}{0}$ and, what's worse, Mathematica tells me the limit is equal to $-\infty$. Where am I going wrong?
| \begin{align*}
&\lim_{x\rightarrow\infty}\frac{1}{\sqrt{x^{2}-4x+1}-x+2}\\
=&\lim_{x\rightarrow\infty}\frac{1}{\sqrt{(x-2)^2-3}-(x-2)}\\
=&\lim_{x-2\rightarrow\infty}\frac{1}{\sqrt{(x-2)^2-3}-(x-2)}\\
=&\lim_{z\rightarrow\infty}\frac{1}{\sqrt{z^2-3}-z}\\
=&\lim_{z\rightarrow\infty}\frac{\frac{1}{z}}{\sqrt{1-\frac{3}{z^2}}-1}
\end{align*}
Note that the absolute value of the last expression is $\infty$ and the denominator is negative as $\sqrt{1-\frac{3}{z^2}}<1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/384471",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Prove $\int_0^{\infty } \frac{1}{\sqrt{6 x^3+6 x+9}} \, dx=\int_0^{\infty } \frac{1}{\sqrt{9 x^3+4 x+4}} \, dx$ Prove that:
$(1)$$$\int_0^{\infty } \frac{1}{\sqrt{6 x^3+6 x+9}} \ dx=\int_0^{\infty } \frac{1}{\sqrt{9 x^3+4 x+4}} \ dx$$
$(2)$$$\int_0^{\infty } \frac{1}{\sqrt{8 x^3+x+7}} \ dx>1$$
What I do for $(1)$ is (something trival):
$$\int_0^{\infty } \frac{1}{\sqrt{6 x^3+6 x+9}} \, dx=\int_0^{\infty } \frac{1}{\sqrt{x^3+36 x+324}} \, dx$$
$$\int_0^{\infty } \frac{1}{\sqrt{9 x^3+4 x+4}} \, dx=\frac{\sqrt3}3\int_0^{\infty } \frac{1}{\sqrt{x^3+4 x+12}} \, dx$$
so it remains to prove that
$$\int_0^{\infty } \frac{1}{\sqrt{x^3+36 x+324}} \, dx=\frac{\sqrt3}3\int_0^{\infty } \frac{1}{\sqrt{x^3+4 x+12}} \, dx$$
Thanks in advance!
| Hint of 1. Substitute $x=\frac{3}{2}y$
But I have no idea about 2...
| {
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"url": "https://math.stackexchange.com/questions/387760",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
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} |
Compute the remainder when $67!$ is divided by $71$. This is how far I've been able to get.
By using Wilson's Theorem:
$$\begin{align}
70! &\equiv -1 \pmod{71} \\
67!(68)(69)(70) &\equiv -1 \pmod{71} \\
67!(68)(69)(-1) &\equiv -1 \pmod{71} \\
67!(68)(69) &\equiv 1 \pmod{71} \\
\end{align}$$
EDIT: Here is how I proceeded using TMM's and Carl Mummert's hints.
$$\begin{align}
&68 \equiv -3 \pmod{71}\\
and\\
&69 \equiv -2 \pmod{71}\\
\end{align}$$
So:
$$\begin{align}
67!(-3)(-2) &\equiv 1 \pmod{71} \\
67! &\equiv 6^{-1} \pmod{71} \\
\end{align}$$
By using the Euclidean algorithm:
$$71 = 6 \cdot 11 + 5$$
$$6 = 5 \cdot 1 + 1$$
$$5 = 1 \cdot 5 + 0$$
Now, going backwards:
$$\begin{align}
1 &= 6 - 5 \\
&= 6 - (71 - 6 \cdot 11) \\
&= 6 + 6(11) - 71 \\
&= 6(1 + 11) - 71(1) \\
&= 6(12) - 71(1)
\end{align}$$
Therefore, $67! \equiv 6^{-1} \equiv 12 \pmod{71}$.
| Hint: $$69 \equiv -2 \pmod{71}, \qquad 68 \equiv -3\pmod{71}.$$
Further hint: $72 = 6 \cdot 12$, so the inverse you need to compute should not be too difficult...
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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"answer_id": 0
} |
Integral of $ \int_{-1}^{1} \frac{x^4}{x^2+1}\,dx $ Any suggestions how to solve it? by parts?
$$ \int_{-1}^{1} \frac{x^4}{x^2+1}dx$$
Thanks!
| $$\dfrac{x^4}{x^2+1}=\dfrac{(x^4+2x^2+1)-2x^2-1}{x^2+1}=\\
=\dfrac{(x^2+1)^2-2x^2-2+1}{x^2+1}=x^2+1-2+\dfrac{1}{x^2+1}=x^2-1+\dfrac{1}{x^2+1}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/390169",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Integrating $\int \left( \frac{\cos\theta}{1+\sin^2\theta} \right)^n \, d\theta$, for $n=2$ and $n=3$ How do you integrate the following functions:
$$\int \left( \frac{\cos\theta}{1+\sin^2\theta} \right)^2 \, d\theta$$ and $$\int \left( \frac{\cos\theta}{1+\sin^2\theta} \right)^3 \, d\theta
$$
respectively?
Note: Initially, I tried integrating the function without the power and obtained the result below.
$$
\int \frac{\cos\theta}{1+\sin^2\theta} \, d\theta = \arctan(\sin\theta)+ C
$$
However, from here it is difficult to proceed. Integration by substitution doesn't seem to work.
How should I go on from here? Any pointers would be greatly appreciated.
| $$ \begin{align*} \int \frac{\cos^2 x}{\left(1 + \sin^2 x\right)^2} \ dx &= \int \frac{\sec^2 x}{\left(\sec^2 x + \tan^2 x\right)^2} \ dx \\ &= \int \frac{1}{\left(2\tan^2x + 1\right)^2}\ d(\tan x) \\ \left(\tan x \mapsto \frac{\tan u}{\sqrt2} \right) &= \frac{1}{\sqrt{2}}\int \frac{\sec^2 u}{\sec^4 u} \ du \\&= \frac{1}{2\sqrt2}\int\cos(2u) + 1 \ du \\ &= \frac{1}{2\sqrt2}\left(\frac{\sin(2u)}{2} + u \right) + C \\ &=\frac{1}{2\sqrt2}\left(\frac{\sin(2\arctan(\sqrt2\tan x))}{2} + \arctan\left(\sqrt{2}\tan x\right)\right)+C\\&= \frac{\tan x}{2\left(2\tan^2x + 1\right)} + \frac{\arctan(\sqrt2 \tan x)}{2\sqrt{2}} + C\end{align*}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/391338",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
$x^4+y^4 \geq \frac{(x^2+y^2)^2}{2}$ I'm doing some exercise to prepare for my multivariable analysis exam. I didn't understand the second part of this question.
Given the function
$$f(x,y)=(x^2+y^2+1)^2 - 2(x^2+y^2) +4\cos(xy)$$
Prove that the taylor polynomial of degree $4$ of $f$ is equal to
$5+x^4+y^4$.
First, $4\cos(xy) = 4 - 2(xy)^2 + 4R_3 $
$(x^2+y^2+1)^2=x^4+2 x^2 y^2+2 x^2+y^4+2 y^2+1$
Therefore: $(x^2+y^2+1)^2 - 2(x^2+y^2)=x^4+2 x^2 y^2+y^4+1$
Therefore: $f(x,y)=x^4+y^4+5+4R_3$
I don't know exactly why I can now conclude that Taylor Polynomial of degree 4 must be $5+x^4+y^4$, but I don't know exactly why.
Now the second question is: $x^4+y^4 \geq \frac{(x^2+y^2)^2}{2}$
New edit
I understand this now thanks to hint of Hagen von Eitzen, thanks !
The third question is:
Determine what kind of stationary point you have in $(0,0)$.
| $$x^4 + y^4 \ge \frac{(x^2+y^2)^2}{2} \\
2x^4+2y^4\ge x^4+2x^2y^2+y^4 \\
x^4-2x^2y^2+y^4 \ge0 \\ (x^2-y^2)^2\ge0$$
Notice that $x^2-y^2\in R$, so all real number with an even exponent are always greater or iqual to zero.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/392544",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Relation between chords length and radius of circle Two chords of a circle, of lengths $2a$ and $2b$ are mutually perpendicular. If the distance of the point at which the chords intersect,from the centre of the circle is $c$($c<$radius of the circle),then find out the radius of the circle in terms of $a,b$ and $c$.Show some short-cut to do it quickly
| Let $P$ be the point where the two chords (and a diameter) meet. Let $h$ (and $k$) be the distance from $P$ to the midpoint of the $2a$ chord (respectively, the $2b$ chord); that is, say $P$ divides the chord into sub-segments of length $a+h$ and $a-h$ (respectively, $b+k$ and $b-k$). Note that $P$ divides a diameter into sub-segments of length $r+c$ and $r-c$ (where $r$ is the radius of the circle); note also that $c$ is the hypotenuse of a right triangle with legs $h$ and $k$: so, $c^2 = h^2 + k^2$.
The Power of a Point principle says that every chord through a particular point of a circle is divided into sub-segments such that the product of the lengths of those sub-segments is a constant (the so-called "power" of the point in question). Thus,
$$(a+h)(a-h) = (b+k)(b-k) = (r+c)(r-c)$$
More succinctly,
$$a^2 - h^2 \;\;=\;\; b^2 - k^2 \;\;=\;\; r^2 - c^2$$
With an eye towards combining an $h^2$ with a $k^2$, I'll add the left-hand and "middle-hand" sides together; their sum is necessarily twice the right-hand side:
$$\begin{align}
( a^2 - h^2 ) + ( b^2 - k^2 ) &= 2 (r^2 - c^2) \\
\implies a^2 + b^2 - ( h^2 + k^2 ) &= 2 r^2 - 2 c^2 \\
\implies a^2 + b^2 - c^2 &= 2 r^2 - 2 c^2 \\
\implies a^2 + b^2 + c^2 &= 2 r^2
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/394257",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
} |
If $2n+1$ and $4n+3$ are prime, then $2n-1$ and $4n+1$ are not when $n>2$ How do you prove that, for $n>2$, if $2n+1$ and $4n+3$ are prime numbers, then $2n-1$ and $4n+1$ are composite numbers?
| When $2n+1$ is prime, $4n+2$ will only have $2$ and $2n+1$ as its prime factor.
From the question, $4n+3$ is also a prime.
As $4n+1,\,4n+2,\,4n+3$ are three consecutive numbers, one of them must be divisible by $3$ and we know that $4n+2$ and $4n+3$ are not, thus we have the conclusion that $4n+1$ is divisible by $3$.
$$\begin{align}
4n+1 \equiv 0 &\pmod 3\\
\implies 4n\equiv-1&\pmod 3\\
\implies n \equiv-1&\pmod 3 [\text{as}\, \gcd(3,4)=1].
\end{align}$$
And as a result, we can work out $2n-1$ and $4n+1$:
$$2(-1)-1 \equiv 0 \pmod 3\\
4(-1)+1 \equiv 0 \pmod 3$$
Hope it helps
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/394411",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Using the hypothesis $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}$ to prove something else Assuming that $$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}$$
Is it possible to use this fact to prove something like:
$$\frac{1}{a^{2013}}+\frac{1}{b^{2013}}+\frac{1}{c^{2013}}=\frac{1}{a^{2013}+b^{2013}+c^{2013}}$$
Just curious. Thanks for the help!
| Well, assuming that $a,b,c\neq 0$, we can see that:
\begin{align}
\frac{1}{a}+\frac{1}{b}+\frac{1}{c}&=\frac{1}{a+b+c}
\\ \\
\frac{ab+bc+ac}{abc}&=\frac{1}{a+b+c}
\\ \\
a^2b+abc+a^2c+ab^2+b^2c+abc+abc+bc^2+ac^2&=abc
\\ \\
(ab+b^2+ac+bc)c+(ab+b^2+ac+bc)a&=0
\\ \\
(ab+b^2+ac+bc)(c+a)&=0
\\ \\
(a+b)(b+c)(c+a)&=0
\end{align}
Now we know that $(a+b)(b+c)(c+a)=0$
$\therefore a=-b\text{ or }b=-c\text{ or }a=-c$.
We can use the fact $a=-b$ to our LHS:
$$\frac{1}{(-b)^{2013}}+\frac{1}{b^{2013}}+\frac{1}{c^{2013}}=-\frac{1}{b^{2013}}+\frac{1}{b^{2013}}+\frac{1}{c^{2013}}=\frac{1}{c^{2013}}$$
$$\text{RHS: }\frac{1}{-b^{2013}+b^{2013}+c^{2013}}=\frac{1}{c^{2013}}=LHS$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/394532",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 0
} |
Finding $y$ value of canonical ellipse. I have an ellipse:
$$
\frac{x^2}{a^2}+\frac{y^2}{b^2}=1
$$
This may be a simple question, but my mind plays tricks on me at the moment;
Which is the most efficient way if I have $x$, $a$ and $b$ and want to find the value of $y$?
Hope someone can help me - thanks in advance :)!
| Given:
$\boxed{\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1}$
We can subtract $\dfrac{x^2}{a^2}$ from both sides:
$\dfrac{x^2}{a^2}-\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1-\dfrac{x^2}{a^2}$
Next multiplying both sides by $b^2$:
$\left(\dfrac{y^2}{b^2}\right)b^2=\left(1-\dfrac{x^2}{a^2}\right)b^2$
Which becomes:
$\dfrac{b^2y^2}{b^2}=b^2-\dfrac{b^2x^2}{a^2}$
The b's on the left hand side cancel leaving us with:
$y^2=b^2-\dfrac{b^2x^2}{a^2}$
We can take the square root of both sides leaving us with our final answer:
$\boxed{y=\pm\sqrt{b^2-\dfrac{b^2x^2}{a^2}}}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/395378",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Calculating $\sqrt{28\cdot 29 \cdot 30\cdot 31+1}$ Is it possible to calculate $\sqrt{28 \cdot 29 \cdot 30 \cdot 31 +1}$ without any kind of electronic aid?
I tried to factor it using equations like $(x+y)^2=x^2+2xy+y^2$ but it didn't work.
| Hint: Use $(x)(x+1)(x+2)(x+3)+1 = (x)(x+3)(x+2)(x+1)+1 =(x^2+3x)(x^2+3x+2)+1= (x^2+3x)^2+2(x^2+3x)+1=(x^2+3x+1)^2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/395962",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 4,
"answer_id": 3
} |
How to evaluate $\sqrt[3]{a + ib} + \sqrt[3]{a - ib}$? The answer to a question I asked earlier today hinged on the fact that
$$\sqrt[3]{35 + 18i\sqrt{3}} + \sqrt[3]{35 - 18i\sqrt{3}} = 7$$
How does one evaluate such expressions? And, is there a way to evaluate the general expression
$$\sqrt[3]{a + ib} + \sqrt[3]{a - ib}$$
| I came across very similar problems while doing Maths Olympiad in high school and had only one approach to it then.
