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If a number is rational, then it has a periodic decimal expression? I have proved the following:
If the decimal expansion of a number is periodic, then it is a rational number.
Now I am trying to prove the converse. For this, I am taking the rational numbers smaller than $1$, that is $\frac{m}{n}$ with $n>m$ because any rational bigger than $1$ can be written as $x$ + $\frac{m}{n}$ with $x\in \Bbb{Z}$. Trying to mimic the process of long division, I write:
$$m=0\cdot n + m$$
Now we multiply $m$ by $10$ and hence:
$$10m=a_1 \cdot n + r_1$$
Now we divide the remainder:
$$r_1=a_2 \cdot n+ r_2$$
And so on. This gives us the decimal expansion: $0,r_1 r_2 r_3...$. Now, by definition $0\leq r_i <n$ and then, $r_i$ divided by $n$ leaves remainder $r_i$. And then for $i\geq 2$ we have that $a_i=0$ and hence:
$$r_i=r_{i+1}$$
Is this correct?
|
EDIT: I'd like to elaborate on something I glossed over in this answer originally - namely, why we can assume that "...we necessarily hit a repeat remainder." The added explanation is included at the bottom of the answer (to keep it readable).
I think you're on the right track. But the idea here is, for $\frac{p}{q}$ with $p < q$, when we take $10p = n_1q + r_1$, there are only finitely many remainders $r < q$ that are possible. We keep using the division algorithm, $10r_1 = n_2q + r_2$, $10r_2 = n_3q + r_3$, etc., until we necessarily hit a repeat remainder$^*$; that is, one with $r_k = p$. Then $p = \frac{n_1q}{10} + \frac{n_2q}{100} + \cdots \frac{n_kq}{10^k} + \frac{p}{10^k}.$
Therefore we have $\frac{p}{q} = \frac{n_1}{10} + \cdots + \frac{n_k}{10^k} + (\frac{1}{10^k}\cdot\frac{p}{q}).$ But then $\frac{p}{q}\cdot(1 - \frac{1}{10^k}) = \frac{n_1}{10} + \cdots + \frac{n_k}{10^k}$, so $$\frac{p}{q} = \left(\frac{1}{1-\frac{1}{10^k}}\right)\left(\frac{n_1}{10} +\cdots + \frac{n_k}{10^k}\right)$$
But $$\left(\frac{1}{1-\frac{1}{10^k}}\right) = \sum_{j = 0}^\infty \left(\frac{1}{10^{k}}\right)^j$$ and so we'll have
\begin{eqnarray}\frac{p}{q} &&= ( 1 + \frac{1}{10^k} + \frac{1}{(10^k)^2} + \cdots) \left(\frac{n_1}{10} +\cdots + \frac{n_k}{10^k}\right) \\ &&= \left(\frac{n_1}{10} +\cdots + \frac{n_k}{10^k}\right) + \left(\frac{n_1}{10^{k+1}} +\cdots + \frac{n_k}{10^{2k}}\right) + \left(\frac{n_1}{10^{2k + 1}} +\cdots + \frac{n_k}{10^{3k}}\right) \cdots \end{eqnarray}
which precisely shows that $\frac{p}{q}$ is a repeating decimal.
$(^*)$ First of all, we can assume $p$ and $q$ are coprime; otherwise before trying to get their decimal values, we could simply cancel all common factors.
When we apply the division algorithm repeatedly, we are computing that $10p \equiv r_1 \bmod(q)$, $10r_1 \equiv r_2 \bmod(q)$, $10r_2 \equiv r_3 \bmod(q)$, etc. This is the same as saying \begin{eqnarray} 10p &&\equiv r_1 \bmod(q), \\10^2p &&\equiv r_2 \bmod(q), \\10^3p &&\equiv r_3 \bmod(q), \\&&... \\10^kp &&\equiv r_k \bmod(q).\end{eqnarray} Then the claim that we reach $r_k = p$ is exactly the claim that for some $k$, $10^kp \equiv p \bmod(q)$.
Now above I noted that the number of possible remainders $r < q$ is finite, so that eventually the repeated use of the division algorithm must give us remainders $r_j = r_k$ with $j > k$. However, we can't assume that either of these remainders is $p$ from the pigeonhole principle.
But we can see that $r_j = r_k$ means that $10^jp \equiv 10^kp \bmod(q),$ so $$(10^{j-k} - 1)(10^k)p \equiv 0 \bmod(q).$$ This means that necessarily we have one of two things: since $q$ is coprime to $p$, either $10^k \equiv 0 \bmod(q)$, or $10^{j-k} \equiv 1 \bmod(q)$. In this case, $10^kp = nq$ for some $k$ and $n$, so we have $\frac{p}{q} = \frac{n}{10^k}$ for $n$ which must be less than $10^k$ (since $p < q$), and now $\frac{n}{10^k}$ is precisely the decimal expansion of $\frac{p}{q}$. (In my original answer, I essentially ignored this possibility.)
The second case is the one where we actually hit a repeat remainder - if $10^{j-k} \equiv 1 \bmod(q)$, then $10^{j-k}p \equiv p \bmod(q)$, and therefore we'll have (after we do it all out with the division algorithm) $p = \frac{n_1q}{10} + \frac{n_2q}{100} + \cdots + \frac{n_{j-k}q}{10^{j-k}} + \frac{p}{10^{j-k}}.$
|
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Finding the global minimum of $x^{n}+x^{n-1}+...+1$ for even $n.$ Inspired by this question, I am curious if there is an asymptotic of the global minimum of the function: $$f_n(x) = x^{2n}+x^{2n-1}+...+x^2+x+1.$$
In the referred question, I showed that $$f_n(x) = x^{2n}+x^{2n-1}+...+x^2+x+1 =x^{2n-2}(x+\dfrac{1}{2})^2+\dfrac{3}{4}x^{2n-4}(x+\dfrac{2}{3})^2+\dfrac{4}{6}x^{2n-6}(x+\dfrac{3}{4})^2+\dots+ \dfrac{n+1}{2n}(x+\dfrac{n}{n+1})^2+\dfrac{n+2}{2n+2}> \dfrac{n+2}{2n+2}.$$
Since the method was completely elementary, I figured this was a loose bound. However, it turned out to be surprisingly close to the actual values for when $n=2,3$ and sharp $n=1.$
When $n=2$, the minimum is $0.673753\approx\dfrac{2}{3}=0.66$ and when $n=3$, the minimum is $0.635\approx \dfrac{5}{8} = 0.625.$
Thus, is it possible to obtain an asymptotic for $\min_{x\in\mathbb{R}}f_n(x)$, as $n\to\infty?$
A closed form solution would be even better, but that just seem hopeless.
|
Claim. The minimum of $f_n$ tends to $\frac12$ when $n \to \infty$.
Let $a_n$ denote the minimum of $f_n(x) = \frac{x^{2n+1}-1}{x-1}$ for $x \in \mathbb{R}$.
For fixed $t>0$ and $n > t$ we also have $$a_n \leq f_n\left(-1 + \frac{t}n\right) = \frac{-1 + (-1 + \frac{t}{n})^{1 + 2 n}}{-2 + \frac{t}{n}} = \frac{1 + (1-\frac{t}{n})^{2n+1}}{2-\frac{t}n} \leq \frac{1+ (1-\frac{t}{n}) e^{-2t}}{2-\frac{t}{n}},$$
because $(1-\frac{t}{n})^n \leq e^{-t}$. This means that for any $t>0$ we have
$$
\frac{1 + \frac{2}{n}}{2+\frac{2}{n}} \leq a_n \leq \frac{1+ (1-\frac{t}{n}) e^{-2t}}{2-\frac{t}{n}}
$$
for sufficiently large $n$. For $n \to \infty$ the lower bound tends to $\frac12$, whereas the upper bound tends to $\frac{1+e^{-2t}}{2}$. Because this holds for any $t$, we find $\lim_{n \to \infty} a_n = \frac12$.
|
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Is it possible to evaluate analytically the following nontrivial triple integral? In a physical mathematical problem, I came across a nontrivial triple integral below obtained upon 3D inverse Fourier transformation. It would be great if someone here could provide with some hints that could help to evaluate analytically
$$
I =\frac{1}{(2\pi)^3} \int_0^{2\pi} \int_0^\infty \int_0^{\pi}
\frac{\sin\theta \sin^2\phi}{a \cos^2 \theta + b \sin^2 \theta} \, e^{ikh\sin\theta\cos\phi} \, \mathrm{d} \theta \, \mathrm{d} k \, \mathrm{d} \phi \, .
$$
where $a$ and $b$ are two positive real numbers.
Using the change of variable $q=\cos\theta$, the latter equation can be written as
$$
I = \frac{1}{(2\pi)^3} \int_0^{2\pi} \int_0^\infty \int_{-1}^{1}
\frac{\sin^2\phi}{(a-b)q^2+b} \, e^{ikh\sqrt{1-q^2}\cos\phi} \, \mathrm{d} q \, \mathrm{d} k \, \mathrm{d} \phi \, .
$$
A guess solution using Maple for some numerical values for $a$ and $b$ is obtained as
$$
I = \frac{1}{4\pi h \sqrt{ab}} \, .
$$
But, is there a way to prove that really? Any idea / feedback is welcome.
Thanks a lot!
Fede
|
Well, for the first integral we have:
$$\mathscr{I}:=\int_0^{2\pi}\int_0^\infty\int_0^\pi\frac{\sin\left(\theta\right)\cdot\sin^2\left(\phi\right)\cdot\exp\left(\sin\left(\theta\right)\cdot\cos\left(\phi\right)\cdot\text{k}\cdot\text{h}\cdot i\right)}{\text{a}\cdot\cos^2\left(\theta\right)+\text{b}\cdot\sin^2\left(\theta\right)}\space\text{d}\theta\space\text{d}\text{k}\space\text{d}\phi\tag1$$
Using Fubini's theorem:
$$\mathscr{I}:=\int_0^{2\pi}\int_0^\pi\int_0^\infty\frac{\sin\left(\theta\right)\cdot\sin^2\left(\phi\right)\cdot\exp\left(\sin\left(\theta\right)\cdot\cos\left(\phi\right)\cdot\text{k}\cdot\text{h}\cdot i\right)}{\text{a}\cdot\cos^2\left(\theta\right)+\text{b}\cdot\sin^2\left(\theta\right)}\space\text{d}\text{k}\space\text{d}\theta\space\text{d}\phi=$$
$$\int_0^{2\pi}\int_0^\pi\frac{\sin\left(\theta\right)\cdot\sin^2\left(\phi\right)}{\text{a}\cdot\cos^2\left(\theta\right)+\text{b}\cdot\sin^2\left(\theta\right)}\int_0^\infty\exp\left(\sin\left(\theta\right)\cdot\cos\left(\phi\right)\cdot\text{k}\cdot\text{h}\cdot i\right)\space\text{d}\text{k}\space\text{d}\theta\space\text{d}\phi\tag2$$
Now, we can use:
$$\int_0^\infty\exp\left(\sin\left(\theta\right)\cdot\cos\left(\phi\right)\cdot\text{k}\cdot\text{h}\cdot i\right)\space\text{d}\text{k}=\frac{i}{\sin\left(\theta\right)\cdot\cos\left(\phi\right)\cdot\text{h}}\tag3$$
when $\Im\left(\sin\left(\theta\right)\cdot\cos\left(\phi\right)\cdot\text{h}\right)>0$
So, we get:
$$\mathscr{I}=\int_0^{2\pi}\int_0^\pi\frac{\sin\left(\theta\right)\cdot\sin^2\left(\phi\right)}{\text{a}\cdot\cos^2\left(\theta\right)+\text{b}\cdot\sin^2\left(\theta\right)}\cdot\frac{i}{\sin\left(\theta\right)\cdot\cos\left(\phi\right)\cdot\text{h}}\space\text{d}\theta\space\text{d}\phi=$$
$$\int_0^{2\pi}\frac{\sin^2\left(\phi\right)\cdot i}{\cos\left(\phi\right)\cdot\text{h}}\int_0^\pi\frac{\sin\left(\theta\right)}{\text{a}\cdot\cos^2\left(\theta\right)+\text{b}\cdot\sin^2\left(\theta\right)}\cdot\frac{1}{\sin\left(\theta\right)}\space\text{d}\theta\space\text{d}\phi=$$
$$\int_0^{2\pi}\frac{\sin^2\left(\phi\right)\cdot i}{\cos\left(\phi\right)\cdot\text{h}}\int_0^\pi\frac{1}{\text{a}\cdot\cos^2\left(\theta\right)+\text{b}\cdot\sin^2\left(\theta\right)}\space\text{d}\theta\space\text{d}\phi\tag4$$
Now, use:
$$\int_0^\pi\frac{1}{\text{a}\cdot\cos^2\left(\theta\right)+\text{b}\cdot\sin^2\left(\theta\right)}\space\text{d}\theta=\frac{\pi}{\sqrt{\text{a}\cdot\text{b}}}\tag5$$
So, we get:
$$\mathscr{I}=\int_0^{2\pi}\frac{\sin^2\left(\phi\right)\cdot i}{\cos\left(\phi\right)\cdot\text{h}}\cdot\frac{\pi}{\sqrt{\text{a}\cdot\text{b}}}\space\text{d}\phi=\frac{\pi\cdot i}{\text{h}\cdot\sqrt{\text{a}\cdot\text{b}}}\int_0^{2\pi}\frac{\sin^2\left(\phi\right)}{\cos\left(\phi\right)}\space\text{d}\phi\tag6$$
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|
find a,b,c where $\sqrt{5^{1/3}-4^{1/3}}=\frac{1}{3}(a^{1/3}+b^{1/3}-c^{1/3})$ Is it ramanujan problems? $$\sqrt{5^{1/3}-4^{1/3}}=\frac{1}{3}(a^{1/3}+b^{1/3}-c^{1/3})$$
find $a,b,c$
Any helps would be appreciated.
|
Note that
\begin{eqnarray*}
\left( \sqrt[3]{\alpha}+\sqrt[3]{\alpha^2 \beta }-\sqrt[3]{\beta^2}\right) ^2 &=& \color{blue}{\sqrt[3]{\alpha^2}}+\color{red}{\alpha\sqrt[3]{\alpha \beta^2}} +\beta\sqrt[3]{\beta}+2 \alpha\sqrt[3]{\beta} \color{red}{-2 \sqrt[3]{\alpha\beta^2}} \color{blue}{-2 \beta\sqrt[3]{\alpha^2}} \\
&=& \color{blue}{(1-2\beta)\sqrt[3]{\alpha^2}}+\color{red}{(\alpha-2)\sqrt[3]{\alpha \beta^2}} +(2 \alpha+\beta)\sqrt[3]{\beta}
\end{eqnarray*}
Now by inspection a solution can be obtained by choosing $\alpha=2$ and $\beta=5$, giving $a=2,b=20,c=25$.
|
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|
Does this integral have a closed-form answer? Does this integral have a closed-form solution?
$$\int\limits_{0}^{\infty}\dfrac{x^4\left(2x^{22}+\left(11\sqrt{5}+11\right)x^{17}-20\right)}{2\left(x^{22}+1\right)\sqrt{4\left(\sqrt{x^{22}+1}+x^{10}+(\sqrt{5}+1)x^{5}\right )+6+2\sqrt{5}}}dx$$
EDIT
Sorry for the lack of context, I was late for a compromise. I saw this integral in my university, in the notes of a professor. I asked him what it was, and as response I got "integrals that computers do not solve."
I plotted the integrand chart on the desmos site. The result was this:
Clearly, the integral is convergent and probably has a value assigned to it.
|
After much effort, I asked my teacher to solve this. Here's his solution:
You need to notice that $$4(\sqrt{x^{22}+1}+x^{10}+(\sqrt{5}+1)x^{5})+6+2\sqrt{5}=4\left (\sqrt{x^{22}+1}+x^{10}+(\sqrt{5}+1)x^{5}+\frac{6+2\sqrt{5}}{4} \right )=4\left (\sqrt{x^{22}+1}+(x^{5}+\varphi)^{2} \right )$$
where $\varphi=\frac{\sqrt{5}+1}{2}$ is the golden ratio. Now the integrand assumes the following form:
$$\int\limits_{0}^{\infty}\dfrac{x^4\left(2x^{22}+\left(11\sqrt{5}+11\right)x^{17}-20\right)}{2\left(x^{22}+1\right)\sqrt{4\left(\sqrt{x^{22}+1}+(x^{5}+\varphi)^{2}\right )}}dx=\int\limits_{0}^{\infty}\dfrac{x^4\left(2x^{22}+\left(11\sqrt{5}+11\right)x^{17}-20\right)}{4\left(x^{22}+1\right)^{\frac{5}{4}}\sqrt{1+\left(\frac{x^{5}+\varphi}{\sqrt[4]{x^{22}+1}} \right)^{2}}}dx$$
Next, subs. $u=\frac{x^{5}+\varphi}{\sqrt[4]{x^{22}+1}}$, therefore
$$du=x^{4}\left[5(x^{22}+1)^{-\frac{1}{4}}-\frac{11}{2}x^{17}(x^{5}+\varphi)(x^{22}+1)^{-\frac{5}{4}} \right]dx$$
After some algebra, the integral will take the following - very simple - form:
$$\int_{0}^{\varphi}\frac{du}{\sqrt{1+u^{2}}}=\boxed{\textrm{arcsinh}(\varphi)}$$
|
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If $a,b,c$ are distinct positive numbers , show that $\frac{a^8 + b^8 + c^8}{a^3 b^3 c^3} > \frac{1}{a} + \frac{1}{b} + \frac{1}{c}$ If $a,b,c$ are distinct positive numbers , show that $$\frac{a^8 + b^8 + c^8}{a^3 b^3 c^3} > \frac{1}{a} + \frac{1}{b} + \frac{1}{c}.$$ I am thinking of Tchebycheff's inequality for this question, but not able to proceed. How do I solve this?
|
The inequality is equivalent to
$$a^8 + b^8 + c^8 > a^2b^3c^3 + a^3b^2c^3+ a^3b^3c^2.$$
Now apply Muirhead's inequality (i.e. $[8,0,0]\geq [3,3,2]$) and note that the equality holds iff $a=b=c$.
|
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Find the maximum and minimum value of $S= \log_{2}a+\log_{2}b+\log_{2}c$. Let $a, b, c \geq 1$ and $a+b+c=4$. Find the maximum and minimum value of $S= \log_{2}a+\log_{2}b+\log_{2}c$.
I found the maximum, it is easy to prove $S_\max = 3\log_{2}\frac{4}{3}$. I think the minimum is $1$ when there are two number are $1$ and the remain is $2$. But I do not know, how to prove it.
|
Let $f(x)=\log_2{x}$ and $a\geq b\geq c$. Hence, $f$ is a concave function.
Also we have: $a\leq2$ and $a+b=4-c\leq3$.
Thus, $(2,1,1)\succ(a,b,c)$ and by Karamata
$$\sum_{cyc}\log_2a\geq\log_22+\log_21+\log_21=1.$$
The equality occurs for $a=2$ and $b=c=1$, which says that $1$ is a minimal value.
In another hand, by AM-GM
$$\sum_{cyc}\log_2a=\log_2abc\leq\log_2\left(\frac{a+b+c}{3}\right)^3=3\log_2\frac{4}{3}.$$
The equality occurs for $a=b=c=\frac{4}{3}$, which says that $\log_2\frac{4}{3}$ is a maximal value.
Done!
|
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Solving equations $a+b+c=5,a^2+b^2+c^2=11,a^3+b^3+c^3=27$
Let $a, b, c $ be real numbers such that $a<b<c$ and satisfying
$$a+b+c=5;$$
$$a^2+b^2+c^2=11;$$
$$a^3+b^3+c^3=27.$$
Prove that $0<a<1<b<2<c<3$.
My understanding :
$(a+b+c)^2 - (a^2+b^2+c^2) = 14$
so, $ab+bc+ca = 7$
$a^2+b^2+c^2-ab-bc-ca = 4$
$(a+b+c)(a^2+b^2+c^2-ab-bc-ca)+3abc=20$
so, $abc = -\frac{7}{3}$
Thus $a, b, c$ are roots of the equation $P(x), x^3-5x^2+7x-\frac{7}{3}=0$.
We have,
$P(0) = -\frac{7}{3} <0$
$P(1) = 1-5+7-\frac{7}{3} >0$
$P(2) = 8-20+14-\frac{7}{3} <0$
$P(3) = 27-45+21-\frac{7}{3} >0$.
by Intermediate value theorem, $P(x)$ has a root in each of the interval $(0,1), (1, 2), (2,3)$.
Since $a<b<c$, we get $0<a<1<b<2<c<3$.
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This is a classical exercise about symmetric functions. By Newton's identities
$$ e_1 = p_1 = 5, $$
$$ 2e_2 = p_1^2 - p_2 = 14, $$
$$ 3e_3 = e_2 p_1 - e_1 p_2 + p_3 = 7 $$
hence $a,b,c$ are roots of the polynomial
$$ p(z)=(z-a)(z-b)(z-c) = z^3-e_1 z^2+e_2 z-e_3 = z^3-5z^2+7z-\frac{7}{3} $$
and the claim follows from the fact that $p(0)<0, p(1)>0, p(2)<0, p(3)>0$.
Numerically the roots lie at
$ \{0.485131\ldots,1.72235\ldots,2.79252\ldots\} $.
To perform the above manipulations in a single step, one may exploit the $\text{EXP}/\text{LOG}$ map explained here and compute the Taylor series at the origin, up to the cubic term, of
$$ \exp\left(-5z+\frac{11}{2}z^2-9z^3\right)=1+5z+7z^2+\frac{7}{3}z^3+\ldots$$
to recover the coefficients of $p(z)$.
|
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$\sqrt{9x^2-16}>3x+1$ I'm trying to solve the following inequality:
$$\sqrt{9x^2-16}>3x+1$$
Here's my attempt:
$\sqrt{9x^2-16}>3x+1$
$\longrightarrow 9x^2-16>9x^2+6x+1$
$\longrightarrow -16>6x+1$
$\longrightarrow x<-\frac{17}{6}$
Now, I need to check the constraints:
$9x^2-16 > 0$
$\longrightarrow (3x)^2 > 4^2$
$\longrightarrow \pm3x > 4$
$\longrightarrow 3x > 4$; $-3x > 4$
$\longrightarrow x > \frac{4}{3}$; $x < -\frac{4}{3}$
Making sure the answer meets the constraints:
$\{(-\infty, -\frac{4}{3})\cup(\frac{4}{3}, \infty)\}\cap (-\infty, -\frac{17}{6}) = (-\infty, -\frac{17}{6})$
So, my answer is $x=(-\infty, -\frac{17}{6})$, however verifying on Wolfram|Alpha results in $x=(-\infty, -\frac{4}{3}]$.
Where, what, and why is wrong with my solution?
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at first it must be $|x|\geq \frac{4}{3}$ if $$x\le -\frac{1}{3}$$ our inequality is true. let $$x>-\frac{1}{3}$$ we can square it:
$$9x^2-16>9x^2+6x+1$$ otr $$x<-\frac{17}{6}$$ together with the conditions $$-\infty<x<-\frac{17}{6}$$ and $$-\frac{1}{3}<x$$ or $$-\frac{1}{3}\geq x$$ and $$x\geq \frac{4}{3}$$ or $$x\le -\frac{4}{3}$$ we get $$-\infty<x\le- \frac{4}{3}$$ as the searched solution set
|
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Find the equation of straight lines through the point $(\dfrac {1}{\sqrt {3}}, 1)$ whose perpendicular distance from the origin is unity. Find the equation of straight lines through the point $\left(\dfrac {1}{\sqrt {3}}, 1\right)$ whose perpendicular distance from the origin is unity.
My Attempt:
Let the equation of line be $ax+by+c=0$.
Its distance from origin is $1$ unit. So,
$$\left|\dfrac {a.0+b.0+c}{\sqrt {a^2+b^2}}\right|=1$$
$$\left|\dfrac {c}{\sqrt {a^2+b^2}}\right|=1$$
What do I solve further?
|
Using the normal form of straight line, let the equation of the line be
$$x \cos \alpha + y \sin \alpha = 1$$
Now since the line passes through $\left ( \frac{1}{\sqrt{3}}, 1 \right )$, we get
$$\frac{\cos \alpha}{\sqrt{3}} + {\sin \alpha} = 1 \implies \alpha = \frac{\pi}{6}
\ \ or \ \ \frac{\pi}{2} $$
Hence we get the equation of line as $\frac{\sqrt{3}x}{2} + \frac{y}{2} = 1$ or $y = 1$.
|
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|
Show that $\frac {1}{\sqrt{5}}[(\frac {1}{x+r_+}) - (\frac {1}{x+r_-}) = \frac {1}{\sqrt{5}x}[(\frac {1}{1-r_{+}x}) - (\frac {1}{1-r_{-}x})] $ I need to manipulate this equation:
$$
\frac {1}{\sqrt{5}}\left(\frac {1}{x+r_+} - \frac {1}{x+r_-}\right)
$$
to show that
$$ \frac {1}{\sqrt{5}}\left(\frac {1}{x+r_+} - \frac {1}{x+r_-}\right) = \frac {1}{\sqrt{5}x}\left(\frac {1}{1-r_{+}x} -\frac {1}{1-r_{-}x}\right)
$$ (*)
where$$r_+=\frac {1+\sqrt{5}}{2} $$and $$r_-=\frac {1-\sqrt{5}}{2}$$
I know that both RHS and LHS of (*) are equal to $\frac{1}{1-x-x^2} $ , but don't know how we get RHS manipulating LHS before someone told us that RHS is also equal to $\frac{1}{1-x-x^2} $
Edit:
Since $$r_+r_-=-1$$, I replace $r_+$ with $r_-$ and $r_-$ with $ r_+$ in the LHS of (*)
So that, I get
$$ \frac {1}{\sqrt{5}}\left(\frac {r_+}{1-r_+x} - \frac {r_-}{1-r_-x}\right)$$ Still, I am not there. Can anybody help me at this step?
