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Complex analysis computation of $Im(f(z)-f(z_0))$ I have a computation that I wanted to carry out and know the answer. My issue is that when I did the computation I did not arrive at the answer that I know to be true. I would really appreciate a fresh set of eyes to comment on my work.
For $f(z)= kz- \sqrt{z}$ we want to compute Im$[f(z)-f\left(\frac{1}{4k^2}\right)]=0$, where $k$ is a real number. What we should obtain is a parabola $x= \frac{1}{4k^2}-k^2y^2$.
My computations
Consider that Im$[f(z)-f\left(\frac{1}{4k^2}\right)]=0$ is the same as Im$[f(z)]=$Im$[f\left(\frac{1}{4k^2}\right)]$.
Then I compute
$f(\frac{1}{4k^2})=k \frac{1}{4k^2} - \sqrt{\frac{1}{4k^2}} =\frac{1}{4k}- \frac{1}{2k}= -\frac{1}{4k}$
Where we know that this is real thus we know the Im$[f\left(\frac{1}{4k^2}\right)]=0$
To compute Im$[f(z)]=0$ we let $z= x+iy$ then we observe
Im $[ k(x+iy) -\sqrt{x+iy}]=0$
Which is equivalent to Im$[k(x+iy)]=$Im$[\sqrt{x+iy}]$. Then to deal with the sqrt I use the exponential definition of complex numbers and see that $r= \sqrt{x^2+y^2}$ and $\theta$ be the corresponding angle. Thus Im$[k(x+iy)]=$Im$[\sqrt{x+iy}]$ is the same as $ky=$Im$[r^{1/2} e^{\theta/2}]$.
Then using Eulers identity to make $e^{\theta/2} =\cos \theta/2+i \sin \theta/2$.
Then we get that
$$ky= r^{1/2} \cos\theta/2.$$
Squaring both sides we obtain
$$k^2y^2= r \cos^2 \theta/2.$$
Using trig we get
$$k^2y^2=\frac{1}{2} r(1+ \cos \theta).$$
Now using that $\cos \theta = \frac{x}{r}$ we get
$$k^2y^2 = \frac{1}{2}(r+ x).$$
We can turn $r$ into $x$ and $y$,
$$k^2y^2= \frac{1}{2}(\sqrt{y^2+x^2} +x).$$
However that is not the answer that is given to be true, i.e. $x= \frac{1}{4k^2}-k^2y^2$. I would really appreciate comments and solutions.
| A mistake I see is that the imaginary part of $r^{1/2}e^{\theta/2}$ is $r^{1/2}\sin\theta/2$ (and not cosine). The change is minor, though: your last equality changes to
$$
k^2y^2= \frac{1}{2}(\sqrt{y^2+x^2} -x).
$$
So
$$
\sqrt{x^2+y^2}=x+2k^2y^2.
$$
Squaring,
$$
x^2+y^2=x^2+4xk^2y^2+4k^4y^4,
$$
so, after subtracting $x^2$ and dividing by $y^2$,
$$
1=4xk^2+4k^4y^2.
$$
Solving for $x$,
$$
x=\frac1{4k^2}-k^2y^2.
$$
| {
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"url": "https://math.stackexchange.com/questions/2213847",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Problem in partial fraction decomposition Consider the following rational function which we want to decompose to partial fractions $$\frac{x}{x^3-3x+2}=\frac{x}{(x-1)(x^2+x-2)}=\frac{A}{x-1}+\frac{Bx+C}{x^2+x-2}$$ Now when we multiply out we get the following expression $$x=A(x^2+x-2)+Bx(x-1)+C(x-1)$$ If we consider $x=1$ we obtain $$1=0$$ What is wrong here?
| $$\frac{x}{x^3-3x+2}=\frac{x}{(x-1)(x^2+x-2)}=\frac{x}{(x-1)(x-1)(x+2)} = \frac{x}{(x-1)^2(x+2)}$$ $$= \frac{A}{x-1}+\frac B{(x-1)^2} + \frac{C}{x+2}$$
$$x = A(x-1)(x+2) + B(x+2) + C(x-1)^2$$
Now, when we test with $x=1$, we get $3B= 1 \iff B=\frac 13$
| {
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"timestamp": "2023-03-29T00:00:00",
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Solve $\sum_{k=0}^{\infty}\frac{1}{1-16k^2}$ I could use some help on calculating this infinite sum: $\sum_{k=0}^{\infty}\frac{1}{1-16k^2}$. Included was that I had to start with a Fourier series for the function $f:\Re \to \Re: x \mapsto \sin(x)$ for $x\in[0, \frac{\pi}{2}[$, so let's start with that.
Let
\begin{eqnarray*}
g(x) = \frac{a_0}{2} + \sum_{k=1}^{\infty}a_k\cos(4kx) + \sum_{k=1}^{\infty}b_k\sin(4kx).
\end{eqnarray*}
This is the Fourier series for $f$.
With $a_k = \frac{4}{\pi}\int_{0}^{\frac{\pi}{2}}\sin(x)\cos(4kx)dx$ and $b_k = \frac{4}{\pi}\int_{0}^{\frac{\pi}{2}}\sin(x)\sin(4kx)dx$. Solving this leads to (or at least I found that): $a_k = \frac{4}{(1-16k^2)\pi}$, $b_k = \frac{16k}{(1-16k^2)\pi}$ for $k\geq1$ and $a_0 = \frac{4}{\pi}$. Bringing this to $g(x)$ gives:
\begin{eqnarray*}
g(x) = \frac{2}{\pi} + \sum_{k=1}^{\infty}\frac{4}{(1-16k^2)\pi}\cos(4kx) + \sum_{k=1}^{\infty}\frac{16k}{(1-16k^2)\pi}\sin(4kx).
\end{eqnarray*}
Since $f(x) \approx g(x)$, we can say that $f(0) = g(0)$. We get
\begin{eqnarray*}
\sin(0) = \frac{2}{\pi} + \sum_{k=1}^{\infty}\frac{4}{(1-16k^2)\pi}\cos(0) + \sum_{k=1}^{\infty}\frac{16k}{(1-16k^2)\pi}\sin(0),
\end{eqnarray*}
this becomes
\begin{eqnarray*}
0 = \frac{2}{\pi} + \sum_{k=1}^{\infty}\frac{4}{(1-16k^2)\pi}.
\end{eqnarray*}
We get
\begin{eqnarray*}
\frac{-2}{\pi} = \frac{4}{\pi}\sum_{k=1}^{\infty}\frac{1}{1-16k^2}
\end{eqnarray*} so
\begin{eqnarray*}
\sum_{k=1}^{\infty}\frac{1}{1-16k^2} = \frac{-1}{2}.
\end{eqnarray*} We need the sum from k = 0. The term $\frac{1}{1-16k^2}$ for k = 0 gives 1, so we add 1 to both sides. This leads to my solution
\begin{eqnarray*}
\sum_{k=0}^{\infty}\frac{1}{1-16k^2} = \frac{1}{2}.
\end{eqnarray*}
However, when approaching this sum numerically and using Wolfram, I find that the sum should be $\frac{4+\pi}{8}$. Could some help and point out where I went wrong with my approach? Thanks in advance
| Maybe it is interesting to see that this series can be easily calculated with the residue theorem. Observing that $$S=\sum_{n\geq0}\frac{1}{1-16n^{2}}=\frac{1}{2}\sum_{n\in\mathbb{Z}}\frac{1}{1-16n^{2}}+\frac{1}{2}$$ and using $$\sum_{n\in\mathbb{Z}}f\left(n\right)=-\sum\left\{ \textrm{Res}\left(\pi\cot\left(\pi z\right)f\left(z\right)\right)\textrm{ at }f\left(z\right)\textrm{'s poles}\right\} $$ and observing that we have poles at $z=\pm\frac{1}{4}$ we get $$S=\color{red}{\frac{\pi}{8}+\frac{1}{2}}$$ as wanted.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Proving $\sec^3\frac{2\pi}{7}+\sec^3\frac{4\pi}{7}+\sec^3\frac{6\pi}{7}=-88$ and $\sec^2\frac{2\pi}{7}+\sec^2\frac{4\pi}{7}+\sec^2\frac{6\pi}{7}=24$
Prove:
$$\begin{align}
\sec^3\frac{2\pi}{7}+\sec^3\frac{4\pi}{7}+\sec^3\frac{6\pi}{7} &=-88 \tag{1} \\[6pt]
\sec^2\frac{2\pi}{7}+\sec^2\frac{4\pi}{7}+\sec^2\frac{6\pi}{7} &=\phantom{-}24 \tag{2}
\end{align}$$
| you do need to know how to work with symmetric polynomials. The basic fact is that the roots of
$$ x^3 + x^2 - 2x-1 $$ are
$$ 2 \cos \frac{2 \pi}{7}, \; \; \; 2 \cos \frac{4 \pi}{7}, \; \; \; 2 \cos \frac{6 \pi}{7}. $$
This tells you the sum $(-1)$, the sum of pairwise products$(-2)$, and the product$(1)$. The observation goes back, essentially, to Gauss. You need to modify by factors of 2,4,8.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2216771",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Can the following integral $I(c): = \int_{|c-1|}^{c+1} \frac{2x\ln(x)}{\pi\sqrt{-x^4 + 2(1+c^2)x^2-(1-c^2)^2}} \: dx$ be computed? Let $c > 0$ be a real number and consider the following integral:
\begin{equation} I(c): = \int_{|c-1|}^{c+1} \frac{2x\ln(x)}{\pi\sqrt{-x^4 + 2(1+c^2)x^2-(1-c^2)^2}} \: dx \end{equation}
I make the following claim:
\begin{align} I(c) = \begin{cases} \ln(c) & c > 1 \\ 0 & 0 < c \leq 1 \end{cases} \end{align}
Indeed, when $c > 1$, a calculation on wolfram alpha will always yield a decimal approximate for $\ln(c)$, and when $0 < c \leq 1$, one will obtain $0$ as the exact result the very same way. In fact, when $c \leq 1$, the corresponding integrand suddenly seems to become symmetric about $x = 1$.