Observe the property for $a,b,c \in \Bbb K$ $$ \text {if } a+b+c=0,\ \text{ then }\ a^3+b^3+c^3=3abc \quad (\ast)$$
This can easily be proved by expanding $(a+b+c)^3$
$$\begin{array}{rll}(a+b+c)^3&=&a^3+b^3+c^3+\underbrace{3a^2b+3a^2c+3b^2c+3b^2a+3c^2a+3c^2b}_X+6abc=0\\X&=&3ab\underbrace{(a+b)}_{=-c}+3bc\underbrace{(b+c)}_{=-a}+3ca\underbrace{(c+a)}_{=-b}=-9abc \\ (a+b+c)^3&=&a^3+b^3+c^3-9abc+6abc\ =\ 0 \ \implies\ (\ast)\end{array}$$
In the problem above you have an expression of the form
$$\sqrt[3]{x+\sqrt y}+\sqrt[3]{x-\sqrt y}=z$$which can also translate to $$\underbrace{\sqrt[3]{x+\sqrt y}}_a+\underbrace{\sqrt[3]{x-\sqrt y}}_b +\underbrace{(-z)}_c=0$$
Applying the property abover
$$\begin{array}{rll}\underbrace{\left(\sqrt[3]{x+\sqrt y}\right)^3+\left(\sqrt[3]{x-\sqrt y}\right)^3}_{2x} +(-z)^3&=&3\underbrace{\left(\sqrt[3]{x+\sqrt y}\right)\left(\sqrt[3]{x-\sqrt y}\right)}_{\text{conjugates}}(-z) \\ \implies \quad 2x-z^3&=&-3z\cdot \sqrt[3]{x^2-y}\end{array}$$
In conclusion the problem resolves to finding the roots
$$z^3-3z\cdot \sqrt[3]{x^2-y} -2x = 0$$
I have also been curious about this, I asked this sometimes ago
| {
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"url": "https://math.stackexchange.com/questions/396915",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 7,
"answer_id": 3
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Solution to second order nonlinear ODE I need to find and exact solution for the following ODEs $$y''=-3y'+2y+2x+3,\qquad y(0)=2$$ $$y(1)=-4+5\exp\left(-3/2+\left(\sqrt{17}\right)/2\right)$$ and $$y''=2y^3-6y-2x^3;$$ $$1\leq x\leq2;$$ $$y(1)=2;$$ $$y(2)=5/2$$
| First equation $y''+3y'-2y=2x+3$ has a solution which is a sum of solution to the homogeneous ODE $y''+3y'-2y=0$ and a particular solution of inhomogeneous problem. By solving the characteristic equation $a^2+3a-2=0$, we can easily check that roots are $\frac{-3\pm\sqrt{17}}{2}$. Therefore, solution to this problem is $C_1e^{\frac{-3+\sqrt{17}}{2}x}+C_2e^{\frac{-3-\sqrt{17}}{2}x}$. Particular solution is $-x-3$ (particular solution has the form $ax+b$; by replacing in equation and solving, we get: $a=-1$, $b=-3$). Solution to the first problem is $C_1e^{\frac{-3+\sqrt{17}}{2}x}+C_2e^{\frac{-3-\sqrt{17}}{2}x}-x-3$. After evaluation $y(0)$, $y(1)$, we get the following sistem
$$C_1+C_2-3=2,$$
$$C_1e^{\frac{-3+\sqrt{17}}{2}}+C_2e^{\frac{-3-\sqrt{17}}{2}}-1-3=-4+5e^{\frac{-3+\sqrt{17}}{2}},$$ from where we get $C_1=5$, $C_2=0$. Final solution is $5e^{\frac{-3+\sqrt{17}}{2}x}-x-3$.
| {
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"url": "https://math.stackexchange.com/questions/397635",
"timestamp": "2023-03-29T00:00:00",
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$\lim_{x\to \pi/2} \;\frac 1{\sec x+ \tan x}$ how to solve it answer is $0$, but $\frac 1{\infty + \infty}$ is indeterminate form
$$\lim_{x \to \pi/2} \frac 1{\sec x + \tan x}$$
| In order to solve this, it would be best to convert the $\sec x$ into $\frac{1}{\cos x}$. and convert $\tan x$ to $\frac{\sin x}{\cos x}$ So we will have
$$\lim_{x\to\frac{\pi}{2}} \frac{1}{\frac{1}{\cos x} + \frac{\sin x}{\cos x}}$$
Now we have to get rid of the fraction in the denominator. So multiply by $\cos x$ so we will have:
$$\lim_{x\to\frac{\pi}{2}} \frac{1}{\frac{\cos x}{\cos x} + \frac{\sin x}{\cos x}}$$
When then is turned into
$$\lim_{x\to\frac{\pi}{2}} \frac{\cos x}{1 + \sin x}$$
Then sub $\pi/2$ and you'll be gold.
$$\lim_{x\to\frac{\pi}{2}} \frac{\cos \frac{\pi}{2}}{1 + \sin \frac{\pi}{2}}$$
And the answer will be $0$. Therefore the limit is $0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/398223",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 2
} |
How to show $x^4 - 1296 = (x^3-6x^2+36x-216)(x+6)$ How to get this result: $x^4-1296 = (x^3-6x^2+36x-216)(x+6)$?
It is part of a question about finding limits at mooculus.
| $$\begin{array}{r|rrrr|r}
& 1 & 0 & 0 & 0 & -1296 \\
-6 & & -6 & 36 & -216 & 1296 \\
\hline
& 1 & -6 & 36 & -216 & 0
\end{array}
$$
This shows that
$$
x^4-1296=(x+6)(x^3-6x^2+36x-216)
$$
Of course you can also use
$$
a^4 - b^4 = (a-b)(a^3 + a^2b + ab^2 + b^3)
$$
with $a=x$ and $b=-6$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/400178",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 3
} |
Probability of no matching pairs of shoes There are 3 different pairs ( i.e. 6 units say aa, bb, cc) of shoes in a lot. Now three person come and pick the shoes randomly (each get 2 units). Let p be the probability that no one is able to wear shoes (i.e. no one gets a correct pair) , then the value of $\frac{13p}{4-p}$ is?
| We use a technique that is definitely overkill for this problem, the Principle of Inclusion/Exclusion. But the idea might be useful for a larger problem, say $5$ people and $5$ pairs of shoes.
Let our people be P, Q, and R. There are $\binom{6}{2}$ ways for P to pick two shoes. And there are $\binom{3}{1}$ ways to pick a matching pair. So the probability that P picks a matching pair is $\frac{\binom{3}{1}}{\binom{6}{2}}$.
This is $\frac{1}{5}$.
If we find the sum $\frac{1}{5}+\frac{1}{5}+\frac{1}{5}$, we will have counted twice the probability P and Q both choose a matching pair, also the same for Q and R, also for P and R. Given that P chose a matching pair, the probability Q does is $\frac{\binom{2}{1}}{\binom{4}{2}}=\frac{1}{3}$.
So the probability P and Q choose a matching pair is $\frac{15}{5}\cdot\frac{1}{3}$. We subtract $3$ times this from $\frac{3}{5}$.
We have subtracted too much. We must add back the probability they all choose a matching pair, which is $\frac{1}{5}\cdot\frac{1}{3}$.
So the probability there is at least one person with a matching pair of shoes is
$$3\cdot\frac{1}{5}-3\cdot \frac{1}{5}\cdot \frac{1}{3}+1\cdot\frac{1}{5}\cdot\frac{1}{3}.$$
This simplifies to $\frac{7}{15}$.
So the probability $p$ of no match is $1-\frac{7}{15}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/401314",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Algebraic expression in its most simplified form I am trying to simplify the algebraic expression:
$$\bigg(x-\dfrac{4}{(x-3)}\bigg)\div \bigg(x+\dfrac{2+6x}{(x-3)}\bigg)$$
I am having trouble though. My current thoughts are:
$$=\bigg(\dfrac{x}{1}-\dfrac{4}{(x-3)}\bigg)\div \bigg(\dfrac{x}{1}+\dfrac{2+6x}{(x-3)}\bigg)$$
$$=\bigg(\dfrac{x(x-3)}{1(x-3)}-\dfrac{4}{(x-3)}\bigg)\div \bigg(\dfrac{x(x-3)}{1(x-3)}+\dfrac{2+6x}{(x-3)}\bigg)$$
$$=\bigg(\dfrac{x(x-3)+(-4)}{(x-3)}\bigg)\div \bigg(\dfrac{x(x-3)+2+6x}{(x-3)}\bigg)$$
$$=\dfrac{x(x-3)+(-4)}{(x-3)}\times \dfrac{(x-3)}{x(x-3)+2+6x}$$
$$=\dfrac{x(x-3)+(-4)(x-3)}{(x-3)x(x-3)+2+6x} $$
$$\boxed{=\dfrac{-4(x-3)}{2(1+3x)} }$$ Which does not appear is not the answer. Am I close? Where exactly did I go wrong? I have tried this question multiple times.
Edit: Figured it out!
$\dfrac{x(x-3) - 4}{x(x - 3) + 2(1 + 3x)}\implies\dfrac{x^2-3x-4}{x^2+3x+2}\implies \dfrac{(x-4)(x+1)}{(x+2)(x+1)}$
$(x+1)$'s cancel leaving us with: $\boxed{\dfrac{x-4}{x+2}}$
| Comment: After the calculation very well described by AWertheim and amWhy, we end up wit $\dfrac{x^2-3x-4}{x^2+3x+2}$.
Note that the calculation is not over. The top and bottom each factor nicely, and there will be some cancellation.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/401571",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
How to solve the irrational system of equations? Solve the system of equations $$\begin{cases} \sqrt{x+2y+3}+\sqrt{9 x+10y+11}=10,&\\[10pt] \sqrt{12 x+13y+14}+\sqrt{28 x+29y+30}=20. \end{cases} $$
| Hint: Try $t=x+y+1$ to get
\begin{cases} \sqrt{t+y+2}+\sqrt{9t +y+2}=10,&\\[10pt] \sqrt{12t+y+2}+\sqrt{28t+y+2}=20. \end{cases}
And solve by squaring.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/402116",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Determine the general solution for $2\cos 2x−5\cos x+2=0$ Determine the general solution for $2\cos 2x−5\cos x +2=0$
my answer I got was : $1.05+n\pi, 4.19+n\pi$
| $$\implies 2(2\cos^2x-1)-5\cos x+2=0\iff \cos x(4\cos x-5) =0$$
If $\cos x=0,x =\frac{(2m+1)\pi}2$ where $m$ is any integer
If $4\cos x-5=0, \cos x=\frac54>1$ but for real $x, -1\le \cos x\le 1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/408613",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Help finding eigenvectors? The given matrix is:
$$
\begin{pmatrix}3 & 1 & 6 \\ 2 & 1 & 0 \\ -1 & 0 & -3\end{pmatrix}\qquad
$$
I got the characteristic polynomial of $$x^3 - x^2 - 5x - 3 = 0$$
which factors down to $$(x+1)^2 * (x-3) = 0$$
I see that it has eigenvalues of -1 and 3.
I know I'm almost there, I plugged in the eigenvalues to $A-\lambda I$ but completely forgot how to find the eigenvectors after this.
When $\lambda$ = 3, I got:
$$\begin{pmatrix}0 & 1 & 6 \\ 2 & -2 & 0 \\ -1 & 0 & 0\end{pmatrix} \begin{pmatrix}x
_1 \\ x_2 \\ x_3\end{pmatrix}=0\qquad$$
and when $\lambda$ = -1, I got:
$$\begin{pmatrix}4 & 1 & 6 \\ 2 & 0 & 0 \\ -1 & 0 & 4\end{pmatrix}\begin{pmatrix}x
_1 \\ x_2 \\ x_3\end{pmatrix}=0\qquad$$
Where do I go from here? Row-reduce the $3x3$ matrices to solve?
| I am going to use the approach you are using so you can see your issues.
We are given: $A = \begin{bmatrix}3 & 1 & 6 \\ 2 & 1 & 0 \\ -1 & 0 & -3\end{bmatrix}$
We set up and and solve: $|A - \lambda I| = 0$, which yields:
$$\left|\begin{matrix}3-\lambda & 1 & 6 \\ 2 & 1-\lambda & 0 \\ -1 & 0 & -3-\lambda\end{matrix}\right| = 0$$
This yields a characteristic polynomial and eigenvalues as:
$$-\lambda^3+\lambda^2+5 \lambda+3 = -(\lambda-3) (\lambda+1)^2 = 0 ~~~\rightarrow ~~~ \lambda_1 = 3, \lambda_{2,3} = -1$$
We have multiplicities of $1$ and $2$ for those eigenvalues.
To find the eigenvectors, we generally solve $[ A - \lambda_i I]v_i = 0$, but since we have a repeated eigenvalue, we may need to change that strategy and find a generalized eigenvalue (I'll let you deal with the details of this and geometric multiplicities).
So, for $\lambda_1 = 3$, we have:
$[A- 3I]v_1 = \begin{bmatrix}0 & 1 & 6 \\ 2 & -2 & 0 \\ -1 & 0 & -6\end{bmatrix}v_1 = 0$
Doing row-reduced-echelon-form (RREF), yields:
$\begin{bmatrix}1 & 0 & 6 \\ 0 & 1 & 6 \\ 0 & 0 & 0\end{bmatrix}v_1 = 0$
Thus, $b = -6c, a = -6c \rightarrow ~~\text{let}~~ c = 1 \rightarrow a = b = -6, v_1 = (-6,-6,1)$.
Repeating this same process for the second eigenvalue, we have as RREF:
$\begin{bmatrix}1 & 0 & 2 \\ 0 & 1 & -2 \\ 0 & 0 & 0\end{bmatrix}v_1 = 0$
This yields an eigenvector of $v_2 = (-2, 2, 1)$.