After showing the equality above I can continue as
$$
\left(\frac {1}{1-r_{+}x}\right) - \left(\frac {1}{1-r_{-}x}\right)= {\sum_{n\ge0}\ r_+^nx^n}-{\sum_{n\ge0}\ r_-^nx^n}
$$
|
From your last step:
$$\frac{1}{\sqrt{5}}\left(\frac{r_{+}}{1 - r_{+}x} - \frac{r_{-}}{1 - r_{-}x}\right)$$
$$= \frac{1}{\sqrt{5}x}\left(\frac{r_{+}x}{1 - r_{+}x} - \frac{r_{-}x}{1 - r_{-}x}\right)$$
$$= \frac{1}{\sqrt{5}x}\left(\left[\frac{1}{1 - r_{+}x} - 1\right] - \left[\frac{1}{1 - r_{-}x} - 1\right]\right)$$
$$= \frac{1}{\sqrt{5}x}\left(\frac{1}{1 - r_{+}x} - \frac{1}{1 - r_{-}x}\right)$$
|
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|
Compute $\int\limits_{-\infty}^{\infty}\frac{1}{(1+x^2)^{n+1}}dx$ via residue calculus. Let $\Gamma_R$ be the semicircle of radius $R$ in the upper half plane. Then,
\begin{align}
\int\limits_{-\infty}^{\infty}\frac{1}{(1+x^2)^{n+1}}dx
&= \lim_{R\to \infty}\int_{\Gamma_R}\frac{1}{(1+z^2)^{n+1}}dz \\
&= 2\pi i \operatorname{Res}\left(\frac{1}{(1+z^2)^{n+1}},i\right)
\end{align}
The pole of the function at $i$ is of order $n+1$, so the residue is computed by
\begin{align}
\operatorname{Res}\left(\frac{1}{(1+z^2)^{n+1}},i\right) &= \frac{1}{n!}\lim_{z\to i}\frac{d^n}{dz^n}\left(\frac{1}{(z+i)^{n+1}}\right) \\
&= \frac{1}{n!}\lim_{z\to i}(-1)^n\frac{(n+1)(n+2)\cdots(2n+1)}{(z+i)^{2n+1}} \\
&=\frac{(2n+1)!}{i2^{2n+1}(n!)^2}
\end{align}
Hence, $$\int_{-\infty}^{\infty}\frac{1}{(1+x^2)^{n+1}}dx = \pi\frac{(2n+1)!}{2^{2n}(n!)^2}$$ The answer provided is $\frac{1\cdot 3\cdot 5\cdots(2n-1)}{2\cdot 4\cdot 6\cdots (2n)}\pi$. How do I manipulate my answer to obtain this answer?
|
Your answer is actually
$$\pi\frac{(2n)!}{2^{2n}(n!)^2}.$$
(Check the derivative.)
Then use
$$ (2n)!=(1 \cdot 3 \cdots (2n-1))\cdot 2^n \cdot (n!)$$ to get
$$\pi\frac{(2n)!}{2^{2n}(n!)^2}=\pi\frac{1 \cdot 3 \cdots (2n-1)}{2^n \cdot (n!)}=\pi\frac{1 \cdot 3 \cdots (2n-1)}{2 \cdot 4 \cdots (2n)}.$$
|
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|
How to prove that $2^0 + 2^1 + 2^2 + 2^3 + \cdots + 2^n = 2^{n+1} - 1$ When I'm reading Computer Systems: A Programmer's Perspective, I met the sum of binary numbers and failed to prove it:
$$ 2^0 + 2^1 + 2^2 + 2^3 + \cdots + 2^n = 2^{n+1} - 1 $$
This might be preliminary knowledge, I'm not good at mathematics, any body could give me a hint?
|
Exploiting the binary representation,
$$111\cdots111_2+1_2=1000\cdots000_2$$ because the carry propagates.
This is exactly
$$2^0+2^1+2^2+\cdots2^{n-2}+2^{n-1}+2^n+1=2^{n+1}.$$
The proof generalizes to other bases. Let $a:=b-1$, then
$$a\cdot111\cdots111_b+1_b= aaa\cdots aaa_b+1_b=1000\cdots000_b,$$ which is a rewrite of
$$(b-1)(b^0+b^1+b^2+\cdots b^{n-2}+b^{n-1}+b^n)+1=b^{n+1}$$ or
$$b^0+b^1+b^2+\cdots b^{n-2}+b^{n-1}+b^n=\frac{b^{n+1}-1}{b-1}.$$
|
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|
Radical problem: $\sqrt{7+2\sqrt{12}}+\sqrt{7-2\sqrt{12}}$ What is the value of
$$\sqrt{7+2\sqrt{12}}+\sqrt{7-2\sqrt{12}}$$
|
With $x=\sqrt{7+2\sqrt{12}}$ and $y=\sqrt{7-2\sqrt{12}}$, we have $x^2+y^2=14$ and $xy=\sqrt{7^2-2^2\cdot12}=1$, so $(x+y)^2=14+2\cdot1=16$ and thus $x+y=4,$ since $x,y$ are positive.
In general, we'd have $$\sqrt{a+b\sqrt{c}}+\sqrt{a-b\sqrt{c}}=\sqrt{2a+2\sqrt{a^2-b^2c}}.$$
|
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|
Why does $a^2x^2+(a^2+b^2-c^2)xy+b^2y^2>0,$ imply $(a^2+b^2-c^2)^2-4a^2b^2<0$? Why does $a^2x^2+(a^2+b^2-c^2)xy+b^2y^2>0,$ imply $(a^2+b^2-c^2)^2-4a^2b^2<0$, if $x, y$ are reals greater than $1$, and $a, b, c$ are positive reals?
A proof with all the math to go from one to the other would be nice.
|
since you added a clause about $x,y > 1,$ some two minutes after first posting the question, the conclusion you require is false. In fact, take any $a,b$ (not both zero) at all and then take $c = 0,$ then for $x,y > 1,$
$$ a^2 x^2 + (a^2 + b^2) xy + b^2 y^2 > 0 $$
however
$$ (a^2 + b^2)^2 - 4 a^2 b^2 = (a^2 - b^2)^2 \geq 0 $$
EXAMPLE WITH NONZERO $c...$ take $a= 20, b = 10, c = 1.$ We find
$$ \color{red}{ 400 x^2 + 499 xy + 100 y^2} $$
is positive whenever both $x,y$ are bigger than one. HOWEVER, the discriminant is
$$ 499^2 - 4 \cdot 400 \cdot 100 = 499^2 - 400^2 = 99 \cdot 899 = 89001 > 0. $$
Furthermore, when we take $$ x = 1, y = -2, $$ we get
$$ 400 - 2 \cdot 499 + 4 \cdot 100 = 400 + 400 - 998 = 800 - 998 = -198. $$
The quadratic form $\langle 400, 499, 100 \rangle$ is positive when both arguments are positive, but not always.
|
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|
How to prove $\ln{\frac{n+1}{n}}\le\frac{2}{n+1},\forall n\in\mathbb{N}^+$? I found an inequality: $$\ln\left(\frac{n+1}{n}\right)\le\frac{2}{n+1},\forall n\in\mathbb{N}^+.$$
I tried induction. It is obvious if $k=1$, when $n=k$, $\ln\sqrt{\left(\frac{k+1}{k}\right)^{k+1}}\le 1$, but bogged down for $n=k+1$:
$$\ln\sqrt{\left(\frac{k+2}{k+1}\right)^{k+2}}\le 1$$
|
Integrate $\frac{1}{x}$ between $n$ and $n+1$: $\ln \frac{n+1}{n} = \ln (n+1) - \ln (n) = \displaystyle\int_n^{n+1} \frac{1}{x} \operatorname{d}x$. As $x\to \frac{1}{x}$ is decreasing, you get $\ln \frac{n+1}{n}\leq \displaystyle\int_n^{n+1} \frac{1}{n} \operatorname{d}x = \frac{1}{n}$.
It is moreover clear that $\frac{1}{n} \leq \frac{2}{n+1}$ for $n\geq 1$, so you get $\ln \frac{n+1}{n} \leq \frac{2}{n+1}$
|
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|
Quadratic equations having a common root. I have a cubic polynomial $f(x)=x^3+px^2+qx+72$ which is divisible by both $x^2+ax+b$ and $x^2+bx+a$ (where a,b,p,q are constants and a$ \neq $b).I have to find the sum of the squares of the the roots of the cubic polynomial.
I tried to attempt it like this.
Since the quadratics divide the polynomial its roots must be the same as the roots of the quadratics.
Let the roots of the first quadratic be A and B and that of second quadratic be D and E.
Now for having 3 roots of the cubic polynomial one root of the 2 quadratics must be common.Then suppose B=D.
Now from the first quadratic $A+B=-a$ and $AB=b$.From second quadratic $B+E=-b$ and $BE=a$.
We need the sum of squares of the roots of the cubic polynomial i.e. $A^2+B^2+E^2$.We have $A^2+B^2+E^2=(A+B+E)^2-2(AB+BE+AE)$.We have $A+B+E=-p$ and $AB+BE+AE=q$.So $A^2+B^2+E^2=-p-2q$.
I don't know how to proceed.
|
The roots are 1, 8, 9. From here you can simply sum the cubes.
Now for the solution:
As you noted your two quadratics have a common root. Let this root be called $y$. Then $y^2 + ay + b = 0$ and $y^2 + by + a = 0$. This is a system of two equations. Solving we get $(a-b)y = (a-b)$. Since $a \neq b$ we conclude $y=1$.
Now since $1$ is a root of $x^2 + ax +b$ we conclude $1+a+b = 0$ or $a+b = -1$.
Furthermore since the quadratics divide the cubic we have $(x-1)(x^3+px^2+qx+72) = (x^2+ax+b)(x^2+bx+a)$. This gives us that $ab = -72$. But now we have two equations for $a,b$. We solve and get that $a = 8$ and $b = -9$ or $a = -9$ and $b = 8$.
We will use $a = 8$ and $b = -9$. From here it is a simple matter to factor the two quadratics and find the roots.
|
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|
Finding solutions for expression to be a perfect square Determine all integers such that $$ n^4- n^2 +64$$ is the square of an integer.
The first two lines of the solution given in the textbook is as below:
Since $n^4-n^2+64>(n^2-1)^2. $ For some non negative integer $k$,
$n^4-n^2+64=(n^2+k)^2$.
I fail to understand what the author tries to say here. Can't this problem be done in another manner?
|
Following the author's solution, since $n^4-n^2+64>(n^2-1)^2$ any $n^4-n^2+64$ which equals a square can be written as $(n^2+k)^2$ for $k$ a non-negative integer (i.e. $k>-1$ in the $(n^2-1)^2$). That means $$n^4-n^2+64 = n^4+2kn^2+k^2$$
and $$n^2=\frac{64-k^2}{(2k+1)}.$$ We need to check for $0\leq k\leq 8$ (since $n^2>0$) to see which values are squares, and find only $k=0, 7$ and $8$ work giving us the solutions $n=0, n=\pm1$, and $n=\pm8$.
|
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|
Arithmetic-geometric mean, prove that $c_n = 4me^{-\ell 2^n+\epsilon_n}$ Let $a$ and $b$ reals with $a > b > 0$. Let $(a_n)$ and $(b_n)$ with $a_0 = a$, $b_0 = b$ and
$$a_{n+1} = \tfrac{a_n+b_n}{2} \quad\text{;}\quad b_{n+1} = \sqrt{a_nb_n}$$
We know that $\lim_{n \rightarrow +\infty} a_n = \lim_{n \rightarrow +\infty} b_n = m$.
Let $c_n = \sqrt{a_n^2-b_n^2}$.
We know that $\lim_{n \rightarrow +\infty} c_n =0$ and that $c_{n+1} \leq \frac{c_n^2}{4m}$.
The question is to prove that there is $\ell > 0$ and $\epsilon_n > 0$ with $\lim_{n \rightarrow +\infty} \epsilon_n = 0$ such as
$$c_n = 4me^{-\ell 2^n+\epsilon_n}$$
An indication is to use $u_n = -2^{-n} \ln(c_n)$ and $\sum (u_{n+1}-u_n)$.
|
Using $c_{n+1} = \frac{c_n^2}{2(a_n+b_n)} = \frac{c_n^2}{4a_{n+1}}$, we get.
\begin{align*}
u_{n+1} & = -2^{-n-1} \ln(c_{n+1}) \\
& = -2^{-n-1} (\ln(c_n)-\ln(4a_{n+1})) \\
& = u_n+2^{-n-1} \ln(4a_{n+1})
\end{align*}
$\ln(4a_{n+1}) = \ln(4m)+\underbrace{\ln(\frac{4a_{n+1}}{4m})}_{\alpha_n}$. The sequence $(\alpha_n)$ is positive and decreasing to zero.
$$ u_n-u_0 = \sum_{k=0}^{n-1} \left( u_{k+1}-u_k \right) = \sum_{k=0}^{n-1} 2^{-k-1} \left( \ln(4m)+\alpha_k \right) $$
Since $\sum_{k=0}^{n-1} 2^{-k-1} = 1-2^{-n}$, we get.
$$ u_n = u_0+\ln(4m)-2^{-n} \ln(4m)+\sum_{k=0}^{n-1} 2^{-k-1} \alpha_k $$
Since $\sum_{k=0}^{n-1} 2^{-k-1} \alpha_k \leqslant \sum_{k=0}^{n-1} 2^{-k-1} \alpha_0 \leqslant \alpha_0 \left( 1-2^{-n} \right)$, we get.
$$ \sum_{k=0}^{+ \infty} 2^{-k-1} \alpha_k = S \quad\text{with}\quad 0 < S \leqslant \alpha_0 $$
Now, the rest.
$$ R_n = \sum_{k=n}^{+\infty} 2^{-k-1} \alpha_k \leqslant \alpha_n \sum_{k=n}^{+\infty} 2^{-k-1} \leqslant 2^{-n}\alpha_n $$
Finally.
\begin{align*}
u_n & = u_0+\ln(4m)-2^{-n} \ln(4m)+S-R_n \\
& = u_0+\ln(4m)+S-2^{-n} \left( \ln(4m)+2^nR_n \right)
\end{align*}
Let $\ell = u_0+\ln(4m)+S$ and $\varepsilon_n = 2^nR_n$.
The last problem is to prove that $\ell > 0$.
As $u_n = -2^{-n} \ln(c_n)$ and $\lim_{n \rightarrow +\infty} c_n = 0$, $u_n > 0$ after some rank and so $\lim_{n \rightarrow +\infty} u_n = \ell \geq 0$. If $\ell = 0$, $c_n = 4me^{\varepsilon_n}$ and $\lim_{n \rightarrow +\infty} c_n = 4m \ne 0$. So $\ell \ne 0$.
|
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|
Representing polynomials as quadratic forms I have the following cubic equation
$$x^3-6x^2+11x-6=(x-1)(x-2)(x-3)=0$$
whose solutions are $x=1,2,3$.
What if the above equation were represented by quadratic form? Let $\mathbf x = \begin{bmatrix} x^2 & x & 1\end{bmatrix}^T$. Can the cubic equation above be represented as follows?
$$x^3-6x^2+11x-6=\begin{bmatrix}x^2&x&1\end{bmatrix}\begin{bmatrix}0&b&c\\d&e&f\\g&h&-6\end{bmatrix}\begin{bmatrix}x^2\\x\\1\end{bmatrix} = \textbf{x}^T \textbf{A} \textbf{x}$$
There are many possibilities to form matrix $A$. For instance the two matrices below
$$\textbf{A}_1 = \begin{bmatrix}0&0.5&0\\0.5&-6&5.5\\0&5.5&-6\end{bmatrix}\quad\quad\quad\textbf{A}_2 = \begin{bmatrix}0&0&0\\1&-3&6\\-3&5&-6\end{bmatrix}$$
Is there any study about finding the homogeneous solution, $\textbf{x}^T \textbf{A} \textbf{x}=0$? Thank you in advance.
|
The simplest solution is to use the 3 homogeneous coordinates $\pmatrix{ x^2 & x & 1}$ and a 3×3 symmetric matrix
$$A x^3 + B x^2 + C x + D = \pmatrix{x^2 \\x \\1}^\top
\begin{bmatrix}
0 & \frac{A}{2} & 0 \\
\frac{A}{2} & B & \frac{C}{2} \\
0 & \frac{C}{2} & D
\end{bmatrix} \pmatrix{x^2 \\x \\1} $$
But the formal solution is actually to use 4 homogeneous coordinates $\pmatrix{ x^3 & x^2 & x & 1}$ and a 4×4 symmetrix matrix
$$A x^3 + B x^2 + C x + D = \pmatrix{x^3 \\ x^2 \\x \\1}^\top \begin{bmatrix}
0 & 0 & 0 & 0\\
0& 0 & \frac{A}{2} & 0 \\
0 & \frac{A}{2} & B & \frac{C}{2} \\
0 & 0 & \frac{C}{2} & D
\end{bmatrix} \pmatrix{x^3 \\ x^2 \\x \\1} $$
The reason is that you have to treat $x^3$ as a separate variable from $x^2$, $x$ and $1$. The coefficients are found by matching and taking successive derivatives of both sides of the expression.
Additional constraints I have imposed (besides the matrix being symmetric) is to make it as diagonal as possible, by making the most off-diagonal terms zero.
|
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An equation for prime numbers $\frac{p-1}{2}(2^p-1)+1=7k^2$
If $2^p-1$ is a Mersenne prime, and $k$ is an integer, then solve
$$\frac{p-1}{2}(2^p-1)+1=7k^2$$
$$$$
If I take modulo $p$ I get $1 \equiv 14k^2 \pmod{p}$.
If I take modulo $q=2^p-1$ I get $1 \equiv 7k^2 \pmod{q}$.
Can I use these to solve the equation? Please help me.
|
This is only a partial answer, but I hope it may help a little. Your evaluations modulo $p$ and $q$ are of course correct, but the argument can be carried a little further. You already know that the case $p=3$ is impossible, so I will assume in what follows that $p > 3$.
It is easily verified that $2^p - 1 \equiv 1 \pmod{6p}$. Thus from the original equation one has $\frac{p-1}{2} + 1 \equiv 7k^2 \pmod{6p}$, or multiplying throughout by $2$,
\begin{equation}
p + 1 \equiv 14k^2 \pmod{12p}.
\end{equation}
Furthermore, the original equation may be written
\begin{equation}
\frac{p - 1}{2} \cdot q + 1 = 7k^2,
\end{equation}
which when $\frac{p-1}{2}$ is even reduces to
\begin{equation}
1 \equiv 7k^2 \pmod{2q},
\end{equation}
making $k$ odd, while when $\frac{p-1}{2}$ is odd it reduces to
\begin{equation}
\frac{p - 1}{2} + 1 \equiv 7k^2 \pmod{2q},
\end{equation}
or, multiplying throughout by 2,
\begin{equation}
p + 1 \equiv 14k^2 \pmod{4q}.
\end{equation}
Third, we can ask under what conditions the left-hand side of the original equation can be divisible by 7, as required. Simply by brute-force exhaustion of all possible cases, it is found that a necessary and sufficient condition is for $p$ to belong to either of the two residue classes $5, 13 \pmod{42}$. Since there are $\phi(42) = 12$ possible residue classes for primes $\pmod{42}$, this condition eliminates $\frac{5}{6}$ of the primes $p$ from consideration.
Finally, it is easily verified that $2^p - 1 \equiv 31 \pmod{48}$. Thus from the original equation one has $31 \cdot \frac{p-1}{2} + 1 \equiv 7k^2 \pmod{48}$, which being multiplied throughout by $62$ gives
\begin{equation}
p + 61 \equiv 50k^2 \pmod{96}.
\end{equation}
Taking into account all possible values of $k^2$ modulo $96$, we see that $p$ can only belong to one of the eight residue classes $5, 11, 35, 37, 43, 53, 67, 85 \pmod{96}$. Since there are $\phi(96) = 32$ possible residue classes for primes $\pmod{96}$, this condition eliminates $\frac{3}{4}$ of the primes $p$ from consideration.
EDIT: Taking into account the observation by Erick Wong, the congruence $1 \equiv 7k^2 \pmod{q}$ in the original posting proves that 7 is a quadratic residue of $q$. Because $p \equiv 1 \pmod{6}$ would imply $q \equiv 15 \pmod{28}$, and 7 is a quadratic non-residue of all such primes $q$, this case is impossible and eliminates all instances above where $p \equiv 1 \pmod{6}$. Thus, $p \equiv 5 \pmod{42}$, and $p \equiv 5, 11, 35, \textrm{or } 53 \pmod{96}$.
Similarly, the congruence $1 \equiv 14k^2 \pmod{p}$ in the original posting proves that 14 is a quadratic residue of $p$. This limits $p$ to the residue classes $1, 5, 9, 11, 13, 25, 31, 43, 45, 47, 51, 55 \pmod{56}$, i.e. to 12 of the $\phi(56) = 24$ possible residue classes.
|
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|
Problems while solving the differential equation. $$x^2\frac{d^2y}{dx^2}+x^2\frac{dy}{dx}-2y=0$$
$x=0$ is a regular singular point.
$$y=\sum_{n=0}^\infty c_nx^{n+r}$$
$$\frac{dy}{dx}=(n+r)\sum_{n=0}^\infty c_nx^{n+r-1}$$
$$\frac{d^2y}{dx^2}=(n+r)(n+r-1)\sum_{n=0}^\infty c_nx^{n+r-2}$$
$$(n+r)(n+r-1)\sum_{n=0}^\infty c_nx^{n+r}+(n+r-1)\sum_{n=1}^\infty c_{n-1}x^{n+r}-2\sum_{n=0}^\infty c_nx^{n+r}=0$$
Taking out a few terms
$$(r)(r-1)c_0x^r+-2c_0x^r+(n+r)(n+r-1)\sum_{n=1}^\infty c_nx^{n+r}+ (n+r-1)\sum_{n=1}^\infty c_{n-1}x^{n+r}-2\sum_{n=1}^\infty c_nx^{n+r}=0$$
The incidal equation is
$$r^2-r-2=0$$
The recurrence formula is,
$$(n+r)(n+r-1)c_n+(n+r-1)c_{n-1}-2 c_n=0$$
Bigger roots
Let $r=r_1=2$
$$(n+2)(n+1)c_n+(n+1)c_{n-1}-2c_n=0$$
$$c_n=\frac{-(n+1)c_{n-1}}{n^2+3n}$$
$$c_1=\frac{-c_0}{2},$$
$$c_2=\frac{-c_{1}}{6}$$
Taking the smaller root, $r=r_2=-1$
$$(n-1)(n-2)c_n+(n-2)c_{n-1}-2c_n=0$$
$$c_n=\frac{(2-n)c_{n-1}}{n^2-3n}$$
How shall I continue this any further? Any help would be appreciated. Can someone hint me on this question.
|
In your equations they are some terms with $n$ in them.
These terms should be in the $\displaystyle{\sum_n}$ , not outside.
Nevertheless, your result is correct which is well. But one can go further.
Case $r=2$ :
$$c_n=\frac{-(n+1)}{n^2+3n}c_{n-1}$$
$$c_n=c_0\prod_{k=1}^n\left(\frac{-(k+1)}{k^2+3k}\right)=
(-1)^n c_0\prod_{k=1}^n\left(\frac{k+1}{k(k+3)}\right)$$
Develop and simplify :
$$c_n=\frac{6(-1)^n(n+1)}{(n+3)!}c_0$$
$$y=\sum_{n=0}^\infty c_nx^{n+2}=6c_0\sum_{n=0}^\infty \frac{(-1)^n(n+1)}{(n+3)!}x^{n+2} = -6c_0\sum_{n=3}^\infty \frac{(-1)^n(n-2)}{n!}x^{n-1}$$
$$y= -6c_0\sum_{n=3}^\infty \frac{(-1)^n}{(n-1)!}x^{n-1}
+12c_0x^{-1}\sum_{n=3}^\infty \frac{(-1)^n}{n!}x^{n}$$
$$y= 6c_0\sum_{n=2}^\infty \frac{(-1)^n}{n!}x^{n}
+12c_0x^{-1}\sum_{n=3}^\infty \frac{(-1)^n}{n!}x^{n}$$
$$y= 6c_0\left(e^{-x}-1+x \right)+12c_0x^{-1}\left(e^{-x}-1+x-\frac{x^2}{2} \right)$$
So, the first family of solutions is obtained :
$$y= 6c_0\left(e^{-x}+\frac{2e^{-x}}{x} -\frac{2}{x}+1\right)$$
Case $r=-1$ :
Proceed on the same manner to obtain the second family of solutions (It is simpler because they are only two terms in the series) :
$$y=c'_0\left(\frac{1}{x}-\frac{1}{2}\right)$$
|
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|
Solving $(1 + i) z - 2 \overline{z} = -11 + 25i$ How do we find the complex number $z$ that satisfies
$$(1 + i) z - 2 \overline{z} = -11 + 25i.$$
How should I start? Am I supposed to express $z$ as $x+yi$?
|
Since both, left hand side and right hand side of
\begin{align*}
(1 + i) z - 2 \overline{z} = -11 + 25i\tag{1}
\end{align*}
represent a complex number, we can equate real and imaginary part of them. We set $z=x+iy$ with real part $x$ and imaginary part $y$.
We obtain
\begin{align*}
\Re\left((1+i)(x+iy)-2(x-iy)\right)&=(x-y)-2x\\
&=-x-y\tag{2}\\
\Im\left((1+i)(x+iy)-2(x-iy)\right)&=(x+y)+2y\\
&=x+3y\tag{3}
\end{align*}
Equating real part (2) and imaginary part (3) with the RHS of (1) we get
\begin{align*}
-x-y&=-11\\
x+3y&=25
\end{align*}
and finally obtain
\begin{align*}
z&=x+iy\color{blue}{=4+7i}
\end{align*}
|
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|
Solve for four values of x where x lies in complex plane. I came across this problem and have to find four roots of this equation. Direct multiplication was of no help and factorizing it is a mess.
I tried to factor it using $a^2+b^2=(a+ib)(a-ib) $
but was of no help.
$$x^2 + (\frac{ax}{x+a})^2= 3a^2$$
Can anyone through in some ideas?
|
Okay I tried to solve this question using two techniques, one is by factorizing and then substitution as below:
$$x^2 +(\frac{ax}{x+a})^2 = 3a^2$$
We can write $ a^2 + b^2$ as $$a^2 + b^2= (a-b)^2 +2ab$$ so $x^2 +(\frac{ax}{x+a})^2$ can be written as $$(x -\frac{ax}{x+a})^2 + 2 * x * \frac{ax}{x+a}$$ $$\implies (\frac{x+ax-ax}{x+a})^2 + 2a*\frac{x}{x+a} = 3a^2 $$ $$\implies (\frac{x}{x+a})^2 + 2a*\frac{x}{x+a} = 3a^2$$
Now substitute $$\frac{x}{x+a} = y$$ $$\implies y^2 + 2a*y= 3a^2 $$$$\implies y^2 + 2a*y-3a^2=0 $$ This becomes $$(y+3a)(y-a)=0 $$
$\implies y=a $ or $y=-3a$
When $y=a$
$$\frac{x^2}{x+a}=a$$ $$\implies x^2-ax-a^2 =0$$ $$\implies x = \frac{a\pm a\sqrt(5)}{2}$$
When $y=-3a$
$$\frac{x^2}{x+a}=-3a$$ $$\implies x^2+3ax+3a^2 =0$$ $$\implies x = \frac{-3a\pm ai\sqrt(3)}{2}$$
|
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|
Let $\ a,b,c$ be integers such that $ \ a^{2} + b^{2} = c^{2}$. If $c$ is divisible by $3$, prove that $a$ and $b$ are both divisible by $3$. Question:
Let $\ a,b,c$ be integers such that $ \ a^{2} + b^{2} = c^{2}$. If $c$ is divisible by $3$, prove that $a$ and $b$ are both divisible by $3$.