Let $c \leq 1$. Then, for every $k \in [0,c)$, the integrand $F_c(x) := \frac{2x\ln(x)}{\pi\sqrt{-x^4 + 2(1+c^2)x^2-(1-c^2)^2}}$ satisfies \begin{equation} F_c(1 + k) = - F_c(1 - k). \end{equation}
These are all quite peculiar equalities, as they are not apparent at all to me, and I am very much interested in how to prove them. Any help is appreciated.
|
Any help is appreciated.
Since $~2x=\big(x^2\big)',\quad\ln x=\dfrac{\ln\big(x^2\big)}2,\quad$ and $x^4=\big(x^2\big)^2,\quad$ substituting $x^2\mapsto x$ seems like the
natural course of action. We then have $I(c)=\dfrac1{2\pi}\displaystyle\int_{(1-c)^2}^{(1+c)^2}\frac{\ln x~\cdot~dx}{\sqrt{-x^2+2(1+c^2)~x-(1-c^2)^2}}.$
Completing the square in the denominator, we have $P(x)=(2c)^2-\Big[x-(1+c^2)\Big]^2.$ Factoring
$2c$ outside the radical, and substituting $\dfrac{x-(1+c^2)}{2c}\mapsto x,~$ we eventually obtain $I(c)=\dfrac{J(c)}{2\pi},$
where $J(c)=\displaystyle\int_{-1}^1\frac{\ln(c^2\pm2cx+1)}{\sqrt{1-x^2}}~dx.~$ Factoring again, and using the basic properties of the
logarithm, we have $J(c)=\displaystyle\int_{-1}^1\frac{\ln(2c)}{\sqrt{1-x^2}}~dx~+~\int_{-1}^1\frac{\ln(x+a)}{\sqrt{1-x^2}}~dx,~$ where $a=\dfrac{1+c^2}{2c}\ge1$
for all $c\ge0.~$ The former is trivial to evaluate, yielding $J_1(c)=\pi\ln(2c).~$ As for the latter, by
differentiating $K(b)=\displaystyle\int_{-1}^1\frac{\ln(a+bx)}{\sqrt{1-x^2}}~dx~$ under the integral sign with regard to b, we are left
with $K'(b)=\displaystyle\int_{-1}^1\frac{x~\cdot~dx}{(a+bx)\sqrt{1-x^2}}=\frac\pi b\bigg(1-\frac a{\sqrt{a^2-b^2}}\bigg),~$ which, after being integrated
back with regards to b, yields $K(1)-K(0)=\displaystyle\int_0^1K'(b)~db=\pi~\Big\{\mid\ln c\mid-\ln(2a)\Big\}.~$ Since
$K(0)=\pi\ln a,~$ it follows that $J_2(c)=K(1)=\pi~\Big\{\mid\ln c\mid-\ln2\Big\},~$ and, by extension, that
$I(c)=0$ for $c\in(0,1],~$ and $I(c)=\ln c,~$ for $c\ge1.$
| {
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"timestamp": "2023-03-29T00:00:00",
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What is the largest power $n$ of 3 such that ${3}^n$ divides $2012^4-2011^4$? What is the largest power $n$ of 3 such that ${3}^n$ divides $2012^4-2011^4$?
[\begin{align}
& \text{My try follows : } \\
& \text{ 201}{{\text{2}}^{\text{4}}}\text{-201}{{\text{1}}^{\text{4}}}= \\
& =\left( \text{201}{{\text{2}}^{\text{2}}}\text{-201}{{\text{1}}^{\text{2}}} \right)\left( \text{201}{{\text{2}}^{\text{2}}}\text{+201}{{\text{1}}^{\text{2}}} \right) \\
& =\left( \text{2012-2011} \right)\left( \text{2012+2011} \right)\left( \text{201}{{\text{2}}^{\text{2}}}\text{+201}{{\text{1}}^{\text{2}}} \right) \\
& =\left( \text{2012+2011} \right)\left( \text{201}{{\text{2}}^{\text{2}}}\text{+201}{{\text{1}}^{\text{2}}} \right) \\
& =\left( \text{4023} \right)\left( \text{201}{{\text{2}}^{\text{2}}}\text{+201}{{\text{1}}^{\text{2}}} \right) \\
& =\text{ }{{\text{3}}^{\text{3}}}\text{*149}\left( \text{201}{{\text{2}}^{\text{2}}}\text{+201}{{\text{1}}^{\text{2}}} \right) \\
& \text{from which i conclude the answer to be 3} \\
& \text{Is my work right ? is there any simpler way ?} \\
& \text{Thanks for your help } \\
\end{align}]
| Just to give another approach, note that $81\cdot25=2025$, so that working mod $81$ (from which we can draw conclusions mod $27$), we have
$$\begin{align}
2012^4-2011^4&\equiv(-13)^4-(-14)^4\\
&\equiv169^2-196^2\\
&\equiv7^2-34^2\\
&=(7-34)(7+34)\\
&=-27\cdot41\mod81\\
&\equiv0\mod27
\end{align}$$
Since $3\not\mid41$, we see that $-27\cdot41\not\equiv0$ mod $81$. Thus $27=3^3$ is the largest power of $3$ that divides $2012^4-2011^4$.
The main advantage of this approach is that it obviates the need to find the highest power of $3$ dividing $4023$ (which I have a hard time doing in my head); it also avoids the need for a side proof that $3$ doesn't divide $2012^2+2011^2$. The main disadvantage is that it requires you to guess (or know) that $81$ is the first power of $3$ that doesn't divide $2012^4-2011^4$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Find locus of intersection of two lines with parameters $\require{cancel}$
I have a triangle
where $$DE || BC, A(a,0), B(0,b), b>0, a>0$$
I need to find the locus of the intersection between $BE$ and $CD$ (in terms of $a$ and $b$).
They didn't tell me more about this parallel line that cuts AB y and AC, therefore I thought I would need to create some variables such that
$$E (e,0) \quad \text{and} \quad D(e,d)$$
From here I get from Tales's theorem that
$$\frac{DE}{BC} = \frac{AE}{AC} \Rightarrow \frac db = \frac{a-e}{a} \Rightarrow d = \frac{b(a-e)}{a}$$
$$\sqrt{a^2-2ae+e^2+d^2} = AD \quad \text{Pythagoras theorem}$$
$$AB = \sqrt{a^2+b^2}$$
$$\tag{$()^2$}\frac{\sqrt{a^2-2ae+e^2+d^2}}{\sqrt{a^2+b^2}} = \frac db$$
$$\frac{a^2-2ae+e^2+d^2}{a^2+b^2} = \frac {d^2}{b^2}$$
$$\frac{a^2-2ae+e^2 + \frac{b^2\left(a^2-2ae+e^2\right)}{a^2}{}}{a^2+b^2} = \frac{\frac{\cancel{b^2}\left(a^2-2ae+e^2\right)}{a^2}}{\frac {\cancel{b^2}}{1}}$$
At the end I would get something like
$$(a^2-2ae+e^2)(a^2+b^2) = a^2(a^2-2ae+e^2) + b^2(a^2-2ae+e^2)$$
Which will not help. I'm doing something wrong. I wanted to find $e$ and $d$ in terms of $a$ and $b$ so I can find the locus I need.
| Parameterize the point $D$ as $(1-t)A+tB=((1-t),bt)$, $t\in[0,1]$. An equation for the line $\overline{CD}$ is then $btx-(1-t)ay=0$. The point $E$ is $((1-t)a,0)$, so an equation for $\overline{BE}$ is $bx+(1-t)ay=(1-t)ab$ and the intersection of these two lines is $x={1-t\over1+t}a$, $y={t\over1+t}b$. From here, it’s a simple matter to eliminate $t$: $x-a=-{2at\over1+t}$, so ${y\over(x-a)}=-{b\over2a}$, therefore the locus of intersections is the line $y=-\frac b{2a}(x-a)$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Find a six-digit perfect square of a particular form – BMO 1993 P1
1993 British Mathematical Olympiad, Round 1, Question 1
Find, showing your method, a six-digit integer $n$ with the following properties:
*
*$n$ is a perfect square,
*the number formed by the last three digits of $n$ is exactly one greater than the number formed by the first three digits of $n$. (Thus $n$ might look like $123124$, although this is not a square.)
I gave it a try, albeit number theory not being a strong suit of mine.
$$\text{n = }\overline{ABCABD}\text{, where } D = C +1$$
$$n = 10^5\times A + 10^4\times B + 10^3\times C + 10^2\times A + 10^1\times B + 10^0\times D \\=A(10^5+10^2) + B(10^4+10) + C(10^3+10^0)+1\qquad\qquad\,$$
$$\\n-1 = (10^3+1)\times(10^2\times A+10^1\times B + 10^0 \times C)\\\frac{n-1}{1001} = \overline{ABC}$$
$\text{The question then becomes:}\\\text{Find } r\in\mathbb{Z^+}\,(r = \overline{ABC})\text{ such that } $
$$n = 1001r + 1\\10^5\le n\le 10^6-1,\,n\in\mathbb{Z^+}$$
This, however, appears to have gotten me nowhere as I have no idea how to solve that linear equation.