Unfortunately, we cannot get another linearly independent eigenvector, so need to get a generalized one, by doing $[A - \lambda I]v_3 = v_2$, so we have:
$\begin{bmatrix}4 & 1 & 6 \\ 2 & 2 & 0 \\-1 & 0 & 2\end{bmatrix}v_3 = \begin{bmatrix} -2 \\ 2 \\ 1 \end{bmatrix}$
After RREF, we arrive at:
$\begin{bmatrix}1 & 0 & 2 \\ 0 & 1 & -2 \\ 0 & 0 & 0\end{bmatrix}v_3 = \begin{bmatrix} -1 \\ 2 \\ 0 \end{bmatrix}$
So, we have: $a = -1 -2c, b = 2 + 2c \rightarrow ~~ \text{let} ~~ c = 0 \rightarrow a = -1, b = 2$, thus $v_3 = (-1,2,0)$
Putting all of this together, we have the eigenvalue/eigenvector pairs:
*
*$\lambda_1 = 3, v_1 = (-6, -6, 1)$
*$\lambda_2 = -1, v_2 = (-2, 2, 1)$
*$\lambda_3 = -1, v_3 = (-1,2,0)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/411104",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Surface integration over section of paraboloid below a plane Let S be the finite portion of the surface $z = 4x^2 + y^2$ cut off by the plane
$z = 8x + 4y - 4$.
Evaluate the surface integral $(x, y, 3z)\,dS$ over the region $S$ where the normal to $S$ points upwards.
I can do this using the divergence theorem, but I don't know what the limits would be for the surface integral. If I parametrise the surface as $r=(\frac{1}{2}r\cos t, r\sin t, r^2)$, then $0\leq t < 2\pi$. But if I substitute the parametrised coordinates into the plane equation then I get $r^2 \leq 4r\cos t + 4r\sin t - 4$. Is this right?
| The plane equation is not right. The points satisfying parametrization you gave live on the paraboloid $z = 4x^2 +y^2$, they only intersect the plane at some points.
Well if you wanna use divergence theorem:
The integration region is as follows:
The bottom is the surface $z=4x^2+y^2$, while it has a cap formed by the plane $z=8x+4y−4$. Equaling the right sides of these two we find the intersecting curve is:
$$
\{(x,y,z): (x-1)^2 + (y-2)^2/2 = 1,\; z=8x+4y−4\}.
$$
Now the cap surface is:
$$
C = \{(x,y,z): (x-1)^2 + (y-2)^2/2 < 1,\; z=8x+4y−4\}.
$$
Its unit normal vector (choose the one pointing upward) is $n_C =(-8/9,-4/9,1/9)$.
The bottom surface is:
$$
S = \{(x,y,z): (x-1)^2 + (y-2)^2/2 < 1,\; z=4x^2+y^2\}.
$$
Let the region bounded by $C$ and $S$ be $V$, $F = (x,y,3z)$, then by divergence theorem:
$$
\int_V \nabla \cdot F\,dV = \int_{\partial V} F\cdot n \,dS = \int_{C} F\cdot n_C \,dS + \int_{S} F\cdot n_S \,dS,
$$
here the normal to $S$ is point downward, if choosing the pointing upward one $n_S = -n$, we have:
$$
\int_{S} F\cdot n_S \,dS = \int_{C} F\cdot n_C \,dS - \int_V \nabla \cdot F\,dV.\tag{1}
$$
Now
$$
F\cdot n_C\bigg|_C = -\frac{8}{9}x - \frac{4}{9}y + \frac{1}{3}z = \frac{2}{9}z - \frac{4}{9},
$$
and we can parametrize the elliptical cap $C$ by
$$\begin{cases} x - 1 = r\cos t, \\ y-2 = \sqrt{2}\,r \sin t,\end{cases}\tag{2}$$
where $0<r<1$, and $t\in (0,2\pi)$, $z = 8(1+r\cos t) + 4(2+ \sqrt{2}\,r \sin t) -4$, the Jacobian $J$ is left for you to compute, then the integral becomes:
$$
\int_{C} F\cdot n_C \,dS = \int^{2\pi}_0\int^1_0\left( \frac{2}{9}\big(8(1+r\cos t) + 4(2+ \sqrt{2}\,r \sin t) -4\big) - \frac{4}{9}\right) |J| \,dr\,dt.\tag{3}
$$
Divergence of $F$ is $5$, so
$$
\begin{aligned}
&\int_V \nabla \cdot F\,dV = 5 \iint_{(x-1)^2 + (y-2)^2/2 < 1}\int^{8x+4y-4}_{4x^2+y^2}\,dz\,dx\,dy
\\
=& 5 \iint_{(x-1)^2 + (y-2)^2/2 < 1} \left( \big(8x+4y-4\big) - \big(4x^2+y^2\big)\right)\,dx\,dy
\\
=& 5 \iint_{(x-1)^2 + (y-2)^2/2 < 1} \left( 2 - 2(x-1)^2 - (y-2)^2\right)\,dx\,dy
\\
=& 5\int^{2\pi}_0\int^1_0 \left( 2 - 2r^2 \cos^2 t - 2r^2 \sin^2 t\right) |J|\,dr\,dt.\tag{4}
\end{aligned}
$$
What left for you to do: Compute the Jacobian from (2), then evaluate (3) and (4) to get (1), the left side of (1) is what you wanna evaluate.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/411414",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
} |
Inequality Of Four Variables Suppose $a,b,c,d$ are real numbers greater than $1$. Prove that
$$abc+abd+acd+bcd-3abcd<1.$$
| The inequality holds because
\begin{align*}
&4\times\left[\,3abcd-(abc+abd+acd+bcd)+1\,\right]\\
\\
=&2(a-1)(b-1)(c-1)(d-1)\\
\\
&+(a+1)(b+1)(c-1)(d-1)+(a+1)(b-1)(c+1)(d-1)\\
&+(a+1)(b-1)(c-1)(d+1)+(a-1)(b+1)(c+1)(d-1)\\
&+(a-1)(b+1)(c-1)(d+1)+(a-1)(b-1)(c+1)(d+1)\\
\\
&+(a+1)(b-1)(c-1)(d-1)\\
&+(a-1)(b+1)(c-1)(d-1)\\
&+(a-1)(b-1)(c+1)(d-1)\\
&+(a-1)(b-1)(c-1)(d+1).
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/412175",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Solve the following in non-negative integers: $3^x-y^3=1$. Solve the following in non-negative integers: $$3^x-y^3=1$$
Of course $(x,y)=(0,0)$ is a trivial solution. After seeing that I proceeded like this:
$$3^x-y^3=1$$$$\implies3^x-1=y^3$$$$\implies2(3^{x-1}+3^{x-2}+ \cdots +3^1+1)=y^3$$$$\therefore2|y$$
So let $y=2k$ for non-negative $k$.
Then $$3^{x-1}+3^{x-2}+ \cdots +3^1+1=4y^3$$ So $4|LHS$ using a mod argument one can easily deduce that this implies that $x$ is even so let $x=2q$.
Then going back to our original expression we have $$9^q-1=8k^3$$$$\implies 9^{q-1}+9^{q-2}+ \cdots +9+1=k^3$$
Then again using mod arguments I deduced that $k \equiv 4 \pmod9$. And after that things started to get more and more yucky. This could be a completely bad approach so I also triad looking at $(y+1)(y^2-y+1)=3^x$ but that didn't lead me anywhere either.
Thanks in advance for any help.
| If $y+1=3^a$ with $a\ge1\iff y>1$
$y^2-y+1=3(3^{2a-1}-3^a+1)$ which can not be power of $3$ unless $2a-1=a\iff a=1$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Algebra simplification in mathematical induction . I was proving some mathematical induction problems and came through an algebra expression that shows as follows:
$$\frac{k(k+1)(2k+1)}{6} + (k + 1)^2$$
The final answer is supposed to be:
$$\frac{(k+1)(k+2)(2k+3)}{6}$$
I walked through every possible expansion; I combine like terms, simplify, factor, but never arrived at the answer.
Could someone explain the steps?
| A good idea for this sort of thing is to use Wolfram Alpha to ensure that the two things are, indeed, equal. In this case, they are, so we can spend some time looking to factor.
$$\begin{align}
\frac{k(k+1)(2k+1)}{6} + (k + 1)^2 &= \frac{k(k+1)(2k+1) + 6(k + 1)^2}{6}\\
&= \frac{(k+1)(k(2k+1) + 6(k + 1))}{6}\\
&= \frac{(k+1)(2k^2 +7k + 6)}{6}\\
&= \frac{(k+1)(k+2)(2k + 3)}{6}\\
\end{align}$$
EDIT: A helpful trick is the one I did in the step from line 1 to line 2: I factored out the $k+1$ immediately, rather than expanding the whole expression. This makes it easier to deal with--most people have more practice factoring quadratics than cubics.
EDIT in response to comment:
$$\begin{align}
\frac{k(k+1)(2k+1) + 6(k + 1)^2}{6} &= \frac{k\color{red}{(k+1)}(2k+1) + 6\color{red}{(k+1)}(k+1)}{6}\\
&=\frac{\color{red}{(k+1)}\Big(k(2k+1) + 6(k+1)\Big)}{6}
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/414184",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
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Evaluate $\int_0^\infty \frac{\log(1+x^3)}{(1+x^2)^2}dx$ and $\int_0^\infty \frac{\log(1+x^4)}{(1+x^2)^2}dx$ Background: Evaluation of $\int_0^\infty \frac{\log(1+x^2)}{(1+x^2)^2}dx$
We can prove using the Beta-Function identity that
$$\int_0^\infty \frac{1}{(1+x^2)^\lambda}dx=\sqrt{\pi}\frac{\Gamma \left(\lambda-\frac{1}{2} \right)}{\Gamma(\lambda)} \quad \lambda>\frac{1}{2}$$
Differentiating the above equation with respect to $\lambda$, we obtain an expression involving the Digamma Function $\psi_0(z)$.
$$\int_0^\infty \frac{\log(1+x^2)}{(1+x^2)^\lambda}dx = \sqrt{\pi}\frac{\Gamma \left(\lambda-\frac{1}{2} \right)}{\Gamma(\lambda)} \left(\psi_0(\lambda)-\psi_0 \left( \lambda-\frac{1}{2}\right) \right)$$
Putting $\lambda=2$, we get
$$\int_0^\infty \frac{\log(1+x^2)}{(1+x^2)^2}dx = -\frac{\pi}{4}+\frac{\pi}{2}\log(2)$$
Question:
But, does anybody know how to evaluate $\displaystyle \int_0^\infty \frac{\log(1+x^3)}{(1+x^2)^2}dx$ and $\displaystyle \int_0^\infty \frac{\log(1+x^4)}{(1+x^2)^2}dx$?
Mathematica gives the values
*
*$\displaystyle \int_0^\infty \frac{\log(1+x^3)}{(1+x^2)^2}dx = -\frac{G}{6}+\pi \left(-\frac{3}{8}+\frac{1}{8}\log(2)+\frac{1}{3}\log \left(2+\sqrt{3} \right) \right)$
*$\displaystyle \int_0^\infty \frac{\log(1+x^4)}{(1+x^2)^2}dx = -\frac{\pi}{2}+\frac{\pi \log \left( 6+4\sqrt{2}\right)}{4}$
Here, $G$ denotes the Catalan's Constant.
Initially, my approach was to find closed forms for
$$\int_0^\infty \frac{1}{(1+x^2)^2(1+x^3)^\lambda}dx \ \ , \int_0^\infty \frac{1}{(1+x^2)^2(1+x^4)^\lambda}dx$$
and then differentiate them with respect to $\lambda$ but it didn't prove to be of any help.
Please help me prove these two results.
| The 2nd evaluation can be obtained from the residue theorem:
*
*Using parity, write the integral as $\displaystyle \frac12\int_{-\infty}^{\infty}\frac{\ln(1+x^4)\,dx}{(1+x^2)^2}$.
*Interpret this as a complex integral and pull the integration contour to, say, $i\infty$. The result will be given by the residue at 2nd order pole at $x=i$ and two integrals of the logarithm jump (equal to $2\pi i$) over the branch cuts emanating from $x=e^{i\pi/4}$ and $x=e^{3i\pi /4}$ in the radial directions.
*For the first contribution, we have
$$2\pi i \cdot\mathrm{res}_{x=i}\frac{\ln(1+x^4)}{(1+x^2)^2}=\frac{\pi}{2}\left(\ln2 -2\right).$$
*The integral over the branch cut $(e^{i\pi/4},e^{i\pi/4}\infty)$ is
$$2\pi i \int_{e^{i\pi/4}}^{e^{i\pi/4}\infty}\frac{dx}{(1+x^2)^2}=\frac{\pi i}{2}\left(\pi-\sqrt{2}-2\arctan e^{i\pi/4}\right),$$
and, similarly, for the second branch cut $(e^{3i\pi/4},e^{3i\pi/4}\infty)$ we find
$$2\pi i \int_{e^{3i\pi/4}}^{e^{3i\pi/4}\infty}\frac{dx}{(1+x^2)^2}=\frac{\pi i}{2}\left(\sqrt{2}-\pi-2\arctan e^{3i\pi/4}\right).$$
*Combining everything, one obtains the answer:
\begin{align}
\frac12\left\{\frac{\pi}{2}\left(\ln2 -2\right)+\pi\, \mathrm{arccoth}\sqrt{2}\right\}=
-\frac{\pi}{2}+\frac{\pi}{4}\ln(6+4\sqrt{2}).
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "31",
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The number of distinct real roots of the folllowing determinant
The number of distinct real roots of determinant $$
\begin{vmatrix}
\csc x & \sec x & \sec x \\
\sec x & \csc x & \sec x \\
\sec x & \sec x & \csc x \\
\end{vmatrix}
=0$$ lies in the interval $\frac{-\pi}{4} \le x \le \frac{\pi}{4}$ is (a) $1$ (b) $2$ (c) $3$ (d) $0$
I tried to solve the determinant and solved it till I got:
$$(\csc x + 2\sec x)(\csc x - \sec x)^2=0$$
How do I get the final answer?Please offer your assistance, Thank you
| We need to solve $\cos(x) + 2\sin(x) = 0$ and $\cos(x) - \sin(x) = 0$. The first equation is solved by setting $u = \cos(x)$ and solving for $u$, i.e.
\begin{align*}
u+2\sqrt{1-u^2} &= 0\\
u &= -2\sqrt{1-u^2} \\
u^2 &= 4(1-u^2) \\
5u^2 &= 4
\end{align*}
and checking which values of $\cos^{-1}(\pm 2/\sqrt{5})$ lie in the specified interval. For the second equation, only $x = \pi/4$ works.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to simplify $\frac{1-\frac{1-x}{1-2x}}{1-2\frac{1-x}{1-2x}}$? $$\frac{1-\frac{1-x}{1-2x}}{1-2\frac{1-x}{1-2x}}$$
I have been staring at it for ages and know that it simplifies to $x$, but have been unable to make any significant progress.