My attempt:
Proof by contradiction:
Assume $c$ is divisible by $3$ and $a$ or $b$ is not divisible by $3$.
Since $c$ is divisible by $3$ we can write $c$ as $ \ c = 3m \implies c^{2} = 9m^{2} \implies 9 | c^{2}$.
Since $a$ and $b$ are not divisible by $3$, $\ a = 3k+1$ and $ \ b = 3n+1$ for some integers $\ k,n.$
Then,
$ a^{2} + b^{2} = (3k+1)^{2} + (3n+1)^{2} = 9k^{2} + 6k +9n^{2} + 6n + 2$.
I am stuck here. I can't find a contradiction. How can I show that $ a^{2} + b^{2} $ is not divisible by $9$.
|
$$n^2\equiv(0,1,-2,4)mod9.$$ It's all!
|
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|
Expanding $(x+1)(x+2)(x+3)(x-1)(x-2)(x-3)$ Is there any trick to multiplying this?: $(x+1)(x+2)(x+3)(x-1)(x-2)(x-3)$?
The brackets are to be eliminated and the result is to be simplified as much as possible.
I have started with this: \begin{align}&(x+1)(x+2)(x+3)(x-1)(x-2)(x-3)\\&= (x^3+2x+3x+x+2+x+3+3x+2x+6)(x^3-2x-3x-x+2-x+3-3x-2x+6)\\&=... \end{align}
But the thing is, that this method takes very long and is prone to include calculation errors.
I'd be thankful for any help!
|
$(x+a)(x-a)=x^2-a^2$ so collect these terms to obtain a third degree polynomial in $x^2$. Solve that as normal.
${(x+1)(x+2)(x+3)(x-1)(x-2)(x-3) \\\qquad=~ (x^2-1)(x^2-4)(x^2-9) \\\qquad=~ x^{2\cdot 3}-(\phantom{1+4+9})x^{2\cdot 2}+(\phantom{4\cdot 9+4+9})x^2-1\cdot 4\cdot 9 \\ \qquad\quad\ddots}$
|
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|
What are the real and imaginary parts of $(x+e^{ix})^{0.5}$? Finding the real and imaginary parts of
$(x+e^{ix})^{n}$ is easy if n is an integer greater than 1. But what if n is a fraction? Suppose I have a function like $(x+e^{ix})^{0.5}$. How do you find the real and imaginary parts of this?
Edit: x is a real number...not complex.
|
If $n$ is not an integer, $(x + e^{ix})^{n}$ is multi-valued.
First write $x + e^{ix} = x + \cos x + i\sin x = e^{i\theta} \sqrt{x^{2} + 2x \cos x + 1}$ for some real $\theta$ (unique up to an additive integer multiple of $2\pi$).
If $n$ is real, then
\begin{align*}
(x + e^{ix})^{n}
&= (x^{2} + 2x\cos x + 1)^{n/2} \exp(in\theta) \\
&= \underbrace{(x^{2} + 2x\cos x + 1)^{n/2} \cos(n\theta)}_{\text{real part}} + i\underbrace{(x^{2} + 2x\cos x + 1)^{n/2} \sin(n\theta)}_{\text{imaginary part}}.
\end{align*}
Generally, letting $\ln$ denote the real branch of the natural logarithm,
\begin{align*}
(x + e^{ix})^{n}
&= (x + \cos x + i\sin x)^{n} \\
&= \exp\bigl[n \log(x + \cos x + i\sin x)\bigr] \\
&= \exp\bigl[n \bigl(\tfrac{1}{2}\ln(x^{2} + 2x \cos x + 1) + i\arg(x + \cos x + i\sin x)\bigr)\bigr] \\
&= \exp\bigl[n \bigl(\tfrac{1}{2}\ln(x^{2} + 2x \cos x + 1) + i\theta\bigr)\bigr].
\tag{1}
\end{align*}
To proceed, write $n = a + bi$ with $a$ and $b \neq 0$ real, multiply out the quantity in square brackets, and use $\exp(u + iv) = e^{u} \cos v + ie^{u} \sin v$ (for $u$ and $v$ real) to simplify (1) to your heart's content.
|
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|
$\lim\limits_{n \rightarrow \infty} \sqrt{n^2+n} -n$?
Calculate
$\displaystyle\lim_{n \to \infty}
\left(\,{\sqrt{\,{n^{2} + n}\,} - n}\,\right)$.
$\displaystyle\lim_{n \to \infty}\left(\,{\sqrt{\,{n^{2} + n}\,} - n}\,\right) =
\infty - \infty$ We have an indeterminate form
So I proceeded to factorize $$\sqrt{n^2+n} -n = \sqrt{ \frac{n^2(n+1)}{n}}-n =n \left[ \sqrt{\frac{n+1}{n}}-1 \right]$$
taking the limit:
$$\lim\limits_{n \rightarrow \infty} n \left[ \sqrt{\frac{n+1}{n}}-1 \right]= \infty \cdot 0$$
indeterminate again
What am i missing? How is the way forward to proceed? Much appreciated
|
Using you approach $$\sqrt{n^2+n} -n = \sqrt{ \frac{n^2(n+1)}{n}}-n =n \left[ \sqrt{\frac{n+1}{n}}-1 \right]=n \left[ \sqrt{1+\frac{1}{n}}-1 \right]$$ Now, use the generalized binomial theorem or Taylor series to get, for small values of $x$, $$\sqrt{1+x}=1+\frac{x}{2}-\frac{x^2}{8}+O\left(x^3\right)$$ $$\sqrt{1+x}-1=\frac{x}{2}-\frac{x^2}{8}+O\left(x^3\right)$$ Make now, since $n$ is large, $x=\frac 1n$ $$\sqrt{1+\frac{1}{n}}-1=\frac{1}{2 n}-\frac{1}{8 n^2}+O\left(\frac{1}{n^3}\right)$$ $$n \left[ \sqrt{1+\frac{1}{n}}-1 \right]=\frac{1}{2}-\frac{1}{8 n}+O\left(\frac{1}{n^2}\right)$$ which shows the limit and how it is approached.
Try with $n=10$ (which is very small, so you will have $\sqrt{110}-10\approx 0.488088$ while the above truncated series would give $\frac{39}{80}=0.487500$.
|
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|
How many 5 letter different words can be formed using the letters AAASSSBB I saw that these sort of questions have been asked before, but not exactly this kind of problem: creating 5 DIFFERENT words, not just all possible combinations.
The words don't need to have any meanings.
I'd appreciate an explanation in addition to the solution.
|
$$
\begin{array}{c|c|c}
\text{Letters} & \text{Calculation} & \text{# Possibilities}\\
\hline\\
AAASS & \displaystyle\binom{5}{3} & 10 \\\\
AAASB & \displaystyle\binom{5}{3} \cdot 2 & 20\\\\
AAABB & \displaystyle\binom{5}{3} & 10\\\\
AASSS & \displaystyle\binom{5}{2} & 10\\\\
AASSB & \displaystyle\binom{5}{2} \cdot \binom{3}{2} & 30\\\\
AASBB & \displaystyle\binom{5}{2} \cdot 3 & 30\\\\
ASSSB & 5\cdot \displaystyle\binom{4}{3} & 20\\\\
ASSBB & 5\cdot \displaystyle\binom{4}{2} & 30\\\\
SSSBB & \displaystyle\binom{5}{3} & 10\\\\
\end{array}
$$
This gives a total of $$\boxed{170}$$
|
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|
Find the curve of intersection between $x^2 + y^2 + z^2 = 1$ and $x+y+z = 0$
Find the curve of intersection between $x^2 + y^2 + z^2 = 1$ and $x+y+z = 0$
My attempt:
*
*$x^2 + y^2 + z^2 = 1$
*$x+y+z=0$
$$(2) \implies z = -(x+y)$$
$$(1) \implies x^2+y^2+(x+y)^2 = 1$$
$$2x^2 + 2y^2 + 2xy = 1$$
This is the curve in the xy-plane. Now if I could get y as a function of x, I could easily parametrize the curve but I am not able to do that.
Is there an easier way to solve the problem?
|
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
\newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
\newcommand{\dd}{\mathrm{d}}
\newcommand{\ds}[1]{\displaystyle{#1}}
\newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
\newcommand{\ic}{\mathrm{i}}
\newcommand{\mc}[1]{\mathcal{#1}}
\newcommand{\mrm}[1]{\mathrm{#1}}
\newcommand{\pars}[1]{\left(\,{#1}\,\right)}
\newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
\newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
\newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
\newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
Find the curve of intersection between $\ds{x^{2} + y^{2} + z^{2} = 1}$ and $\ds{x + y + z = 0}$.
\begin{align}
1 & = x^{2} + y^{2} + \pars{-x - y}^{2} =
2x^{2} + 2y^{2} + 2xy
\\[5mm] & =
2\bracks{{x + y \over 2} + {x - y \over 2}}^{2} +
2\bracks{{x + y \over 2} - {x - y \over 2}}^{2} +
2\bracks{{x + y \over 2} + {x - y \over 2}}\bracks{{x + y \over 2} - {x - y \over 2}}
\\[5mm] & =
{3 \over 2}\pars{x + y}^{2} + {1 \over 2}\pars{x - y}^{2}.
\\ & \text{which is}\ identically\ satisfied\ \text{by}\qquad
\left\{\begin{array}{rcrcr}
\ds{x} & \ds{+} & \ds{y} & \ds{=} & \ds{{\root{6} \over 3}\cos\pars{\theta}}
\\
\ds{x} & \ds{-} & \ds{y} & \ds{=} & \ds{\root{2}\sin\pars{\theta}}
\end{array}\right.
\end{align}
$$
\bbx{x =
{\root{6} \over 6}\,\cos\pars{\theta} +
{\root{2} \over 2}\,\sin\pars{\theta}\,,\qquad
y = {\root{6} \over 6}\,\cos\pars{\theta} -
{\root{2} \over 2}\,\sin\pars{\theta}}
$$
|
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|
How can i find the least number n that can be presented as product of a*b Which the least number n can we imagine in product n = a∙b like k ways? Products a∙b and b∙a is one of the way, where all numbers is natural (1≤ k ≤50)
I tried to loop from 1 to 1000000 and with each number perform the following:
looping from 1 to square root of that number and if number%i==0 then add to one variable and after reaching the square root check if var is equal to k if yes then print i,but this solution took too much time
|
Suppose $$n=2^{n_1}\cdot 3^{n_2}\dots \cdot p_r^{n_r}$$ where $p_r$ is the $r^{th}$ prime. The number of divisors of $n$ $$d(n)=(n_1+1)(n_2+1)\dots (n_r+1)$$
The number of factorisations is half of this unless $n$ is a square in which case it is half $d(n)+1$.
To find a number having $k$ factorisations, reverse this process. Look at factorisations of $2k$ and allocate the largest factors to the smallest primes. Check also factorisations of $2k-1$.
Suppose $k=2$ we need $d=2k=4$ or $d=2k-1=3$.
If $d=4$ we have either
(1) the single factor $4$ which gives exponent $4-1=3$ and choose the lowest prime, $2^3=8$ with factorisations $1\cdot 8$ and $2\cdot 4$
(2) The factors $2\cdot 2$ which gives two exponents equal to $2-1=1$. The lowest possibility is $2^13^1=6$ and factorisations $1\cdot 6$ and $2\cdot 3$
If $d=3$ we have just the factor $3$ and exponent $3-1=2$ which gives $2^2=4$ with $1\cdot 4$ and $2\cdot 2$
This tells us that the possible lowest numbers having two factorisations are $4, 6, 8$ - obviously $4$ is the one.
With $k=6$ we have $d=12$ or $d=11$
$d=12$ factorises
(1) as $12$ giving $2^{11}=2048$
(2) $6\cdot 2$ giving $2^5\cdot 3=96$
(3) $4\cdot 3$ giving $2^3\cdot 3^2=72$
(4) $3\cdot 2 \cdot 2$ giving $2^2\cdot 3\cdot 5=60$
$d=11$ gives just $2^{10}=1024$
Clearly $60$ is the number you want.
In this last case, how do we count the factors of $60$?
Well with the prime $2$ we have three possible factors $1,2,4$
With the prime $3$ we get $1,3$
With the prime $5$ we get $1,5$
We need to choose one factor associated with each of the primes, hence $3\times 2\times 2=12$ possibilities.
How do we know we got all the possibilities for $12=2^2\cdot 3^1$. Well this gives $d=6$ and $k=3$ for two factor factorisations, and there is just the fourth possibility $2\cdot 2\cdot 3$ - we got them all.
|
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Continuity- function Let $f : R \to R $ be a function which is continuous at $0$ and $f(0)=1$ . Also assume that f satisfies the following relation for all $x$ $$f(x)-f\left(\frac{x}{2}\right)=\frac{3x^2}{4}+x$$ Find $f(3)$
[I have tried it solve. I will be grateful indeed if someone can review the following solution and made comments wherever necessary ]
Ans: $f(0)=1=\lim_{x\to 0^+}f(x)=\lim_{x\to 0^-} f(x).........................(1) $
$f(x)-f\left(\frac{x}{2}\right)=\frac{3x^2}{4}+x$..........................$(2)$
$f(x)=f\left(\frac{x}{2}\right)+\frac{3x^2}{4}+x=f\left(\frac{x}{4}\right)+\left(\frac{3}{4}\frac{x^2}{2^2}\right)+3\frac{x^2}{4} +x .......$.(Using (2) and putting $\frac{x}{2}$ in place of $x$)
$=f\left(\frac{x}{4}\right)+\frac{3}{4}\left(x^2+\frac{x^2}{2^2}\right)+ \left(x+\frac{x}{2}\right)=f\left(\frac{x}{8}\right)+\frac{3}{4}\left(x^2+\frac{x^2}{2^2}+\frac{x^3}{2^3}\right)+\left(x+\frac{x}{2}+\frac{x}{4}\right)$
Continuing this way, for an infinitely large positive no. $n$ we have
$f(x)=f\left(\frac{x}{2^n}\right)+\frac{3}{4}x^2\left(1+\frac{1}{2^2}+\frac{1}{2^4}+..........+\frac{1}{2^{2n-2}}\right)+x\left(1+\frac{1}{2}+\frac{1}{2^2}+...............+\frac{1}{2^{n-1}}\right)$
Now , putting $x=3$ in the last eqn. and considering $n\to \infty$ , $$f\left(\frac{x}{2^n}\right)\to f(0)=1$$Assuming the two sums in the R.H.S. of $(A)$ as infinite G.P. (Am I correct here ?)
$$f(3)=f(0)+\frac{3}{4}.3^2\left(1+\frac{1}{2^2}+\frac{1}{2^4}+.........\right)+3.\left(1+\frac{1}{2}+\frac{1}{2^2}+.............\right)=1+\frac{27}{4}.\frac{4}{3}+3.2=1+9+6=16$$
|
hint
$$f (3)-f (\frac {3}{2})=\frac {3^3}{2^2}+3$$
$$f (\frac {3}{2})-f (\frac {3}{2^2})=\frac {3^3}{2^4}+\frac {3}{2} $$
...
$$f (\frac {3}{2^n})-f (\frac {3}{2^{n+1}})=\frac {3^3}{2^{2n+2}}+\frac {3}{2^n} $$
sum and $n\to +\infty $.
You will find
$$f (3)=f (0)+\frac {27}{4}\frac {1}{1-\frac {1}{4}}+3\frac {1}{1-\frac {1}{2}} $$
$$=16$$
|
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|
Solve the Diophantine equation $x^6 + 3x^3 + 1 = y^4$
Find all pairs $(x, y)$ of integers, such that $x^6 + 3x^3 + 1 = y^4$.
My solution:
Claim: The pair $(0, 1)$ are the only solutions.
Proof. Suppose there exists other solutions for $y \gt 1$ and $x \gt 0$, then I shall show that such pairs are impossible if $x$ and $y$ are integers. Let us Factorize the given equation as follows.
$x^6 + 3x^3 = y^4 - 1$
$x^6 + 3x^3 = (y^2)^2 - 1^2$
$x^6 + 3x^3 =(y^2 - 1)(y + 1)(y^2 + 1)$
$x^3(x^3+ 3) = (y - 1)(y + 1)(y^2 + 1) . . . (*)$
Because the numbers $(y - 1), (y + 1)$, and $(y^2 + 1)$ are all distinct, it follows that their products can never be a cube, hence, from $(*)$, we obtain the system.
$(y - 1)(y + 1)(y^2 + 1) = x^3 + 3$, and
$x^3 = 1$,
Which is equivalent to:
$(y - 1)(y + 1)(y^2 +) = 4 . . .(**)$
Since $(y - 1) \gt 0$, then $y$ is minimum if and only if $y = 2$ which clearly does not satisfy $(**)$. Hence we must have $x = 0$ and $y = 1$.
Please, Is there any mistake in my solution?.
|
$x^6+3x^3+1-y^4=0$ is a quadratic equation.
Thus, there is an integer $n$ for which $$3^2-4(1-y^4)=n^2$$ or
$$n^2-4y^4=5$$ or
$$(n-2y^2)(n+2y^2)=5$$
and we have four cases only.
|
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|
Trigonometry limit's proof: $\lim_{x\to0}\frac{\sin(x)+\sin(2x)+\cdots+\sin(kx)}{x}=\frac{k(k+1)}{2}$ How to prove that $$\lim_{x\to0}\frac{\sin(x)+\sin(2x)+\cdots+\sin(kx)}{x}=\frac{k(k+1)}{2}$$
I tried to split up the fraction and multiple-divide every new fraction with its $x$ factor but didn't work out.
ex: $$\lim_{x\to 0}\frac{\sin(2x)}{x} = \lim_{x\to 0}\frac{\sin(2x)\cdot 2}{2\cdot x}=2$$
|
You are so close. Note that
\begin{align*}
\frac{\sin(x)+\sin(2x)+\cdots+\sin(kx)}{x}
&= \frac{\sin(x)}{x}+\frac{\sin(2x)}{x}+\cdots+\frac{\sin(kx)}{x} \\
&= \frac{\sin(x)}{x}+2\frac{\sin(2x)}{2x}+\cdots+k\frac{\sin(kx)}{kx} \\
&\to 1 + 2 + \cdots + k \\
&= \frac{k(k+1)}{2}
\end{align*}
as $x\to0$.
I suspect you may not have been able to finish because you didn't recognize the identity
$$
1 + 2 + \cdots + k = \frac{k(k+1)}{2}.
$$
This identity has a very cute proof.
Set $S:=1+2+\cdots+k$. Adding
\begin{align*}
1 + 2 + \cdots + k &= S \\
k + (k-1) + \cdots + 1 &= S \\
\end{align*}
gives
\begin{align*}
\underbrace{(k+1)+(k+1)+\cdots+(k+1)}_{k\ \text{times}} = 2S. \\
\end{align*}
Therefore $k(k+1)=2S$ and consequently
$$1+2+\cdots+k = S = \frac{k(k+1)}{2}.$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Proof that $n!$ is divisible by $(n+1)^2$ for $n=xy+x+y$ I noticed the other day that for $n>8, n\in\mathbb{N}$, the factorial $n!$ seems to be divisible by $(n+1)^2$ when $n$ can be written in the form $xy+x+y$ (where $x,y\geq1$ and $\in\mathbb{N}$). Some examples:
*
*$n=14=2 \times 4+2+4$ (i.e. $x=2$, $y=4$), and we have $14!|15^2$
*$n=15=3 \times 3+3+3$ (i.e. $x=y=3$), and we have $15!|16^2$,
*$n=19=3 \times 4+3+4$ (i.e. $x=3$, $y=4$), and we have $19!|20^2$,
*...
This seems to hold for the first $15$ values I checked, so it seemed natural to try to prove it for all $n=xy+x+y$, however I could not find a proof. Here is my attempt:
Plugging $xy+x+y$ into the expressions above and noticing that $xy+x+y+1=(x+1)(y+1)$, we need to prove that
$$
(xy+x+y)!|(x+1)^2(y+1)^2,
$$
or
$$
\Gamma((x+1)(y+1))|(x+1)^2(y+1)^2.
$$
It is easy to prove that
$$
\Gamma((x+1)(y+1))|(x+1)(y+1)
$$
however I am struggling to prove the same for $(x+1)^2(y+1)^2$. Any hints?
|
$\color{Green}{\text{Question}}$ :
Find all integers $n$;
such that $\color{Green}{(n+1)! \mid n!} $ .
$\color{Blue}{\star \ \ \ \ \text{condition}}$ :
Now suppose that $n+1$ can be written
in the form $n+1=ab$;
with $2 < a,b$.
Also assume that one of the following occures:
*
*If $a= b$; then $5 \leq a=b$.
*If $a=2b$; then $5 \leq b $.
*If $2a=b$; then $5 \leq a $.
With the condition as in
$\color{Blue}{\star}$ ;
we claim that
the assertion is $\color{Blue}{\text{true}}.$
Let $n, a, b$ to be fixed as above in $\color{Blue}{\star}$ .
Notice that if we let $x:=a-1$ and $y:=b-1$
then one can easilly see that:
$$n=xy+x+y.$$
$\color{Blue}{\text{Remark(I)}}$ :
With the notation as in $\color{Blue}{\star}$;
*
*If $a=b$ , with $5 \leq a$
then we have:
$$b < 2b < 3b < 4b < ab=n+1 \Longrightarrow b < 2b < 3b <4b \leq n=ab-1 \ ; $$
so we can conclede that:
$$(b)(2b)(3b)(4b)=24(ab)^2=24(n+1)^2 \mid n! ;$$
which implies that: $(n+1)^2 \mid n! \ \ $ .
*
*If $a=2b$ ; with $5 \leq b$
then we have:
$$b < 2b < 3b < 4b < ab=n+1 \Longrightarrow b < 2b < 3b <4b \leq n=ab-1 \ ; $$
so we can conclede that:
$$(b)(2b)(3b)(4b)=6(ab)^2=6(n+1)^2 \mid n! ;$$
which implies that: $(n+1)^2 \mid n! \ \ $ .
*
*If $2a=b$ ; with $5 \leq a$
then we have:
$$a < 2a < 3a < 4a < ba=n+1 \Longrightarrow a < 2a < 3a <4a \leq n=ba-1 \ ; $$
so we can conclede that:
$$(a)(2a)(3a)(4a)=6(ab)^2=6(n+1)^2 \mid n! ;$$
which implies that: $(n+1)^2 \mid n! \ \ $ .
*
*If $a \neq b$ and $a \neq 2b$ and $2a \neq b$ ;
then we have:
$$b < 2b < ab=n+1 \Longrightarrow b < 2b \leq n=ab-1 \ ; $$
$$a < 2a < ab=n+1 \Longrightarrow a < 2a \leq n=ab-1 \ ; $$
so we can conclede that:
$$(b)(2b)(a)(2a)=4(ab)^2=4(n+1)^2 \mid n! ;$$
which implies that: $(n+1)^2 \mid n! \ \ $ .
Remark(II): Integer $N$
can not be written
in the form $N=ab$;
with $2 < a,b$;
if and only if
$N$ is prime or twice a prime.
Remark(III):
We can split the integres in three cases:
*
*$\color{Red}{ n+1 = \text{prime numbers and twice of prime numbers;
for which the assertion is wrong!} }.$
*$\color{Purple}{ n+1= a.a
\ \ \
\text{for}
\ \ \
a \in \{ 1, 2, 3, 4 \} }.$
This case has only four numbers;
which can you check out that
the assertion is $\color{Blue}{\text{true}}$ only for
$\color{Blue}{n+1=16}$ and
$\color{Blue}{n+1= 1}$ .
*$\color{Purple}{ n+1= a.(2a)
\ \ \
\text{for}
\ \ \
a \in \{ 1, 2, 3, 4 \} }.$
This case has only four numbers;
which can you check out that
the assertion is $\color{Blue}{\text{true}}$ only for
$\color{Blue}{n+1=18}$ and
$\color{Blue}{n+1=32}$ .
$\color{Green}{\text{Remark(IV)}}$ :
Except for the following cases;
for which the assertion is
$\color{Red}{\text{wrong}}$
1.$\color{Red}{ n+1 = \text{prime numbers and twice of prime numbers
} }.$
2.$\color{Red}{ n+1= 4, 9, 2, 8 } .$
for all the other cases
the assertion is
$\color{Blue}{\text{true}}$ .
$\color{Red}{\text{Remark(V)}}$ :
Let $p$ to be an odd prime number;
then we have:
$p^2 \color{Red}{\nmid} (2p-1)! \ \ . $
For example let $p$ to be an odd prime number;
and let's define:
$$n:=2p-1
\ \ \
\text{and}
\ \ \
x:=1
\ \ \
\text{and}
\ \ \
y:=p-1.$$
Notice that we have: $n=xy+x+y$
We calaim that $(2p)^2 \color{Red}{\nmid} (2p-1)!$.
[
Suppose on contrary
that $(2p)^2 \mid (2p-1)!$ ,
on the otherhand notice that
$p^2 \mid (2p)^2$.
So we must have
$p^2 \mid (2p-1)!$
;
which has an obvious contradiction
with the above $\color{Red}{\text{Remark(V)}}$.