SIDE NOTE:
Using Python, I found that $r\in\{{183,328,528,715}\}$.
Therefore, $n\in\{{183184,328329,528529,715716}\}$.
However, I would like to see a mathematical solution.
| Hint The condition $n = 1001 r + 1$ does not express one of our conditions, namely that $n$ is a square, say, $m^2$ for some positive integer $m$. In terms of $m$, our condition is $m^2 = 1001 r + 1$, and the key observation is that this equation can be rearranged and then factored:
$$(m + 1) (m - 1) = 1001 r .$$
Now, both sides are factorizations of the same integer. The prime factorization of $1001$ is $1001 = 7 \cdot 11 \cdot 13$, so one or two of these three primes are factors of $m + 1$ and the other(s) are factors of $m - 1$. (If all three were factors of the same term, we would have $m \pm 1 = 1001$, and in both cases here $m^2$ has seven digits.)
If we take the case that $7, 11$ are factors of $m + 1$ and $13$ is a factor of $m - 1$, then we have$$\left\{\begin{array}{rcl}m + 1 &=& 7 \cdot 11 s \\m - 1 &=& 13 t\end{array}\right.$$ for some integers $m, s, t$; $r$ is just $r = s t$. Eliminating $m$ gives $$77 s = 13 t + 2$$ for some $s, t$, and reducing modulo $13$ and solving gives $s \equiv 11 \pmod {13}$, so the smallest positive solution is $s = 11$; substituting gives that the the corresponding value is $t = 65$, so $r = st = 715$ (reproducing one of the given answers) and hence $n = 715716$. The next smallest positive solution is $s = 24$, but this corresponds to a four-digit $m$, and hence to an $m^2$ of at least seven digits, so $r = 715$ is the only solution from this case. Proceeding similarly for the other five cases should yield the other three solutions.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Integrate: $\int (x^2+a^2)^{-3/2} \cdot dx$
Integrate: $\int (x^2+a^2)^{-3/2} \cdot dx$
My Approach:
$\int (x^2+a^2)^{-3/2} \cdot dx$
$\int (x^2+a^2)^{-3/2} \cdot d(a^2+x^2)\cdot \frac{dx}{d(x^2+a^2)}$
But this doesn't give the right answer. I showed this to my friend and he said $d(x^2+a^2)$ is not possible which makes sense since you can't take a small element of the form $(x^2+a^2)$. How can I then solve this integration without using trigonometry?
| Here is the correct solution. Note for anything of the form $({x^2+a^2)^{n/2}}$ for n odd, consider using the following trigonometric substituion.
So, let $x=atan(u) \ (or \ x=asinh(u)$ works aswell$)$ $\Rightarrow dx=(a)sec^2(u)du$
Thus we have:
$$\int \frac{1}{(x^2+a^2)^{3/2}}dx=\int \frac {asec^2u}{(a^2tan^2(u)+a^2)^{3/2}}du=\frac {a}{a^3}\int \frac {sec^2u}{(tan^2(u)+1)^{3/2}}du$$
$$=\frac {1}{a^2}\int \frac {sec^2u}{(tan^2(u)+1)^{3/2}}du=\frac {1}{a^2}\int \frac {sec^2u}{(sec^2(u))^{3/2}}du=\frac {1}{a^2}\int \frac {1}{sec(u)}du=\frac {1}{a^2}\int cos(u)du$$
Finally, we have:
$$=\frac{1}{a^2}sin(u)+c=\frac{1}{a^2}sin(arctan(\frac xa))+c=\frac{x}{a^2 \sqrt{a^2+x^2}}+c$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2222429",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Maclaurin Series : $f(x) = 3\sin^2(x)$ Use a Maclaurin series in this table to obtain the Maclaurin series for the given function.
$$f(x) = 3\sin^2(x)$$
[Hint: Use $\sin^2(x) = \frac{1}{2}\cdot(1-\cos(2x))$]
Series Sum from $0$ to $\infty$ = ?
| Alternatively, consider that:
$f(x) = 3\sin^2(x) \implies f'(x) = 6\sin(x)\cos(x) = 3\sin(2x)$
The series for $f'(x)$, we can easily find:
$\displaystyle f'(x) = 3\cdot\sum_{k=0}^{\infty} \frac{(-1)^{k}2^{2k + 1}x^{2k + 1}}{(2k + 1)!}$
Thus $\displaystyle f(x) = \int f'(x) dx = \int 3\cdot\sum_{k=0}^{\infty} \frac{(-1)^{k}2^{2k + 1}x^{2k + 1}}{(2k + 1)!}$
$f(x) \displaystyle = 3\cdot\sum_{k=0}^{\infty} \frac{(-1)^{k}2^{2k + 1}x^{2k + 2}}{(2k
+ 2)!}$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Kindly provide the stepwise solution for this integral. $$\int_{-b}^b\int_{-a}^a\frac{1}{(x^2+y^2+h^2)^{3/2}}dxdy$$
First I put x=$\sqrt{y^2+h^2}tan\theta$ and arrive at:
$$\int_{-b}^b\frac{2a}{(y^2+h^2)(y^2+h^2+a^2)^{1/2}}dy$$
Can you please tell me what to do after this?
| If we substitute
$$y=\sqrt{a^2+h^2}\tan u$$
$$dy=\sqrt{a^2+h^2}\sec^2u \text{ }du$$
We get the integral
$$\int \frac{2a(\sqrt{a^2+h^2}\sec^2u)}{((a^2+h^2)\tan^2u+h^2)((a^2+h^2)\tan^2u+a^2+h^2)^{1/2}}du$$
Remembering that $\tan^2x+1=\sec^2x$, we get
$$\int \frac{2a\sqrt{a^2+h^2}\sec^2u}{((a^2+h^2)\sec^2u -h^2)((a^2+h^2)\sec^2u)^{1/2}}du$$
Now you can simplify quite a bit.
Hope this helps!
| {
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Which of the following circles has the greatest number of points of intersection with the parabola $x^{2} = y + 4$ Which of the following circles has the greatest number of points of intersection with the parabola $x^2 = y + 4$?
(A) $x^2 + y^2=1$
(B) $x^2 + y^2=2$
(C) $x^2 + y^2=9$
(D) $x^2 + y^2=16$
(E) $x^2 + y^2=25$
I tried to solve every equation by substituting the value of $x^2$ in it, but is this the smartest way to solve this question?
The question is supposed to be answered in no more than $2.5$ minutes
| Eliminate $x$ between them leading to quadratic equation
$$ x^2+y^2= R^2,\, x^2= y+4 ,\, \rightarrow y^2+ y + 4- R^2 =0 $$
If its discriminant is $>0$ then real roots occur when
$$ R>\frac{\sqrt15} {2} \approx 1.9365$$
Repeated roots occur when $ R = \dfrac{\sqrt15}{2} $
But graphically repeated roots occur also for
$$ R= 4 $$
EDIT 1:
For a comprehensive algebraic ( analytical geometrical) procedure (above is incomplete),
we next consider symmetry along y- axis, $x=0$
$$ y=-4, x=0, \rightarrow \, R=4 $$
which supplies another tangent point.
So the complete interval to be considered for maximum real roots cutting situation is:
$$ (\dfrac{\sqrt15}{2}<R < 4 )$$
| {
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"timestamp": "2023-03-29T00:00:00",
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$\int_0^1 \frac{1}{1-t} ( \frac{1}{\sqrt{1-4 x}}-\frac{1}{\sqrt{1-4 t x}}) \, dt=\frac{2}{\sqrt{1-4x}}\log (\frac{1+\sqrt{1-4x}}{2\sqrt{1-4x}})$ How to prove
$$\int_0^1 \frac{1}{1-t} \bigg( \frac{1}{\sqrt{1-4 x}}-\frac{1}{\sqrt{1-4 t
x}}\bigg) \, dt=\frac{2}{\sqrt{1-4x}}\log \bigg(\frac{1+\sqrt{1-4x}}{2\sqrt{1-4x}}\bigg)$$
| This is a weird one. Put $t = \frac{1}{4x}(1-u^2)$. The limits become $1$ and $\sqrt{1-4x}$. Then
$$ \frac{dt}{1-t} = \frac{2u \, du}{1-4x-u^2}, $$
while $1/\sqrt{1-4tx} = 1/u$. But $1-4x-u^2 = (\sqrt{1-4x}+u)(\sqrt{1-4x}-u)$, so we can cancel part of this with the numerator of
$$\frac{1}{\sqrt{1-4x}}-1/u = \frac{u-\sqrt{1-4x}}{u\sqrt{1-4x}},$$
and we end up with the easily integrable
$$ \frac{2}{\sqrt{1-4x}}\int_{\sqrt{1-4x}}^1 \frac{du}{\sqrt{1-4x}+u}, $$
which gives
$$ \frac{2}{\sqrt{1-4x}} \left( \log{\left(1+\sqrt{1-4x}\right)} - \log{\left(2\sqrt{1-4x}\right)} \right) $$
as expected.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove by Induction $1\cdot 2+2\cdot 5+3\cdot 8+4\cdot 11+...+ n(3n-1) = n^2(n+1)$ Prove by induction that the following equality holds true for all n that's an element of a natural number.
$$1\cdot 2+2\cdot 5+3\cdot 8+4\cdot 11+...+ n(3n-1) = n^2(n+1)$$
My work:
Base Case: $n = 1$
l.s = 2
r.s = 2
True
Induction Hypothesis: Assume for some $k$ that's an element of a natural number, $$1\cdot 2+2\cdot 5+3\cdot 8+4\cdot 11+...+ k(3k-1) = k^2(k+1)$$
Now show that, $$1\cdot 2+2\cdot 5+3\cdot 8+4\cdot 11+...+ (k+1)(3(k+1)-1) = (k+1)^2((k+1)+1)$$
$$(k+1)(3k +2) = (k+1)^2(k+2)
$$
$$3k^2 + 5k+ 2 = k^3 + 4k^2 + 5k + 2
$$
$$0 = k^3 + 4k^2-3k^2 + 5k - 5k + 2 - 2
$$
$$0 = k^3 + k^2
$$
by Induction Hypothesis, $k^3 + k^2 = k(3k-1)$
I know you're not supposed to start off with what you are trying to prove/show but if I reverse this whole process, wouldn't that be a correct proof? Is there a faster way?