I have tried doing $(\frac{1-x}{1-2x})(\frac{1+2x}{1+2x})$ but that doesn't help as it leaves $\frac{1+x-2x^2}{1-4x^2}$ any ideas?
| $$\frac{1-\frac{1-x}{1-2x}}{1-2\frac{1-x}{1-2x}}$$
(take L.C.M. in numerator and denominator)
$$\dfrac{\dfrac{{(1-2x)}-{(1-x)}}{1-2x}}{\dfrac{(1-2x)-2{(1-x)}}{1-2x}}$$
$$\dfrac{\dfrac{1-2x-1+x}{1-2x}}{\dfrac{1-2x-2+2x}{1-2x}}$$
$$\dfrac{\dfrac{-x}{1-2x}}{\dfrac{-1}{1-2x}}$$
since $$\dfrac{\dfrac ab}{\dfrac cd}=\dfrac ab\times\dfrac {d}{c}$$
so
$${\dfrac{-x}{1-2x}}\times{\dfrac{1-2x}{-1}}$$
$$x$$
| {
"language": "en",
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"source": "stackexchange",
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Prove that $\{x_1=1,\,x_{n+1}=\frac {x_n}2+\frac 1{x_n} \}$ converges when $n \to \infty$ I want to prove that the sequence defined by $\{x_1=1,\,x_{n+1}=\frac {x_n}2+\frac 1{x_n} \}$ has a limit.
By evaluating the sequence I notice that the sequence is strictly monotonically decreasing starting from $x_2=1.5$.
It seems to suggest itself to prove that the sequenced is bounded by $1\le\big( x_n \big)_{n\ge1} \le 1.5$ and to prove that it is strictly monotonically decreasing starting at $x_2$ which would imply convergence.
How would I proceed and could one prove the existence of the limit without first evaluating the values of the sequence to see how the sequence behaves?
| A. $x_n>0$ , $n\in \mathbb{N}$.
B.
We will show that $\{x_2, x_3, \ldots, x_n, \ldots \}$ is monotonically non-increasing.
$$
x_n-x_{n+1} = x_n - \frac{x_n}{2} - \frac{1}{x_n} = \frac{x_n}{2} - \frac{1}{x_n} =
\frac{x^2_n-2}{2x_n}, \qquad n>1.
$$
But $x^2_n-2>0$, when $n>1$, because
$$
x^2_n-2 = \left( \dfrac{x_{n-1}}{2}+\dfrac{1}{x_{n-1}} \right)^2 - 2 =
\dfrac{x^2_{n-1}}{4} + 1 + \dfrac{1}{x^2_{n-1}} - 2 =
$$
$$
\dfrac{x^2_{n-1}}{4} - 1 + \dfrac{1}{x^2_{n-1}} =
\left( \dfrac{x_{n-1}}{2}-\dfrac{1}{x_{n-1}} \right)^2 \geqslant 0.
$$
So,
$$x_n-x_{n+1} \geqslant 0.$$
Proved.
C. Since the sequense has the limit (denote $X$) $-$ as bounded and non-increasing sequence, then
$$
X = \frac{X}{2}+\frac{1}{X};
$$
$$
X^2 = \frac{X^2}{2}+1;
$$
$$
\frac{X^2}{2}=1;
$$
$$
X = \sqrt{2}.
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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derive the Maclaurin series by using partial fractions. derive the Maclaurin series for the function
$(x^3+x^2+2x-2)/(x^2-1)$ by using partial fractions and a known Maclaurin series.
question.
how can I use partial fractions in this case?
Is this case the special one?
plz provide me a hint. thanks.
| Using polynomial long division:
$$\frac{(x^3+x^2+2x-2)}{(x^2-1)}=x + 1 + \frac{3x-1}{x^2-1}$$
For partial fraction decomposition, think of it as the inverse of finding a common denominator for two fractions: first factor the denominator:
$$x + 1 + \frac{3x-1}{x^2-1} = x + 1 + \frac{3x-1}{(x - 1)(x+ 1)}$$
Then we split the fraction into the sum of two partial fractions, setting up the decomposition to solve for the constants $A$ and $B$:
$$x + 1 + \frac{3x-1}{(x - 1)(x+ 1)} = \frac A{x-1} + \frac B{x + 1}$$
Now, we know that $$A(x+1) + B(x - 1) = (A + B)x + (A - B) = 3x - 1 $$ $$\iff A+ B = 3 \;\; \text{and}\;\; A - B = -1$$
So solving for the system of equations in $A, B$: $$A = B - 1 \implies 2B - 1 = 3 \iff B = 2 \implies A = 1$$
Using these values, we have that $$\frac{(x^3+x^2+2x-2)}{(x^2-1)}=x + 1 + \frac 1{x-1} + \frac 2{x + 1}$$
With practice, partial fraction decomposition becomes easier (as with any skill), and there are "short-cuts" you'll pick up along the way that can simplify the process.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Upper bound for expression involving logarithms Let $N = 2^p$ for some $p \in \mathbb{N}$. Find the smallest upper bound for $\frac{N}{2}\log\left(\frac{N}{2}\right) + \frac{N}{4}\log\left(\frac{N}{4}\right) + \ldots + 1$
I guess I could first rewrite this to $\frac{2^p}{2}\log\left(\frac{2^p}{2}\right) + \frac{2^p}{4}\log\left(\frac{2^p}{4}\right) +\ldots+ 1$ and then to
$2^{p-1}\log(2^{p-1}) + 2^{p-2}\log(2^{p-2}) +\ldots+1$ but I still don't know how I should proceed.
All help appreciated.
Edit: Now I though also writing it to the form $(p-1)2^{p-1} + (p-2)2^{p-2} + ... + 1 \Leftrightarrow \sum_{i=1}^{log N} (p-i)2^{p-i}$ but I still feel like ...... ;-( (yes, the log is base 2, sorry, forgot to mention that)
| Fix $k = \log 2 = 1$. Then expression yields
$$\begin {eqnarray}
S & = & \frac {N} {2} (\log N - k) + \frac {N} {4} (\log N - 2k) + \cdots + \frac {N} {2^{p - 1}} (\log N - (p - 1)k) + 1 \nonumber \\
& = & N \log N (\frac {1} {2} + \frac {1} {4} + \cdots + \frac {1} {2^{p - 1}}) - N (\frac {1} {2} + \frac {2} {4} + \cdots + \frac {p - 1} {2^{p - 1}}) + 1 \nonumber \\
& > & N \log N - 2 \log N - 2N + 1 \nonumber \\
& = & p 2^p - 2^p - 2p + 1.
\end {eqnarray}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the values of the constants in the following identity $2x^3+3x^2-14x-5=(ax+b)(x+3)(x+1)+C$ I'm working through identities but I can't figure out how to get further than multiplying out the above to get :
$$2x^3+3x^2-14x-5=2ax^3+3ax^2+3ax+bx^2+3bx+bx+3b+C$$
can someone give me a hint on what to do next?
| By expanding and grouping like terms together, we obtain:
$$ \begin{align*}
2x^3+3x^2-14x-5&=(ax+b)(x+3)(x+1) +C\\
2x^3+3x^2-14x-5&=(ax+b)(x^2+4x+3) +C\\
2x^3+3x^2-14x-5&=(ax)(x^2+4x+3)+b(x^2+4x+3) +C\\
2x^3+3x^2-14x-5&=(ax^3+4ax^2+3ax)+(bx^2+4bx+3b) +C\\
2x^3+3x^2-14x-5&=(a)x^3+(4a+b)x^2+(3a+4b)x+(3b+C)
\end{align*} $$
Now let's compare coefficients. In particular, focus on the coefficients of $x^3$, $x^2$, and $x^0$ (the constant term). From this, we obtain:
$$ \begin{align*}
x^3 &: \boxed{2=a} \\
x^2 &: 3 = 4a+b \implies b = 3-4a=3-4(2) \implies \boxed{b=-5} \\
x^0 &: -5=3b+C \implies C=-5-3b=-5-3(-5) \implies \boxed{C=10}
\end{align*} $$
| {
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Find $x,y,z \in \mathbb Q$ such that $x + \frac 1y, y + \frac 1z, z+ \frac 1x \in \mathbb Z$
Find $x,y,z \in \mathbb Q$ such that: $$x + \frac 1y, y + \frac 1z, z+
\frac 1x \in \mathbb Z$$
Here is my thinking:
$$x + \frac 1y, y + \frac 1z, z+ \frac 1x \in \mathbb Z\\ \implies \left ( x + \frac 1y\right ) \left ( y + \frac 1z\right )\left ( z + \frac 1x\right )\in \mathbb Z \\ \iff \frac 1 {xyz} + xyz + \left( x+\frac 1y \right) + \left( y+\frac 1z \right) + \left( z+\frac 1x \right) \in \mathbb Z \\ \iff xyz + \frac 1 {xyz} \in \mathbb Z \\ \iff |xyz| = 1 \text{ (I proved it, easily)}$$
Case 1:
$xyz=1 \implies \exists a,b,c \in \mathbb Z$ such that: $x = \frac ab,y=\frac bc,z=\frac ca$
$$\implies \frac {a+c} b = x+\frac 1y \in \mathbb Z$$
and so on.
Now, I am stuck. Do you thing I am going a right way ?
| Let's start with your observation, that $|xyz| = 1$, and therefore $xyz = \pm 1$. If we take any solution $(x, y, z)$ with $xyz = -1$ and change it to $(-x, -y, -z)$, it is still a solution and we have $(-x)(-y)(-z) = 1$, so it's enough to consider the solutions with $xyz = 1$ and obtain the rest by flipping all the signs. So we have $x = \frac{a}b$, $y = \frac{b}c$, and $z = \frac{c}a$ for some integers $a, b, c$. Without loss of generality, let |a| ≤ |b| ≤ |c|. We actually can't make the assumption $|a| \le |b| \le |c|$ without loss of generality! But we can let $c$ be the largest of the three in magnitude, which means that $|c| \ge |b|$ and $|c| \ge |a|$. The other solutions will be obtained by cyclically rotating the three.
Then, consider the fact that (as $y + \frac1z$ is an integer) $c$ divides $(a + b)$ and therefore $|c|$ divides $|a+b|$. As $|a| \le |c|$ and $|b| \le |c|$, we have $|a + b| \le 2|c|$, with equality holding only when $|a| = |c|$ and $|b| = |c|$ (in which case we have the solutions $|x| = |y| = |z| = 1$).
Otherwise, $|a+b|$ is strictly less than $2|c|$, and so either $|a + b| = 0$, or $|a + b| = |c|$.
If $|a + b| = 0$ so that $b = -a$, then the fact that $a$ divides $b+c = c-a$ means that $c$ is a multiple of $a$, say $c = ka$. This is always a solution, and gives $x = \frac{a}{b} = -1$, $y = \frac{b}{c} = \frac{-1}{k}$, and $z = \frac{c}{a} = k$.
The other case is that $|a + b| = |c|$, so $a + b = \pm c$. We can consider these separately as two cases:
*
*$a + b = -c$. This happens to be fine for any pair of nonzero integers $(a, b)$, and gives the family of solutions $x = \frac{a}{b}$, $y = \frac{b}{c} = \frac{-b}{a+b}$, and $z = \frac{c}{a} = \frac{-a-b}{a}$.
*$a + b = c$. In this case, the fact that $b$ divides $(c + a) = 2a + b$ means that $b$ divides $2a$, and similarly the fact that $a$ divides $(b + c) = a + 2b$ means that $a$ divides $2b$. So $|b|$ is one of $|a|/2$, $|a|$, or $2|a|$. Let's consider all three separately, in descending order:
*
*
*If $|b| = 2|a|$, then $b = 2a$ (as $b = -2a$ gives $c = a + b = -a$, which contradicts our assumption that $|c| \ge |b|$). So $c = a + b = 3a$, and this gives the solution $x = \frac{a}{b} = \frac12$, $y = \frac{b}{c} = \frac23$, $z = \frac{c}{a} = 3$.
*
*
*If $|b| = |a|$, then we can ignore the case $b = -a$ as we already considered it earlier, and look at $b = a$. This gives $c = 2a$, and indeed a solution $x = \frac{a}b = 1$, $y = \frac{b}{c} = \frac12$, and $z = \frac{c}{a} = 2$.
*
*
*If $|b| = |a|/2$, then $a = 2b$ (as $a = -2b$ gives $c = a + b = -b$, contradicting our assumption that $|c| \ge |a|$). So $c = a + b = 3b$, and this gives the solution $x = \frac{a}{b} = 2$, $y = \frac{b}{c} = \frac13$, and $z = \frac{c}{a} = \frac32$.
To summarize, all the solutions $(x, y, z)$ are (note that all the below are distinct):
*
*$(1, 1, 1)$,
*$(2, 1, \frac12)$,
*$(\frac12, \frac23, 3)$,
*$(-k, -1, \frac1k)$ for any nonzero integer $k$,
*$(\frac{a}{b}, \frac{-b}{a+b}, \frac{a+b}{-a})$, for any pair of nonzero integers $a$ and $b$,
and the solutions obtained by changing any $(x, y, z)$ above to $(y, z, x)$, $(z, x, y)$, $(-x, -y, -z)$, $(-y, -z, -x)$, or $(-z, -x, -y)$.
| {
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"timestamp": "2023-03-29T00:00:00",
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How can $\left({1\over1}-{1\over2}\right)+\left({1\over3}-{1\over4}\right)+\cdots+\left({1\over2n-1}-{1\over2n}\right)+\cdots$ equal $0$?
How can $\left({1\over1}-{1\over2}\right)+\left({1\over3}-{1\over4}\right)+\cdots+\left({1\over2n-1}-{1\over2n}\right)+\cdots$ equal $0$?