]
|
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"timestamp": "2023-03-29T00:00:00",
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|
Evaluate the given limit: Evaluate the given limit:
$$\lim_{x\to 1} \dfrac {1+\cos \pi x}{\tan^2 \pi x}$$
My Attempt:
$$=\lim_{x\to 1} \dfrac {1+\cos \pi x}{\dfrac {\sin^2 \pi x}{\cos^2 \pi x}}$$
$$=\lim_{x\to 1} (1+\cos \pi x) \times \dfrac {\cos^2 \pi x}{\sin^2 \pi x}$$
$$=\lim_{x\to 1} (1+\cos \pi x) \cos^2 \pi x (\dfrac {\pi x}{\sin \pi x} \times \dfrac {1}{\pi^2 x^2})$$
|
Make the problem simpler using $x=y+1$ and later $\pi y=z$ $$\lim_{x\to 1} \dfrac {1+\cos (\pi x)}{\tan^2 (\pi x)}=\lim_{y\to 0} \dfrac {1-\cos (\pi y)}{\tan^2 (\pi y)}=\lim_{z\to 0} \dfrac {1-\cos (z)}{\tan^2 (z)}$$ and consider either equivalents $$\cos(z)\sim 1-\frac{z^2}2 \qquad \text{and}\qquad\tan(z)\sim z \implies \tan^2(z)\sim z^2$$ to conclude.
If you want more than the limit itself, use Taylor series
$$\cos(z)=1-\frac{z^2}{2}+\frac{z^4}{24}+O\left(z^6\right)$$
$$\tan(z)=z+\frac{z^3}{3}+O\left(z^5\right)$$
$$\tan^2(z)=z^2+\frac{2 z^4}{3}+O\left(z^6\right)$$ which make
$$\dfrac {1-\cos (z)}{\tan^2 (z)}=\dfrac {\frac{z^2}{2}-\frac{z^4}{24}+O\left(z^6\right) }{z^2+\frac{2 z^4}{3}+O\left(z^6\right) }=\frac{1}{2}-\frac{3 z^2}{8}+O\left(z^4\right)$$ which shows the limit and how it is approached.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Prove that for every real number $x$, if $|x − 3| > 3$ then $x^2 > 6x$. This is Velleman's exercise 3.5.10:
Prove that for every real number $x$, if $|x − 3| > 3$ then $x^2 > 6x$.
And here's my proof of it:
Proof. Suppose $|x − 3| > 3$. We now consider two cases:
Case 1. $x - 3 \ge 0$. Then $|x − 3| = x - 3$, so we have $x - 3 > 3$, and therefore $x > 6$. Since $x > 6$ and positive, then multiplying the inequality by $x$, we get $x^2 > 6x$.
Case 2. $x - 3 < 0$. Then $|x − 3| = 3 - x$, so we have $3 - x > 3$, and therefore $x < 0$. Subtracting 3 from both sides of the inequality $x < 0$, we get $x - 3 < -3$. Multiplying the inequality by ($x - 3$), we get $x^2 - 6x + 9 > -3x + 9$ and therefore $x^2 > 3x$.
Since by one of the cases we have $x^2 > 6x$ then $|x − 3| > 3$ $\Rightarrow$ $x^2 > 6x$.
Is my proof valid?
Thanks in advance.
|
Your proof is incomplete: you should show in BOTH cases that $x^2>6x$. So in case 2, when $x<0$, at the end, it suffices to sya that "and therefore $x^2>3x>6x$".
However, there is a much shorter way that I warmly recommend. Since $|x − 3|$ and $3$ are both non-negative, the inequality $|x − 3| > 3$ is equivalent to $|x-3|^2>3^2$. Hence $$0<(x-3)^2-9=(x^2-6x+9)-9=x^2-6x$$
and we are done.
|
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|
How to find linear dependency relationship
The Question I am attempting to answer gives vectors of,
(2,1,4,3) (2,3,2,3) (4,11,-1,6) and asks for an explicit dependency relationship.
I have found them to be linearly dependent. (found the determinant = 0)
How do i then find a dependency relationship?
Looking at this explanation, I understand up until the "which implies..." in the final line.
If someone could explain it that would be much appreciated.
Thanks, JD
|
Look for the relationship between the three vectors:
$$
xu+yv+zw=
x
\left[\begin{matrix}
2\\
1\\
4\\
3
\end{matrix}\right]
+
y
\left[\begin{matrix}
2\\
3\\
2\\
3
\end{matrix}\right]
+
z
\left[\begin{matrix}
4\\
11\\
-1\\
6
\end{matrix}\right]
=
\left[\begin{matrix}
2 & 2 & 4\\
1 & 3 & 11\\
4 & 2 & -1\\
3 & 3 & 6\\
\end{matrix}\right]
\left[\begin{matrix}
x\\
y\\
z\\
\end{matrix}\right]
=
\left[\begin{matrix}
0\\
0\\
0\\
0
\end{matrix}\right].
$$
Writing the matrix in augmented form, and then performing Gaussian elimination gives
$$
\left[\begin{array}{ccc|c}
1 & 0 & \frac{-5}{2} & 0\\
0 & 1 & \frac{9}{2} & 0\\
0 & 0 & 0 & 0\\
0 & 0 & 0 & 0\\
\end{array}\right],
$$
where we see that there is one free variable, $z$. Two relationships are obtained from this matrix, namely
$$x=\frac{5z}{2},y=-\frac{9z}{2}.$$
Return to the first equation mentioned, and plug in these two relationships.
$$xu+yv+zw=\left(\frac{5z}{2}\right)u-\left(\frac{9z}{2}\right)v+zw=z\left(\frac{5}{2}u-\frac{9}{2}v+w\right)=0.$$
Notice that we can choose $z$ as we please to obtain a relationship between these three vectors. Divide $z$ out. Then we have
$$\frac{5}{2}u-\frac{9}{2}v+w=0,$$
and we can solve for $w$, obtaining a relationship between the vectors. For completeness, it is
$$w=-\frac{5}{2}u+\frac{9}{2}v.$$
|
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|
Show that if $a+b+c=0$, $2(a^4 + b^4+ c^4)$ is a perfect square
Show that for $\{a,b,c\}\subset\Bbb Z$ if $a+b+c=0$ then $2(a^4 + b^4+ c^4)$ is a perfect square.
This question is from a math olympiad contest.
I started developing the expression $(a^2+b^2+c^2)^2=a^4+b^4+c^4+2a^2b^2+2a^2c^2+2b^2c^2$ but was not able to find any useful direction after that.
Note: After getting 6 answers here, another user pointed out other question in the site with similar but not identical content (see above), but the 7 answers presented include more comprehensive approaches to similar problems (e.g. newton identities and other methods) that I found more useful, as compared with the 3 answers provided to the other question.
|
A systematic way doing this is using Newton's identifites.
Let $p_k = a^k + b^k + c^k$ for $k = 1, 2, 3, 4$ and
$$\begin{align}
s_1 &= a + b + c\\
s_2 &= ab+bc+ca\\
s_3 &= abc
\end{align}$$
be the elementary symmetric polynomials associated with $a, b, c$.
Newton's identities tell us:
$$\require{cancel}\begin{array}{rlrlrlrlrl}
p_1 &-& s_1 &&&&&= 0\\
p_2 &-&
\cancelto{ 0}{\color{grey}{s_1 p_1}}
&+& 2s_2 &&&= 0\\
p_3 &-&
\cancelto{ 0}{\color{grey}{s_1 p_2}}
&+&
\cancelto{ 0}{\color{grey}{s_2 p_1}}
&- &3s_3 &= 0\\
p_4 &-&
\cancelto{ 0}{\color{grey}{s_1 p_3}}
&+& s_2 p_2 &-&
\cancelto{ 0}{\color{black}{s_3 p_1}}
&= 0
\end{array}
$$
When $a + b + c = 0$, $s_1 = 0$ and $1^{st}$ equation $p_1 - s_1 = 0$ tell us $p_1 = 0$.
Substitute back into $2^{nd}$ and $4^{th}$ equations lead to
$$\begin{cases}
p_2 = -2s_2,\\
p_4 = -s_2 p_2
\end{cases}
\quad\implies\quad
2(a^4+b^4+c^4) = 2p_4 = -2s_2 p_2 = p_2^2 = (a^2+b^2+c^2)^2$$
|
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|
find a and b such that the limit will exist and find the limit I am given this question $$\lim_{x\rightarrow 0} \frac{e^{\sqrt{1+x^2}}-a-bx^2}{x^4}$$
And asked to find the $a$ and $b$ such that the limit exists and also to compute the limit
My solution:
Let $t$=$\sqrt{1+x}$.
Then the Maclaurin polynomial is : $$\sqrt{1+x}=1+\frac{x}{2}-\frac{x^2}{8}+\mathcal{O}(x^3)$$
Now plugging in $x^2$ for x we get $$\sqrt{1+x^2}=1+\frac{x^2}{2}-\frac{x^4}{8}+\mathcal{O}(x^6)$$
and from the common Maclaurin polynomial we have that $e^t=1+t+\frac{t^2}{2}+ \mathcal{O}(t^3)$. Plugging in $\sqrt{x^2+1}$ for $t$ we get:
$$e^{1+\frac{x^2}{2}-\frac{x^4}{8}+\mathcal{O}(x^6)
}$$
which in turn is:
$$e^{1+\frac{x^2}{2}-\frac{x^4}{8}+\mathcal{O}(x^6)
}=1+\left(1+\frac{x^2}{2}-\frac{x^4}{8}+\mathcal{O}(x^6)\right)+\mathcal{O}(x^6)$$
hence we have:
$$\lim_{x\rightarrow 0} \frac{1+\left(1+\frac{x^2}{2}-\frac{x^4}{8}+\mathcal{O}(x^6)\right)+\mathcal{}O(x^6)-a-bx^2}{x^4}$$
And i argue that $a=2, b=1/2$ and the $\lim=-1/8 $
However the book disagrees with me and argues that the they should be $a=e, b=e/2$ and limit $=0$
i mean i can see how theyd done it, by not expanding $e^t$ but by only expanding $\sqrt{x^2+1}$ however what i dont understand how come we didnt get the same values or at least the same value for the limit?
|
Hint. When you write
$$
e^{1+\frac{x^2}{2}-\frac{x^4}{8}+\mathcal{O}(x^6)
}=1+\left(1+\frac{x^2}{2}-\frac{x^4}{8}+\mathcal{O}(x^6)\right)+\mathcal{O}(x^6)
$$ it is wrong, since it is not right that, as $x \to 0$,
$$
\left(1+\frac{x^2}{2}-\frac{x^4}{8}+\mathcal{O}(x^6)\right)^m=\mathcal{O}(x^6),\qquad m=2,3,\cdots.
$$
Please look at
$$
\lim_{x \to 0}\left(1+\frac{x^2}{2}-\frac{x^4}{8}+\mathcal{O}(x^6)\right)^m.
$$
You can't neglect the previous quantities.
Do you see the issue?
|
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|
$P\left(x\right)=x^{100}+x^{50}-2x^4-x^3+x+1$, $\frac{P(x)}{x^3+x}$ remainder Given the polynomial:
$P\left(x\right)=x^{100}+x^{50}-2x^4-x^3+x+1$
What is the remainder of $\frac{P(X)}{x^3+x}$?
I don't think the long division is efficient the way to go, and the remainder theorem doesn't seem to be applicable here as $x^3+x$ is not linear. Could I have some hints on how to approach this? Thank you.
|
The remainder is
$$\frac{x+2}{x^2+1}\:+\:\frac 1x\qquad\text{or}\quad2x^2+2x+1\;\;\text{respectively}\\[3ex]$$
obtained by taking a lazy approach: Feeding the command
apart((x**100+x**50-2*x**4-x**3+x+1)/(x**3+x))
into http://live.sympy.org/ ,
thus asking Python to go through the long division in short time,
yields
$$\frac{P(x)}{x(x^2+1)}\;=\;\sum_{m=48}^{24}(-1)^mx^{2m+1}-2x-1+\frac 1x+\frac{x+2}{x^2+1}\\[3ex]$$
Note that the $\,\sum\,$ equals $(x^{100}+x^{50})/(x^3+x)=x^{49}(x^{50}+1)/(x^2+1)\,$, whence no contribution to the remainder.
... did I fail to address the issue in question?
|
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|
Is this true $n!\leq(\frac{5n+7}{12})^n,n∈N$? Is the following inequality true?
For all $n\in \Bbb N$ prove that:
$$n!\leq\left(\frac{5n+7}{12}\right)^n.$$
I know the answer,but I want to see other people how to prove the problem.
In my proof I used $\frac{5n+7}{12}=\frac{\frac{n+1}2+\frac{n+2}3}2\geq \sqrt{\frac{(n+1)(n+2)}6}$
$=\sqrt{\frac{1}{n}\left(\frac{n(n+1)(n+2)}{6}\right)}$
$=\sqrt{\frac{1}{n}\sum_{k=1}^{n}k(n-k+1)}$
$\ge\sqrt{\sqrt[n]{(n!)^2}}=\sqrt[n]{n!}$.
|
By AM-GM
$$\frac{1\cdot n+2(n-1)+...+n\cdot1}{n}\geq\sqrt[n]{(n!)^2}$$ or
$$\left(\sqrt{\frac{(n+1)(n+2)}{6}}\right)^n\geq n!.$$
Thus, it remains to prove that
$$\frac{5n+7}{12}\geq\sqrt{\frac{(n+1)(n+2)}{6}},$$
which is $$(n-1)^2\geq0.$$
Done!
$$1\cdot n+2(n-1)+...+n\cdot1=\sum_{k=1}^nk(n-k+1)=$$
$$=(n+1)\sum_{k=1}^nk-\sum_{k=1}^nk^2=(n+1)\cdot\frac{n(n+1)}{2}-\frac{n(n+1)(2n+1)}{6}=$$
$$=\frac{n(n+1)}{6}\cdot(3n+3-2n-1)=\frac{n(n+1)(n+2)}{6}.$$
|
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|
Polynomials$ P(x^2)=(x+1)P(x)$ Find all polynomials $P(x) \in\mathbb{R}[x]$ satisfying $$ P(x^2)=(x+1)P(x)$$
Please check my work :
Let $r$ be root of $P(x)$ so $P(r)=0$
Substitute $x=r$, $ P(r^2)=(r+1)P(r)$ so $P(r^2)=0$, i.e., if $r$ is root then $r^2$ is also root.
Substitute $x=-1$, so $P(1)=0$, i.e., if $r$ is root then $r^2$ is also root.
If $ \mid\; r\mid >1$ then $ \mid\; r^4\mid > \mid\; r^2\mid > \mid\; r\mid $ so $P(x)$ has infinitely many roots.
If $ \mid\; r\mid <1$ then $ \mid\; r^4\mid < \mid\; r^2\mid < \mid\; r\mid $ so $P(x)$ has infinitely many roots.
Hence $ \mid\; r\mid =1$ so $P(x)=c, \;c(x-1), \;c(x+1), \;c(x-1)(x+1)$
By checking, my answer is $P(x)=c(x-1)$
Edited work :
If deg$(P(x))=0$, then $P(x)=0$.
Assume deg$(P(x))=k \not= 0$ so deg$(P(x^2))=2k$ and deg$((x+1)P(x))=k+1$
so $2k=k+1$ we have $k=1$ so $P(x)=ax+b$
so $P(x^2)=ax^2+b=(x+1)(ax+b)$, we obtain $a=-b$
$P(x)=ax+b=ax-a=a(x-1)$
Answer : $P(x)=a(x-1)$, $P(x)=0$.
|
Suppose $P(x)$ has a root other than $1$ and $-1$ (say $r$). Then $r^2$ is also a root of the polynomial, i.e. $r^4$ is also a root of the polynomial which leads $P$ has infinitely many roots, which is not possible.
Then only roots $P$ can have are $1$ and $-1$. It is easy to see that $1$ is a root of $P$.
$c\neq 0$
Let $P(x)=c(x-1)(x+1)\Rightarrow P(x^2)=\color{red}{c(x^2-1)(x^2+1)}=(x+1)P(x)=\color{green}{c(x-1)(x+1)^2}$, which is not possible, since degree of the red coloured polynomial is $\color{red}{4}$ and degree of the green coloured polynomial is $\color{green}{3}$.
Let $P(x)=c(x-1)\Rightarrow P(x^2)=c(x^2-1)=(x+1)P(x)=(x+1)c(x-1)=c(x^2-1)$
Hence $P(x)$ is constant or $c(x-1)\space\space\space\blacksquare$
|
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|
Proving that for $\{ a,b\} \subset \Bbb{R^{+}}$; $a+b=1 \implies a^2+b^2 \ge \frac{1}{2}$
For $\{ a,b\} \subset \Bbb{R^{+}}$; $a+b=1 \implies a^2+b^2 \ge
\frac{1}{2}$
I'm trying to prove this in the following way, but I'm not sure if it's correct. Could anyone please check it and see if it's okay?
$a+b=1 \implies (a+b)^2 = 1^2 = 1 \implies (a+b)-(a+b)^2 = 1-1 =0$ (1)
$(a-b)^2 \ge 0$
So by (1) we have:
$(a-b)^2 \ge (a+b)-(a+b)^2$
$(a^2-2ab+b^2) \ge (a+b) - (a^2+2ab+b^2)$
$(a^2-2ab+b^2) + (a^2+2ab+b^2) \ge (a+b) $
$ a^2+a^2+b^2+b^2+2ab-2ab \ge (a+b)$
$2(a^2+b^2) \ge (a+b)$
$2(a^2+b^2) \ge 1$
$(a^2+b^2) \ge \frac{1}{2} $
$\blacksquare$
|
Follows directly from Hölder's inequality on a finite measure space (with dual exponents $(2,2)$): $1 = (a+b)^2 \le (a^2+b^2)(1^2+1^2)$.
|
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|
Show that, for any two integers $a$ and $b$, at least one of the expressions $a^3, b^3, a^3+b^3$ or $a^3-b^3$ will be divisible by $7$. This is a very interesting word problem that I came across in an old textbook of mine:
Show that, for any two integers $a$ and $b$, at least one of the expressions $a^3, b^3, a^3+b^3$ or $a^3-b^3$ will be divisible by $7$.
So I know it's got something to do with Euler's totient function and the Euler-Fermat theorem, which yields the shortest, simplest proofs, but other than that, the textbook gave no hints really and I'm really not sure about how to approach it. Any guidance hints or help would be truly greatly appreciated. Thanks in advance.
|
If $a$ or $b$ is a multiple of 7 then it's obvious, else $a^6\equiv b^6 \equiv 1 \pmod 7$ and $(a^3+b^3)(a^3-b^3)=a^6-b^6\equiv 0 \pmod 7$. Since 7 is prime it divides $a^3+b^3$ or $a^3-b^3$.
|
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|
Solve $\frac{1}{x}<\frac{x}{2}<\frac{2}{x}$. First I consider the case $\frac{1}{x}<\frac{x}{2}$, manipulating to $\frac{(x-\sqrt{2})(x+\sqrt{2})}{2x}<0\Leftrightarrow x < -\sqrt{2}.$ Did this by subtracting $\frac{x}{2}$ from both sides and writing everything with a common denominator.
Second case to consider is $\frac{x}{2}<\frac{2}{x}\Leftrightarrow\frac{(x-2)(x+2)}{2x}<0\Leftrightarrow x<-2.$
Since both of these inequalities have to be satisfied simultaneously, one can combine them to get $x<-2.$ Correct answer is $x\in(\sqrt{2},2)$
|
If $x>0$, multiply by $x$ to get
$$1<\frac{x^2}{2}<2\iff 2<x^2 <4\implies \sqrt 2<x<2$$
If $x< 0$ multiply by $x$ and
$$1>\frac{x^2}{2}>2\iff 2>x^2>4\implies \text{false}$$
|
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|
Simple logarithmic differentiation Given the equation $\frac{dy}{dx} = \frac{-x}{y}$
How can you solve for $\frac{d^2y}{d^2x}$ using logarithmic differentiation?
Here is my work:
$ln(\frac{dy}{dx}) = ln(\frac{-x}{y})$
$ln(\frac{dy}{dx}) = ln(-x) - ln(y)$
(Take derivative of both sides)
$(\frac{d^2y}{d^2x})/(\frac{dy}{dx}) = \frac{1}{x} - \frac{1}{y}$
(Then solve for $\frac{d^2y}{d^2x}$)
$\frac{d^2y}{d^2x} = \frac{\frac{dy}{dx}}{x} - \frac{\frac{dy}{dx}}{y}$
(Then plug in $\frac{-x}{y}$ for $\frac{dy}{dx}$)
$\frac{d^2y}{d^2x} = \frac{-x}{yx} + \frac{x}{y^2}$
And simplify to get:
$\frac{d^2y}{d^2x} = \frac{-1}{y} + \frac{x}{y^2}$
$ = \frac{(x-y)}{y^2}$
However, when I solve the derivative using the quotient rule, I get:
$\frac{x\frac{-x}{y} - y}{y^2}$
$= \frac{\frac{-x^2}{y}-y}{y^2}$
What am I doing wrong?
|
That equation can be rewritten with another notation like y'y = -x
simply claculate the dervative of that expression
$$y"y+y'²= -1 $$
where y'= dy/dx and y" = d²y/dx²
Rewriting this last equation with your notation
$$d²y/dx²=-(x²+y²)/y³$$
|
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|
If the constant term of binomial expansion $ (2x-\frac{1}{x})^{n}$ is$-160$ ,then $n$ equal to? As I know the expansion binomial expression of $ (2x-\frac{1}{x})^{n}$ .But I don't know that for which $-160$ in $n$th term of given expression.
Please help me to solve this.
|
As usual, the binomial expansion helps:
$$
\left(2x - \frac 1x\right)^n = \sum_{k=0}^n (-1)^k\binom nk\frac{1}{x^k}(2x)^{n-k} = \sum_{k=0}^n \binom nk (-1)^k2^{n-k}x^{n-2k}
$$
Now, when does the constant term appear? Precisely when $n=2k$, as one sees above. Also, if $n$ is odd, then you see that the constant term is zero, so $n$ must be even.
Then, the constant itself is $(-1)^k\binom nk 2^{n -k } = \binom n{\frac n2} 2^{\frac n2}$.
Now, this is equal to $-160$, as you say. We note that $160$ is a multiple of $5$, hence $n \geq 5$ must happen, otherwise $\binom nk$ cannot be a multiple of $5$. Also, $k$ is odd, as the coefficient is negative, hence $n = 6,10,...$ As we notice, $n=6$ gives $\binom 63 \times 8 = 20 \times 8 = 160$. Hence, the answer is $n = 6$.
Almost as if to prove a point:
$ \left(2x - \frac 1x\right)^6 = 64 x^6 + \frac 1{x^6} - 192 x^4 - \frac{12}{x^4} + 240 x^2 + \frac{60}{x^2} - 160$.
|
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|
Proving that $\{x\in\Bbb{R}\mid 1+x+x^2 = 0\} = \varnothing$ without the quadratic formula and without calculus I'm asked to prove that $\{x\in\Bbb{R}\mid 1+x+x^2 = 0\} = \varnothing$ in an algebra textbook.
The formula for the real roots of a second degree polynomial is not introduced yet. And the book is written without assuming any prior calculus knowledge so I can't prove this by finding the minimum and the limits as x approaches $\infty \text{ and} -\infty $.
So there has to be a simple algebraic proof involving neither the quadratic formula nor calculus but I'm stuck.
Here are some things I thought:
Method 1:
$1+x+x^2 = 0 \iff 1+x+x^2+x = x$
$\iff x^2+2x+1 = x$
$\iff (x+1)^2 = x $
And here maybe prove that there is no x such that $(x+1)^2 = x$ ???
Method 2:
$1+x+x^2 = 0$
$\iff x^2+1 = -x$
By the trichotomy law only one of these propositions hold: $x=0$ or
$x>0$ or $x<0$.
Assuming $x=0$:
$x^2+1= 0^2+1 = 0 +1 = 1$
$-x = - 0 = 0$
And $1\neq 0$
Assuming $x>0$:
$x>0 \implies -x < 0$
And $x^2+1 \ge 1 \text{ } \forall x$
With this method I have trouble proving the case $x<0$:
I thought maybe something like this could help but I'm not sure:
$x<0 \implies -x=|x|$
$x^2 = |x|^2$
And then prove that there is no x such that $|x|^2 + 1 = |x|$??
Can anyone please help me? Remember: No calculus or quadratic formula allowed.
|
To pursue your ideas:
"And here maybe prove that there is no $x$ such that $(x+1)^2=x$ ???"
If $x > 0$ then $x + 1 > x$ and $x + 1 > 1$ so $(x+1)^2 = (x+1)(x+1) > x *1 = x$.
If $x = 0$ then $(x+1)^2 = 1 \ne 0 = x$
If $x < 0$ then $(x+1)^2 \ge 0 > x$.
"With this method I have trouble proving the case x<0"
If $x > 0$ then $x^2 > 0; x> 0; 1 > 0$ so $1 + x + x^2 > 0$.
If $x =0$ then $1 + x + x^2 = 1 > 0$.
So $x < 0$.
"And then prove that there is no x such that $|x|^2+1=|x|$"
$|x|^2 \ge 0$ so $|x|^2 + 1 \ge 1$ so if $|x|^2 + 1 = |x|$ then $|x| \ge 1$.
If $|x| = 1$ then $|x|^2 + 1 = 2 \ne 1 = |x|$.
If $|x| > 1$ then $|x|^2 > |x|$ so $|x|^2 + 1 > |x| + 1 > |x|$.
====
Basically the first thing to notice is that if $1 + x + x^2 = 0$ then $1>0; x^2 \ge 0$ so $1 + x^2 = -x \ge 1 > 0$ so $x$ is negative.
The next thing to notice is $1 + x^2 = |x|$ So $x^2 < |x|$ so $|x| < 1$. But that contradicts $|x| = x^2 + 1 \ge 1$.
But another approach is:
$1 + x +x^2 = 0$
$x(x + 1) = -1$ so $x$ and $x+1$ must be opposite signs.
So either $x > 0$ and $x + 1 < 0$ which is impossible or $x < 0$ and $x +1 > 0$.
But if $x< 0 < x+1$ the $|x| < 1$ and and $|x+1| < 1$. So $|x(x+1)| < 1$ which is a contradiction.
|
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|
$|x^2-5x+2|\leq 4$
Solve the following inequality.
$$|x^2-5x+2|\leq 4.$$
I know how to interpret $|x-a|$ as the distance between $x$ and $a$ along the $x$-axis, but how does one interpret an absolute value of $|ax^2+bx+c|$?