Thanks
| Your proof is incredibly confusing (not least due to lack of proper type-setting).
As you know, you really should just start on one side and try to get to the other side:
$$\sum_{n=1}^{k+1}(n(3n-1)) = $$
$$\sum_{n=1}^{k}(n(3n-1)) + (k+1)(3(k+1)-1)= $$ (Inductive Hypothesis)
$$k^2(k+1) + (k+1)(3(k+1)-1)= $$
$$(k+1)(k^2 + 3(k+1) -1) =$$
$$(k+1)(k^2 +3k+2)=$$
$$(k+1)(k+1)(k+2)=$$
$$(k+2)(k + 1)^2$$
So it is really only in the second to last step that I was looking at the goal ($(k+2)(k + 1)^2$) that told me I had to factor out a $k+2$, but other than that I just worked from left to right
| {
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Limit of $S_n = \frac{1}{2n + 1} + \frac{1}{2n + 3} + ... + \frac{1}{4n - 1}$ when $n\to\infty$ $$S_n = \frac{1}{2n + 1} + \frac{1}{2n + 3} + ... + \frac{1}{4n - 1}$$
the task is to find $$\lim_{n \to \infty} S_n$$
I've tried different ways, but all I could do is to make an estimation that the limit is somewhere between 0.5 and 1, but that's not the exact answer.
Manually doing fist elements gives
$$\frac{1}{3}, \quad \frac{1}{5} + \frac{1}{7}, \quad \frac{1}{7} + \frac{1}{9} + \frac{1}{11}, \quad ...$$
I'm trying to restate it as
$$\lim_{n \to \infty} \bigg[ \sum_{i = 1}^{\infty} \frac{1}{2i - 1} - \sum_{i = 1}^{n} \frac{1}{2i - 1} - \sum_{i = 4n}^{\infty} \frac{1}{2i - 1} \bigg]$$
so that I cut off the beginning of the series and it's tail.
| Consider the following inequalities:
$$ \begin{align}
\frac{1}{2n+1+2x} &\le \frac{1}{2n+1} \le \frac{1}{2n-1+2x}, \quad 0\le x\le 1, \quad (1) \\
\frac{1}{2n+1+2x} &\le \frac{1}{2n+3} \le \frac{1}{2n-1+2x}, \quad 1\le x\le 2 , \quad (2)\\
\frac{1}{2n+1+2x} &\le \frac{1}{2n+5} \le \frac{1}{2n-1+2x}, \quad 2\le x\le 3 , \quad (3)\\
\frac{1}{2n+1+2x} &\le \frac{1}{2n+7} \le \frac{1}{2n-1+2x}, \quad 3\le x\le 4 , \quad (4)\\
... \quad & \le \quad ... \quad \le \quad... \\
\frac{1}{2n+1+2x} &\le \frac{1}{2n+2n-1} \le \frac{1}{2n-1+2x}, \quad n-1\le x\le n. \quad (5)
\end{align} $$
Integrating each inequality over the corresponding interval, side by side.
For example:
$$ \begin{align}
\int_0^1 (1) dx \Rightarrow \int_0^1 \frac{dx}{2n+1+2x} &\le \frac{1}{2n+1} \le \int_0^1 \frac{dx}{2n-1+2x}, \quad (1') \\
\int_1^2 (2) dx \Rightarrow \int_1^2 \frac{dx}{2n+1+2x} &\le \frac{1}{2n+3} \le \int_1^2 \frac{dx}{2n-1+2x}, \quad (2') \\
\int_2^3 (3) dx \Rightarrow \int_2^3 \frac{dx}{2n+1+2x} &\le \frac{1}{2n+5} \le \int_2^3 \frac{dx}{2n-1+2x}, \quad (3') \\
... \quad & \le \quad ... \quad \le \quad... \\
\int_{n-1}^n (5) dx \Rightarrow \int_{n-1}^n \frac{dx}{2n+1+2x} &\le \frac{1}{2n+2n-1} \le \int_{n-1}^n \frac{dx}{2n-1+2x}. \quad (5')
\end{align} $$
Summing $(1')$ through $(5')$, side by side:
$$ \begin{align}\int_0^n \frac{dx}{2n+1+2x} &\le S_n \le \int_0^n \frac{dx}{2n-1+2x} \quad (6)\\
\frac12 \ln|2n+1+2x|_0^n &\le S_n \le \frac12 \ln |2n-1+2x|_0^n \\
\frac12 \ln\left|\frac{4n+1}{2n+1}\right| &\le S_n \le \frac12 \ln \left|\frac{4n-1}{2n-1}\right|, \quad (7)
\end{align}$$
Thus, from $(7)$:
$$ \lim_{n\to\infty} S_n = \frac{\ln 2}{2} $$
| {
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"source": "stackexchange",
"question_score": "2",
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Prove this $(ab)^2+(bc)^2+(ca)^2\le ab+bc+ac$
Let$a$, $b$ and $c$ be non-negative numbers such that
$$a^2+b^2+c^2=a+b+c.$$
Show that
$$(ab)^2+(bc)^2+(ca)^2\le ab+bc+ac.$$
I tried the uvw's technique and BW and more but without some success.I think can use C-S solve it?
| Another way:
$$\sum_{cyc}(ab-a^2b^2)=\frac{1}{2}\sum_{cyc}(2ab-2a^2b^2)=$$
$$=\frac{1}{2}\sum_{cyc}(a^2+2ab-a^4-2a^2b^2+a^4-a^2)=\frac{1}{2}\sum_{cyc}(a^4-a^2)=$$
$$=\frac{1}{2}\sum_{cyc}(a^4-a^2-2(a^2-a))=\frac{1}{2}\sum_{cyc}(a^4-3a^2+2a)=\frac{1}{2}\sum_{cyc}a(a+2)(a-1)^2\geq0.$$
Done!
| {
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Analyzing quadratic forms $x^2-3xy+y^2$, $2xy+yz-3xz$ and $x^2+y^2+2xy-xt+2yt$. Positive, negative or indefinite?
By means of successive coordinate changes, write each one of the
quadratic forms below as a sum of terms of the type $\pm u^2$ and
decide wich ones are positive, negative or indefinite:
$$A(x,y) = x^2-3xy+y^2$$
$$B(x,y,z) = 2xy+yz-3xz$$
$$C(x,y,z,t) = x^2+y^2+2xy-xt+2yt$$
For $A$ I did $A(x,y) = x^2-3xy+y^2 = x^2-2xy+y^2-xy = (x-y)^2-xy$ which is indefinite.
For $B(x,y) = 2xy+yz-3xz$ I remembered that $(x+y+z)^2 = x^2 + 2 x y + y^2 + 2 x z + 2 y z + z^2$ so $2xy+yz-3xz = (x+y+z)^2 -x^2-y^2-z^2-yz+xz$ which won't help anything.
For $C(x,y,z,t)$ there's too much terms, how should I do it?
| If you want to work directly with the quadratic expressions, you should just "complete" the square. For $C$, we have
$$ C(x,y,z,t) = x^2 + y^2 + 2xy - xt + 2yt = \left( x + y - \frac{1}{2}t \right)^2 - \frac{1}{4}t^2 + 3yt = \\ \left( x + y - \frac{1}{2}t \right)^2 - \left(\frac{1}{2}t - 3y \right)^2 - 9y^2.$$
I've manipulated the expression to get the terms $x^2,y^2,2xy, -xt$ as part of the expansion of the first square term which resulted in an "error" of the form $-\frac{1}{4}t^2 - yt$. Then I tried to get the terms $-\frac{1}{4}t^2, 3yt$ as part of the expansion of the second square, etc.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to compute the determinant of $D_n$? Let
\begin{align*}
D_n = \begin{pmatrix}
a & 0 & b & 0 & 0 & \cdots & b & 0 \\
0 & a & 0 & b & 0 & \cdots & 0 & b \\
b & 0 & a & 0 & b & \cdots & 0 & 0 \\
0 & b & 0 & a & 0 & \ddots & 0 & 0 \\
0 & 0 & b & 0 & a & \ddots & b & 0 \\
\vdots & \vdots & \vdots & \ddots & \ddots & \ddots & \vdots & \vdots \\
b & 0 & 0 & 0 & b & \cdots & a & 0 \\
0 & b & 0 & 0 & 0 & \cdots & 0 & a
\end{pmatrix}.