Let
$$\begin{align*}x &= \frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n} + \cdots\\
y &= \frac{1}{1} + \frac{1}{3} + \frac{1}{5} + \cdots + \frac{1}{2n-1} + \cdots\\
z &= \frac{1}{2} + \frac{1}{4} + \frac{1}{6} + \cdots + \frac{1}{2n} + \cdots \end{align*}$$
so we have
$$x = y + z.$$
However, $x = 2\cdot z$, so $y$ = $z$ or
$$\frac{1}{1} + \frac{1}{3} + \frac{1}{5} + \cdots + \frac{1}{2n-1} + \cdots = \frac{1}{2} + \frac{1}{4} + \frac{1}{6} + \cdots + \frac{1}{2n} + \cdots$$
This looks ok if I interpret it as
$$\frac{1}{1} = \left (\frac{1}{2} - \frac{1}{3} \right ) + \left (\frac{1}{4} - \frac{1}{5} \right ) + \left (\frac{1}{6} - \frac{1}{7} \right ) + \cdots + \left (\frac{1}{2n} - \frac{1}{2n+1} \right ) + \cdots$$
However, it's a bit weird if I write it as
$$\left (\frac{1}{1} - \frac{1}{2} \right ) + \left (\frac{1}{3} - \frac{1}{4} \right ) + \left (\frac{1}{5} - \frac{1}{6} \right ) + \cdots + \left (\frac{1}{2n-1} - \frac{1}{2n} \right ) + \cdots = 0.$$
How can a sum of positive numbers equal $0$?
| Answering only "How can a sum of only positive numbers equal 0?" : look at $1^2 + 2^2 + 3^2 + 4^2 + 5^2+ ... = 0$ which is a meaningful result if the series is understood as zeta-series.
| {
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Why is $\frac{1}{\frac{1}{X}}=X$? Can someone help me understand in basic terms why $$\frac{1}{\frac{1}{X}} = X$$
And my book says that "to simplify the reciprocal of a fraction, invert the fraction"...I don't get this because isn't reciprocal by definition the invert of the fraction?
| $$y=\frac1{\frac 1 x} $$
$$y'(x)=\left(\frac1{\frac 1 x}\right)' = -\left(\frac 1 {\left(\frac 1x\right)^2}\right)\left({\frac 1 x}\right)' =
\frac 1 {\left(\frac 1x\right)^2} \cdot {\frac 1 {x^2}} = \frac {y^2(x)}{x^2}$$
So we have that $$x^2dy = y^2dx\\
\int \frac{dy}{y^2} = \int \frac{dx}{x^2}\\
-\frac{1}{y} = -\frac 1 x + C\\$$
Let's take a look at $y(1)$. $\frac 1 1 = 1$, this is already explained in a more common problem here:
Why is $n$ divided by $n$ equal to $1$?
So $y(1)=\frac{1}{\frac{1}{1}} =\frac 1 1 = 1$.
Note that I lost one possible solution, $y(x)=0$, by dividing by $y$. But since $y(1)=1$, it isn't really the solution.
Again: $y(1)=1$, so $~-\frac 1 1 = -\frac 1 1 + C ~~\Rightarrow~~ C=0$. Then $\frac {1} {y} = \frac {1}{x} \Rightarrow x=y$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Show that $\frac 1 2 <\frac{ab+bc+ca}{a^2+b^2+c^2} \le 1$
If $a,b,c$ are sides of a triangle, then show that $$\dfrac 1 2 <\dfrac{ab+bc+ca}{a^2+b^2+c^2} \le 1$$
Trial: $$(a-b)^2+(b-c)^2+(c-a)^2 \ge 0\\\implies a^2+b^2+c^2 \ge ab+bc+ca $$ But how I prove $\dfrac 1 2 <\dfrac{ab+bc+ca}{a^2+b^2+c^2}$ . Please help.
| Adding the triangle inequalities:$$a^{2}>(b-c)^{2}\\b^{2}>(a-c)^{2}\\c^{2}>(b-a)^{2}$$
We get $$a^{2}+b^{2}+c^{2}>2a^{2}+2b^{2}+2c^{2}-2ab-2bc-2ca$$
Now divide by $a^{2}+b^{2}+c^{2}$ to get,$$1>2-2\left(\frac{ab+bc+ca}{a^2+b^2+c^2}\right)\\\rightarrow \frac{ab+bc+ca}{a^2+b^2+c^2}>\frac{1}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/424367",
"timestamp": "2023-03-29T00:00:00",
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Explanation of this step in a modular arithmetic problem
The multiplicative inverse of $5$ is $7$, when using mod $34$.
$$\begin{align*}
5\cdot x&=3\\[0.1in]
7\cdot 5\cdot x &=7\cdot 3\\[0.1in]
1\cdot x &=7\cdot 3\\[0.1in]
x&=21
\end{align*}$$
I don't understand this part:
$$\begin{align*}
7\cdot 5\cdot x &=7\cdot 3\\[0.1in]
1\cdot x &=7\cdot 3
\end{align*}$$
How is 7*5*x the same as 1*x?
| Since $5x = 3 \pmod{34}$, you can multiply both sides of this equation by $7$ (the inverse of $5$ to obtain $7 \cdot 5x = 7 \cdot 3 \pmod{34}$. (The reason why this works is because $a=b \pmod{n}$ implies $ca = cb \pmod{n}$, and you should check this yourself if you haven't.)
Therefore since
$$
7 \cdot 5 = 35 = 34 + 1 = 0 + 1 = 1 \pmod{34},
$$
we obtain
$$
x = 1 \cdot x = 7 \cdot 5 \cdot x = 7 \cdot 3 = 21 \pmod{34}.
$$
Hope that helps,
| {
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Recursion problem help The following are the teachers example problems. The issue is that I don't understand the exact steps they took to go from $f(0)$ to $f(1)$ to $f(2)$ to $f(3)$. What I'm asking here is if someone could be so kind to show me how the answers for $f(0)\ldots f(3)$ are derived for each problem.
Let $f$ be defined as follows: $f(0) = 3$
$f(n) = (−1)^nf(n−1) + 4n$ for $n\ge1$.
Find $f(7)$
The correct answer is $3$
$f(0) = 3$
$f(1) = 1$
$f(2) = 9$
$f(3) = 3$
$f(4) = 19$
$f(5) = 1$
$f(6) = 25$
$f(7) = 3$
Let $f$ be defined as follows:
$f(0) = 1$
$f(n+1) = f(n) + n$ for $n≥0$
Find $f(6)$
The correct answer is $22$
$f(0) = 1$
$f(1) = 2$
$f(2) = 4$
$f(3) = 7$
$f(4) = 11$
$f(5) = 16$
$f(6) = 22$
Let $f$ be defined as follows:
$f(0) = 4$
$f(1) = 3$
$f(n) = f(n−1) + 3f(n−2)$ for $n≥2$.
Find $f(6)$
The correct answer is $348$
$f(0) = 4$
$f(1) = 3$
$f(2) = 15$
$f(3) = 24$
$f(4) = 69$
$f(5) = 141$
$f(6) = 348$
| You are given $f(0)=3$ and $$\tag1f(n)=(-1)^nf(n)+4n\quad\text {for }n\ge 1.$$
To compute $f(1)$, you let $n=1$ in equation $(1)$, i.e. $f(1)=(-1)^1f(1-1)+4\cdot 1$. Since you knwo $f(09=3$, this gives you $f(1)=(-1)^1\cdot 3+4=1$. To compute $f(2)$, you let $n=2$ in $(1)$, which gives you $f(2)=(-1)^2f(1)+4\cdot 2 = 1\cdot 1+8=9$. You should be able to veify the rest now.
Especially, do you see why we need two "starting values" in the last example? The first you get, letting $n=2$ in $f(n)=f(n-1)+3f(n-2)$, is $f(2)=f(1)+3f(0)=3+3\cdot 4=15$ so it's good that both $f(0)$ and $f(1)$ wer explicitly given, after that $f(3)=f(2)+3f(1)=15+3\cdot 3=24$.
| {
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Find area of the surface obtained by rotating curve around x-axis? I got a curve $y=a \cdot cosh \frac{x}{a}$ where $|x|\le b$. The task is to find area of the surface obtained by rotating curve around x-axis.
Here is my solution. Unfortunately the result is not identical with the result of the textbook. Would you please look at this and tell me where is my mistake? Maybe there is a mistake in my textbook. Thanks
My attempt at a solution:
Formula for area of the surface obtained by rotating curve:
$$S=2\pi \int_{a}^{b}f(x)\sqrt{1+(f'(x))^2}\,dx$$
I calculated $(f'(x))$:
$$f'(x)=\frac{d}{dx}(a\cdot \cosh\frac{x}{a})=\ldots =\sinh\frac{x}{a}$$
so: $(f'(x))^2=\sinh^2 \frac{x}{a}$
Then I put it into the formula for area of the surface obtained by rotating curve:
$$S=2\pi \int_{a}^{b}a\cdot \cosh\frac{x}{a}\sqrt{1+\sinh^2 \frac{x}{a}}\,dx=2\pi a \int_{a}^{b} \cosh\frac{x}{a}\sqrt{1+\sinh^2 \frac{x}{a}}dx$$
I made 1st substitution $u=\frac{x}{a} $ and $du=\frac{1}{a}dx $ then I obtained:
$$2\pi a^2 \int_{a}^{b} \cosh(u)\sqrt{1+\sinh^2 (u)}\,du $$
I made 2nd substitution $v=\sinh(u)$ and $dv=\cosh(u)du$ then I got:
$$2\pi a^2\int_{\sinh a}^{\sinh b}\sqrt{1+v^2}dv$$
To solve previous integral I need to calculate this integral $\int_{}^{}\frac{1}{\cos^3\varphi }=\int_{}^{}\sec^3\varphi $:
$$\int_{}^{}\sec^3\varphi =\frac{1}{2} \left( \sec\varphi \cdot \tan\varphi +\ln\right|sec\varphi +\tan\varphi \left|\right)$$
Then I returned back to $2\pi a^2\int_{\sinh a}^{\sinh b}\sqrt{1+v^2}\,dv$ and I made 3rd substitution: $v=\tan\varphi $ so $dv=\sec^2\varphi d\varphi$ then I got:
$$2\pi a^2\int_{\sinh a}^{\sinh b}\sqrt{1+v^2}\,dv=2\pi a^2\int_{...}^{...}\sqrt{\sec^2\varphi }\sec^2\varphi d\varphi=2\pi a^2\int_{...}^{...}\sec^3 \varphi d\varphi$$
Now we can see that we just calculated the value of $\int_{...}^{...}\sec^3 \varphi d\varphi$ so:
$$S=2\pi \int_{a}^{b}a\cdot \cosh\frac{x}{a}\sqrt{1+\sinh^2 \frac{x}{a}}\,dx=2\pi a^2\int_{\sinh a}^{\sinh b}\sqrt{1+v^2}\,dv=2\pi a^2\int_{...}^{...}\sec^3 \varphi d\varphi=2\pi a^2\frac{1}{2}\left[ \sec\varphi \cdot \tan\varphi +\ln\left|sec\varphi +\tan\varphi \right|\right]_{...}^{...}$$
Then I made a substitution back from $\varphi $ to $x$ , I obtained:
$$S=2\pi a^2\frac{1}{2}\left[ \sec\varphi \cdot \tan\varphi +\ln\left|\sec\varphi +\tan\varphi \right|\right]_{...}^{...}=2\pi a^2\frac{1}{2}\left[ v\sqrt{1+v^2}+\ln\left|\sqrt{1+v^2}+v\right|\right]_{...}^{...}=$$
$$=\pi a^2 \left[ \sinh\frac{x}{a}\sqrt{1+\sinh^2\frac{x}{a}}+\ln \left| \sqrt{1+\sinh^2\frac{x}{a}}+\sinh\frac{x}{a} \right|\right]_{a}^{b}=$$
$$=\pi a^2 \left[ \sinh\frac{x}{a}\cosh\frac{x}{a}+\ln \left| \cosh\frac{x}{a}+\sinh\frac{x}{a} \right|\right]_{a}^{b}=$$
$$=\pi a^2 \left[ \frac{1}{2}\sinh\frac{2x}{a}+\ln \left| e^{\frac{x}{a}}\right|\right]_{a}^{b}=\pi a^2 \left( \frac{1}{2}\sinh\frac{2b}{a}+\ln \left| e^{\frac{b}{a}}\right| - \frac{1}{2}\sinh 2 -\ln \left| e \right| \right)$$
| Hint: Use whe well-known hyperbolic identity
$$\cosh^2 x-\sinh^2x=1\iff 1+\sinh^2x=\cosh^2x$$
Use the above and you'll save about 80% of the work you did...:)
| {
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"timestamp": "2023-03-29T00:00:00",
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Induction and convergence of an inequality: $\frac{1\cdot3\cdot5\cdots(2n-1)}{2\cdot4\cdot6\cdots(2n)}\leq \frac{1}{\sqrt{2n+1}}$ Problem statement:
Prove that $\frac{1*3*5*...*(2n-1)}{2*4*6*...(2n)}\leq \frac{1}{\sqrt{2n+1}}$ and that there exists a limit when $n \to \infty $.
, $n\in \mathbb{N}$
My progress
LHS is equivalent to $\frac{(2n-1)!}{(2n)!}=\frac{(2n-1)(2n-2)(2n-3)\cdot ....}{(2n)(2n-1)(2n-2)\cdot ....}=\frac{1}{2n}$ So we can rewrite our inequality as:
$\frac{1}{2n}\leq \frac{1}{\sqrt{2n+1}}$ Let's use induction:
For $n=1$ it is obviously true. Assume $n=k$ is correct and show that $n=k+1$ holds.
$\frac{1}{2k+2}\leq \frac{1}{\sqrt{2k+3}}\Leftrightarrow 2k+2\geq\sqrt{2k+3}\Leftrightarrow 4(k+\frac{3}{4})^2-\frac{5}{4}$ after squaring and completing the square. And this does not hold for all $n$
About convergence: Is it not enough to check that $\lim_{n \to \infty}\frac{1}{2n}=\infty$ and conclude that it does not converge?
| It boils down to showing
$$\frac{2k+1}{2k+2} \leqslant \frac{\sqrt{3k+1}}{\sqrt{3k+4}}$$
for $k \geqslant 2$. Squaring that, we need to show
$$\left(1-\frac{1}{2k+2}\right)^2 = 1 - \frac{1}{k+1} + \frac{1}{4(k+1)^2} \leqslant 1 - \frac{3}{3k+4}.$$
From then on, it should not be difficult.
| {
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"url": "https://math.stackexchange.com/questions/431234",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Upper Bound of Logarithm For $1\leq x < \infty$, we know $\ln x$ can be bounded as following:
$\ln x \leq \frac{x-1}{\sqrt{x}}$.
Then what is the upper bound of $\ln x$ for following condition?