For the inequality $|x^2-5x+2|\leq 4$ I factored the LHS and got $$\left|\left(x+\frac{\sqrt{7}-5}{2}\right)\left(x+\frac{5-\sqrt{7}}{2}\right)\right|\leq 4.$$
I don't know how to case-divide the LHS.
|
Hints:
*
*$$|x^2-5x+2| \le 4 \iff -4 \le x^2-5x+2 \le4 \iff \frac{1}{4} \le x^2-5x+\frac{25}{4} \le \frac{33}{4}.$$
*$$x^2-5x +\frac{25}{4}=\left(x-\frac{5}{2}\right)^2.$$
*$$a \le y^2 \le b \iff \sqrt{a} \le y \le \sqrt{b} \quad \text{or} \quad -\sqrt{b} \le y \le -\sqrt{a}.$$
|
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|
Finding the minimum value of $a^2+b^2+c^2$
Let $a$, $b$ and $c$ be $3$ real numbers satisfying $2 \leq ab+bc+ca$. Find the minimum value of $a^2+b^2+c^2$.
I've been trying to solve this, but I don't really know how to approach this. I thought of $(a+b+c)^2 = a^2+b^2+c^2 + 2(ab+bc+ca)$, but that gives me $a+b+c$, which is unknown. How can I solve this?
|
Notice that by Inequality of arithmetic and geometric means we know that:
$$
2ab \leq a^2+b^2;
\\
2ac \leq a^2+c^2;
\\
2bc \leq b^2+c^2;
$$
so we can conclude that:
$$
2\left(ab+ac+bc\right)
\leq
2\left(a^2+b^2+c^2\right)
\Longrightarrow
\\
\ \
\left(ab+ac+bc\right)
\leq
\ \
\left(a^2+b^2+c^2\right)
\Longrightarrow
\\
\ \ \ \ \
\ \ \ \ \
\ \ \ \ \
\ \ \ \
2
\leq
\ \
\left(a^2+b^2+c^2\right)
.
$$
Note that this in-equlality is sharp for $a=b=c=\sqrt{\dfrac{2}{3}}$;
for which one can see the value of $a^2+b^2+c^2$ is equal to $2$.
|
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|
Find all integers $n>1$ such that$\frac{2^n+1}{n+1}$ is an integer. Find all integers $n>1$ such that $\frac{2^n+1}{n+1}$ is an integer.
Second attempt : I'd like to know if it's correct or not. Thank you.
Since $\frac{2^n+1}{n+1}$ for all integers $n>1$ is equivalent to $\frac{2^{n-1} \;+1}{n}$ for all integers $n>2$,
we will find all integers $n>2$ such that $\frac{2^{n-1}+1}{n}$ is an integer instead.
Let $n=\displaystyle\prod_{i=1}^l p_i^{k_i}$, where $k_i \geq 1$, $p_i \in \text{prime}$.
Let $p_j$ be prime such that $p_j\mid n$ and $p_j$ has minimum $v_2(p_j-1)$.
Since $n \mid 2^{n-1}+1$, so $n\in$ odd, then $p_j\in$ odd.
so $v_2(p_j-1) \geq 1$, i.e., there exist $r_j, m_j \in \mathbb{N}$ and $m_j \in \text{odd}$ such that $p_j-1=2^{r_j}\cdot m_j$
we have $p_j \equiv 1 (\bmod{2^{r_j}})$
Since $v_2(p_j-1)$ is minimum and $n\in$ odd, so $n \equiv 1 (\bmod{2^{r_j}})$
then we can write $n$ in the form, $n = 2^mt+1$, where $m, t \in \mathbb{N}$ and $m\geq r_j$
so $n \equiv -1 (\bmod{2^m})$
$-1 \equiv 2^{n-1} \equiv 2^{ 2^m \cdot t}\equiv 2^{ 2^m\cdot t\cdot m_j} (\bmod{p_j})$, since $m_j$ is odd,
$-1 \equiv 2^{(2^{r_j} \cdot m_j) 2^{m-r_j}\;\cdot t} (\bmod{p_j})$
$-1 \equiv 2^{(p_j-1) 2^{m-r_j}\;\cdot t} (\bmod{p_j})$
$-1 \equiv 1 (\bmod{p_j})$, then $p_j=2$, contradicts that $p_j\in$ odd.
Therefore, there is no such $n$.
|
HINT: The following procedure is a sort of descent proving the impossibility of solution.
It is obvious that the denominator cannot be even so $n$ should be even and the quotient (supposing it is integer) should be odd.
$$2^{2n}+1=(2n+1)(2k+1)\iff2^{2n}=4kn+2(k+n)\iff2^{2n-1}=2kn+k+n$$ It follows that $k$ and $n$ have the same parity. We have to consider the two possible cases which give
$$►(1)\text{ both odd: } 2^{2n-1}==2^3k_1n_1+(2+1)(k_1+n_1)+2^2$$
$$►(2)\text{ both even: } 2^{2n-1}=2^3k_1n_1+2(k_1+n_1) $$
$(1)$ and $(2)$ give respectively $$2^{n-2}=2^2k_1n_1+(2+1)(k_1+n_1)+2\\2^{n-2}=2^2k_1n_1+k_1+n_1$$
Both cases require $k_1$ and $n_1$ have the same parity.
Iterating the procedure we get in each step $k_i$ and $n_i$ have the same parity
which leads, after sufficient iterations, to an equality in which an integer is equal to another greater integer. Contradiction.
$$\text{ A concise example}$$ $$(2^6+1)=(6+1)(2k+1)=14k+6+1\iff2^6=14k+6\iff2^5=7k+3\\ 2^5=7k+3\Rightarrow2^5=7(2k_1+1)+3=14k_1+10\iff2^4=7k_1+5\\2^4=7k_1+5\Rightarrow2^4=7(2k_2+1)+5=14k_2+12\iff2^3=7k_2+6\text{
absurde }.$$
|
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|
Question about parametric divisibility relations Let $a,b,c,d$ be pairwise relatively prime positive integers such that the divisibility relations
$$
(kc-d) \mid \bigl((2k^2+1)b^2-2kab-(k^2-1)a^2\bigr)
$$
and
$$
(c-kd) \mid \bigl((k^2+2)b^2-2kab+(k^2-1)a^2\bigr)
$$
hold for every nonzero integer $k$.
QUESTION: Can any of $a,b,c,d$ be determined or characterized (even partially) from this information alone?
If it helps, I’m trying to prove $c=d=1$.
|
\begin{cases}
\begin{align}
(2k^2+1)b^2-2kab-(k^2-1)a^2 &= (kc-d)q, \\
(k^2+2)b^2-2kab+(k^2-1)a^2 &= (c-kd)r.
\end{align}
\end{cases}
\begin{align}
a &= ((k+1)s+(k-1)p)((k-1)p^2+2(k^3-k^2-1)ps+(k+1)(k^2+2)s^2), \\[0.5em]
b &= ((k+1)s+(k-1)p)(-(k-1)^2p^2+2k(k-1)ps+k^2(k^2-1)s^2, \\[0.5em]
c &= (2k^4-5k^3+7k^2-5k+1)p^2+(4k^4+2k^3-4k^2-2k-4)ps \\
&\hspace{2em} +(k^5+k^4+k^3+3k^2+4k+2)s^2, \\[0.5em]
d &= (k^4-k^3+3k^2-5k+2)p^2+(6k^4-4k^3-2k^2-4)ps \\
&\hspace{2em} +(k^6-k^5-k^4+3k^3+4k^2+4k+2)s^2, \\[0.5em]
q &= (k-1)^3p^4-2(2k+1)(k-1)^2sp^3-2k(k^2-1)(2k^2-2k-1)p^2s^2 \\
&\hspace{2em} -2(k^2-2)(k+1)^2ps^3+(k^4-2k^3-2)(k+1)^3s^4, \\[0.5em]
r &= -((k-1)p^2-2kps+(k+1)(k^2+2)s^2)((k+1)s+(k-1)p)^2.
\end{align}
|
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|
Induction: Bounds on sum of inverses of first $n$ square roots I am trying to show by induction that $2(\sqrt{n}-1)<1+\dfrac{1}{\sqrt{2}}+\cdots+\dfrac{1}{\sqrt{n}}<2\sqrt{n}$.
For the upper bound, I have that if $1+\dfrac{1}{\sqrt{2}}+\cdots+\dfrac{1}{\sqrt{n}}<2\sqrt{n}$, then $1+\dfrac{1}{\sqrt{2}}+\cdots+\dfrac{1}{\sqrt{n}}+\dfrac{1}{\sqrt{n+1}}<2\sqrt{n}+\dfrac{1}{\sqrt{n+1}}=$
$2\sqrt{n}+\dfrac{2}{2\sqrt{n+1}}<2\sqrt{n}+\dfrac{2}{\sqrt{n}+\sqrt{n+1}}=2\sqrt{n}+2(\sqrt{n+1}-\sqrt{n})=2\sqrt{n+1}$.
However, despite hours of trying, I cannot figure out how to show that $2(\sqrt{n+1}-1)<1+\dfrac{1}{\sqrt{2}}+\cdots+\dfrac{1}{\sqrt{n}}+\dfrac{1}{\sqrt{n+1}}$.
|
$$\begin{array}{cl}
& 1+\dfrac{1}{\sqrt{2}}+\cdots+\dfrac{1}{\sqrt{n}} \\
>& \dfrac2{1+\sqrt2}+\dfrac{2}{\sqrt{2}+\sqrt3}+\cdots+\dfrac{2}{\sqrt{n}+\sqrt{n+1}} \\
=& \dfrac{2(\sqrt2-1)}{(\sqrt2+1)(\sqrt2-1)}+\dfrac{2(\sqrt3-\sqrt2)}{(\sqrt{3}+\sqrt2)(\sqrt3-\sqrt2)}+\cdots+\dfrac{2(\sqrt{n+1}-\sqrt n)}{(\sqrt{n+1}+\sqrt{n})(\sqrt{n+1}-\sqrt n)} \\
=& 2(\sqrt2-1)+2(\sqrt3-\sqrt2)+\cdots+2(\sqrt{n+1}-\sqrt n) \\
=& 2\sqrt{n+1}-2 \\
\end{array}$$
|
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|
Finding $\int\frac{x}{\sqrt{3-2x-x^2}} dx$. I was looking for the integral of
$$\int\frac{x}{\sqrt{3-2x-x^2}} dx$$
My work:
$$\int \frac{x}{\sqrt{3-2x-x^2}} dx = \int \frac{x}{\sqrt{(3)+(-2x-x^2)}} dx $$
$$ = \int \frac{x}{\sqrt{(3)-(2x+x^2)}} dx $$
$$ = \int \frac{x}{\sqrt{(3)-(1+2x+x^2) +1}} dx $$
$$\int \frac{x}{\sqrt{3-2x-x^2}} dx = \int \frac{x}{\sqrt{(4)-(x+1)^2}} dx $$
Then I often remember this integral $\frac{u}{\sqrt{a^2 - x^2}} du$. So I modified the above integral to
look like the integral $\frac{u}{\sqrt{a^2 - x^2}}$.
$$\int \frac{x}{\sqrt{3-2x-x^2}} dx = \int \frac{x+1-1}{\sqrt{(4)-(x+1)^2}} dx $$
$$\int \frac{x}{\sqrt{3-2x-x^2}} dx = \int \frac{x+1}{\sqrt{(4)-(x+1)^2}} dx + \int \frac{-1}{\sqrt{(4)-(x+1)^2}} dx$$
I recognized the the last integral $\int \frac{-1}{\sqrt{(4)-(x+1)^2}} dx$ has the form $\int \frac{1}{\sqrt{a^2-u^2}} du$, where $a =2$ and $u = x+1$.
It's corresponding integral would be $\arcsin \left( \frac{u}{a}\right) + c$.
Evaluating $\int \frac{-1}{\sqrt{(4)-(x+1)^2}} dx$, it would be $-\arcsin \left( \frac{x+1}{2}\right)$
Here's the problem: I couldn't find the integral of $\int \frac{x+1}{\sqrt{(4)-(x+1)^2}} dx, $ because my Table of Integral doesn't show what is the
integral of $\frac{u}{\sqrt{a^2 - x^2}} du$.
How to evaluate the integral of $\frac{x}{\sqrt{3-2x-x^2}} dx$ properly?
|
Hint: You have done till here
$$\int \frac{x}{\sqrt{3-2x-x^2}} dx = \int \frac{x}{\sqrt{(4)-(x+1)^2}} dx$$
now let $x+1=2\sin t$ and simplify!
|
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|
How to calculate the determinant of this n by n matrix? Find the determinant of this n by n matrix.
$$
\begin{pmatrix}
0 & x_1 & x_2 & \cdots& x_k \\
x_1 & 1 & 0 & \cdots & 0 \\
x_2 & 0 & 1& \cdots & 0 \\
\vdots& \vdots& \vdots& \ddots & \vdots\\
x_k & 0 & 0 & \cdots& 1 \\
\end{pmatrix}
$$
where, $$ k=n-1 $$.
I am new to matrices and determinants, but this is what I did:
I developed the determinant using the second column:
$$
(-1)^2*x_1
\begin{pmatrix}
x_1 & 0 & \cdots & 0 \\
x_2 & 1& \cdots & 0 \\
\vdots& \vdots& \ddots& \vdots \\
x_k & 0 & \cdots& 1 \\
\end{pmatrix} + (-1)^3 *1
\begin{pmatrix}
0 & x_2 & \cdots & x_k \\
x_2 & 1& \cdots & 0 \\
\vdots& \vdots& \ddots& \vdots \\
x_k & 0 & \cdots& 1 \\\end{pmatrix}
$$
the first determinant is triangular, so its equal to $ x_1 $ but this is where I got stuck. I don't know what to do with the second determinant. Any help is appriciated. Thanks
|
You should just factorize:\begin{pmatrix}
0 & x_1 & x_2 & \cdots& x_k \\
x_1 & 1 & 0 & \cdots & 0 \\
x_2 & 0 & 1& \cdots & 0 \\
\vdots& \vdots& \vdots& \ddots & \vdots\\
x_k & 0 & 0 & \cdots& 1 \\
\end{pmatrix} = \begin{equation}\begin{pmatrix}
-1 & x_1& \cdots &x_k\\
0 &1 &\cdots &0\\
\vdots &&1\\
0&&&1
\end{pmatrix}\begin{pmatrix}
\sum x_i^2 & 0 &\cdots&\\
x_1&1&&\\
\vdots&&1&&\\
x_k&&&1
\end{pmatrix}
\end{equation}
and then \begin{equation}
\det\begin{pmatrix}
0 & x_1 & x_2 & \cdots& x_k \\
x_1 & 1 & 0 & \cdots & 0 \\
x_2 & 0 & 1& \cdots & 0 \\
\vdots& \vdots& \vdots& \ddots & \vdots\\
x_k & 0 & 0 & \cdots& 1 \\
\end{pmatrix}=\det\begin{pmatrix}
-1 & x_1& \cdots &x_k\\
0 &1 &\cdots &0\\
\vdots &&1\\
0&&&1
\end{pmatrix}\det\begin{pmatrix}
\sum x_i^2 & 0 &\cdots&\\
x_1&1&&\\
\vdots&&1&&\\
x_k&&&1
\end{pmatrix}
\end{equation}
\begin{equation}
=-1\cdot\sum x_i^2=-\sum x_i^2
\end{equation}
|
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|
Show $\sqrt[4]{2}$ on horizontal axis It is very basic question ,to show $\sqrt[2]{2}$ on $x$ axis .we do like below
but,how can we show $\sqrt[4]{2}$ on $x$ axis ?(like we do for $\sqrt2)$
I am thankful if you guide me .
$\bf{Remark}:$ may be a question for many people .
|
The following steps will work . . .
\begin{align*}
1.\;\,&\text{Construct a segment of length$\,\sqrt{2}$.}\\[4pt]
2.\;\,&\text{Construct a segment $AB$ of length$\,\sqrt{2} + {\small{\frac{1}{4}}}$.}\\[4pt]
3.\;\,&\text{Construct a circle with $AB$ as a diameter.}\\[4pt]
4.\;\,&\text{Construct a point $P$ on $AB$ such that $PB$ has length$\,\sqrt{2} - {\small{\frac{1}{4}}}$.}\\[4pt]
5.\;\,&\text{Construct a circle with center $B$, and radius $BP$.}\\[4pt]
6.\;\,&\text{Let $C$ be one of the two points where the circles intersect.}\\[10pt]
\end{align*}
\begin{align*}
\text{Then}\;\;|AC|^2 &= |AB|^2 - |BC|^2
\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\,
\\[4pt]
&=\left(\sqrt{2} + {\small{\frac{1}{4}}}\right)^2 -\left(\sqrt{2} - {\small{\frac{1}{4}}}\right)^2\\[4pt]
&=
\left(2+{\small{\frac{1}{2}}}\sqrt{2}+{\small{\frac{1}{16}}}\right)
-
\left(2-{\small{\frac{1}{2}}}\sqrt{2}+{\small{\frac{1}{16}}}\right)\\[4pt]
&=\sqrt{2}\\[8pt]
\text{hence}\;\;|AC|\;&= \sqrt[4]{2},\;\text{as required.}\\[4pt]
\end{align*}
|
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|
Partial Differentiation Ex 2.2 Eng Mathematics by H.K. Das Test for continuity when
$f(x,y) = \frac{x^3\,y^3}{x^3+y^3}$ when $x\neq0, y\neq0$ and f(x,y)=0 when $x=0,y=0$.
|
for $$x>0,y>0$$ we have $$x^3+y^3\geq 2(xy)^{3/2}$$ thus we obtain
$$\frac{x^3y^3}{x^3+y^3}\le \frac{x^3y^3}{2(xy)^{3/2}}=\frac{1}{2}(xy)^{3/2}->0$$
|
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|
let : $\forall n \in \mathbb{N}$ : $x_n=\sqrt{1+(1+1/n)^2}+\sqrt{1+(1-1/n)^2}$ then : $\displaystyle{\sum_{i=1}^{20}}\dfrac{1}{x_i}=?$ let : $\forall n \in \mathbb{N}$ : $x_n=\sqrt{1+(1+1/n)^2}+\sqrt{1+(1-1/n)^2}$
then :
$$\sum_{i=1}^{20}\dfrac{1}{x_i}=?$$
my try :
$$\dfrac{1}{\sqrt{5}}+\dfrac{2}{\sqrt{5}+\sqrt{13}}+...$$
But boring .
|
I notice that:
$$\begin{align}\frac{1}{\sqrt{1+(1+1/n)^2}+\sqrt{1+(1-1/n)^2}} &= \frac{\sqrt{1+(1+1/n)^2}-\sqrt{1+(1-1/n)^2}}{(1+(1+1/n)^2)-(1+(1-1/n)^2)}\\
&= \frac{\sqrt{1+(1+1/n)^2}-\sqrt{1+(1-1/n)^2}}{(1+1/n)^2-(1-1/n)^2}\\
&= \frac{\sqrt{1+(1+1/n)^2}-\sqrt{1+(1-1/n)^2}}{4/n}\\
&= \frac14\left(n\sqrt{1+(1+1/n)^2}-n\sqrt{1+(1-1/n)^2}\right)\\
&= \frac14\left(\sqrt{n^2+(n+1)^2}-\sqrt{n^2+(n-1)^2}\right)
\end{align}$$
That last expression provides for a telescoping sum:
$$\frac14\left(\left(\sqrt{1^2+2^2} - \sqrt{1^2+0^2}\right) + \left(\sqrt{2^2+3^2} - \sqrt{2^2+1^2}\right) + \cdots + \left(\sqrt{19^2+20^2} -
\sqrt{19^2+18^2}\right) + \left(\sqrt{20^2+21^2} -
\sqrt{20^2+19^2}\right)\right)$$
which collapses down to:
$$\frac14\left(\sqrt{20^2+21^2} - \sqrt{1^2+0^2}\right)$$
|
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|
How to integrate $\int \frac {e^y}{y} dy$? The question is to evaluate $$\iint_R \frac {x}{y} e^y dx dy$$ where R is the region bounded by $0 \leq x \leq 1$ and $x^2 \leq y \leq x$.
So i write it as $$\int_0^1 \int_{x^2}^{x} \frac{x}{y} e^y dy dx$$.
The thing is, how do i evaluate $I=\int_{x^2}^{x} \frac{1}{y} e^y dy$? I tried integration by parts but failed. Then i tried to use infinite series.
Use $$e^x=\sum_{n=0}^{\infty} \frac{x^n}{n!}$$
we get \begin{align}
I & =\int_{x^2}^{x} \frac{1}{y}+1+\frac{y}{2!}+\frac{y^2}{3!}+\cdots dy \\
&=ln(\frac{1}{x})+(x-x^2)+\frac{1}{4}(x^2-x^4)+\frac{1}{18}(x^3-x^6)+ \cdots
\end{align}
Now get back to $\int_0^1 I dx$. We need to find $P=\int_0^1 ln(\frac{1}{x})$ and $Q=\int_0^1 (x-x^2)+\frac{1}{4}(x^2-x^4)+\frac{1}{18}(x^3-x^6)+ \cdots$
integrate by parts, we find $P=0$.
But for Q,we have $\left((\frac{1}{2}x^2-\frac{1}{3}x^3)+\frac{1}{4}(\frac{1}{3}x^3-\frac{1}{5}x^5)+ \cdots \right)_0^1$
Can we simplify the infinite sum in the bracket as a simple result?
|
There is a theorem that says when you are given double integrals, you must flip them. Well, not always, but in this case you want to do that.
Reparametrize your region $R$ as $R = \{(x,y) | \sqrt{y} \le x \le y, 0 \le x \le 1\}.$
Then by Fubini, we can write
$$\int_0^1 \int_{x^2}^x \frac{x}{y}e^y dydx = \int_0^1 \int_{\sqrt{y}}^y \frac{x}{y}e^y dxdy.$$
This is much easier to integrate. You end up having to integrate $e^y$ and $ye^y$. The former is immediate, the latter an integration by parts.
|
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|
A floor function equation I want to solve below equation analytically $$\lfloor \frac{2x+1}{3}\rfloor +\lfloor \frac{\lfloor4x\rfloor+2}{3}\rfloor=1$$ but I have no idea to start .
Implicit: I solve it by graphing and the solution was $x\in[\frac {1}{4},1)$
I am thankful for any idea in advance .
I try to use $x=n+p \\0\leq p<1 $ but can't go further $$\lfloor \frac{2n+2p+1}{3}\rfloor +\lfloor \frac{\lfloor4n+4p\rfloor+2}{3}\rfloor=1\\0\leq 2p<2 \\0\leq 4p<4 $$
|
Use the fact that for each $x$ we have:
$$x-1<[x]\leq x$$
So $$ \frac{2x+1}{3}+ \frac{\lfloor4x\rfloor+2}{3}-2<\lfloor \frac{2x+1}{3}\rfloor +\lfloor \frac{\lfloor4x\rfloor+2}{3}\rfloor \leq \frac{2x+1}{3}+ \frac{\lfloor4x\rfloor+2}{3}$$
So $$ \frac{2x+1}{3}+ \frac{4x+1}{3}-2<1 \leq \frac{2x+1}{3}+ \frac{4x+2}{3}$$
So So $$ \frac{6x+2}{3}-2<1 \leq \frac{6x+3}{3}$$
Thus: $6x-4<3\leq 6x+3$ and so $0\leq x<{7\over 6}$ and this is not the end...
|
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|
For which $k \in \mathbb{R}$ the vector $(1,k^2-k,k) \in Im(f)$ where $f$ is a linear map? Let $f:\mathbb{R^3}\to \mathbb{R^3} $ the map defined by $A_h$
$$
A_h= \begin{pmatrix}
2 & 1 & -1 \\
1 & 2 & 1 \\
-1 & 1 & h \\
\end{pmatrix}
$$
where $h \in \mathbb{R}$. $f(x,y,z)=(x-y-2z,x+2y+z,3y+3z)$
I have found the value of $h$ for which $\dim(\ker(f))>0$ which is $h=2$.
Then a basis of $Im(f)$ is $\{(1,2,1),(-1,1,2)\}$.
$$
A_2= \begin{pmatrix}
2 & 1 & -1 \\
1 & 2 & 1 \\
-1 & 1 & 2 \\
\end{pmatrix}
$$
I have to determine for which $k \in \mathbb{R}$ the vector $v=(1,k^2-k,k) \in Im(f)$ (with $h=2$).
My attempt
I tried to find $k$ such that $v$ is equal to the linear combination of the two vectors of $Im(f)$, but it does not work.
I also tried to find $k$ such that $v$ is equal to $(x-y-2z,x+2y+z,3y+3z)$ and I got $k=1+\sqrt{2}$ or $k=1-\sqrt{2}$, but I think I am wrong.
What is the right approach?
|
$$a(1,2,1) + b(-1,1,2) = (1,k^2-k,k)$$
Separating components:
$$\begin{cases}
a - b &=& 1 &(1) \\
2a + b &=& k^2-k &(2) \\
a+2b &=& k &(3)
\end{cases}$$
$(1)+(3)$ gives:
$$2a+b=k+1$$
Comparing it with $(2)$ gives:
$$k^2-k=k+1$$
Solving the quadratic equation gives:
$$k=\dfrac{2\pm\sqrt{8}}{2}=1\pm\sqrt2$$
We're basically done here.
Let $k=1+\sqrt2$.
$(3)-(1)$ gives:
$$3b=k-1=\sqrt2$$
Thus:
$$\begin{cases}
a &=& \dfrac{\sqrt2}{3}+1\\
b &=& \dfrac{\sqrt2}3
\end{cases}$$
Checking:
$$\begin{cases}
a - b &=& 1 \\
2a + b &=& 2+\sqrt2 \\
a+2b &=& 1+\sqrt2
\end{cases}$$
Let $k=1-\sqrt2$.
$(3)-(1)$ gives:
$$3b=k-1=-\sqrt2$$
Thus:
$$\begin{cases}
a &=& -\dfrac{\sqrt2}{3}+1\\
b &=& -\dfrac{\sqrt2}3
\end{cases}$$
Checking:
$$\begin{cases}
a - b &=& 1 \\
2a + b &=& 2-\sqrt2 \\
a+2b &=& 1-\sqrt2
\end{cases}$$
|
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|
How to solve $\cos({r^{-4}}\cos 4\theta - {4\theta}) = 0$ for $\theta$ and $r$? Problem Statement
How does one solve the trigonometric equation
$$\cos({r^{-4}}\cos 4\theta - {4\theta}) = 0$$ for $\theta$ and $r$?
My Try
Taking the inverse cosines of both sides, I get
$${r^{-4}}\cos 4\theta - {4\theta} = \cos^{-1}(0) = \frac{\pi}{2}+k\pi.$$
Rearranging the last equation gives
$$\frac{\cos 4\theta}{4 \theta} - r^4 = \frac{{r^4}\left(\frac{\pi}{2}+k\pi\right)}{4 \theta}.$$
This is where I get stuck.