\end{align*}
In $D_n=(d_{ij})$, $d_{ii} = a$ for all $1 \leq i \leq n$, $d_{ij} = b$ for $|j-i|=2$ and $d_{n-1,1} = d_{n,2} = d_{1,n-1} = d_{2,n}=b$ for $n \geq 6$.
It is easy to compute $|D_1| = a$, $|D_2| = a^2$, $|D_3| = a (a^2 - b^2)$, $|D_4| = (a^2 - b^2)^2$, $|D_5| = a^5 - 3a^3 b^2 + 2 a b^4$. How to compute $|D_n|$?
Thanks to Semiclassical pointing that $n \geq 6$, $D_n$ is a circulant matrix .
| This is a circulant matrix with associated polynomial $f(x)=a+bx^2+bx^{n-2}$; the definition, as well as the result used below, can be found in the Wikipedia link. The eigenvectors/eigenvalues of a circulant matrix can be written down explicitly, and from the latter the determinant can be found. In the present case, we specifically have
$$|D_n|=\prod_{j=0}^{n-1}f(e^{2\pi i j/n})=\prod_{j=0}^{n-1}(a+be^{4\pi i j/n}+be^{2\pi ij(n-2)/n})=\prod_{j=0}^{n-1}(a+2b\cos(4\pi j/n))$$ where the fact that $e^{2\pi i j}=1$ for any integer $j$ has been used to simplify the exponentials.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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A point is reflected in the three sides of a triangle; when do the lines connecting the reflected points to corresponding vertices concur?
Let $P$ be a point in $\triangle ABC$, and let $P_A$ be its reflection about $\overline{BC}$, and similarly define $P_B$ and $P_C$. When do $\overleftrightarrow{AP_A}, \overleftrightarrow{BP_B},$ and $\overleftrightarrow{CP_C}$ concur?
They appear to concur when $P$ is the orthocenter, the incenter (no idea why though), and the circumcenter (and some other points) (also not sure why it happens here).
| I'll derive a condition based on the distances from $P$ to the edges of $\triangle ABC$. For simplicity, I'll assume that $P$ lies in the interior of the triangle; some tweaks to the intermediate may be required otherwise, but the final condition is (probably) universal.
Let the midpoints of $\overline{PP_A}$, $\overline{PP_B}$, $\overline{PP_C}$ be $D$, $E$, $F$, respectively, and define
$$d := |\overline{PD}| \qquad e := |\overline{PE}| \qquad f := |\overline{PF}|$$
Note that, because $\square AEPF$, $\square BFPD$, $\square CDPE$ each has two right angles, we have
$$\angle EPF = \pi - A \qquad \angle DPF = \pi - \angle B \qquad \angle DPE = \pi - \angle C$$
Let's coordinatize, placing $P$ at the origin and $D$ along the positive $x$-axis at $D = (d,0)$. Then
$$\begin{align}
E &= e \left(\; \cos(\phantom{-}\angle DPE), \sin(\phantom{-}\angle DPE) \;\right) = e \left(\;-\cos C, \phantom{-}\sin C\;\right) \\
F &= f \left(\; \cos(-\angle DPF), \sin(-\angle DPF)\;\right) = f\left(-\cos B, -\sin B\;\;\right)
\end{align}$$
The sides of the triangle are perpendicular to $\overline{PD}$, $\overline{PE}$, $\overline{PF}$ at $D$, $E$, $F$, and we derive
$$\overleftrightarrow{BC}: x = d \qquad \overleftrightarrow{CA}: x \cos C - y \sin C = -e \qquad \overleftrightarrow{AB}: x \cos B + y \sin B = -f$$
so that
$$\begin{align}
A &= \frac{1}{\sin A}\left(\;-e \sin B - f \sin C, e \cos B - f \cos C\right) \\[8pt]
B &= \left(d, -d \cot B - f \csc B \right)\\[8pt]
C &= \left(d, \phantom{-}d \cot C + e \csc C \right)
\end{align}$$
Writing $D^\prime$ for the point where $\overleftrightarrow{AP_A}$ meets $\overleftrightarrow{BC}$, we can find
$$D^\prime = \left(\;d\;,\; \frac{d \left( e \cos B - f \cos C \right)}{
2 d \sin A + e \sin B + f \sin C}\;\right)$$
Also,
$$\begin{align}
\overrightarrow{BD^\prime} = D^\prime - B =
\left(\;0\;,\; \frac{(f + 2 d \cos B) ( d \sin A + e \sin B + f \sin C )}{\sin B\;(
2 d \sin A + e \sin B + f \sin C)}\;\right) \\[8pt]
\overrightarrow{D^\prime C} = C - D^\prime =
\left(\;0\;,\; \frac{(e + 2 d \cos C) ( d \sin A + e \sin B + f \sin C )}{\sin C\;(
2 d \sin A + e \sin B + f \sin C)}\;\right) \\[8pt]
\end{align}$$
Conveniently, the $y$-components of these vectors are exactly the signed distances used in the corresponding ratio from Ceva's Theorem:
$$\frac{|BD^\prime|}{|D^\prime C|} = \frac{\sin C\;(f+2d \cos B)}{\sin B\;(e+2d \cos C)}$$
With the appropriate variants of the above for the other sides of the triangle, Ceva's Theorem gives (after canceling sines, clearing fractions, and combining terms) this condition for concurrency:
$$\begin{align}
0 &= d ( e^2 - f^2 ) ( \cos A - 2 \cos B \cos C ) \\
&+ e ( f^2 - d^2 ) ( \cos B - 2 \cos C \cos A ) \tag{$\star$}\\
&+ f ( d^2 - e^2 ) ( \cos C - 2 \cos A \cos B )
\end{align}$$
Right away, we see that the incenter satisfied this condition, as $d = e = f = \text{inradius}$. Less-obviously, if $P$ is the circumcenter and $r$ the circumradius, then
$$d = r \cos A \qquad e = r \cos B \qquad f = r \cos C$$
and $(\star)$ holds. Also, for $P$ the orthocenter (and $r$ still the circumradius),
$$d = 2 r \cos B \cos C \qquad e = 2 r \cos C \cos A \qquad f = 2 r \cos A \cos B$$
satisfying $(\star)$. Exploration of other cases, and/or conversion of $(\star)$ into a better form, is left as an exercise to the reader.
It's worth pointing-out some geometric interpretations of expressions used above. For instance, from the Law of Sines (with $r$ the circumradius), we have
$$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2 r$$
so that
$$ d \sin A = \frac{d a}{2r} = \frac{1}{r}|\triangle PBC| \qquad e\sin B = \frac{1}{r}|\triangle APC| \qquad f \sin C = \frac{1}{r} |\triangle ABP|$$
and
$$ d \sin A + e \sin B + f \sin C = \frac{1}{r}|\triangle ABC| \qquad 2d\sin A + e \sin B + f \sin C = \frac{1}{r}|\square ABP_AC|$$
Moreover, if $E^{\prime\prime}$ and $F^{\prime\prime}$ are the feet of the perpendiculars from $P_A$ to $\overleftrightarrow{PE}$ and $\overleftrightarrow{PF}$, respectively, then $\angle P_APE^{\prime\prime} = \angle C$ and $\angle P_APF^{\prime\prime} = \angle B$, so that
$$|EE^{\prime\prime}| = e + 2d \cos C \qquad |FF^{\prime\prime}| = f + 2d \cos B$$
These facts may somehow provide a coordinate-free path to $(\star)$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Is it a coincidence that the 12th Fibonacci number is 12 squared? The first $12$ Fibonacci numbers are:
$$\begin{array}{|c|c|c|c|c|c|c|c|c|c|c|c|c|c|}
\hline
n & 0& 1& 2& 3& 4& 5& 6& 7& 8 & 9& 10& 11& 12\\
\hline
f_{n} & 0& 1& 1& 2& 3& 5& 8& 13& 21& 34& 55& 89& 144\\
\hline
\end{array}$$
When you get to $f_{12}$ you find it is equal to $144$.
What strikes me here is the following:
\begin{align*}
f_{12} &= 144\\
12^2 &=144.
\end{align*}
Not only is $f_{12}$ equal to $144$, but so is $12^2$. Here, $n^2= f_n$.
This is the only example as $n^2 > f_n$ is true for all $1<n<12$ (works for $0$ and $1$) and $n^2< f_n$ is true for all $n > 12$.
So is this just a coincidence?
| This is a very nice manifestation of the Strong Law of Small Numbers formulated by Richard K. Guy:
"There aren't enough small numbers to meet the many demands made of them."