$2\leq x <\infty$
| To get your first upper bound we may start with the expansion :
$$\ln(1+x)=x - \frac{x^2}2 + \frac{x^3}3 - \frac{x^4}4 + O(x^5)$$
with the upper bound :
$$\frac x{\sqrt{1+x}}=x - \frac{x^2}2 + \frac{3x^3}8 - \frac{5x^4}{16} + O(x^5)$$
To stay in the same spirit you may use the expansion :
$$\ln\left(1+\frac x2\right)=\frac x2 - \frac{x^2}8 + \frac{x^3}{24} - \frac{x^4}{64} + O(x^5)$$
and more generally (for $a$ positive) :
$$\ln\left(1+\frac xa\right)=\frac xa - \frac{x^2}{2a^2} + \frac{x^3}{3a^3} - \frac{x^4}{4a^4} + O(x^5)$$
we could use the same formula for the upper bound (so that the initial proof remains valid) :
$$\frac {\frac xa}{\sqrt{1+\frac xa}}= \frac xa - \frac{x^2}{2a^2} + O(x^3)$$
Let's use $\ln\left(1+\frac xa\right)=\ln(a+x)-\ln(a)$ and $z:=a+x$ to get :
$$\ln(z)\le\ln(a)+\frac {\frac za-1}{\sqrt{1+\frac za-1}}$$
or for $x\ge a$ and $a>0$ :
$$\ln(x)\le\ln(a)+\frac{x-a}{\sqrt{ax}}$$
(as obtained by coffeemath...)
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Trig substitution $\int x^3 \sqrt{1-x^2} dx$ $$\int x^3 \sqrt{1-x^2} dx$$
$x = \sin \theta $
$dx = \cos \theta d \theta$
$$\int \sin^3 \theta d \theta$$
$$\int (1 - \cos^2 \theta) \sin \theta d \theta$$
$u = \cos \theta$
$du = -\sin\theta d \theta$
$$-\int u^2 du$$
$$\frac{-u^3}{3} $$
$$\frac{\cos^3 \theta}{3}$$
With the triangle trick I get:
$$\frac{-\sqrt{1-x^2}^3}{3}$$
This is wrong but I am not sure where I went wrong.
| What about by integration by parts? It looks pretty simple:
$$u=x^2\;,\;\;u'=2x\\v'=x\sqrt{1-x^2}\;,\;\;v=-\frac13(1-x^2)^{3/2}$$
and thus
$$\int x^2\cdot x\sqrt{1-x^2}\,dx=-\frac13x^2(1-x^2)^{3/2}+\frac23\int x(1-x^2)^{3/2}dx=$$
$$-\frac13x^2(1-x^2)^{3/2}-\frac2{15}(1-x^2)^{5/2}+C$$
Note: We used above the following:
$$\int x(1-x^2)^k\,dx=-\frac12\int (-2x\,dx)(1-x^2)^k=-\frac12\frac{(1-x^2)^{k+1}}{k+1}\;\ldots$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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} |
the value of $\lim\limits_{n\rightarrow\infty}n^2\left(\int_0^1\left(1+x^n\right)^\frac{1}{n} \, dx-1\right)$ This is exercise from my lecturer, for IMC preparation. I haven't found any idea.
Find the value of
$$\lim_{n\rightarrow\infty}n^2\left(\int_0^1 \left(1+x^n\right)^\frac{1}{n} \, dx-1\right)$$
Thank you
| By integration by parts,
\begin{align*}
\int_{0}^{1} (1 + x^{n})^{\frac{1}{n}} \, dx
&= \left[ -(1-x)(1+x^{n})^{\frac{1}{n}} \right]_{0}^{1} + \int_{0}^{1} (1-x)(1 + x^{n})^{\frac{1}{n}-1}x^{n-1} \, dx \\
&= 1 + \int_{0}^{1} (1-x) (1 + x^{n})^{\frac{1}{n}-1} x^{n-1} \, dx
\end{align*}
so that we have
\begin{align*}
n^{2} \left( \int_{0}^{1} (1 + x^{n})^{\frac{1}{n}} \, dx - 1 \right)
&= \int_{0}^{1} n^{2} (1-x) (1 + x^{n})^{\frac{1}{n}-1} x^{n-1} \, dx.
\end{align*}
Let $a_{n}$ denote this quantity. By the substitution $y = x^{n}$, it follows that
\begin{align*}
a_{n}
&= \int_{0}^{1} n \left(1-y^{1/n}\right) (1 + y)^{\frac{1}{n}-1} \, dy
= \int_{0}^{1} \int_{y}^{1} t^{\frac{1}{n}-1} (1 + y)^{\frac{1}{n}-1} \, dtdy
\end{align*}
Since $0 \leq t (1 + y) \leq 2$ and $ \int_{0}^{1} \int_{y}^{1} t^{-1}(1+y)^{-1} \, dtdy < \infty$, an obvious application of the dominated convergence theorem shows that
\begin{align*}
\lim_{n\to\infty} a_{n}
= \int_{0}^{1} \int_{y}^{1} \frac{dtdy}{t(1+y)}
&= - \int_{0}^{1} \frac{\log y}{1+y} \, dy \\
&= \sum_{m=1}^{\infty} (-1)^{m} \int_{0}^{1} y^{m-1} \log y \, dy
= \sum_{m=1}^{\infty} \frac{(-1)^{m-1}}{m^2}
= \frac{\pi^2}{12}.
\end{align*}
| {
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"timestamp": "2023-03-29T00:00:00",
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Differentiate the following w.r.t. $\tan^{-1} \left(\frac{2x}{1-x^2}\right)$ Differentiate : $$ \tan^{-1} \left(\frac {\sqrt {1+x^2}-1}x\right) \quad w.r.t.\quad \tan^{-1} \left(\frac{2x}{1-x^2}\right) $$
| Putting $x=\tan\theta,$
$$\quad \tan^{-1} \left(\frac{2x}{1-x^2}\right) $$
$$=\quad \tan^{-1} \left(\frac{2\tan\theta}{1-\tan^2\theta}\right) $$
$$=\quad \tan^{-1}(\tan2\theta)=n\pi+2\theta=n\pi+2\tan^{-1}x $$ where $n$ is any integer
$$\frac {\sqrt {1+x^2}-1}x=\frac {\sqrt {1+\tan^2\theta}-1}{\tan\theta}=\frac {\sqrt {1+\tan^2\theta}-1}{\tan\theta}=\frac{1-\cos\theta}{\sin\theta}=\frac{1-\frac{1-\tan^2\frac{\theta}2}{1+\tan^2\frac{\theta}2}}{\frac{2\tan\frac{\theta}2}{1+\tan^2\frac{\theta}2}}=\tan\frac{\theta}2$$
$$\implies\tan^{-1} \left(\frac {\sqrt {1+x^2}-1}x\right) \quad=\tan^{-1} \left(\tan\frac{\theta}2 \right)=m\pi+\frac{\theta}2=m\pi+\frac12\tan^{-1} x $$ where $m$ is any integer
So, $$\frac{d \tan^{-1} \left(\frac {\sqrt {1+x^2}-1}x\right) \quad}{d \tan^{-1} \left(\frac{2x}{1-x^2}\right)}=\frac{d(m\pi+\frac12\tan^{-1} x )}{d (n\pi+2\tan^{-1}x )}=\frac{\frac{d(m\pi+\frac12\tan^{-1} x )}{d(\tan^{-1} x)}}{\frac{d (n\pi+2\tan^{-1}x)}{d(\tan^{-1} x})}$$
| {
"language": "en",
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"source": "stackexchange",
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Find all real numbers such that $\sqrt{x-\frac{1}{x}} + \sqrt{1 - \frac{1}{x}} = x$ Find all real numbers such that
$$\sqrt{x-\frac{1}{x}} + \sqrt{1 - \frac{1}{x}} = x$$
My attempt to the solution :
I tried to square both sides and tried to remove the root but the equation became of 6th degree.Is there an easier method to solve this?
| Multiply with $\sqrt{x-1/x}-\sqrt{1-1/x}$ to get
$$x-1=(x-\frac1x)-(1-\frac1x) =x\left(\sqrt{x-1/x}-\sqrt{1-1/x}\right)$$
so $$1-\frac1x = \sqrt{x-1/x}-\sqrt{1-1/x}$$
and by adding
$$ x+1-\frac1x =2\sqrt{x-\frac1x}$$
Now with $z=x-\frac1x$, this is simply
$$z+1=2\sqrt z,$$
a quadratic in $\sqrt z$ with (double) solution $\sqrt z=1$, so $z=1$ and we need to solve
$$ x-\frac1x=1$$
i.e. $$ x^2-x-1=0.$$
This has solutions $x=\frac{1\pm\sqrt 5}2$, but only the positive value is possible.
| {
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"timestamp": "2023-03-29T00:00:00",
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For a convex function, the average value lies between $f((a+b)/2)$ and $(f(a) + f(b))/2$ Suppose that $f\in C^2$, $f''(x)\geq 0$ $\,\,\,\forall x \in [a,b]$. I want to show that $$\frac{1}{2}(b-a)(f(a)+f(b))\leq \int_a^bf(t)\,dt\leq (b-a)f\left(\frac{a+b}{2}\right).$$If we divide by $b-a$, we see that the left term is less than the right term by definition of convexity, and it remains to show that the average value of the function lies between $f\left(\frac{a+b}{2}\right)$ and $\frac{1}{2}(f(a)+f(b))$.
The mean value theorem for integrals implies that the average value is attained at some point $c\in (a,b)$. But it's not clear to me why $f(c)$ should lie in between two other points. Perhaps there's another theorem about integration we should apply. Any ideas?
| Here is an answer that doesn't assume $f$ is differentiable.
$f$ is convex iff $\frac{f(x)-f(y)}{x-y}$ is non-decreasing in both $x$ and $y$. If $f$ is convex, $f$ is continuous.
Assume $a\lt b$ and let $c=\frac{a+b}{2}$.
Since $\dfrac{f(x)-f(c)}{x-c}$ is non-decreasing,
$$
D^-=\sup_{x\lt c}\frac{f(x)-f(c)}{x-c}
\le\inf_{x\gt c}\frac{f(x)-f(c)}{x-c}=D^+\tag{1}
$$
Let $D=\frac{D^-{+}D^+}{2}$.
Inequality $(1)$ says
$$
f(x)-f(c)\ge D(x-c)\tag{2}
$$
Integrate $(2)$ over $[a,b]$ to get
$$
\begin{align}
\int_a^bf(x)\,\mathrm{d}x-(b-a)f(c)
&\ge D\left[\frac12x^2-cx\right]_a^b\\
&=D\left[\frac12(b^2-a^2)-\frac{a+b}2(b-a)\right]\\[6pt]
&=0\tag{3}
\end{align}
$$
For any $x\in[a,b]$, the convexity of $f$ implies
$$
\frac{f(x)-f(a)}{x-a}\le\frac{f(x)-f(b)}{x-b}\tag{4}
$$
therefore, since $(x-a)(x-b)\le0$, $(4)$ is equivalent to
$$
(x-b)(f(x)-f(a))\ge(x-a)(f(x)-f(b))\tag{5}
$$
Subtract $xf(x)$ from both sides of $(5)$ and integrate to get
$$
\frac12(b^2-a^2)(f(b)-f(a))+(bf(a)-af(b))(b-a)
\ge(b-a)\int_a^bf(x)\,\mathrm{d}x\tag{6}
$$
Divide both sides of $(6)$ by $b-a$ and regroup to get
$$
\begin{align}
\int_a^bf(x)\,\mathrm{d}x
&\le\frac12(a+b)(f(b)-f(a))+(bf(a)-af(b))\\
&=\frac12(b-a)(f(a)+f(b))\tag{7}
\end{align}
$$
Bringing $(3)$ and $(7)$ together yields
$$
(b-a)f\left(\frac{a+b}{2}\right)
\le\int_a^bf(x)\,\mathrm{d}x
\le\frac12(b-a)(f(a)+f(b))\tag{8}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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system of equations solving for positive $a,b,c$ i need help
i need to find positive number $a,b,c$ solving this system of equations?
$$(1-a)(1-b)(1-c)=abc$$
$$a+b+c=1$$
I found that $0<a,b,c<1$ and I try to solve it by try $(1-a)=a$, $(1-b)=b$ and $(1-c)=c$ and got that
$a=0.5,b=0.5,c=0.5$ but it contradicts the second equations.
Can someone help me?
| by expanding the left hand side you will have
$$1-(a+b+c)+ab+bc+ac-abc=abc
$$
using the second equality we will have:
$$2abc=ab+bc+ca
$$
And using the substitution method we can replace $a$ by $1-(b+c)$ then we have:
$$2bc(1-(b+c))=(1-(b+c))(b+c)+bc
$$
$$\Rightarrow 2bc-2b^2c-2bc^2=b+c-b^2-c^2-2bc+bc
$$
$$\Rightarrow 2b^2c+2bc^2+b+c-b^2-c^2-3bc=0 \qquad s.t. \quad 0 \leq b,c \leq 1
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $\sin 10^\circ \sin 20^\circ \sin 30^\circ=\sin 10^\circ \sin 10^\circ \sin 100^\circ$? $\sin 10^\circ \sin 20^\circ \sin 30^\circ=\sin 10^\circ \sin 10^\circ \sin 100^\circ$
This is a competition problem which I got from the book "Art of Problem Solving Volume 2". I'm not sure how to solve it because there's just so many possibilities of using most of the trig identities that I don't know which path to take. One way I tried is canceling out the $\sin 10$ on both sides:
$\sin 20^\circ \sin 30^\circ= \sin 10^\circ \sin 100^\circ$
$2 \sin 10^\circ \cos 10^\circ \sin 30^\circ=\sin 10^\circ \sin 100^\circ$
$2 \sin 80^\circ \sin 30^\circ= \sin 100^\circ$
$2 \sin 80^\circ \sin 30^\circ= \sin 30^\circ \cos 70^\circ+\sin 70^\circ \cos 30^\circ$
$2 \sin 80^\circ \sin 30^\circ=\sin 20^\circ \sin 30^\circ+\sin 70^\circ \sin 60^\circ$
$2 \sin 80^\circ \sin 30^\circ=\sin 20^\circ \sin 30^\circ + 2\sin 70^\circ \sin 30^\circ \cos 30^\circ$
$2 \sin 80^\circ \sin 30^\circ=\sin 20^\circ \sin 30^\circ + 2\sin 70^\circ \sin 30^\circ \sin 60^\circ$
$1=\large \frac {\sin 20^\circ}{2 \sin 80^\circ}+ \frac {\sin 70^\circ \sin 60^\circ}{\sin 80^\circ}$
Now I set $\sin^2 \theta=\large \frac {\sin 20^\circ}{2 \sin 80^\circ}$ and $\cos^2 \theta =\large \frac {\sin 70^\circ \sin 60^\circ}{\sin 80^\circ}$
I solves for $\sin \theta$ and $\cos \theta$ and set up a right triangle, which then led me to the equation
$\sin 70^\circ \sin 60^\circ=\sin 80^\circ-\frac 12 \sin 20^\circ$
Another approach I tries is from the beginning having $\sin 10^\circ \sin 20^\circ \sin 30^\circ=\sin^2 10^\circ \sin 100^\circ$ and the using the power-reducing formula, but that also got me nowhere. Any help is appreciated. Thanks.