QUERY
Of course, I know that I can rewrite the last equation as
$$\cos 4\theta = {r^4}\left(4\theta + \frac{\pi}{2} + k\pi\right).$$
Unfortunately, I do not know what to do or what approach to take past this point. Should I use iteration to approximate the solution(s), if any? Or is it (logically) possible to prove that there do not exist any solutions to this equation?
Thanks!
Added September 15 2017
Actually, I have
$$r = (x^2 + y^2)^{1/2} := f(x, y)$$
and
$$\theta = \tan^{-1}\left(\frac{y}{x}\right) := g(x,y).$$
Does that help?
|
Notice that, from the equation
$$\cos 4\theta = {r^4}\left(4\theta + \frac{\pi}{2} + k\pi\right),$$
and the equation
$$r = (x^2 + y^2)^{1/2}$$
I get that $r \geq 0$, so that
$$\cos 4\theta \geq 0.$$
This last inequality implies that
$$-\frac{\pi}{2} \leq 4\theta \leq \frac{\pi}{2}$$
which implies that
$$-\frac{\pi}{8} \leq \theta \leq \frac{\pi}{8}.$$
(Note that, when $r = 0$, then $\theta = -\pi/8, \pi/8$.)
We now bound $r$ from above, given that
$$r^4 = \frac{\cos 4\theta}{4\theta + \frac{\pi}{2} + k\pi}.$$
An upper bound for $\cos 4\theta$ is given by $1$. However, there is no lower bound for $4\theta + \frac{\pi}{2} + k\pi$, since $k \in \mathbb{Z}$.
Thus, there is no global maximum for
$$\frac{\cos 4\theta}{4\theta + \frac{\pi}{2} + k\pi}.$$
Hence, $r^4$ is not bounded from above.
Consequently, we conclude that the equation
$$\cos 4\theta = {r^4}\left(4\theta + \frac{\pi}{2} + k\pi\right)$$
is solvable when
$$\frac{-\pi}{8} \leq \theta \leq \frac{\pi}{8}.$$
|
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|
Prove $ \forall n \ge 4$, $n^{3} + n < 3^{n}$
Prove that $\forall n \ge 4$, $n^{3} + n < 3^{n}$
My attempt: Base case is trivial.
Suppose $ \ n \ge 4$, $n^{3} + n < 3^{n}$. Then,
$$ (n+1)^{3} + (n+1) = n^3 + n + 3n^2 + 3n + 1 +1 < 3^{n} + 3n^2 + 3n + 2 \\< 3\cdot 3^{n} + 3n^2 + 3n +2.$$
Not sure how to get rid of $3n^2 + 3n +2$.
|
Hint:
$$ 3n^2+3n+2<n^3+n<3^n. $$
|
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|
Expectation of $Y := \min\{\frac{a}2, X\}$ if $X$ is uniform on $(0,a)$ The density of X is given by
$
f(x)=
\begin{cases}
\frac{1}{a}&\, 0 \leq x < a\\
0&\, otherwise\\
\end{cases}
$
Y := min{$\frac{a}{2}$, X} so Y belongs to [0, $\frac{a}{2}$]
Based on this solution, $E(Y) = \frac{3a}{8}$. Why is $E(Y) = \frac{3a}{8}$?
My Work
$
F_Y(y)=
\begin{cases}
\frac{y}{a}&\, 0 \leq y < \frac{a}{2}\\
1&\, y \geq \frac{a}{2}\\
\end{cases}
$
$
(F_Y(y))' = f_Y(y)=
\begin{cases}
\frac{1}{a}&\, 0 \leq y < \frac{a}{2}\\
0&\, y \geq \frac{a}{2}\\
\end{cases}
$
$
E(Y) = \int_{0}^{\frac{a}{2}} yf(y) dy = \int_{0}^{\frac{a}{2}} \frac{y}{a} dy = \frac{a}{8}
$
|
A different approach would be:
$$EY=E\left[Y|X<\frac{1}{2}a\right]+E\left[Y|X\ge\frac{1}{2}a\right]= \int_0^{\frac{1}{2}a}x\frac{1}{a}dx= 0.125a+ 0.5a=\frac{3}{8}a$$
It relays on the properties of conditional expectations.
|
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|
Solving $y' = y^{2} + 1$ if $y(0) = 1$ The solution to the initial-value problem $y' = y^{2} + 1$ with $y(0)=1$ is $y = \tan(x + \frac{\pi}{4})$. I would like to show that this is the correct solution in a way that is analogous to my solution to differential equation $y' = y - 12$.
Solution to Differential Equation
If $y' = y - 12$,
\begin{equation*}
\frac{y'}{y - 12} = 1 ,
\end{equation*}
\begin{equation*}
\left(\ln(y - 12)\right)' = 1 ,
\end{equation*}
\begin{equation*}
\ln(y - 12) = t + C
\end{equation*}
for some constant $C$. If $C' = e^{C}$,
\begin{equation*}
y = C' e^{t} + 12 .
\end{equation*}
|
The equation can be written as
$$\frac {y'}{y^2+1}=1 $$
and after integration
$$\arctan (y)=x+C $$
for $x=0$, it gives
$$\arctan (1)=0+C=\frac {\pi}{4} $$
thus
$$y=\tan (x+\frac {\pi}{4}) .$$
|
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|
$\arctan{x}=\arccos{2x}.$ First, let's determine some domains. By definition, it follows that $\arccos:[-1,1]\rightarrow[0,\pi]$ and $\arctan:\mathbb{R}\rightarrow\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$. Since the domain of $\arctan$ is the entire reals, the restricting factor here should be the domain of arccos. This means that $2x\in[-1,1]\Rightarrow x\in\left[-\frac{1}{2},\frac{1}{2}\right]$.
I now take cosine of both sides and get $$\cos{(\arctan{x})}=\cos{(\arccos{2x})}=2x.$$
I use the formular $\cos{a}=\frac{1}{\sqrt{1+\tan^2{a}}},$ and rewrite as
$$\frac{1}{\sqrt{1+\tan^2{(\arctan{x})}}}=\frac{1}{\sqrt{1+x^2}}=2x.$$
Squaring both sides, simplifying and setting $t=x^2$ gives the quadratic equation
$$t^2+t-\frac{1}{4}=0\Longleftrightarrow \left\{
\begin{array}{rcr}
t_1 & = & -\frac{1}{2}-\frac{\sqrt{2}}{2} <0 \\
t_2 & = & -\frac{1}{2}+\frac{\sqrt{2}}{2} >0\\
\end{array}
\right.$$
Since $t_1<0$, it is a false root since a square of a real number $x$ can not become a negative real number $t$. So the final two roots in terms of $x$ are
$$\left\{\begin{array}{rcr}
x_1 & = & +\sqrt{-\frac{1}{2}+\frac{\sqrt{2}}{2}} = \sqrt{\frac{1}{2}(\sqrt{2}-1)} \\
x_2 & = & -\sqrt{-\frac{1}{2}+\frac{\sqrt{2}}{2}} = -\sqrt{\frac{1}{2}(\sqrt{2}-1)} \\
\end{array}
\right.$$
Clearly, both $x_1,x_2\in[-\frac{1}{2},\frac{1}{2}]$. Only $x_1$ is correct however. I did something wrong setting up the domain didin't I? I have a feeling that the correct domain for the solutions is $[0,\frac{1}{2}],$ I don't know how to see it.
|
Since $x \in \left[\frac{1}{2}, \frac{1}{2}\right]$ we have $\arctan x \in \left[-\arctan\frac{1}{2}, \arctan\frac{1}{2}\right] \subseteq \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$, so $\cos(\arctan x) \ge 0$.
So $\cos(\arctan x) = 2x$ implies $x \geq 0$, which discards the negative solution $x_2$.
|
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Calculating the limit $\lim\limits_{x \to 0^+} \frac{\sqrt{\sin x}-\sin\sqrt{ x}}{x\sqrt{x}}$
Calculate $$\lim\limits_{x \to 0^+} \dfrac{\sqrt{\sin x}-\sin \sqrt{x}}{x\sqrt{x}}$$
without use Taylor serie and L'Hôpital.
$$\lim\limits_{x \to 0^+} \dfrac{\sqrt{\sin x}-\sin \sqrt{x}}{x\sqrt{x}}\cdot\dfrac{\sqrt{\sin x}+\sin \sqrt{x}}{\sqrt{\sin x}+\sin \sqrt{x}}=\lim\limits_{x \to 0^+} \dfrac{\sin x-\sin^2\sqrt{x}}{x\sqrt{x}(\sqrt{\sin x}+\sin \sqrt{x})}$$
now what ?
|
$$\mathrm L =\lim\limits_{x \to 0^+} \dfrac{\sqrt{\sin x}-\sin \sqrt{x}}{x\sqrt{x}}= \lim\limits_{x \to 0^+} \dfrac{(\sqrt{\sin x}- \sqrt{x}) -(\sin \sqrt{x} - \sqrt{x})}{x\sqrt{x}} \\= \overbrace{\lim\limits_{x \to 0^+}\dfrac{(\sqrt{\sin x}- \sqrt{x})}{x\sqrt{x}}}^{\large \rm L^\prime} - \overbrace{\lim\limits_{x \to 0^+}\dfrac{(\sin \sqrt{x} - \sqrt{x})}{x\sqrt{x}}}^{\large \rm L^{\prime\prime}}$$
Let $y = \sqrt{x}$
Then $${\rm L^{\prime\prime}} = \lim\limits_{x \to 0^+}\dfrac{(\sin \sqrt{x} - \sqrt{x})}{x\sqrt{x}} = \lim\limits_{x \to 0^+}\dfrac{(\sin y - y)}{y^3} = -\dfrac 16$$
For the other limit,
$${\rm L^{\prime}}=\lim\limits_{x \to 0^+}\dfrac{(\sqrt{\sin x}- \sqrt{x})}{x\sqrt{x}} =\lim\limits_{x \to 0^+}\dfrac{(\sin x- x)}{x^3}\dfrac{x\sqrt{x}}{(\sqrt{\sin x} + \sqrt{x})}= \dfrac {-1}6\lim\limits_{x \to 0^+}\dfrac{x}{\left(\sqrt{\dfrac{\sin x}{x}} + 1\right)} = 0$$
Hence our limit is $\mathrm L = \dfrac {1}{6}$.
I used $$\lim_{x \to 0} \dfrac{\sin x - x}{x^3} = \dfrac{-1}6$$
Are all limits solvable without L'Hôpital Rule or Series Expansion ?
|
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"timestamp": "2023-03-29T00:00:00",
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|
Computing $\int_0^1 \sin(3\pi x)\sin(2\pi x) \mathrm{d}x$
How do I show
$$\int_0^1 \sin(3\pi x)\sin(2\pi x) \mathrm{d}x=0.$$
I tried using
$$\sin (x) \sin (y) = \cos ( x \pm y ) \mp \cos (x) \cos (y) $$
Then the integral becomes
$$\int_0^1\cos(3\pi x)\cos(2\pi x) \mathrm{d}x$$
Which doesn't help me. I also tried using Taylor series expansion of sinus but I don't see how it is helpful since we have the product of two infinite sums.
|
Using integration by parts twice,
$\begin{align}J&=\int_0^1 \sin\left(3\pi x\right)\sin\left(2\pi x\right)\,dx\\
&=\left[-\frac{1}{2\pi}\sin\left(3\pi x\right)\cos\left(2\pi x\right)\right]_0^1+\frac{1}{2\pi}\int_0^1 \sin\left(3\pi x\right)\cos\left(2\pi x\right)\, dx\\
&=\frac{1}{2\pi}\int_0^1 \sin\left(3\pi x\right)\cos\left(2\pi x\right)\, dx\\
&=\left[\frac{1}{(2\pi)^2}\sin\left(3\pi x\right)\sin\left(2\pi x\right)\right]_0^1-\frac{1}{(2\pi)^2}J\\
&=-\frac{1}{(2\pi)^2}J
\end{align}$
Since $\displaystyle 1+\frac{1}{(2\pi)^2}\neq 0$ then $J=0$.
|
{
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Summation involving double factorials How can we evaluate exact infinite sum of the following?
$${ \sum_{r=1}^{\infty}\frac{(2r)!!}{(2r+3)!!}}$$
where
$k!! = \begin{cases} 2\cdot4\cdot6 . . . k, & \text{if $k$ is even} \\ 1\cdot3\cdot5 . . . k, & \text{if $k$ is odd} \end{cases}$
What approach do we need to solve such type of summations?
My attempt:
\begin{align}
\sum_{r=1}^{\infty}\frac{(2r)!!}{(2r+3)!!}
&= \sum_{r=1}^{\infty}\frac{(2r)!!\cdot (2r+2)!!}
{(2r+3)!!(2r+2)!!} \\
&= \sum_{r=1}^{\infty}\frac{2^r\cdot (r)!\cdot 2^{r+1}\cdot (r+1)!}{(2r+3)!} \\
&= \sum_{r=1}^{\infty}\frac{2^{2r+1}\cdot (r)!\cdot (r+1)!}{(2r+3)!}
\end{align}
I also tried using the generalized binomial theorem but it doesn't seem to follow the pattern.
EDIT: I suspect it might be solved using telescopic method, but can't figure out how.
|
Starting with
My attempt:
\begin{align}
S = \sum_{r=1}^{\infty}\frac{(2r)!!}{(2r+3)!!}
&= \sum_{r=1}^{\infty}\frac{(2r)!!\cdot (2r+2)!!}
{(2r+3)!!(2r+2)!!} \\
&= \sum_{r=1}^{\infty}\frac{2^r\cdot (r)!\cdot 2^{r+1}\cdot (r+1)!}{(2r+3)!} \\
&= \sum_{r=1}^{\infty}\frac{2^{2r+1}\cdot (r)!\cdot (r+1)!}{(2r+3)!}
\end{align}
One can continue as follows:
\begin{align}
S &= \sum_{r=1}^{\infty}\frac{2^{2r+1}\cdot (r)!\cdot (r+1)!}{(2r+3)!} \\
&= 2 \, \sum_{r=1}^{\infty} \frac{\Gamma(r+1) \, \Gamma(r+2)}{ \Gamma(2r + 4) } \, 4^{r} \\
&= \frac{2}{3!} \, \sum_{r=1}^{\infty} \frac{(1)_{r} \, (2)_{r}}{4^{r} \, (2)_{r} \, \left(\frac{5}{2}\right)_{r} } \, 4^{r} \\
&= \frac{1}{3} \, \sum_{r=1}^{\infty} \frac{(1)_{r} \, (1)_{r}}{r! \, \left(\frac{5}{2}\right)_{r} } = 2 \left[ {}_{2}F_{1}\left(1, 1; \frac{5}{2}; 1\right) -1 \right] \\
&= \frac{1}{3} \left[\frac{\Gamma(5/2) \, \Gamma(1/2)}{\Gamma^{2}(3/2)} - 1 \right] = \frac{2}{3}.
\end{align}
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find the range of function We have the function \begin{equation*}f(x)=-\frac{(x-3)^2}{x+1}\end{equation*}
I want to determine the domain and the range of the function.
The root of the deniminator is $x=-1$. Therefore, the domain is $D_f=\mathbb{R}\setminus\{-1\}=(-\infty, -1)\cup (-1,+\infty )$.
Does it holds that \begin{equation*}W_f=f(D_f)=f\left ((-\infty, -1)\cup (-1,+\infty )\right )=f\left ((-\infty , -5)\cup (-5,-1) \cup (-1,3) \cup (3,+\infty)\right )\end{equation*} ?
If this is correct, then we have to know the monotonicity of $f$ at each of these intervalls.
We have that at $(-\infty , -5)$ the function is decreasing, at $(-5,-1)$ the function is incresing, at $(-1,3)$ the function is increasing and at (3,+\infty)$ the function is decreasing.
We have the following:
*
*\begin{align*}f\left ((-\infty , -5)\right )= \left (16, +\infty\right ) \end{align*}
*\begin{align*}f\left ((-5,-1)\right )= \left (16, +\infty\right )\end{align*}
*\begin{align*}f\left ((-1,3)\right )= \left (-\infty, 0\right )\end{align*}
*\begin{align*}f\left ((3,+\infty)\right ) = \left (-\infty, 0\right )\end{align*}
Therefore the range is
\begin{equation*}W_f= \left (-\infty, 0\right )\cup \left (16, +\infty\right ) \end{equation*}
$$$$
Is everything correct? Or do we have to determine the range in an other way?
|
$$-\frac{(x-3)^2}{x+1}=\frac{-x^2+6x-9}{x+1}=\frac{-x^2+6x+7-16}{x+1}=7-x-\frac{16}{x+1}.$$
Thus, for $x<-1$ by AM-GM we obtain
$$f(x)=8-x-1-\frac{16}{x+1}\geq8+2\sqrt{(-x-1)\left(-\frac{16}{x+1}\right)}=16$$
and for $x>-1$ by AM-GM again we obtain:
$$f(x)=8-x-1-\frac{16}{x+1}\leq8-2\sqrt{(x+1)\cdot\frac{16}{x+1}}=0.$$
Since $\lim\limits_{x\rightarrow-1^+}f(x)=-\infty$ and $\lim\limits_{x\rightarrow-1^-}f(x)=+\infty$ and $f$ is a continuous function
on all interval of the domain, we got the answer:
$$(-\infty,0]\cup[16,+\infty).$$
|
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|
Finding all values such that column vector is a linear combination Question:
For which values(s) of $ \ a$ is the column $ \ c = \begin{bmatrix} a\\a^2\\0\\a+1\end{bmatrix}$ a linear combination of the columns of,
$ \ x = \begin{bmatrix} 1\\1\\1\\1\end{bmatrix}$, $\ y = \begin{bmatrix} 1\\0\\-1\\0\end{bmatrix}$, $ \ z= \begin{bmatrix} 0\\1\\0\\1\end{bmatrix}$.
My attempt:
$ \ c = \begin{bmatrix} a\\a^2\\0\\a+1\end{bmatrix} = \ r\begin{bmatrix} 1\\1\\1\\1\end{bmatrix} + s \begin{bmatrix} 1\\0\\-1\\0\end{bmatrix} + t \begin{bmatrix} 0\\1\\0\\1\end{bmatrix} $
We know its a linear combination $\iff $ $ \ r+s = a, r+t = a^2, r-s = 0, r+t = a+1$ . I am not sure how to find all values of $ \ a$ now.
|
$\require{cancel}$
Using your four obtained equations, you know that $r=s$ and so $2r=a.$ You then have $t=a^2-\frac{a}{2}.$ Inserting all that into the last equation, you get
$$\cancel{\frac{a}{2}}+a^2-\cancel{\frac{a}{2}}=a+1,$$
the solutions of which are $\frac{1\pm\sqrt{5}}{2}.$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Determine the derivative of $\arctan$ function Find $f'(x)$ for the function:
$f(x)= \arctan(\frac{a+x}{1-ax}))$ , $a\in R$
So this is what I've done:
$f(x) = \arctan x$
$f'(x) = \frac{1}{1+x^2}$
$x= \frac{a+x}{1-ax}$
$f'(x) = \frac{1}{1+\frac{(a+x)^2}{(1-ax)^2}}$
$f'(x) = \frac{(1-ax)^2}{(1-ax)^2+(a+x)^2}$
Is this correct?
|
$$\arctan\left(\frac{x+a}{1-ax}\right)=\arctan x+\arctan a$$
(give or take a multiple of $\pi$), so
$$\frac d{dx}\arctan\left(\frac{x+a}{1-ax}\right)=\frac1{1+x^2}.$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
$\frac{a^3}{b^2} + \frac{b^3}{c^2} + \frac{c^3}{a^2} \geq 3 \, \frac{a^2 + b^2 + c^2}{a + b + c}$ Proposition
For any positive numbers $a$, $b$, and $c$,
\begin{equation*}
\frac{a^3}{b^2} + \frac{b^3}{c^2} + \frac{c^3}{a^2}
\geq 3 \, \frac{a^2 + b^2 + c^2}{a + b + c} .
\end{equation*}
I am requesting an elementary, algebraic explanation to this inequality. (I suppose the condition for equality is that $a = b = c$.) I am not familiar with symmetric inequalities in three variables. I would appreciate any references.
|
By C-S and Holder we obtain:
$$\sum_{cyc}\frac{a^3}{b^2}=\sum_{cyc}\frac{a^5}{a^2b^2}\geq\frac{\left(a^{\frac{5}{2}}+a^{\frac{5}{2}}+a^{\frac{5}{2}}\right)^2}{\sum\limits_{cyc}a^2b^2}=$$
$$=\frac{\left(a^{\frac{5}{2}}+a^{\frac{5}{2}}+a^{\frac{5}{2}}\right)^2(a+b+c)}{(a+b+c)\sum\limits_{cyc}a^2b^2}\geq\frac{(a^2+b^2+c^2)^3}{(a^2b^2+a^2c^2+b^2c^2)(a+b+c)}\geq$$
$$\geq\frac{3(a^2b^2+a^2c^2+b^2c^2)(a^2+b^2+c^2)}{(a^2b^2+a^2c^2+b^2c^2)(a+b+c)}=\frac{3(a^2+b^2+c^2)}{a+b+c}.$$
Done!
|
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|
Is the limit $\lim_{(x,y) \to \infty} \frac{x+2y}{x^2 - 2xy + 2y^2}$ zero? I have this limit:$$\lim_{(x,y) \to \infty} \frac{x+2y}{x^2 - 2xy + 2y^2}$$
At first sight seems that limit equals 0. But WolframAlpha says that there is no limit. I tried to prove it. I considered cases $y = kx$, and so on. I never got to find subsequence, whose has limit $\neq 0$.
I think there is a problem in denominator. When $x\rightarrow \infty$ and $y \rightarrow \infty$ we got there $\infty - \infty + \infty$. It's unclear what to do with it and how to find necessary subsequence.
Maybe i'm on the wrong way to solve it. Please, give me a tip.
|
$$
\begin{align}
(x-y)^2+y^2
&=x^2-2xy+2y^2\\
&=(2-\phi)x^2+(\phi-1)x^2-2xy+\phi y^2+(2-\phi)y^2\\
&=(2-\phi)\left(x^2+y^2\right)+\left(\sqrt{\phi-1}\,x+\sqrt{\phi}\,y\right)^2\\
&\ge(2-\phi)\left(x^2+y^2\right)\\
&=\frac1{\phi^2}\left(x^2+y^2\right)
\end{align}
$$
Therefore,
$$
\begin{align}
\left|\frac{x+2y}{(x-y)^2+y^2}\right|
&\le\phi^2\frac{|x+2y|}{x^2+y^2}\\
&\le\frac{3\phi^2}{\sqrt{x^2+y^2}}
\end{align}
$$
|
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|
Find $\operatorname{ord}_{17} (a)$ for all $a = 1, \ldots, 16$ The $\operatorname{ord}_{m} (a)$ is defined as the least positive integer $x$ satisfying $a^x \equiv 1 \mod m$.
To find $\operatorname{ord}_{17} (a)$ for all $a = 1, ..., 16,$ I know I can go through each numbers from 1 to 16 one by one (and have gotten 1, 8, 16, 4, 16, 16, 16, 8, 8, 16, 16, 16, 4, 16, 8, 2), but is there any trick to it without having to go through them one by one?
|
So here are some tricks which might help to reduce the work.
If $a\neq \pm 1$ and the order of $a$ is even, then $-a(\equiv 17-a)$ will have the same order as $a$.
The numbers for which $a^r\equiv 1 \bmod m$ form a group. Therefore $a$ and $a^{-1}$ have the same order. Here you have $18\equiv 1 \bmod 17$ and $18=2\times 9=3\times 6$ so that $9$ has the same order as $2$ and $6$ has the same order as $3$. Also $35 \equiv 1$ so you can do the same trick with $5$ and $7$.
Further, if $a$ has even order, $a^2$ will have half the order of $a$.
So $1$ has order $1$ and $-1$ has order $2$.
$2$ has the same order as $15$ and $9$ and $8$
$3$ has the same order as $14$ and $6$ and $11$
$4$ has the same order as $13$ and $4^2=16\equiv -1$ so $4$ has order $4$.
$5$ has the same order as $7, 12, 10$
So we can combine these as follows. $2^2=4$ has order $4$ so $2$ has order $8$ (clearly a factor of $8$ and can't be less).
$3^2=9$ has order $8$, so $3$ has order $16$ (again easy to check can't be less).
$5^2=25\equiv 8$ has order $8$ so $5$ has order $16$ too.
This is made somewhat easier by the fact that $17-1=16$ has only the prime factor $2$, and squaring is easy. Tricks are fine for small cases, but a more systematic approach wins out in general.
|
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|
Equation of a straight line passing through passing through a point and equally inclined to two other lines
Find the equation of the straight line passing through the point $(4,5)$ and equally inclined to the lines $3x= 4y+7$ and $5y=12x+6$.
I know that the equation of the bisector is given by:
$\dfrac{a_1x+b_1y+c_1}{\sqrt{a_1^2+b_1^2}}$=$\pm$$\dfrac{a_2x+b_2y+c_2}{\sqrt{a_2^2+b_2^2}}$
but I am facing real difficulty in finding which sign I should choose and why?
the answers are: $9x-7y=1 $ and $7x+9y=73$
Here is a Desmos graph of the above equation.
|
You have TWO lines which form equal angles with the given straight lines
$3x-4y-7=0;\;12x-5y+6=0$
$\dfrac{3x-4y-7}{\sqrt{3^2+4^2}}=\pm\dfrac{12x-5y+6}{\sqrt{12^2+5^2}}$
$\dfrac{3x-4y-7}{5}=\pm\dfrac{12x-5y+6}{13}$
$13(3x-4y-7)=\pm 5(12x-5y+6)$
$99 x-77 y-61=0;\;21 x+27 y+121=0$
$y=\dfrac{9 x}{7}-\dfrac{61}{77};\;y=-\dfrac{7 x}{9}-\dfrac{121}{27}$
If the wanted lines gave to pass through the point $(4,5)$ then their equation is
$y-5=m(x-4)$
where $m_1=\dfrac{9}{7};\;m_2=-\dfrac{7}{9}$
Hope this helps
|
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|
$f(x)=?$ if we have $f\left(\frac{x}{x^2+x+1}\right)=\frac{x}{x^2-x+1}$ $f(x)=?$
If we have $$f\left(\frac{x}{x^2+x+1}\right)=\frac{x}{x^2-x+1}$$ to fractions are very similar. I don't have an idea to find $f(x)$. Can someone show me a clue ?
|
Let $y = \frac x{x^2+x+1}$ and $z = \frac x{x^2-x+1}$. Then we have
\begin{align}
y(x^2+x+1) = x &\implies y(x^2-x+1)=x -2xy\\
&\implies y=z(1-2y)\\
&\implies z=\frac y{1-2y}\\
\end{align}
which gives us $f(y) = \frac{y}{1-2y}$.
|
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|
If $\alpha,a,b$ are integers and $b\neq-1$, then prove that, if $\alpha$ satisfies the equation $x^2+ax+b+1=0$, $a^2+b^2$ must be composite.