A more striking manifestation of the same "law" of small numbers is this:
$$
\begin{array}{|c|c|c|c|c|c|c|c|c|c|c|c|c|c|}
\hline
n & 0& 1& 2& 3& 4& 5& 6& 7& 8 & 9& 10& 11& 12\\
\hline
\mbox{Fibonacci }
f_{n} & 0& 1& 1& 2& 3& 5& 8& 13& 21& 34& 55& 89& 144\\
\hline
\hline
\mbox{# squares in }[e^{n-1},e^n)
& 0& 1& 1& 2& 3& 5& 8& 13& 21& 36& 58& 96& 159\\
\hline
\end{array}
$$
The last line is OEIS A306486: the number of squares in the interval $[e^{n-1},e^n)$, which happens to be the same as the Fibonacci number $f_n$ for $0\le n\le 8$. Note that both $f_n$ and A306486($n$) grow about as fast as certain geometric progressions; the common ratio is $\phi\approx1.618\ldots$ for the former and $\sqrt{e}\approx1.6487\ldots$ for the latter.
| {
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Better way of finding $\int {x^2 + 1 \over x^4 + x^2 + 1} dx$.
$$\int {x^2 + 1 \over x^4 + x^2 + 1} dx$$
By separating partial fractions,
$${Ax + B \over x^2 - x + 1} + {Cx + D \over x^2 + x + 1} = {x^2 + 1 \over x^4 + x^2 + 1} \\ \implies (Ax + B)(x^2 + x + 1) + (Cx + D)(x^2 - x + 1) = x^2 + 1$$
I get $$\begin{cases} A = -1/2 \\
B = 1/2\\
C = 1/2 \\
D = 1/2\end{cases}$$
For which the integrand becomes
$${1 \over 2}\int {1 - x\over x^2 - x + 1} dx + {1\over2} \int {x + 1\over x^2 + x + 1} dx$$
Now these two are easy enough to solve but still very tedious, not to mention that partial fractions was also very tedious.
Is there a less cumbersome way to solve this ? I tried to change the integrand of form $1 + 1/x^2$ so that I substitute $u = 1 - 1/x$ but was unsuccessful .
| Notice that if you plug $x=i$ into $$(Ax+B)(x^2+x+1)+(Cx+D)(x^2-x+1)=x^2+1$$
You get $$Bi-A+C-Di=(C-A)+(B-D)i=0$$
or $C=A,B=D$ so
$$(Ax+B)(x^2+x+1+x^2-x+1)=2(Ax+B)(x^2+1)=x^2+1$$
From this it's easy to see that $A=C=0$ and $B=D=1/2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2240614",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
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} |
Simple inequality $\left|\frac{3x+1}{x-2}\right|<1$
$$\left|\frac{3x+1}{x-2}\right|<1$$
$$-1<\frac{3x+1}{x-2}<1$$
$$-1-\frac{1}{x-2}<\frac{3x}{x-2}<1-\frac{1}{x-2}$$
$$\frac{-x+1}{x-2}<\frac{3x}{x-2}<\frac{x-3}{x-2}$$
$$\frac{-x+1}{x-2}<\frac{3x}{x-2}<\frac{x-3}{x-2} \text{ , }x \neq 2$$
$${-x+1}<{3x}<{x-3} \text{ , }x \neq 2$$
$${-x+1}<{3x} \text{ and } 3x<{x-3} \text{ , }x \neq 2$$
$${1}<{4x} \text{ and } 2x<{-3} \text{ , }x \neq 2$$
$${\frac{1}{4}}<{x} \text{ and } x<{\frac{-3}{2}} \text{ , }x \neq 2$$
While the answer is
$${\frac{1}{4}}>{x} \text{ and } x>{\frac{-3}{2}}$$
| We can not multiply by $x-2$ even if it is $\ne0$
In fact the inequality sign remains valid only if $x-2>0$
$$-1<\dfrac{3x+1}{x-2}\iff0<\dfrac{3x+1-(x-2)}{x-2}=\dfrac{x+3}{x-2}$$
So, we need $(x-2)(x+3)>0$
Now if $(x-a)(x-b)>0;a\le b$ we can prove either $x>b$ or $x<a$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2242064",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to solve the equation $x^2y-y^2 = e^ax^2+x^2$ in terms of $x$? Please help me solve the following equation in terms of $x$, I have difficulty with this:
$$x^2y-y^2 = e^ax^2+x^2$$
Edit with my solution:
Rewrite the equation
$$0 = y^2 - x^2y + (e^a+1)x^2$$
Using the quadratic formula we obtain
$$y = \frac{x^2±\sqrt{x^4-4(e^a+1)x^2}}{2}$$
and simplify we obtain
$$y = \frac{x^2}{2} \pm \frac{x\sqrt{x^2-4e^a-4}}{2}$$
| Rewrite your equation as
$$y^2 - (x^2)y + (e^ax^2+x^2) = y^2 -by +c$$
where $b$ and $c$ correspond to the terms in brackets. Then apply the standard technique to solve a quadratic equation.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2245263",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
What is $1+x+x^2+x^3...$? What is the difference between these two series?
$$
\begin{align}
1+x+x^2+x^3+...+x^n+\mathcal O(x^{n+1})&=\frac{1}{1-x}\\
\\
1+r+r^2+r^3+...+r^{n-1}&=\frac{r^n-1}{r-1}\\
\end{align}
$$
I can't wrap my head around it; they both start with $1+x+x^2+x^3...$? Of course the first is the maclaurin series. But other than that, why don't they both yield the same result?
Thanks!
| Beware that "$=$" is not a symmetric relation in the presence of asymptotic notation. So while it is true that
$$\frac{1}{1-x} = 1+x+x^2+x^3+...+x^n+O(x^{n+1})$$
as $x\to 0$, we do not usually say that
$$1+x+x^2+x^3+...+x^n+O(x^{n+1})=\frac{1}{1-x}$$
because the left-hand side of that can be a lot of things other than $\frac{1}{1-x}$.
Some consider the conventional use of $=$ in these contexts to be misguided and misleading, and would rather we write
$$\frac{1}{1-x} \in 1+x+x^2+x^3+...+x^n+\mathcal O(x^{n+1})$$
whose right-hand side is interpreted as a set of functions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2245655",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Why is $\int \frac{1}{\sqrt{y}\sqrt{1 - y}} dy = \frac{2\sqrt{y - 1}\sqrt{y} \log(\sqrt{y - 1} + \sqrt{y})}{\sqrt{(-(y - 1) y)}} $? Fairly self-explanatory question title. Why is
$$\int \frac{1}{\sqrt{y}\sqrt{1 - y}} dy = \frac{2\sqrt{y - 1}\sqrt{y} \log(\sqrt{y - 1} + \sqrt{y})}{\sqrt{-(y - 1)}\sqrt{y}}\ ? $$
I'm assuming you have to use substition, but I'm not sure how.
edit: $$ y \in (0,1) $$
| Is the answer correct? Notice $y\in (0,1)$, as both $y> 0$, $1-y>0$. But why answer has $\sqrt{y-1}$?
Let $y=\sin ^2x$, $x\in (0,\frac{\pi}{2})$
$$\int \frac{1}{\sqrt{y}\sqrt{1 - y}} dy = \int \frac{2\sin x \cos x}{\sin x\cos x} dx =2x +C=2\arcsin \sqrt{y}+C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2246475",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluating $\int_C\frac{z+1}{z^2-2z}dz$, where $C$ is the circle $|z|=3$
Evaluate the contour integral $\int_C\frac{z+1}{z^2-2z}dz$ using Cauchy's residue theorem, where $C$ is the circle $|z|=3$.
I see that the function has 2 singularities, at 0 and 2, so I need to find the residue of each. By examining the Laurent series, I have the following:
$$f(z)=\left(\frac{z+1}{z}\right)\left(\frac{1}{z-2}\right)=\left(\frac{z+1}{z^2}\right)\left(\frac{1}{1-2/z}\right)=\left(\frac{1}{z}+\frac{1}{z^2}\right)\left(\frac{1}{1-2/z}\right)$$ and therefore $$f(z)=\left(\frac{1}{z}+\frac{1}{z^2}\right)\left(1-\frac{2}{z}+\frac{4}{z^2}-\frac{8}{z^3}+\cdots\right)=\frac{1}{z}-\frac{1}{z^2}+\frac{2}{z^3}-\frac{4}{z^4}+\cdots$$
so the residue at 0 is 1.
Similarly, $$f(z)=\left(\frac{z+1}{2(z-2)}\right)\left(\frac{1}{1+(z-2)/2}\right)=\left(\frac{1}{2}+\frac{3}{2(z-2)}\right)\left(\frac{1}{1+(z-2)/2}\right)$$ and so $$f(z)=\left(\frac{1}{2}+\frac{3}{2(z-2)}\right)\left(1-\frac{z-2}{2}+\frac{(z-2)^2}{4}-\frac{(z-2)^3}{8}+\cdots\right)$$
Thus $$f(z)=\frac{3}{2(z-2)}-\frac{1}{4}+\frac{1}{8}(z-2)-\frac{1}{16}(z-2)^2+\cdots$$
and so the residue at 2 is $\frac{3}{2}$.
So I think $\int_Cf(z)dz=2\pi i(1+\frac{3}{2})= 5\pi i$, but that's not what the book is telling me - the book says the answer should be $2\pi i$. What am I doing wrong?
| You can calculate the partial fraction decomposition to avoid calculating the Laurent's series :
$$\int_C\frac{z+1}{z^2-2z}dz=\frac32\int_C\frac{1}{z-2}dz-\frac12\int_C\frac{1}{z}dz$$
Then we apply Cauchy formula :
$$\int_C\frac{1}{z-2}dz=2\pi i=\int_C\frac{1}{z}dz$$
Or with Cauchy residue's theorem, $\frac{1}{z-2}$ is holomorphic inside $C$ except in $z=2$, so its Laurent series around $2$ is $\frac{1}{z-2}+\sum_{i=0}^\infty0z^i=\frac{1}{z-2}$. So the residue of $f$ around $2$ is $1$, so $\int_C\frac{1}{z-2}dz=2\pi i$. You can do the same with other integral.