| With $s=\sin 10^\circ$ and $c=\cos 10^\circ$, we have $\sin 20^\circ =\sin(2\cdot 10^\circ)=2sc$ and $\sin100^\circ=c$ and $\sin 30^\circ =\frac12$, hence
$$ \sin10^\circ\sin20^\circ\sin30^\circ = s\cdot2sc\cdot \frac12=ssc=\sin10^\circ\sin10^\circ\sin100^\circ$$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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Indefinite integral $\int{\frac{dx}{x^2+2}}$ I cannot manage to solve this integral:
$$\int{\frac{dx}{x^2+2}}$$
The problem is the $2$ at denominator, I am trying to decompose it in something like $\int{\frac{dt}{t^2+1}}$:
$$t^2+1 = x^2 +2$$
$$\int{\frac{dt}{2 \cdot \sqrt{t^2-1} \cdot (t^2+1)}}$$
But it's even harder than the original one. I also cannot try partial fraction decomposition because the polynomial has no roots. Ho to go on?
| $$
\frac{dx}{x^2+2} = \frac{dx}{2\left(\frac{x^2}{2} + 1\right)} =\frac{dx}{2\left(\left(\frac{x}{\sqrt{2}}\right)^2+1\right)} = \frac{dx/\sqrt{2}}{\sqrt{2}\left(\left(\frac{x}{\sqrt{2}}\right)^2+1\right)} = \frac{1}{\sqrt{2}}\cdot\frac{du}{u^2+1}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Expressing a combination of sine and cosine as a single cosine Prove that:
$$\dfrac{\sqrt2}2 \cos \omega t - \dfrac{\sqrt2}2 \sin \omega t = \cos \left(\omega t + \dfrac\pi4\right)$$
Obviously, if we are evaluating the right side of the equation, it would a easier. It would only take the following steps:
$cos \left(\omega t + \dfrac\pi4\right) = \cos \omega t\cos\dfrac\pi4 - \sin \omega t\sin\dfrac\pi4 = \dfrac{\sqrt2}2 \cos \omega t - \dfrac{\sqrt2}2 \sin \omega t$
Thus, it was proven.
But when I was evaluating the left side of the equation, I was having a hard time. Any idea on how or what approach I should do?
Thanks a lot.
| If you want to prove it as if you didn't know what it actually should be equal to, then here what people usually do. Assume, you have the expression $$a \cos x + b \sin x$$ and you want to find what it is equal to. The very first step is to notice the if you pull some factor outside of the brackets, so the whole expression becomes $$c (a_1 \cos x + b_1 \sin x)$$, so that $$a_1^2 + b_1^2 = 1$$ and $$0 \le |a_1|, |b_1| \le 1,$$ i.e. you can substitute $a_1 = \cos t$ and $b_1 = \sin t$.
$$
a_1 = \frac ac; \qquad b_1 = \frac bc
$$
and now just use condition above
$$
\left ( \frac ac \right )^2 + \left (\frac bc \right )^2 = 1
$$
from which you can find that the factor is
$$
c = \sqrt{a^2 + b^2}
$$
and therefore
$$
a \cos x + b \sin x = \sqrt{a^2 + b^2} \left ( \frac a{\sqrt{a^2 + b^2}} \cos x + \frac b{\sqrt{a^2 + b^2}} \sin x\right ) = \\
= \sqrt{a^2 + b^2} \left ( \cos t \cos x + \sin t \sin x \right ) = \sqrt{a^2 + b^2} \cos (x - t)
$$
where
$$
\cos t = \frac a{\sqrt{a^2 + b^2}} \\
\sin t = \frac b{\sqrt{a^2 + b^2}}
$$
Now, just substitute your numbers.
$$
c = \sqrt{\frac 12 + \frac 12} = 1
$$
so no need to factor anything out, since coefficients already make unity. So, all you need to do is finding some angle that corresponds to those coefficients, namely
$$
\sin t = \frac {\sqrt 2}2 \\
\cos t = -\frac {\sqrt 2}2
$$
you can easily see, that $t = -\frac \pi 4$ is one of those angles. At the same time, you could've taken any angle of the form $-\frac \pi 4 + 2 \pi n$.
So, angle is found, then substitute it to the final answer
$$
\frac {\sqrt 2}2 \cos \omega t - \frac {\sqrt 2}2 \sin \omega t = \cos \left ( \omega t - \left (-\frac \pi 4 \right )\right ) = \cos \left ( \omega t + \frac \pi 4 \right )
$$
| {
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The value of $w$ also has a max error of $p\%$ Suppose $\frac{1}{w}=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}$ where each variable $x,y,z$ can be measured with a max error of $p\%$
Prove that the calculated value of $w$ also has a max error of $p\%$
I guess I need to take its derivative i.e $-\frac{1}{w^2}dw=-\frac{1}{x^2}dx-\frac{1}{y^2}dy-\frac{1}{z^2}dz$
After there, what do I need to do?
I know this question maybe seem nonsense. But please help me to solve. Thanks!:)
| From this line
$$-\frac{1}{w^2}dw=-\frac{1}{x^2}dx-\frac{1}{y^2}dy-\frac{1}{z^2}dz$$
The percentage error of $w$ is
$$p_w=100\frac{dw}{w}\%$$
and likewise for the other variables, therefore
$$\frac{1}{w}p_w=\frac{1}{x}p_x+\frac{1}{y}p_y+\frac{1}{z}p_z$$
So assuming a maximum error of $p\%$ on the independent variables we have that
$$\frac{1}{w}p_w\leq\frac{1}{x}p+\frac{1}{y}p+\frac{1}{z}p=\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)p=\frac{1}{w}p$$
| {
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Domain of $\sqrt{1+\frac1x}$ I solved this as follows:
$1+\frac1x \ge 0$
$\frac1x \ge -1$ (subtract $1$ from both sides)
$1 \ge -x$ (multiply $x$ to both sides, cancel out $x$ from bottom of left side)
$-1 \le x$ (multiply both sides by $-1$, change sign)
$x \ge -1$
Another way to solve is:
$1+\frac1x \ge 0$
$\frac{(x+1)}x \ge 0$
$x+1 \ge 0$
$x \ge -1$
The domain should be $x\ge -1$, not including $0$. However, I see from WolframAlpha and searching online that the domain is $x\le -1$ or $x>0$. Why is my solution wrong?
| Here's a better way at looking at this:
$$\begin{align*}
1+\frac{1}{x} &\ge 0 \\
x+1 &\ge 0 \tag{multiply both sides by $x$ when $x >0$} \\
&\textrm{or} \\
x+1 &\le 0 \tag{multiply both sides by $x$ when $x < 0$}
\end{align*}$$
In the first case, we restrict $x>0$, so every permissible value of $x$ satisfies the inequality.
In the second case, we get $x \le -1$, as expected.
| {
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The no. of values of k for which $(16x^2+12x+39) + k(9x^2 -2x +11)$ is perfect square is: I wanted to know, how can i determine the no. of values of k for which $(16x^2+12x+39) + k(9x^2 -2x +11)$ is a perfect square.($x \in R$)
I have tried, since $x$ is real the discriminant must be $\geq 0$.
$D = 4(6-k)^2 -4(16+9k)(11k+39) \geq 0.$
which gives $k \in [-4,-1.5]$. But how can i determine the values where the question will be a perfect square.
Any help appreciated.
Thanks.
| HINT:
Let $(16x^2+12x+39) + k(9x^2 -2x +11)=(ax+b)^2$
$\implies (16+9k)x^2+(12-2k)x+3+11k=a^2x^2+2abx+b^2$
Comparing the coefficients of $x^2,x,x^0$
we have $a^2=16+9k,2ab=12-2k\implies ab=6-k, b^2=3+11k$
$\implies (6-k)^2=(ab)^2=(16+9k)(3+11k)$ which is a Quadratic Equation in $k$ on re-arrangement
| {
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Number Theory divisibilty How can I check if $$12^{2013} + 7^{2013}$$ is divisible by $19$?
Also, how can I format my questions to allow for squares instead of doing the ^ symbol.
| First we prove two theorems that we will use
If $a,\;b,\;c\;\in \mathbb{N}$ with $a\neq0$ and $x,y\in\mathbb{N} $such that $a\mid b$ and $a\mid c$, then, $a\mid bx+cy$;
Show:$a\mid b$ and $a\mid c$ implies that there $m,n\in\mathbb{N}$ such that $b=a\cdot m$ and $c=a\cdot n$;$$bx+cy=am\cdot x+an\cdot y=a(mx+ny)\Longrightarrow a\mid bx+cy \;\;\;\Box$$Now we have to prove the theorem we use to solve this issue, and in this test we use the theorem proved above
Are $a,b,n\in\mathbb{N}$ with $a+b\neq0$ we have $a+b\mid a^{2n+1}+b^{2n+1}$
Show: Have demonstrated induction $n=0$ the statement is true because $a+b\mid a^{2\cdot 0+1}+b^{2\cdot0+1}\Longrightarrow a+b\mid a+b$; hypothesis: $a+b\mid a^{2n+1}+b^{2n+1}$;$$a^{2(n+1)+1}+b^{2(n+1)+1}=a^{2n+3}+b^{2n+3}=a^{2n+1+2}+b^{2n+1+2}=a^2a^{2n+1}+b^2b^{2n+1}=$$$$=a^2a^{2n+1}+b^2b^{2n+1}\underbrace{-b^2a^{2n+1}+b^2a^{2n+1}}_{=0}=a^2a^{2n+1}-b^2a^{2n+1}+b^2b^{2n+1}+b^2a^{2n+1}=$$$$=a^{2n+1}(a^2-b^2)+b^2(a^{2n+1}+b^{2n+1})$$we know that $a+b\mid a^2-b^2$ because $a^2-b^2=(a+b)(a-b)$; our hypothesis gives us that $a+b\mid a^{2n+1}+b^{2n+1}$ and the theorem shown above assures us that $a+b\mid a^{2n+1}(a^2-b^2)+b^2(a^{2n+1}+b^{2n+1})$ thus showing that the theorem is valid for $n +1$, then, goes for any $n\in\mathbb{N}\;\;\Box$
Question: Show that $19\mid 12^{2013}+7^{2013}$
Using the theorem proved above, we have easily that $19=12+7$ and $$12+7\mid 12^{2013}+7^{2013}$$because $2013=2\cdot1006+1$ and $1006\in\mathbb{N}$
| {
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Integration $\int \frac{\sqrt{x^2-4}}{x^4}$ Problem :
Integrate $\int \frac{\sqrt{x^2-4}}{x^4}$
I tried : Let $x^2-4 =t^2 \Rightarrow 2xdx = 2tdt$
$\int \frac{\sqrt{x^2-4}}{x^4} \Rightarrow \frac{t^3 dt}{\sqrt{t^2+4}(t^4-8t+16)}$
But I think this made the integral too complicated... please suggest how to proceed.. Thanks..
| HINT:
As the radical contains $x^2-4,$
put $x=2\sec\theta$
$$\implies \int\frac{\sqrt{x^2-4}}{x^4}dx=\int \frac{2|\tan\theta|}{\sec^4\theta} 2\sec\theta\tan\theta d\theta=4\cdot \text{sign}(\tan\theta)\int \sin^2\theta \cos \theta d\theta$$
Putting $\sin\theta=u,$
$$4\text{sign}(\tan\theta)\int \sin^2\theta \cos \theta d\theta$$
$$=4\text{sign}(\tan\theta)\int u^2 du=\frac43\cdot\text{sign}(\tan\theta)\cdot u^3+K$$
$$=\frac43\cdot\text{sign}(\tan\theta)\cdot \sin^3\theta+K$$ where $K$ is an arbitrary constant of indefinite integral
Now, $\sin\theta=\pm\sqrt{1-\cos^2\theta}=\pm\sqrt{1-\left(\frac2x\right)^2}=\pm\frac{\sqrt{4-x^2}}x$
Now observe that the sign$(\sin\theta)=$ sign$(\tan\theta)\cdot$sign$\left(\frac2x\right)$
| {
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Simplifying compound fraction: $\frac{3}{\sqrt{5}/5}$ I'm trying to simplify the following:
$$\frac{3}{\ \frac{\sqrt{5}}{5} \ }.$$
I know it is a very simple question but I am stuck. I followed through some instructions on Wolfram which suggests that I multiply the numerator by the reciprocal of the denominator.
The problem is I interpreted that as:
$$\frac{3}{\ \frac{\sqrt{5}}{5} \ } \times \frac{5}{\sqrt{5}},$$
Which I believe is:
$$\frac{15}{\ \frac{5}{5} \ } = \frac{15}{1}.$$
What am I doing wrong?
| One thing that helps me organize my thoughts, is to convert both numerator and denuminator to fractions as follows.
$$\begin{align}
\frac{3}{\frac{5}{\sqrt{5}}}&=\frac{\frac{3}{1}}{\frac{5}{\sqrt{5}}}=\frac{\frac{3}{1}}{\frac{5}{\sqrt{5}}}\cdot\frac{\frac{\sqrt{5}}{5}}{\frac{\sqrt{5}}{5}}=\frac{\frac{3}{1}\cdot\frac{\sqrt{5}}{5}}{\frac{5}{\sqrt{5}}\cdot\frac{\sqrt{5}}{5}}=\frac{\frac{3}{1}\cdot\frac{\sqrt{5}}{5}}{1}=\frac{3}{1}\cdot\frac{\sqrt{5}}{5}=\text{etc.}\\
\end{align}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Power series for $(1+x^3)^{-4}$ I am trying to find the power series for the sum $(1+x^3)^{-4}$ but I am not sure how to find it. Here is some work:
$$(1+x^3)^{-4} = \frac{1}{(1+x^3)^{4}} = \left(\frac{1}{1+x^3}\right)^4 = \left(\left(\frac{1}{1+x}\right)\left(\frac{1}{x^2-x+1}\right)\right)^4$$
I can now use
$$\frac{1}{(1-ax)^{k+1}} = \left(\begin{array}{c} k \\ 0 \end{array}\right)+\left(\begin{array}{c} k+1 \\ 1 \end{array}\right)ax+\left(\begin{array}{c} k+2 \\ 2 \end{array}\right)a^2x^2+\dots$$
on the $\frac{1}{1+x}$ part but I am not sure how to cope with the rest of formula.
| This would not be the best answer but may help if you are interested in computing the first terms of the Taylor expansion.