Let $\alpha,a,b$ be integers such that $b\neq-1$. Assume that $\alpha$ satisfies the equation $x^2+ax+b+1=0$. Prove that the integer $a^2+b^2$ must be composite.
$\alpha=\frac{-a\pm\sqrt{a^2-4(b+1)}}{2}$is an integer. But manipulating this expression is leading me to nowhere. Please help.
|
Let the roots be $r,s$.
By hypothesis, at least one of $r,s$ is an integer.
By Vieta's formulas
\begin{align*}
r + s &= -a\\[4pt]
rs &= b + 1\\[4pt]
\end{align*}
From the first of the above equations, since one of $r,s$ is an integer, and $a$ is an integer, $r,s$ must both be integers.
From the second of the above equations, since $b \ne -1$, it follows that $r,s$ are both nonzero.
\begin{align*}
\text{Then}\;
a^2+b^2&=(r+s)^2 + (rs-1)^2\\[4pt]
&=(r^2 + 2rs + s^2) + (r^2s^2 - 2rs + 1)\\[4pt]
&=r^2s^2 + r^2 + s^2 + 1\\[4pt]
&=(r^2+1)(s^2+1)\\[4pt]
\end{align*}
Since $r,s$ are nonzero integers, it follows that $a^2+b^2$ is composite.
|
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|
Beta Gamma functions How to convert this integral into Beta Gamma function?
Any hints will be appreciated.
$$\int_0^{b} x \sqrt{b^3-x^3} dx$$ Thanks
|
Set $x =bt$ so that, $dx =bdt$
$$I:=\int_0^{b} x \sqrt{b^3-x^3} dx = \int_0^{1} b^2t \sqrt{b^3-b^3t^3} dt = b^{7/2}\int_0^{1} t (1-t^3)^{1/2} dt$$
Setting $u=t^3~~$ i.e $~~t= u^{1/3}~~$ yields $dt =\frac{1}{3}u^{-2/3}du~~~$ therefore,
$$I= b^{7/2}\int_0^{1} t (1-t^3)^{1/2} dt = \frac{1}{3}b^{7/2}\int_0^{1} u^{-1/3} (1-u)^{1/2} du\\=\frac{1}{3}b^{7/2}\int_0^{1} u^{2/3-1} (1-u)^{3/2-1} du\\=\frac{1}{3}b^{7/2} B\left(\frac{2}{3},\frac{3}{2}\right)$$
But
$$B\left(\frac{2}{3},\frac{3}{2}\right) = \frac{\Gamma\left(\frac{2}{3}\right)\Gamma\left(\frac{3}{2}\right)}{\Gamma\left(\frac{2}{3}+\frac{3}{2}\right)} =\frac{\Gamma\left(\frac{2}{3}\right)\Gamma\left(1+\frac{1}{2}\right)}{\Gamma\left(2+\frac{1}{6}\right)}=\frac{\frac{1}{2}\Gamma\left(\frac{2}{3}\right)\Gamma\left(\frac{1}{2}\right)}{(1+\frac{1}{6})\frac{1}{6}\Gamma\left(\frac{1}{6}\right)}\\=
\color{red}{\sqrt{\pi}}\frac{18}{7}\frac{\Gamma\left(\frac{2}{3}\right)}{\Gamma\left(\frac{1}{6}\right)}$$
Since$$\Gamma\left(\frac{1}{2}\right) =\sqrt{\pi}$$
Conclusion
$$\color{blue}{\int_0^{b} x \sqrt{b^3-x^3} dx =\color{red}{\sqrt{\pi}}b^{\frac{7}{2}}\frac{6}{7}\frac{\Gamma\left(\frac{2}{3}\right)}{\Gamma\left(\frac{1}{6}\right)} }$$
|
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|
How to find a formula for the sum up to $n$ terms of the sum $1+11+111+11111...$ using $1+2x+3x^2+4x^3...$ I'm trying to find the formula for the sum up to $n$ terms of the sum $1+11+111+11111...$ using $1+2x+3x^2+4x^3...=\frac{1-x^n}{(1-x)^2}-\frac{nx^n}{1-x}$. It's easy to find the formula if we expressed the sum as $1+(1+10)+(1+10+100)+(1+10+100+1000)...$ however when I try to express the sum as $1+2x+3x^2+4x^3...$ I can't find a constant value for $x$. Can anyone guide me on how I can express $1+11+111+11111...$ as $1+2x+3x^2+4x^3...$.
|
$1 + 11 + 111 + .... = n + (n-1)10 + (n-2)10^2 + ....+2*10^{n-2} + 10^{n-1}$
$= 10^{n-1}(n*10^{-n} + (n-1)*10^{-n+1} + (n-2)10^{-n+2} + .....+2*10 +1)$
$= 10^{n-1}(1 + 2*\frac 1{10} + ..... + (n-2)*\frac 1{10} + (n-1))$
$= 10^{n-1}[\frac{1-\frac{1}{10}^n}{(1-\frac{1}{10})^2}-\frac{n\frac{1}{10}^n}{1-\frac{1}{10}}]$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Solving $x_{n} - 3x_{n-1} = -8$ with $n\geq 1$ and $x_0 = 2$ I tried two methods which gave different answers:
Method 1:
$$x_{n} - 3x_{n-1} = -8 \\ x_n = 3(3x_{n-2} - 8) - 8 \\ = 3^2 x_{n-2} -8 ( 1+3) \\ = 3^3 x_{n-3} - 8(1+3+3^2) \\ = 3^n x_{0} - 8(1+3+3^2 + \ldots + 3^{n-1}) \\ = 2\times 3^n - 8\left(\frac{3^n - 1}{3-1}\right)\\ = 2\times 3^n - 4(3^n - 1) \\ = -2\times 3^n +4.$$
Method 2:
Solving the homogenous equation using $x_n = r^n$,
$$r^n - 3r^{n-1} = 0 \\ \implies r = 3$$
So the homogenous solution is $h_n = a\times 3^n$ for some $a\in\mathbb{R}$.
With the initial condition, $h_n = 2\times 3^n$.
Also, by guessing the particular solution with $x_n = C$,
$$C - 3C = -8 \implies C = 4$$.
So, the final solution is
$$x_n = 2\times 3^n + 4.$$
I'm unsure why the two methods differ. It looks obvious that the first one gives the correct result.
|
Here's another method. Convert the recurrence to a generalized Fibonacci form as follows:
$$
x_n=Ax_{n-1}+B,\quad x_{_0} ~\text{given}\\
x_{n-1}=Ax_{n-2}+B\\
x_n-x_{n-1}=Ax_{n-1}-Ax_{n-2}\\
x_n=(A+1)x_{n-1}-Ax_{n-2},\quad x_{_1}=Ax_{_0}+B
$$
The characteristic roots for this equation are always
$$\alpha,\beta=A,1$$
|
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|
Calculate $\iint_D x dxdy$ using polar coordinates Using polar coordinates, I want to calculate $\iint_D x dxdy$, where $D$ is the disk with center $(2,3)$ and radius $2$.
$$$$
I have done the following:
We have $D=\{(x,y)\mid (x-2)^2+(y-3)^2\leq 4\}$.
We use $(x,y)=(r\cos \theta, r\sin \theta)$.
From the inequality $$(x-2)^2+(y-3)^2\leq 4\Rightarrow x^2-4x+4+y^2-6y+9\leq 4 \Rightarrow x^2+y^2-4x-6y\leq -9$$ we get $$r^2\cos^2\theta+r^2\sin^2\theta-4r\cos\theta-6r\sin\theta\leq -9 \Rightarrow r^2-r(4\cos\theta-6\sin\theta)+9\leq 0$$
To find for which values of $r$ that inequality is true, we have to find first the roots of $r^2-r(4\cos\theta-6\sin\theta)+9=0$.
The roots are $$2\cos \theta+3\sin\theta\pm \sqrt{12\cos\theta\sin\theta-5\cos^2\theta}$$ Therefore, we get the inequality $r^2-r(4\cos\theta-6\sin\theta)+9\leq 0$ for $$2\cos \theta+3\sin\theta-\sqrt{12\cos\theta\sin\theta-5\cos^2\theta}\leq r\\ \leq 2\cos \theta+3\sin\theta+\sqrt{12\cos\theta\sin\theta-5\cos^2\theta}$$ or not?
So, at the integral do we use these limits for $r$ ? And what about $\theta$ ? Does it hold that $0\leq \theta\leq 2\pi$ ?
|
Hint: Use substitution
\begin{cases}
x=2+r\cos\theta,\\
y=3+r\sin\theta,\\
\end{cases}
then
$$\iint_D xdxdy=\int_0^{2\pi}\int_0^2r(2+r\cos\theta)\,dr\,d\theta$$
|
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|
Transform term for induction proof Could you help me to mathematically show that these two terms are the same (they are). This is the last (but probably the most important :( ) step of an induction proof.
$$First: \frac{(n+1)(n+2)(2(n+1)+7)}{6}$$
$$Second: \frac{n(n+1)(2n+7)+6(n+1)(n+3)}{6}$$
Thank you! :)
|
A (hopefully) more elegant variant:
Once you've factored out $\dfrac{n+1}6$, there remains
*
*$n(2n+7)+6(n+3)=n\bigl(2(n+3)+1\bigr)+6(n+3)=\color{red}{2(n+3)^2+n}$.
*$(n+2)\bigl(2(n+1)+7\bigr)=\bigl((n+3)-1\bigr)\bigl(2(n+3)+3\bigr)=2(n+3)^2+(3-2)(n+3)-3=\color{red}{2(n+3)^2+n}.$
|
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|
How to find $\lim_{x\to1}\left(\tan\frac{\pi x}{4}\right)^{\tan\frac{\pi x}{2}}$?
Find
$$\lim_{x\to1}\left(\tan\frac{\pi x}{4}\right)^{\tan\frac{\pi x}{2}}$$
My attempt: ON THE basis of This post
$$\lim_{x\to1}\tan\frac{\pi x}{4} =1,\quad \lim_{x\to1}\tan\frac{\pi x}{2}=\infty$$
$$\implies\lim_{x\to1}\left(\tan\frac{\pi x}{4}\right)^{\tan\frac{\pi x}{2}}= e^{\lim_{x\to1}\left[\tan\frac{\pi x}{4}-1\right]\tan\frac{\pi x}{2}}$$
Now I need to solve $\lim_{x\to1}\left[\tan\frac{\pi x}{4}-1\right]\tan\frac{\pi x}{2}$, but I don't know how to go on.
|
$$\lim_{x\rightarrow1}\left(\tan\frac{\pi x}{4}\right)^{\tan\frac{\pi x}{2}} = \lim_{x \to 1}e^ {{\tan\frac{\pi x}{2}}\ln(\tan\frac{\pi x}{4})}$$
Now we will check
\begin{align*}
\lim_{x \to 1} {{\tan\frac{\pi x}{2}}\ln\left(\tan\frac{\pi x}{4}\right)}
&\stackrel{\text{(L'Hôpital's rule)}}{=}
\lim_{x \to 1} \frac{-\frac{\pi}{4}\tan^2(\frac{x\pi}{2})\cos^2(\frac{x\pi}{2})}{\cos^2(\frac{x\pi}{4})\tan(\frac{x\pi}{4})} \\&= \frac{\frac{-1}{2}}{\frac{1}{2}} = -1
\
\end{align*}
and therefore:
$\lim_{x \to 1} e^{{{\tan\frac{\pi x}{2}}\ln(\tan\frac{\pi x}{4})}} = e^{-1}$
|
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|
If $a+b+c=0$ prove that $ \frac{a^2+b^2+c^2}{2} \times \frac{a^3+b^3+c^3}{3} = \frac{a^5+b^5+c^5}{5} $
If $a+b+c=0$, for $a,b,c \in\mathbb R$, prove
$$ \frac{a^2+b^2+c^2}{2} \times \frac{a^3+b^3+c^3}{3} = \frac{a^5+b^5+c^5}{5} $$
I've tried squaring, cubing, etc. the $a+b+c=0$, but I've just dug myself in.
Is there a more elegant way to prove this, or a way to find the right "trick"?
|
The theory for this type of questions are Newton's Identities.
For a cubic polynomial
$$\begin{align}ax^3+bx^2+cx+d\end{align}\\$$
Vieta's formulas for the roots $\alpha_1, \alpha_2, \alpha_3$ are:
$$\begin{align}a(\alpha_1 + \alpha_2 + \alpha_3) + b=0\tag{V}\\
a(\alpha_1\alpha_2 + \alpha_2\alpha_3 + \alpha_3\alpha_1) - c=0\\
a\alpha_1\alpha_2\alpha_3 + d=0\end{align}$$
Newton's Identities for the power sums $P_i:=\alpha_1^i+\alpha_2^i+\alpha_3^i$ are:
$$\begin{align}
aP_1 ~+ ~&b&=0\tag{N1}\\
aP_2 ~+ ~&bP_1 +~ 2c&=0\tag{N2}\\
aP_3 ~+ ~&bP_2 +~ cP_1 +~ 3d&=0\tag{N3}\\
aP_4 ~+ ~&bP_3 +~ cP_2 +~ dP_1&=0\tag{N4}\\
aP_5 ~+ ~&bP_4 +~ cP_3 +~ dP_2&=0\tag{N5}\\
\end{align}$$
If $\alpha_1 + \alpha_2 + \alpha_3=0$ then from $(V)$ it follows that $b=0$. Using $(N5)$
$$\begin{align}
aP_5 +~ cP_3 +~ dP_2&=0\\
\end{align}$$
multiply by $6$
$$\begin{align}
6aP_5 +~ 3\cdot2cP_3 +~ 2\cdot3 dP_2&=0\\
\end{align}$$
using $(N2)$ and $(N3)$
$$\begin{align}
6aP_5 -~ 3\cdot(aP_2+bP_1)P_3 -~ 2\cdot(aP_3+bP_2+cP_1)P_2=\end{align}$$
$$\begin{align}
6aP_5-3aP_2P_3 -~ 2aP_3P_2=0\\
\end{align}$$
If $a\neq 0$ then
$$\begin{align}
6P_5=5P_2P_3\\
\end{align}$$
|
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|
$\lim_{n\to \infty} F(k)=\frac{(1^{k}+2^{k}+3^{k}+.....+n^{k})}{(1^{2}+2^{2}+3^{2}+.....+n^{2})(1^{3}+2^{3}+3^{3}+.....+n^{3})}$ Find F(5) and F(6) Find the value of F(5) & F(6).It is given that
$$F(k)= \lim_{n\to \infty} \frac{(1^{k}+2^{k}+3^{k}+.....+n^{k})}{(1^{2}+2^{2}+3^{2}+.....+n^{2})(1^{3}+2^{3}+3^{3}+.....+n^{3})}$$ where 'k' is a Natural number
My approach Denominator is $\frac{(n)^{3}*(n+1)^{3}*(2n+1)}{24}$
I dont know how to proceed from here as the Numerator has power in k. I also tried definite integral formula but not able to solve it
|
For $f$ increasing function we have that $ \int \limits_{a-1}^{b} f(s)ds \leq \sum \limits_{s=a}^{b} f(s) < \int \limits_{a}^{b+1} f(s) ds$
So $ \lim \limits_{n \to \infty} \frac{\int \limits_{0}^{n} s^k ds}{\sum \limits_{s=1}^{n} s^2 \sum \limits_{s=1}^{n} s^3}< F(k) <\lim \limits_{n \to \infty} \frac{\int \limits_{1}^{n+1} s^k ds}{\sum \limits_{s=1}^{n} s^2 \sum \limits_{s=1}^{n} s^3}$
Which is $ \lim \limits_{ n \to \infty} \frac{24 n^{k-2}}{(k+1) (n+1)^3 (2 n+1)} \leq F(k) \leq \lim \limits_{n \to \infty } \frac{24 \left((n+1)^{k+1}-1\right)}{(k+1) n^3 (n+1)^3 (2 n+1)}$
So, $0= \lim \limits_{ n \to \infty} \frac{24 n^{5-2}}{(5+1) (n+1)^3 (2 n+1)}\leq F(5) \leq \lim \limits_{n \to \infty } \frac{24 \left((n+1)^{5+1}-1\right)}{(5+1) n^3 (n+1)^3 (2 n+1)} = 0$
And, $\frac{12}{7} = \lim \limits_{ n \to \infty} \frac{24 n^{6-2}}{(6+1) (n+1)^3 (2 n+1)} \leq F(6) \leq \lim \limits_{n \to \infty } \frac{24 \left((n+1)^{6+1}-1\right)}{(6+1) n^3 (n+1)^3 (2 n+1)} = \frac{12}{7}$
|
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Stokes' Theorem verified problem
For the given vector field
$$\vec{H(r)} = rcos( \phi - \frac{\pi}{4} ) \vec{ a_{r} } +sin \phi \vec{ a_{ \phi }} $$
a) Calculate line integral of $\vec{H(r)} $ over the close path $ \Gamma $ with corners at ABCD on xy-plane shown in Figure.
b) Confirm the result of the line integral by using Stokes' theorem.-
i solved but I couldn't verified, I don't understand where I am doing wrong.
My solution:
$\vec{H(r)} = rcos( \phi - \frac{\pi}{4} ) \vec{ a_{r} } +sin \phi \vec{ a_{ \phi }} $
a) $ \int_A^B \vec{H(r)}.\vec{ a_{r}} .dr |_{\phi = \frac{ \pi }{4}} + \int_B^C \vec{H(r)}.r.\vec{ a_{ \phi }} d \phi |_{r=1} +\int_C^D \vec{H(r)}\vec{ a_{ r }} dr |_{ \phi = \frac{ 3\pi }{4}} +\int_D^A \vec{H(r)}r\vec{ a_{ \phi }} d \phi |_{r=2}$
then,
$\int_{r=2}^1 rcos(\phi - \frac{ \pi }{4} ) |_{ \phi = \frac{ \pi }{4}} + \left(\int_{\frac{ 3\pi }{2}}^\frac{ \pi }{4} sin\phi.d \phi.r |_{r=2}] + \int_{\frac{ 3\pi }{4}}^\frac{ 3\pi }{2} sin \phi. d \phi. r ) |_{ r=2}\right ) + \left( \int_{r=1}^2 rcos(\phi - \frac{ \pi }{4} ) |_{ \phi = \frac{ 3\pi }{4}} \right ) + [\int_{\frac{ 3\pi }{2}}^\frac{ 3\pi }{4} sin\phi.d\phi.r |_{r=1} + \int_{\frac{ \pi }{4}}^\frac{ 3\pi }{2} sin\phi.d\phi.r |_{r=1}] $
$\therefore $
= $ \frac{ r^2 }{2} |_{ 2}^1 + \left (2(-cos\phi)|_{ \frac{ 3\pi }{2}}^\frac{ \pi }{4} + 2(-cos\phi)|_{ \frac{ 3\pi }{4}}^\frac{ 3\pi }{2} \right ) + \left ( (-cos\phi)|_{ \frac{ 3\pi }{2}}^\frac{ 3\pi }{4} + (-cos\phi)|_{ \frac{ \pi }{4}}^\frac{ 3\pi }{2} \right )$
=$-\frac{ 3 }{2} - \sqrt{2}$
b) $\bigtriangledown \times \vec{H} = \vec{a_z} \left ( {\frac{ sin \phi }{r}} - sin( \frac{ \pi }{4}- \phi) \right )$
$\int_{S} \bigtriangledown \times \vec{H}.d\vec{s} = rd \phi dr \left ( {\frac{ sin \phi }{r}} - sin( \frac{ \pi }{4}- \phi) \right )$ $\Rightarrow $
$ = \left( \int_{r=1}^2 \int_{\frac{ 3\pi }{2}}^ \frac { \pi }{4}sin \phi d \phi dr - \int_{r=1}^2 \int_{\frac{ 3\pi }{2}}^ \frac { \pi }{4} sin (\frac{ \pi }{4} - \phi) r d \phi dr \right ) + \left( \int_{r=1}^2 \int_{\frac { 3\pi }{4}}^\frac{ 3\pi }{2} sin \phi d \phi dr - \int_{r=1}^2 \int_{\frac { 3\pi }{4}}^\frac{ 3\pi }{2} sin (\frac{ \pi }{4} - \phi) r d \phi dr \right )$
then,
$ \left( (-cos\phi)|_{ \frac{ 3\pi }{2}}^\frac{ \pi }{4}.r|_{1}^2 - cos(\frac{ \pi }{4} - \phi) |_{ \frac{ 3\pi }{2}}^\frac{ \pi }{4}.\frac{ r^2 }{2} |_{ 1}^2 \right ) + \left( (-cos\phi)|_{ \frac{ 3\pi }{4}}^\frac{ 3\pi }{2}.r|_{1}^2 - cos(\frac{ \pi }{4} - \phi) |_{ \frac{ 3\pi }{4}}^\frac{ 3\pi }{2}.\frac{ r^2 }{2} |_{ 1}^2 \right )$
$=-\frac{ 3 }{2} - \sqrt{2}$
|
Solution:
$ \int_{r=2}^1 r.cos(\phi-\frac{ \pi }{4}) .dr |_{\phi = \frac{ \pi }{4}} + \int_{\phi=\frac{ \pi }{4}}^\frac{ -5\pi }{4} sin\phi.d\phi.r |_{r=1} + \int_{\phi=\frac{ 3\pi }{4}}^\frac{ 9\pi }{4} sin\phi.d\phi.r |_{r=2} $
$=-\frac{ 3 }{2} - \sqrt{2}$ Line integral boundaries:
B $\Rightarrow $ C $\frac{ \pi }{4}, \frac{ -5\pi }{4}$
D $\Rightarrow $ A $\frac{ 3\pi }{4}, \frac{ 9\pi }{4}$
stokes' is verified, boundaries: $ \phi= \frac{ 3\pi }{4} \Rightarrow \frac{ 3\pi }{2}$ and $ \phi= \frac{ 3\pi }{2} \Rightarrow \frac{ \pi }{4}$
|
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|
Find $n\in\mathbb N$ so that: $\sqrt{1+5^n+6^n+11^n}\in\mathbb N$ Find $n\in\mathbb N$ so that: $\sqrt{1+5^n+6^n+11^n}\in\mathbb N$
My attempt led me to have $\quad n=2k+1:\quad k\in\mathbb N$
The expression under square root is odd, so the square root's value is also odd.
I assumed $\sqrt{1+5^n+6^n+11^n}=2a+1:\quad a\in\mathbb N \Rightarrow$
$5^n+6^n+11^n=(2a+1)^2-1 \Rightarrow 5^n+6^n+11^n=4a(a+1)$
Then the expression on the left must be divisible by 8
$5^n\equiv 5\pmod 8$ if $n$ is odd, $5^n\equiv 1\pmod 8$ if $n$ is even.
$6^n\equiv 0\pmod 8$ if $n\ge3$.
$11^n\equiv 3\pmod 8$ if $n$ is odd, $11^n\equiv 1\pmod 8$ if $n$ is even.
Then $n$ must be odd and $\ge3$ for the expression to be divisible by 8.
I'm stuck here, would anyone give a hand please?
|
You could do also like this (for kids in elementary school).
$5^a$ always ends with $5$
$6^b$ always ends with $6$
$11^c$ always ends with $1$
So $1+5^a+6^b+11^c$ always end with 3 and there for it can not be perfect square.
|
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|
Prove convergence of the series Help please to prove the convergence: $$\sum_{n=1}^{\infty}\left(\frac{1}{\sin \frac{1}{n}}-n\cos\frac1n\right)\cos \frac{\pi n(n-1)}2$$
It can be proved with Dirichlet's test, but there are come problems with $$\left(\frac{1}{\sin \frac{1}{n}}-n\cos\frac1n\right)$$ monotone. Other steps in this way are clear. Or I should use another test?
|
Let $\frac{1}{n}=x$.
Hence,
$$\lim_{n\rightarrow\infty}\left(\frac{1}{\sin \frac{1}{n}}-n\cos\frac1n\right)=\lim_{x\rightarrow0}\left(\frac{1}{\sin{x}}-\frac{\cos{x}}{x}\right)=$$
$$\lim_{x\rightarrow0}\frac{x-\frac{1}{2}\sin2x}{x\sin{x}}=\lim_{x\rightarrow0}\frac{1-\cos2x}{\sin{x}+x\cos{x}}=$$
$$=\lim_{x\rightarrow0}\frac{2\sin^2x}{x\left(\frac{\sin{x}}{x}+\cos{x}\right)}=0.$$
Since with $$\sum_{n=1}^{\infty}\cos\frac{\pi n(n-1)}{2}$$ you know what to do, by the Dirichlet's Test we'll done if we'll prove that $a_{n+1}\leq a_{n}$, where
$$a_n=\frac{1}{\sin \frac{1}{n}}-n\cos\frac1n$$ or $\lim\limits_{x\rightarrow0^+}f'(x)>0$, where
$$f(x)=\frac{1}{\sin{x}}-\frac{\cos{x}}{x},$$
which is easy.
Indeed, $$\lim_{x\rightarrow0^+}f'(x)=\lim_{x\rightarrow0^+}\left(-\frac{\cos{x}}{\sin^2x}-\frac{-x\sin{x}-\cos{x}}{x^2}\right)=$$
$$=\lim_{x\rightarrow0^+}\left(\frac{\sin{x}}{x}+\frac{\cos x(\sin^2x-x^2)}{x^2\sin^2x}\right)=$$
$$=1+\lim_{x\rightarrow0^+}\frac{\sin{x}-x}{x^3}\lim_{x\rightarrow0^+}\frac{\frac{\sin{x}}{x}+1}{\frac{\sin^2x}{x^2}}=$$
$$=1+2\lim_{x\rightarrow0^+}\frac{\cos{x}-1}{3x^2}=1-\frac{1}{3}\lim_{x\rightarrow0^+}\frac{\sin^2\frac{x}{2}}{\frac{x^2}{4}}=\frac{2}{3}>0$$
and we are done!
|
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Calculate a triple integral I want to draw the space $D=\{x,y,z)\mid z\geq 0, x^2+y^2\leq 1, x^2+y^2+z^2\leq 4\}$ and calculate the integral $\iiint_D x^2\,dx\,dy\,dz$.