So :
$$\int_C\frac{z+1}{z^2-2z}dz=2\pi i$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2248563",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Determine the unique vector x in the row space for A, for which Ax = b Given the Vector $b = (9, -6, 3)^T$
$A =\begin{pmatrix}
1 & 2 & 0 & 1\\
-1 & -1 & -1 & 0 \\
1 & 0 & 2 &-1
\end{pmatrix}$
I know that the row space is $ \begin{pmatrix} 1 & 0 & 2& -1\end{pmatrix}$ , $ \begin{pmatrix} 0 & 1 & -1& 1\end{pmatrix}$
Not really sure how to proceed.
| As far as I can see, this system of equations doesn't even a have a solution, for we have:
$$
\begin{align}
\left[
\begin{array}{cccc|c}
1&2&0&1&6\\
-1&-1&-1&0&-6\\
1&0&2&-1&3
\end{array}
\right]
&\iff
\left[
\begin{array}{cccc|c}
1&2&0&1&6\\
0&1&-1&1&0\\
0&-2&2&-2&-3
\end{array}
\right]
\iff
\left[
\begin{array}{cccc|c}
1&2&0&1&6\\
0&1&-1&1&0\\
0&0&0&0&-3
\end{array}
\right]
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2251271",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
A question regarding Czarnowski's cat function The Czarnowski cat function $c(x, n) = \sin^n(x) + \cos(x)$ gives the so-called Czarnowski's cats. There are infinitely many of these cats, and legend has it that Czarnowski created a clever mathematical naming scheme to give them all unique names.
The problem: for Czarnowski's principal cat (named Mruk, the one centered at the origin) find the area between the cats corresponding to $n = 18$ and $n = 26$.
Some background. In 1896, Polish mathematician Żurzysław Czarnowski (born 1875) discovered the cat function while doodling on his note pad during a particularly boring talk by his teacher, Belgian mathematician Per Hansel. It is rumored that one day Czarnowski presented the cat function to Hansel, who studied it and noticed that for very large $n$ the cats turn into devils. It is rumored that Hansel was so terrified and distraught by this discovery that he committed suicide. This prompted Czarnowski to find an optimal cat, i.e. a value $n$ for which the cats are "most cat-like and least devil-like". It is rumored that this $n$ lies somewhere between 18 and 26, and that if we can find the area between the curves corresponding to "cat 18" and "cat 26" we will have half of the solution.
| We seek a reduction formula for integrands of the form $$\sin^{2n} x.$$ Integration by parts with the choice $$u = \sin^{2n-1} x, \quad du = (2n-1) \sin^{2n-2} x \cos x \, dx, \\ dv = \sin x \, dx, \quad v = -\cos x, $$ gives
$$\begin{align*} I_n (x) &= \int \sin^{2n} x \, dx = -\sin^{2n-1} x \cos x + (2n-1) \int \sin^{2n-2} x \cos^2 x \, dx \\
&= -\sin^{2n-1} x \cos x + (2n-1) \int \sin^{2n-2} (1 - \sin^2 x) \, dx \\
&= -\sin^{2n-1} x \cos x + (2n-1) I_{n-1}(x) - (2n-1)I_n(x).
\end{align*}$$
Consequently, $$I_n(x) = \frac{1}{2n} \left( -\sin^{2n-1} x \cos x + (2n-1) I_{n-1}(x) \right).$$ On the interval $[0,\pi]$, we then have $$I_n = \int_{x=0}^\pi I_n(x) \, dx = \frac{2n-1}{2n} I_{n-1}.$$ And for $n = 0$, we trivially have $$I_0(x) = 1, \quad I_0 = \pi.$$ It follows that $$I_n = \pi \prod_{k=1}^n \frac{2k-1}{2k} = \pi \prod_{k=1}^n \frac{(2k-1)(2k)}{(2k)^2} = \frac{(2n)!}{4^n (n!)^2} \pi = \binom{2n}{n}\frac{\pi}{4^n}.$$ Therefore, since $c(x,n+2) \le c(x,n)$, $$\begin{align*} \int_{x=-\pi}^\pi c(x,18) - c(x,26) \, dx
&= 2 \int_{x=0}^\pi \sin^{18} x - \sin^{26} x \, dx \\
&= 2(I_{9} - I_{13}) \\
&= 2\pi \left(\binom{18}{9}\frac{1}{4^{9}} - \binom{26}{13} \frac{1}{4^{13}}\right) \\
&= \frac{\pi}{2^{25}}\left(\binom{18}{9} 4^4 - \binom{26}{13}\right) \\
&= \frac{255765 \pi }{4194304}.
\end{align*}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2252633",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Error of approximation for Euler's constant Is there a simple way to show this bound for Euler’s constant:
$$ \frac{1}{2n+2} < \sum_{k=1}^n\frac{1}{k} - \ln(n) - \gamma < \frac{1}{2n}$$
For proving convergence, I showed that $\sum_{k=1}^n 1/k - \ln(n)$ is bounded and monotonic, using the property
$$\frac{1}{k+1} < \int_k^{k+1} x^{-1} dx <\frac{1}{k}$$
This showed
$$\frac{1}{n} < \sum_{k=1}^n \frac{1}{k} - \ln(n) < 1$$
But this does not help prove the error bounds.
Thanks
| Use convexity.
Since $f(x) = 1/x$ is both decreasing and convex we have for $k \leqslant x \leqslant k+1$,
$$\frac{f(k+1) - f(k+2)}{k + 2 - (k+1)} \leqslant \frac{f(x) - f(k+1)}{k + 1 - x} \leqslant \frac{f(k) - f(k+1)}{k + 1 - k} $$
Hence,
$$\tag{*}\left(\frac{1}{k+1} - \frac{1}{k+2}\right)(k + 1 - x) \leqslant \frac{1}{x}- \frac{1}{k+1} \leqslant \left(\frac{1}{k} - \frac{1}{k+1} \right) (k+1 - x)$$
Note that
$$\int_{k}^{k+1}(k + 1 - x) \, dx = \left.-\frac{1}{2} (k+1 -x)^2\right|_k^{k+1} = \frac{1}{2}$$
Integrating (*) over $[k,k+1]$ and summing from $k = n$ to $N-1$ we get
$$\frac{1}{2(n+1)} - \frac{1}{2(N+1)}\leqslant \sum_{k=n}^{N-1} \int_{k}^{k+1} \frac{dx}{x}- \sum_{k=n}^{N-1} \frac{1}{k+1} \leqslant \frac{1}{2n} - \frac{1}{2N},$$
which reduces to
$$\frac{1}{2(n+1)} - \frac{1}{2(N+1)}\leqslant \sum_{k=1}^n\frac{1}{k} -\log n - \left(\sum_{k=1}^N \frac{1}{k}- \log N\right) \leqslant \frac{1}{2n} - \frac{1}{2N}$$
Then take the limit as $N \to \infty$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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If $\alpha$ and $\beta$ are the roots of $x^2-2x+4=0$ then the value of $\alpha^6+\beta^6$ is
If $\alpha$ and $\beta$ are the roots of $x^2-2x+4=0$ then the value
of $\alpha^6+\beta^6$ is
I know that, here, $\alpha\beta=4$ and $\alpha + \beta = 2$ and use that result to find $\alpha^2 + \beta^2$ using the expansion of $(a+b)^2$
But how to find $\alpha^6+\beta^6$ ?
| From $\alpha^2-2\alpha+4=0$ we get that for $n\geq 0$ we have $\alpha^{n+2}-2\alpha^{n+1}+4\alpha^n=0$, and we have the same relation for $\beta$. Putting $u_n=\alpha^n+\beta^n$ and adding, we get that $u_{n+2}-2u_{n+1}+4u_n=0$ for all $n$. We have $u_0=2$, $u_1=2$, and now it is easy to compute $u_6$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2256142",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Prove that there are infinite sets of $(a, b, c)$ such that $ab + 1$, $ac + 1$ and $bc + 1$ are perfect square Prove that there are infinite sets of integer numbers $(a, b, c)$ that $ab + 1$, $ac + 1$ and $bc + 1$ are perfect square. ($a$, $b$, $c$ are different numbers)
Any hints how to prove the statement? I'm trying to find a way to turn that into a Pell's equation
and then prove it has infinite answers.
| If $a=4n$ , $b=n+1$ and $c=n-1$ then
\begin{eqnarray*}
ab+1&=&(2n+1)^2 \\
bc+1&=& n^2 \\
ca+1&=&(2n-1)^2
\end{eqnarray*}
EDIT : for Ovi.
We $ab+1,bc+1,ca+1$ are perfect squares. So there exist $x,y,z$ such that
\begin{eqnarray*}
z^2=ab+1 \\
x^2=bc+1 \\
y^2=ca+1
\end{eqnarray*}
Now subtract these equations pairwise (difference of two squares factorise) and we have
\begin{eqnarray*}
(z-x)(z+x)=b(a-c) \\
(z-y)(z+y)=a(b-c) \\
(y-x)(y+x)=c(a-b)
\end{eqnarray*}
At this point I totally guessed to equate the factors pairwise (there is no justification for this). Amazingly this gives
\begin{eqnarray*}
x=\frac{-a+b+c}{2} \\
y=\frac{a-b+c}{2} \\
z=\frac{a+b-c}{2}
\end{eqnarray*}
Now substitute this formula for $z$ into $z^2=ab+1$ & after some algebra, we have
\begin{eqnarray*}
a^2+b^2+c^2-2(ab+bc+ca)=4.\\
a^2-2a(b+c)+(b-c)^2-4=0
\end{eqnarray*}
Now consider this as a quadratic in $a$ ... Its disciminant (over 4) needs to be a perfect square ...
\begin{eqnarray*}
\Delta=(b+c)^2- \left( (b-c)^2-4 \right) =4bc+4=(2n)^2.