Just use the fact that $\frac{1}{1-z}= \sum_{n=0}^\infty z^n $ for $|z|<1$ and the binomial formula, i.e.
\begin{align}
\frac{1}{(1+x^3)^4} &= \frac{1}{1+4x^3+6x^6+4x^9+x^{12}} \\
&= \sum_{n=0}^\infty (-1)^n (4x^3+6x^6+4x^9+x^{12})^n.
\end{align}
| {
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$\int \dfrac {\sqrt{x+1}} {x^{7/2}} dx$ without using trigonometry? $$\int \dfrac {\sqrt{x+1}} {x^{7/2}} dx$$
Is there any way to find the answer without using trigonometry, like this?
Hint by Parth Thakkat:
$$\int \dfrac {\sqrt{x+1}} {x^{7/2}} dx$$
$$ = \int \dfrac {\sqrt{x+1}} {x^{1/2}} \cdot \dfrac{dx} {x^{3}}$$
$$ = \int \sqrt{1+\dfrac 1 x} \cdot \dfrac{dx} {x^{3}}$$
Take $t^2 = 1 + \dfrac 1 x$
and note that $ \dfrac 1 x = t^2 - 1$
More Hint:
$t^2 = 1 + \dfrac 1 x$
$\implies 2tdt = -\dfrac 1 {x^2}dx$
Next
$$ = \int t dx/x^3$$
Since $dx = -2tx^2 dt$
$$= \int t(-2t)(t^2-1) dt$$
$$ = -2\int t^4-t^2 dt$$
$$ = -2(\dfrac{t^5} {5} - \dfrac{t^3}{3} + C$$
C = constant
$$ = -\dfrac{2}{5}(1+\dfrac{1}{x})^\dfrac{5}{2}+\dfrac{2}{3}(1+\dfrac{1}{x})^\dfrac{3}{2} + C$$
Thanks alot :)
| $$\int \sqrt{x+1}/x^{7/2} dx=\int\frac{1}{x^3} \sqrt{1+\frac{1}{x}} dx$$
$1+\frac{1}{x}=t$ then $\frac{1}{x^2}dx=-dt $ and $\frac{1}{x}={t-1}$
$$\int\frac{1}{x^3} \sqrt{1+\frac{1}{x}} dx=\int\frac{1}{x} \sqrt{1+\frac{1}{x}} \frac{dx}{x^2}=-\int(t-1)\sqrt {t}dt=\int (t^{1/2}-t^{3/2})dt$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Partial Fractions I here would like to clear my doubt on the question below:
$$\frac{1}{x(x-1)(x-2)}\;,$$
that is, we want to bring it into the form:
$$\frac{A}{x}+\frac{B}{x-1}+\frac{C}{x-2}\;,$$
in which the unknown parameters are $A,B$, and $C$. Multiplying these formulas by $x(x − 1)(x − 2)$ turns both into polynomials, which we equate:
$$A(x-1)(x-2) + Bx(x-2) + Cx(x-1) = 1\;,$$
or, after expansion and collecting terms with equal powers of $x$:
$$(A+B+C)x^2 - (3A+2B+C)x + 2A = 1\;.$$
At this point it is essential to realize that the polynomial $1$ is in fact equal to the polynomial $0x^2 + 0x + 1$, having zero coefficients for the positive powers of $x$. Equating the corresponding coefficients now results in this system of linear equations:
$$\left\{\begin{align*}
&A+B+C = 0\\
&3A+2B+C = 0\\
&2A = 1\;.
\end{align*}\right.$$
Solving it results in:
$$A = \frac{1}{2},\, B = -1,\, C = \frac{1}{2}\;.$$
So from my solving I had different values of $A,B$, and $C$ which gave me:
$$\left\{\begin{align*}
&A=\frac12\\
&B= 2\\
&C= -\frac52\;.
\end{align*}\right.$$
Can someone please tell me if these answers are correct because when I substitute these values into equation $A+B+C= 0$, it still gave me a zero.
But this time for the $2$nd equation, instead of $3A+2B+C= 0$, I used $-3A+2B+C= 0$, which then by substituting the values of $A, B$, and $C$ I had, also gave me a zero. Only $A= \frac12$ was the same as obtained from $2A= 1$.
Does this mean that the values that I have obtained for $A, B$, and $C$ are also correct? Kindly can someone please give a clear explanation to this?
Many thanks.
| No, your values are not correct: they do not satisfy the equation $3A+2B+C=0$. The equation $-3A+B+C=0$ has nothing to do with the problem, so numbers obtained by using it simply aren’t relevant. I suspect that you got the equation $-3A+B+C=0$ by making a sign error in collecting the coefficients of $x$ in $A(x-1)(x-2)+Bx(x-2)+Cx(x-1)$, e.g., by converting $-(3A+2B+C)$ to $-3A+2B+C$ instead of to $-3A-2B-C$.
| {
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Maximum and minimum function on area Find maximum and minimum value of function $f(x,y) = 3x+14y$ on $ \left\{ (x,y): 3x^4 + xy + y^4 =6\right\} $.
I will grateful for hints and yours help.
| What could be said about this "skewed superellipse" $ \ 3x^4 \ + \ xy \ + \ y^4 \ = \ 6 \ $ is that since the curve has symmetry about the origin [if a point $ \ (x,y) \ $ lies on the curve, so does $ \ (-x,-y) \ $ ], we should expect that extrema are located at corresponding points in opposing quadrants. user64494 finds, by implicit differentiation of the curve equation, the slope of a tangent line to be $ \ \frac {12x^3+y}{4y^3+x} \ $ , and from the linear function to be extremized, the "useful" equation
$$\frac {12x^3 \ + \ y} {4y^3 \ + \ x} \ \ = \ \ \frac{3}{14} \ \ . $$
As the OP, Thomas, indicated that this was an exam problem, I wondered whether some indication was given that the student was to locate a rational point on this curve in answering the question, since the time would be severely (hopelessly?) limited to find real zeroes of the related polynomial without a computational aid. If it were given that the coordinates of a rational point have the same denominator, so that $ \ x \ = \frac{a}{c} \ \ , \ \ y \ = \frac{b}{c} \ \ , $ we would be able to write
$$\frac {12 \left( \frac{a}{c} \right)^3 \ + \ \frac{b}{c} } {4\left( \frac{b}{c} \right)^3 \ + \frac{a}{c} } \ \ = \ \ \frac {12 a^3 \ + \ bc^2 } {4 b^3 \ + \ ac^2 } \ \ = \ \ \frac{3}{14} \ \ . $$
We might then begin a search with $ \ c \ = \ 2 \ $ (much as we do in "hunting" rational zeroes of a polynomial), giving us
$$ \ \frac {12 a^3 \ + \ 4b } {4 b^3 \ + \ 4a } \ \ = \ \ \frac{3}{14} \ \ . $$
Since we need $ \ a \ $ and $ \ b \ $ to be integers, we may require that $ \ 6 a^3 \ + \ 2b \ \ = \ \ 3k \ $ and $ \ 2 b^3 \ + \ 2a \ \ = \ \ 14k \ \ , $ with $ \ k \ $ being a positive integer. It takes only a little effort to find $ \ 6 · 1^3 \ + \ 2·3 \ \ = \ \ 12 \ \ = \ \ 3·4 \ $ ; we then note that $ \ 14 · 4 \ \ = \ \ 56 \ \ = \ \ 2 ·3^3 \ + \ 2·1 \ \ \ \ . $
This gives us the desired ratio $ \ \frac{12}{56} \ \ = \ \ \frac{3}{14} \ \ . $ As all of the exponents in the slope expression are odd, the signs of both $ \ a \ $ and $ \ b \ $ can both be changed without altering the sign of the ratio. This then yields two rational points $ \ \left( \ \frac{1}{2} \ , \ \frac{3}{2} \ \right) $ and $ \ \left( \ -\frac{1}{2} \ , \ -\frac{3}{2} \ \right) \ \ , $ as expected from the symmetry of the curve. The maximum and minimum of $ \ f(x,y) \ = \ 3x \ + \ 14y \ $ occur, respectively, for those two points as $ \ \frac{45}{2} \ $ and $ \ -\frac{45}{2} \ $ . A representative graph is shown below. A calculation of this sort may have been possible, with the clues mentioned, in the available time.
$$ $$
I can't help feeling that this is one of those exam problems, if it is correctly reproduced here, that the problem-poser had not worked out in advance to estimate how much work would be necessary. (I've had to reject many prospective "interesting" exam problems after seeing the amount of solution-writing that would be called for...) An optimization problem for an exam would be rather more tractable if the curve were a "rotated ellipse", say, $ \ 3x^2 \ + \ xy \ + \ y^2 \ = \ C \ $ .
| {
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find a solution of 9x = 24 (mod 21) I need help finding a solution of $9x\equiv {24}\pmod {21}$.
Here is what I tried, but it's wrong.
mod x is the positive value of x. mod $21 = 21.$
$9x\equiv {24}\pmod {21}$.
$9x = 24*21$
$x = 24*21/9 = 56$
| Since $21=3\cdot 7$ and $(3,7)=1$, we have:
$$9x\equiv 24\pmod{21}\Leftrightarrow 9x\equiv 24\pmod{3}\wedge9x\equiv 24\pmod{7}$$
Now, $9x\equiv 24\pmod{3}\Leftrightarrow 0x\equiv 0\pmod{3}$, which is always true and
$$9x\equiv 24\pmod{7}\Leftrightarrow 2x\equiv 3\pmod{7}\Leftrightarrow 4\cdot2x\equiv 4\cdot 3\pmod{7}\Leftrightarrow x\equiv 5\pmod{7}$$
| {
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Error in approximation of nonzero root to $x^2=\sin x$ using Taylor's cubic polynomial. I have successfully obtained the root's approximation $r=\sqrt{15}-3$ as I'm supposed to as following:$$\begin{align}
\displaystyle f(x)=\sin x &= x - \frac{x^3}{6} + E\\
x^2 &\approx x - \frac{x^3}{6}\\
r^2 &= r - \frac{r^3}{6}\\
r &= \sqrt{15}-3
\end{align}$$
Given $r=\sqrt{15}-3<0.9$, show that $|\sin r-r^2|<\frac{1}{200}$. I have tried but of no avail like this:
$$
\displaystyle f(r)=\sin r = r - \frac{r^3}{6} + E\\
|\sin r - r^2| = |\sin r-(r-\frac{r^3}{6})|=|E|=E\\
E>0\text{ because } r>a=0\text{ and } f^4(x)=\sin x > 0 \text{ on }(0,\pi)\text{ containing }r\\
\text{Because }f^4(r)=\sin r <\sin 0.9\\
\therefore E<\sin0.9\frac{r^4}{4!}\approx 0.01895639019\\
\frac{1}{200}=0.005\\
$$
| The series for $\sin x$ is an alternating series. So for numbers $x$ near $r$, we have $\sin x\lt x-\frac{x^3}{3!}+\frac{x^5}{5!}$.
Since $r^2=r-\frac{r^3}{3!}$, the absolute value of the difference is $\lt \frac{(0.9)^5}{5!}$. This brings is a little below $\frac{1}{200}$. Using $\sqrt{15}-3$ instead of $0.9$ brings us more cleanly under $\frac{1}{200}$.
The Lagrange remainder term when we cut off the expansion of $\sin x$ after the (zero) term in $x^4$ yields the same result.
| {
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How do you do partial fraction $\frac2{x^2 - 8}$? How do you do partial fraction on $\frac2{x^2 - 8}$ ? Or are there other method of doing? I tried trig substitution but could not get the answer.
| It's very simple:
$$\frac{2}{x^2-8} = \frac{2}{(x+2\sqrt 2)(x-2\sqrt 2)} = \frac{\frac{1}{4} \sqrt 2}{x-2\sqrt 2}-\frac{\frac{1}{4} \sqrt 2}{x+2\sqrt 2}$$
Thus, for example, $$\int \frac{2}{x^2-8} dx = \frac{1}{4} \sqrt 2 \int \frac{dx}{x-2\sqrt 2} - \frac{1}{4} \sqrt 2 \int \frac{dx}{x+2\sqrt 2}$$
| {
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Average of all 6 digit numbers that contain only digits $1,2,3,4,5$ How do I find the average of all $6$ digit numbers which consist of only digits $1,2,3,4$ and $5$?
Do I have to list all the possible numbers and then divide the sum by the count? There has to be a more efficient way, right?
Thank you!
| The generating function of these numbers is
$$f(x) = \prod_{q=0}^5
\left(x^{10^q} + x^{2 \times 10^q}+ \cdots + x^{5 \times 10^q}\right).$$
Hence the number of values is given by
$$f(1) = \prod_{q=0}^5 5 = 5^6,$$
(this part is trivial but it confirms that we have the right generating function).
Similarly, the sum of these numbers is given by
$$\left.\frac{d}{dx} f(x)\right|_{x=1} \\
= \left.\prod_{q=0}^5
\left(x^{10^q} + x^{2 \times 10^q}+ \cdots + x^{5 \times 10^q}\right) \\ \times
\sum_{q=0}^5 \frac{10^q \times x^{10^q-1} + 2\times 10^q \times x^{2 \times 10^q-1}+ \cdots + 5\times 10^q \times x^{5 \times 10^q-1}}
{x^{10^q} + x^{2 \times 10^q}+ \cdots + x^{5 \times 10^q}}
\right|_{x=1}$$
which is $$5^6 \frac{1}{5} \sum_{q=0}^5 10^q \times(1+2+\cdots+5),$$
so the average is
$$\frac{\left.\frac{d}{dx} f(x)\right|_{x=1}}{f(1)} =
\frac{1}{5} \times 15 \times \sum_{q=0}^5 10^q =
3\times 111111 = 333333.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/459880",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 6,
"answer_id": 5
} |
Sum of integer parts of different numbers I have the sum of all these integer parts of different numbers
$$
\lfloor 1\rfloor + \lfloor \sqrt{2} \rfloor + \lfloor \sqrt{3} \rfloor + \dots + \lfloor \sqrt{15} \rfloor
$$
I don't have any idea to solve this exercise.
| More generally, for nonnegative integer $n$, $\lfloor \sqrt{x} \rfloor = n$ if
$n \le \sqrt{x} < n+1$, i.e. $n^2 \le x < (n+1)^2$. Since $(n+1)^2 - n^2 = 2n+1$,
there are $2n+1$ integers $x$ for which this is true. Thus
$$\lfloor \sqrt{1} \rfloor + \lfloor \sqrt{2}\rfloor + \ldots + \lfloor \sqrt{(n+1)^2 - 1}\rfloor = \sum_{j=1}^n (2 j + 1) j = \dfrac{n(n+1)(4n+5)}{6}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/460366",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
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