How can we draw that space?
About the integral I have done the following:
We have that $x^2+y^2+z^2\leq 4\Rightarrow z^2\leq 4-x^2-y^2 \Rightarrow -\sqrt{4-x^2-y^2}\leq z\leq \sqrt{4-x^2-y^2}$. Since $z\geq 0$ we get that $0\leq z \leq \sqrt{4-x^2-y^2}$.
It holds that $4-x^2-y^2\geq 0$, i.e. $x^2+y^2\leq 4$ since $x^2+y^2\leq 1$.
We have that $x^2+y^2\leq 1 \Rightarrow y^2\leq 1-x^2 \Rightarrow -\sqrt{1-x^2}\leq y \leq \sqrt{1-x^2}$.
It must hold that $1-x^2\geq 0 \Rightarrow x^2\leq 1 \Rightarrow -1\leq x\leq 1$.
Therefore $D$ can be written also in the following form: $$D=\{(x,y,z)\mid -1\leq x\leq 1, -\sqrt{1-x^2}\leq y \leq \sqrt{1-x^2}, 0\leq z\leq \sqrt{4-x^2-y^2}\}$$
Therefore, we get \begin{align*}\iiint_D x^2 \,dx\,dy\,dz&=\int_{-1}^1 \left (\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}\left (\int_0^{\sqrt{4-x^2-y^2}}x^2 \ dz\right ) \ dy\right ) \ dx \\ & =\int_{-1}^1 \left (\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}x^2\sqrt{4-x^2-y^2} \ dy\right ) \ dx \\ & =\int_{-1}^1 x^2\left (\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}\sqrt{4-x^2-y^2} \ dy\right ) \ dx \end{align*}
Is everything correct so far? How could we continue? How could we calculate the inner integral?
$$$$
EDIT:
So to calculate the inner integral $\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}\sqrt{4-x^2-y^2}dy$ we do the following:
We set $y=\sqrt{4-x^2}\cdot \sin t$, then $dy=\sqrt{4-x^2}\cdot \cos t \ dt$.
If $y=-\sqrt{1-x^2}$ then $t=\arcsin \left (-\sqrt{\frac{1-x^2}{4-x^2}}\right )$ and if $y=\sqrt{1-x^2}$ then $t=\arcsin \left (\sqrt{\frac{1-x^2}{4-x^2}}\right )$.
Therefore we get the following:
\begin{align*}\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}\sqrt{4-x^2-y^2}dy&=\int_{\arcsin \left (-\sqrt{\frac{1-x^2}{4-x^2}}\right )}^{\arcsin \left (\sqrt{\frac{1-x^2}{4-x^2}}\right )}(4-x^2)\cdot \cos^2t \ dt \\ & = (4-x^2)\cdot\int_{\arcsin \left (-\sqrt{\frac{1-x^2}{4-x^2}}\right )}^{\arcsin \left (\sqrt{\frac{1-x^2}{4-x^2}}\right )} \frac{1+\cos (2t)}{2} \ dt \\ & = \frac{4-x^2}{2}\cdot\left [ t+\frac{\sin (2t)}{2} \right ]_{\arcsin \left (-\sqrt{\frac{1-x^2}{4-x^2}}\right )}^{\arcsin \left (\sqrt{\frac{1-x^2}{4-x^2}}\right )} \\ & = \frac{4-x^2}{2}\cdot\left [ \arcsin \left (\sqrt{\frac{1-x^2}{4-x^2}}\right )-\arcsin \left (-\sqrt{\frac{1-x^2}{4-x^2}}\right ) \\ +\frac{1}{2}\sin \left (2\arcsin \left (\sqrt{\frac{1-x^2}{4-x^2}}\right )\right )-\frac{1}{2}\sin \left (\arcsin \left (-\sqrt{\frac{1-x^2}{4-x^2}}\right )\right )\right ]\end{align*}
Is everything correct so far? How could we continue?
|
$D$ is a cylinder inside a hemisphere. Or, a cylinder with a spherical cap.
Your set-up for the integral is correct.
You could proceed with a trig substitution $y = \sqrt {4-x^2} \sin \theta$
$\int \int x^2 (4-x^2) \cos^2 \theta \ d\theta\ dx$
Or you could convert to polar coordinates (which is also a trig substitution of sorts, but it would be to both x and y simultaneously.)
$\int_0^{2\pi} \int_0^1 r^2\cos^2\theta \sqrt {4-r^2} (r\ dr)\ d\theta$
Update:
It looks to me like the path in polar coordinates will be the easier one to take.
$\int_0^{2\pi} \cos^2 \theta \ d\theta\int_0^1 r^3 \sqrt {4-r^2} \ dr\\
u = 4-r^2; du = -2r \ dr\\
\int_0^{2\pi} \cos^2 \theta \ d\theta\int_3^4 \frac 12 (4u-u^3) \ du\\
(\frac 12 \theta + \frac 12\sin\theta\cos\theta)|_0^{2\pi}\frac {\pi}{2} (2u^2-\frac 14 u^4)|_3^4\\
\frac {\pi}{2}(\frac {81}{4} - 18)\\
\frac {9\pi}{8}\\
$
Suppose we don't translate into polar
$\int_{-1}^1 \int_{-\sqrt{1-x^2}}^{\sqrt {1-x^2}} x^2 \sqrt {4-x^2-y^2}\ dy\ dx\\
y = \sqrt {4-x^2} \sin\theta\\
dy = \sqrt {4-x^2} \cos \theta\\
\displaystyle\int_{-1}^1 \int_{-\arcsin \sqrt{\frac {1-x^2}{4-x^2}}}^{\arcsin \sqrt{\frac {1-x^2}{4-x^2}}} x^2 (4-x^2)\cos^2\theta \ d\theta\ dx\\
\displaystyle\int_{-1}^1 x^2 (4-x^2)\frac 12(\theta+\sin\theta\cos\theta)|_{-\arcsin \sqrt{\frac {1-x^2}{4-x^2}}}^{\arcsin \sqrt{\frac {1-x^2}{4-x^2}}} dx\\
\displaystyle\int_{-1}^1 x^2 (4-x^2)(\arcsin \sqrt{\frac {1-x^2}{4-x^2}}+\frac {\sqrt {1-x^2}}{\sqrt{4-x^2}}\frac {\sqrt 3}{\sqrt {4-x^2}})\ dx\\
\displaystyle\int_{-1}^1 x^2 (4-x^2)\arcsin \sqrt{\frac {1-x^2}{4-x^2}}+x^2\sqrt {3(1-x^2)} dx\\
$
Now it looks like I need to so some integration by parts.
|
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|
Given $x\odot y=x+y+xy$, prove $x\odot x\odot\ldots\odot x= (1+x)^n -1$
Given $x\odot y=x+y+xy$, prove that
$$\underbrace{x\odot x\odot\ldots\odot x}_{n\text{ times}}= (1+x)^n -1$$ for all $n\in \Bbb{N}$ and $x \in\Bbb R\setminus\{-1\}$.
I have tried to use the binomial theorem on $(1+x)^n$ but was unable to simplify the nested function. Also tried mathematical induction but that didn't work out. The nested function is still stumping me as I haven't figured out how to generalize it for $n$ terms. Also some information on how to deal with nested functions would be greatly appreciated.
|
By induction:
$$x_1 = x$$
$$x_n = x \odot x_{n-1} = x+x_{n-1} + xx_{n-1} \Rightarrow \\
x_n = (1+x)x_{n-1} + x.$$
Then:
$$x_n = (1+x)\left[(1+x)x_{n-2} + x\right] + x = \\
= (1+x)^2x_{n-2} + (1+x)x + x = \\
= (1+x)^3x_{n-3} + (1+x)^2x + (1+x)x + x \Rightarrow\\
x_n = (1+x)^{h}x_{n-h}+x\sum_{k=0}^{h-1} (1+x)^k.$$
For $h=n-1$:
$$x_n = (1+x)^{n-1}x_{1}+x\sum_{k=0}^{n-2} (1+x)^k \Rightarrow \\
x_n = (1+x)^{n-1}x + x\frac{1-(1+x)^{n-1}}{1-(1+x)} \Rightarrow \\
x_n = (1+x)^{n-1}x - (1-(1+x)^{n-1}) \Rightarrow \\
x_n = (1+x)^{n-1}(1+x) - 1 \Rightarrow\\
x_n = (1+x)^n - 1.$$
|
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|
Find the general solution to non homogeneous linear equation The question:
State the general solutuon of:
$$5x_1-2x_2+4x_3=5$$
My attempt: I tried following closely an example provided for homogeneous equations without any luck:
$$
x=\begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix}=
\begin{bmatrix}2x_2-4x_3+5\\x_2\\x_3\end{bmatrix}=
x_2\begin{bmatrix}2\\1\\0\end{bmatrix}
+x_3\begin{bmatrix}-4\\0\\1\end{bmatrix}
+\begin{bmatrix}5\\1\\1\end{bmatrix}
$$
And after this step I get stuck since I don't know whether is correct and should proceed with the solution space (and also how to find the solution space for non homogenous equations)
As for who might be concerned, the question is homework related.
It would be nice to know why I got downvoted so I can improve my question
Based on the feedback received from @Isham the system should have been written as follows:
$$
x=\begin{bmatrix}5x_1\\x_2\\x_3\end{bmatrix}=
\begin{bmatrix}2x_2-4x_3+5\\x_2\\x_3\end{bmatrix}=
\begin{bmatrix}2/5x_2-4/5x_3+1\\x_2\\x_3\end{bmatrix}=
x_2\begin{bmatrix}2/5\\1\\0\end{bmatrix}
+x_3\begin{bmatrix}-4/5\\0\\1\end{bmatrix}
+\begin{bmatrix}1\\0\\0\end{bmatrix}
$$
|
Consider the pattern:
For
$$5x_1-2x_2+4x_3=5$$
then $x_{3} = \frac{1}{4} \, (5 - 5 \, x_{1} + 2 \, x_{2})$
and
\begin{align}
x &=\begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix}=
\begin{bmatrix} x_{1} \\ x_2 \\ \frac{1}{4} \, (5 - 5 \, x_{1} + 2 \, x_{2}) \end{bmatrix}
= \frac{1}{4} \, \begin{bmatrix} x_{1} \\ x_2 \\ 5 - 5 \, x_{1} + 2 \, x_{2} \end{bmatrix} \\
&= \frac{x_1}{4} \, \begin{bmatrix} 1 \\ 0 \\ -5 \end{bmatrix}
+ \frac{x_2}{4} \, \begin{bmatrix} 0 \\ 1 \\ 2 \end{bmatrix}
+ \frac{5}{4} \, \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}.
\end{align}
|
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|
If $a,b,c,a^2+b^2+c^2$ are primes, then $a$ or $b$ or $c$ is equal to $3$ Given primes $a,b,c$ such that $a^2+b^2+c^2$ is prime, then $3\in\{a,b,c\}$.
Tested for $a,b,c<500$.
|
If $p\ge 5 $ is a prime then it is always of the form $6n\pm 1 ~, n \ge 1$.
Let $a,b,c$ be primes $\ge 5$. Then we've ;
$$(6n_1\pm 1)^2+(6n_2 \pm 1)^2+(6n_2 \pm 1)^2 = \text{prime}$$
$$36(n_1^2+n_2^2+n_3^2)+12(\pm n_1\pm n_2 \pm n_3)+3 =\text{prime}$$
But in the above equation, LHS is clearly divisible by $3$ and hence not a prime. Therefore our assumption was false. Atleast one of $a,b,c$ is equal to $3$. Also, any of $a,b$ and $c$ can not be $2$ because that will lead the sum $a^2+b^2+c^2$ to be even.
|
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|
$4y^2+y=3x^2+x$ implies that $x-y$ is a perfect square. I need a help to prove the following statement. (Sorry for my bad english).
If $x,y\in\mathbb{Z}$ are solutions of $4y^2+y=3x^2+x$, then $x-y$ is a perfect square.
I've tried to rewrite the equation as $\cfrac{y^2}{x-y}=3(x+y)+1 $ and conclude that $x-y\mid y^2$, but I do not think that's useful. I found two solutions: $(x,y)=(2,-2)$ and $(x,y)=(30,26)$. If I could find a recursive way to build more solutions, I would probably show that $x-y\in\mathcal{Q}:=\big\{ n^2\mid n\in \mathbb{N} \big\}$.
Thanks to everyone which will help me to solve (prove) it.
|
The equation
$$
4Y^2 + Y = 3X^2 + X
$$.
Let me proof why $X - Y = c$.
where $c$ is a perfect square, note: the values $X$ and $Y$ here are the points $( X, Y )$ on the graph of the equation.
$4Y^2+Y = 3X^2+X$.
make $Y$ the subject of the formula there.
$Y = \frac{-1 \pm \sqrt{48X^2+16X+1}}{8}$.
Now say $X-Y = c$.
So that.
$
\begin{align}
X - \frac{-1 \pm \sqrt{48X^2+16X+1}}{8} = c\\
8X - \frac{-1 \pm \sqrt{48X^2+16X+1}}{1} = 8c\\
8X + 1 \mp \sqrt{48X^2+16X+1} = 8c\\
8X + 1 - 8c = \pm \sqrt{48X^2+16X+1}\\
(8X+1-8c)^2 = 48X^2+16X+1\\
1-16c+64c^2+16X-128cX+64X^2 = 48X^2+16X+1\\
16X^2-128cX+64c^2-16c = 0
\end{align}
$.
This equation would give us the value of the $X$ coordinate for which $X-Y = c$.
$X^2-8cX+4c^2-c = 0$.
solving for $X$ shows that.
$X = \frac{8c \pm \sqrt{64c^2-4(4c^2-c)}}{2}$.
which can be further depressed to.
$X = 4c \pm \sqrt{12c^2+c}$.
if we had put $c = 4$, we would get coordinate point $X = 2 \text{and} 30$ as solution.
the determinant here is $12c^2+c$, were $c$ was defined to be a perfect square, let's assume we are looking for integers values of $X$.
Then $12c^2+c = c(12c+1)$ must also be a perfect square, then the number $(12c+1)$ must be a perfect square also, but as far I can say, this worked only when $c = 4$ where $(12×4+1) = 49$, and $49$ is also a perfect number.
Let's recall that $X^2-8cX+4c^2-c = 0$.
now we want to see the value of $X$ that can make $c$ a perfect square.
put as $4c^2-(8X+1)c+X^2 = 0$.
$c = \frac{8X+1 \pm \sqrt{48X^2+16X+1}}{8}$.
one of these condition is that $(48X^2+16X+1)$ must be a perfect square.
we are still limited to only two values that can fit this $X = 2 \text{and} 30$
|
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|
How to solve a xor equation? I have the following equations I need to solve for A and B:
2A ⊕ B = 7
A ⊕ B = 14
Note:
All variables are unsigned 32-bit integers
⊕ is the XOR operator for 32-bit integers.
|
XOR is the same as addition in the finite field $F_2$. Thus if we assume a solution exists with $A<8$ and $B<16$ (reasonable given the magnitudes of the right-hand sides) you can rewrite your equations as
\begin{align*}
\left[\begin{array}{cccc}0 & 1 & 0 & 0\\0 & 0& 1 & 0\\0 & 0& 0& 1\\0 & 0& 0&0\end{array}\right]\mathbf{A} + I_{4\times 4}\mathbf{B} &\equiv \left[\begin{array}{c}0\\1\\1\\1\end{array}\right]\\
I_{4\times 4}\mathbf{A} + I_{4\times 4}\mathbf{B}&\equiv\left[\begin{array}{c}1\\1\\1\\0\end{array}\right]
\end{align*}
in $F_2$, with the vector $\mathbf{A}$ containing the bits of $A$ (from most to least significant).
We thus have an $8\times 8$ linear system over $F_2$, which you can row-reduce to get
$$\left[\begin{array}{cccccccc}1 & 0 &0 &0 & 1 & 0 & 0 & 0\\
0 & 1 & 0 & 0 & 0 & 1 & 0 & 0\\
0 & 0 & 1 & 0 & 0 & 0 & 1 & 0\\
0 & 0 & 0 & 1 & 0 & 0 & 0 & 1\\
0 & 0 &0 &0 &1 & 1 & 0 &0\\
0 & 0 &0 &0 &0 & 1 & 1 &0\\
0 & 0 &0 &0 &0 & 0 & 1 &1\\
0 & 0 &0 &0 &0 &0 &0 &1
\end{array}\right]\left[\begin{array}{c}\mathbf{A}\\\mathbf{B}\end{array}\right] \equiv \left[\begin{array}{c}1\\1\\1\\0\\0\\1\\0\\1\\1\end{array}\right]$$
and finally back-substitute to get
$$\mathbf{A} = \left[\begin{array}{c}0\\1\\1\\1\end{array}\right], \mathbf{B}=\left[\begin{array}{c}1\\0\\0\\1\end{array}\right].$$
But wait, as pointed out in the comments, the answer isn't unique... that's right, notice that at the beginning we assumed that both $A$ and $B$ had at most 4 bits, and that $A<8$. If we allow larger $A$ or $B$ we get larger linear systems (which will be rank-deficient).
|
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|
Find $k$ such that $f(k)$ is Minimum Find $k$ such that
$$f(k)=\int_{0}^{4} |4x-x^2-k|dx$$ is Minimum
I splitted the Modulus in to two cases:
$1.$ if $4x-x^2-k \ge 0$ Then
$$f(k)=\int_{0}^{4} (4x-x^2-k)dx=\frac{32}{3}-4k$$
$2.$ if $4x-x^2-k \lt 0$ Then
$$f(k)=4k-\frac{32}{3}$$
But How to minimize $f(k)$ which is linear in $k$?
|
We need to separate it into three cases :
*
*Case 1 : If $k\ge 4$, then we have $4x-x^2-k=-(x-2)^2+4-k\le 0$, so $$f(k)=\int_0^4 (x^2-4x+k)dx=4k-\frac{32}{3}$$So, $f(k)\ge f(4)=\frac{16}{3}$ for $k\ge 4$.
*Case 2 : If $0\lt k\lt 4$, then $0\lt 2-\sqrt{4-k}\lt 2+\sqrt{4-k}\lt 4$ where $x=2\pm\sqrt{4-k}$ are roots of $4x-x^2-k$, so we have $$\begin{align}\small f(k)&=\int_{0}^{2-\sqrt{4-k}}(x^2-4x+k)dx+\int_{2-\sqrt{4-k}}^{2+\sqrt{4-k}}(-x^2+4x-k)dx+\int_{2+\sqrt{4-k}}^{4}(x^2-4x+k)dx\\\\&=\frac 43\left(2(4-k)^{3/2}+3k-8\right)\end{align}$$
and
$$f'(k)=4\left(1-\sqrt{4-k}\right)$$ from which we see that $f(k)$ is decreasing for $0\lt k\lt 3$ and increasing for $3\lt k\lt 4$ to have $f(k)\ge f(3)=4$ for $0\lt k\lt 4$.
*Case 3 : If $k\le 0$, then $4x-x^2-k=x(4-x)-k\ge 0$, so $$f(k)=\int_0^4(-x^2+4x-k)dx=\frac{32}{3}-4k$$So, $f(k)\ge f(0)=\frac{32}{3}$ for $k\le 0$.
Therefore, the minimum of $f(k)$ is attained when $\color{red}{k=3}$.
|
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|
Non unique factorization in $\mathbb{Z}_5$
Prove $3X^3+4X^2+3=(X+2)^2(3X+2)=(X+2)(X+4)(3X+1)$ in $\mathbb{Z}_5$.
I see that $2$ is a root (since $-24+16+3=-5$) , then $3X^3+4X^2+3=(X+2)(3X^2)$, and also $-4$ (since -$64\cdot3+16\cdot4+3=-125$) and although $x=-1/3$ is not in $\mathbb{Z}_5$, because it also works as a root then $3x+1$ is a nother factor.
Then it should be $3X^3+4X^2+3=(X+2)(X+4)(3X+1)$ but the product of the right side gives $3X^3+24X^2+30X+8=3X^3+4X^2+3+5(4X^2+6X^2+1)=3X^3+4X^2+3.$
But why is it $3X^3+4X^2+3=(X+2)^2(3X+2)$? It seems $(X+2)^2(3X+2)=(X^2+2X+4)(3X+2)=3X^4+8X^2+10X+8=3X^4+3X^2+3+5(X+2X+1)=3X^4+3X^2+3$ which is not the initial polynomial.
Also, $\mathbb{Z}_5$ is UFC domain, wouldn't the double factorization imply a contradiction?
|
No contradiction.
In a UFD (such as $\mathbb{Z}_5[x]$), the factorization is unique, only up to unit factors.
Note that $2$ and $3$ are units in $\mathbb{Z}_5$, hence are also units in $\mathbb{Z}_5[x]$.
Also, we have $6=1$.
Then since
$$x+4 = 6x+4 = 2(3x+2)$$
and
$$3x+1 = 3x + 6 = 3(x+2)$$
the factors match up, one-to-one, up to unit factors.
Another way to see it is as follows . . .
\begin{align*}
&(x+2)(x+4)(3x+1)\\[4pt]
=\;&(x+2)(x+4)(3x+1)(6)&&\text{[since $6=1$]}\\[4pt]
=\;&(x+2)\bigl(3(x+4)\bigr)\bigl(2(3x+1)\bigr)\\[4pt]
=\;&(x+2)(3x+12)(6x+2)\\[4pt]
=\;&(x+2)(3x+2)(x+2)&&\text{[since $12=2$ and $6=1$]}\\[4pt]
=\;&(x+2)^2(3x+2)\\[4pt]
\end{align*}
which shows how we can convert one of the factorizations directly into the other, by multiplying the original factors by selected unit factors whose product is $1$ (e.g., by $2$ and $3$).
|
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|
Factorising $(4 + 3i)z^2 + 26iz + (-4+3i)$? Quadratic formula attempt included. I'm trying to factorise $(4 + 3i)z^2 + 26iz + (-4+3i)$.
I tried to use the quadratic formula to factorise $(4 + 3i)z^2 + 26iz + (-4+3i)$, where $b = 26i$, $a = (4 + 3i)$, $c = (-4+3i)$. This gives us the roots $z = \dfrac{-4i - 12}{25}$ and $z =-4i - 3$.
But $\left( z + \dfrac{4i + 12}{25} \right) \left( z + 4i + 3 \right) \not= (4 + 3i)z^2 + 26iz + (-4+3i)$.
So I did what we do when using the quadratic formula with real numbers and have that $a \not= 0, 1$: $(4 + 3i)\left( z + \dfrac{4i + 12}{25} \right) \left( z + 4i + 3 \right)$. But, as far as I can tell, $(4 + 3i)\left( z + \dfrac{4i + 12}{25} \right) \left( z + 4i + 3 \right) \not= (4 + 3i)z^2 + 26iz + (-4+3i)$.
So what is the correct way to go about this?
I would greatly appreciate it if people could please take the time to explain this.
|
The solutions of the equation $(4+3i)z^2+26iz+(3i-4)=0$ are $z=-3-4i$ and $z=\dfrac{-3-4i}{25}$. Try using the Factor Theorem now, it should work.
|
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|
Prove or disprove : $\sqrt{1 + \sqrt{4+\sqrt{16+\sqrt{256...}}}} = \sqrt{2+\sqrt{5}}$ Edit: My initial question was regarding the expressions:
$\sqrt{1 + \sqrt{4+\sqrt{16+\sqrt{64... + \sqrt{4^{n}}}}}}$
And in general:
$\sqrt{1 + \sqrt{k+\sqrt{k^2+\sqrt{k^3... + \sqrt{k^{n}}}}}}$
And their limits, however, my working out was wrong. So, if someone can explain how to find the values of the above two expressions, that would be great!
This is just something I've noticed playing around with the radicals, and I honestly don't have much idea on how to prove it. These were my ideas:
Let $\sqrt{1 + \sqrt{4+\sqrt{16+\sqrt{256... + \sqrt{4^{2^n}}}}}} = A_n$
My logic was as follows:
$A_n = \sqrt{1 + \sqrt{4+\sqrt{16+\sqrt{256... + \sqrt{4^{2^n}}}}}}$
$A_n^2 = 1 + \sqrt{4+\sqrt{16+\sqrt{256... + \sqrt{4^{2^n}}}}}$
$A_n^2 = 1 + 2\sqrt{1+\frac{1}{4}\sqrt{16+\sqrt{256... + \sqrt{4^{2^n}}}}}$
$A_n^2 = 1 + 2\sqrt{1+\sqrt{1+\frac{1}{16}\sqrt{256+... + \sqrt{4^{2^n}}}}}$
$A_n^2 = 1 + 2\sqrt{1+\sqrt{1+\sqrt{1+... + \sqrt{1}}}}$
$A_{\infty}^2 = 1 + 2\sqrt{1+\sqrt{1+\sqrt{1+... }}} = 1 + 2\phi = 2 + \sqrt{5}$
$A_{\infty} = \sqrt{2 + \sqrt{5}}$
Then, rather than this specific case, how would we evaluate:
$L = \sqrt{1 + \sqrt{k+\sqrt{k^2+\sqrt{k^4+\sqrt{k^8+...}}}}}$
Would we also get $L = \sqrt{\sqrt{k}\phi + 1}$?
And what about expressions such as:
$\sqrt{1 + \sqrt{2+\sqrt{3+...}}}$
Thanks for any answers and guidance!
|
Let $\phi_n:=\sqrt{1+\sqrt{1+\cdots+\sqrt{1}}}$ ($n$ square roots); note that $\phi_n\to\phi$, the golden ratio.
Let $a_{n,r}:=\sqrt{k^{2^{r-n+1}}+\sqrt{\cdots+\sqrt{k^{2^r}}}}$.
For induction in $n$, assume that $a_{n,r}=k^{2^{r-n}}\phi_n$.
Then $$a_{n+1,r}=\sqrt{k^{2^{r-n}}+a_{n,r}}=\sqrt{k^{2^{r-n}}+k^{2^{r-n}}\phi_n}=k^{2^{r-n-1}}\phi_{n+1}$$
The initial case $n=0$ is trivially true.
Thus $\sqrt{k+\sqrt{\cdots}}=\lim_{r\to\infty}a_{r+1,r}=\lim_{r\to\infty}\sqrt{k}\phi_{r+1}=\sqrt{k}\phi$.
Hence $\sqrt{1+\sqrt{k+\cdots}}=\sqrt{1+\sqrt{k}\phi}$.
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/2492189",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
}
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Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.