\end{eqnarray*}
& the above solution follows.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2256701",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Analytical evaluation of double integral I'm looking to evaluate the following integral:
$$\int^a_{-a}\int^a_{-a}\frac{\sqrt{x^2+y^2}}{4a^2} \,\mathrm{d}x \mathrm{d}y,$$
where $a> 0$.
Mathematica gives the following answer:
$$ \frac{a}{3} (\sqrt{2} + \sinh^{-1}(1)).$$
This goes well beyond my basic calculus 101 training.
Is anybody able to give me a step-by-step analytical solution?
Regards
| $$
\frac 1 {4a^2} \int^a_{-a}\int^a_{-a} \sqrt{x^2+y^2} \, dx \, dy = \text{what?}
$$
$$
\int_{-a}^a (\text{an even function}) = 2\times\int_0^a (\text{the same function}), \text{ so}
$$
\begin{align}
& \int_{-a}^a \sqrt{x^2+y^2} \ dx = 2\int_0^a \sqrt{x^2+y^2}\ dx = 2\int_0^{\arctan a} \sqrt{y^2\tan^2\theta + y^2} (\sec^2\theta\,d\theta) \\[10pt]
= {} & 2|y| \int_0^{\arctan a} \Big(\sec\theta\Big) (\sec^2\theta\,d\theta) = 2|y| \int_0^{\arctan a} \sec^3\theta\,d\theta.
\end{align}
This is one of the more challenging among freshman-level integrals. This article tells us how to find that
$$
\int \sec^3\theta\,d\theta = \frac 1 2 \sec\theta\tan\theta + \frac 1 2 \log|\sec\theta+\tan\theta| + C.
$$
So we need
$$
\left[ \frac 1 2 \sec\theta\tan\theta + \frac 1 2 \log|\sec\theta+\tan\theta| \right]_0^{\arctan\theta}.
$$
From trigonometry we have
\begin{align}
\tan\arctan a & = a, \\[10pt]
\sec\arctan a & = \sqrt{1+a^2}.
\end{align}
Maybe you can do the rest. (?)
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "7",
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Find the sum of all values of $ f ( 2017 ) $ given $ f ^ { f ( a ) } ( b ) f ^ { f ( b ) } ( a ) = f ( a + b ) ^ 2 $.
Let $ f :\mathbb N \to \mathbb N $ be an injective function such that
$$ f ^ { f ( a ) } ( b ) f ^ { f ( b ) } ( a ) = f ( a + b ) ^ 2 $$
for all $ a , b \in \mathbb N $. Let $ S $ be the sum of all possible values of $ f ( 2017 ) $. Find $ S \mod 1000 $. Here
$$ f ^ k ( n ) = \underbrace { f \Big( \dots f \big( f } _ { k \text{ times} } ( n ) \big) \dots \Big) \text . $$
So I put $ a = b $ and found out that $ f ^ { f ( a ) } ( a ) = f ( 2 a ) $.
Since $ f $ is injective, $ f ^ { f ( a ) - 1 } ( a ) = 2 a $. I found out that $ f ( 1 ) = 2 $.
Now $ f ( 2 ) = 3 $ by putting value of $ a = 1 $ in initial equation and solving for $ f ( b ) = 3 $. It seems by guessing that $ f ( n ) = n + 1 $ but can't the function be periodic after certain interval and is unbounded and repeat its value? It can still be injective.
| It's easy to see that $ f ( n ) = n + 1 $ satisfies the functional equation
$$ f ^ { f ( a ) } ( b ) f ^ { f ( b ) } ( a ) = f ( a + b ) ^ 2 \text , \tag 0 \label 0 $$
since then for every positive integers $ m $ and $ n $, one will have $ f ^ m ( n ) = n + m $. It can be shown that this is the only injective solution to \eqref{0} that maps positive integers to positive integers, and thus the only possible value for $ f ( 2017 ) $ is $ 2018 $, which shows $ S = 2018 $ and hence $ S \mod 1000 = 18 $. To see this, first note that $ f $ cannot take the value $ 1 $, because if $ f ( n ) = 1 $ for some positive integer $ n $, then letting $ a = b = n $ in \eqref{0}, you get $ f ( 2 n ) = 1 $, which by injectivity of $ f $ yields $ n = 2 n $, which can't happen. Next, let $ b = a $ in \eqref{0} to get
$$ f ^ { f ( a ) } ( a ) = f ( 2 a ) \text , $$
and by injectivity,
$$ f ^ { f ( a ) - 1 } ( a ) = 2 a \text , \tag 1 \label 1 $$
as you've done yourself. We can use \eqref{1} to inductively prove
$$ f ^ { \sum _ { m = 1 } ^ n f \left( 2 ^ { m - 1 } a \right) - n } ( a ) = 2 ^ n a $$
for every nonnegative integer $ n $ (note that if $ n = 0 $, the left-hand side gives $ f ^ 0 ( a ) $ that means $ a $ itself), which in particular shows that $ \left\{ f ^ i ( a ) \right\} _ { i = 0 } ^ \infty $ is infinite. As a consequence, we can see that if for some nonnegative integers $ i $ and $ j $ we have $ f ^ i ( a ) = f ^ j ( a ) $, then we must have $ i = j $.
We can now show that $ f ( 1 ) = 2 $. If that's not the case, we must have $ f ( 1 ) = n + 2 $ for some positive integer $ n $. Then we have $ f ^ { n + 1 } ( 1 ) = 2 $ by putting $ a = 1 $ in \eqref{1}, and letting $ a = f ^ n ( 1 ) $ in \eqref{1} we get
$$ 2 f ^ n ( 1 ) = f ^ { f \left( f ^ n ( 1 ) \right) - 1 } \big( f ^ n ( 1 ) \big) = f ^ { f ^ { n + 1 } ( 1 ) - 1 } \big( f ^ n ( 1 ) \big) = f ^ { 2 - 1 } \big( f ^ n ( 1 ) \big) = f ^ { n + 1 } ( 1 ) = 2 \text . $$
But this means $ f \left( f ^ { n - 1 } ( 1 ) \right) = 1 $, which is impossible.
Then, we show that $ f ( 2 ) = 3 $. To see this, first note that if for some positive integer $ n $ and some positive integer $ k \ne 1 $ we have $ f ( n ) = 2 k $, letting $ a = k $ in \eqref{1} we get $ f ^ { f ( k ) - 1 } ( k ) = f ( n ) $, and as we must have $ f ( k ) > 2 $, we get $ f \left( f ^ { f ( k ) - 3 } ( k ) \right) = f ( n ) $. By injectivity of $ f $, we conclude that there is a positive integer $ m $ such that $ f ( m ) = n $. Since by putting $ a = 1 $ and $ b = 2 $ in \eqref{0} and using \eqref{1} we have
$$ f ( 3 ) ^ 2 = f ^ { f ( 1 ) } ( 2 ) f ^ { f ( 2 ) } ( 1 ) = f ^ 2 ( 2 ) f ^ { f ( 2 ) - 1 } \big( f ( 1 ) \big) = f ^ 2 ( 2 ) f ^ { f ( 2 ) - 1 } ( 2 ) = 4 f ^ 2 ( 2 ) \text , $$
it follows that there is a positive integer $ k $ such that $ f ( 3 ) = 2 k $. As $ f ( 3 ) \ne f ( 1 ) = 2 $, we must have $ k \ne 1 $, and thus there is a positive integer $ m $ with $ f ( m ) = 3 $. We know that $ m \ne 1 $. $ m $ cannot be equal to $ 3 $ either, because then by \eqref{1} we would have $ 6 = f ^ { f ( 3 ) - 1 } ( 3 ) = f ^ 2 ( 3 ) = 3 $. If $ m > 3 $, we can let $ a = m - 3 $ and $ b = 3 $ in \eqref{0} and see that
$$ 9 = f ( m ) ^ 2 = f ^ { f ( m - 3 ) } ( 3 ) f ^ { f ( 3 ) } ( m - 3 ) \text . $$
As non of $ f ^ { f ( m - 3 ) } ( 3 ) $ and $ f ^ { f ( 3 ) } ( m - 3 ) $ can be equal to $ 1 $, we must have $ f ^ { f ( m - 3 ) } ( 3 ) = f ^ { f ( 3 ) } ( m - 3 ) = 3 $. But
$ f ^ { f ( m - 3 ) } ( 3 ) = 3 = f ^ 0 ( 3 ) $ leads to $ f ( m - 3 ) = 0 $, which is impossible. Thus the only possible case is $ m = 2 $, as desired.
At last, we use strong induction on $ n $ to prove that $ f ( n ) = n + 1 $ for every positive integer $ n \ge 3 $. Note that as $ n \ge 3 $, if we let $ a = \left\lfloor \frac { n + 1 } 2 \right\rfloor $ and $ b = \left\lceil \frac { n + 1 } 2 \right\rceil $, then $ a $ and $ b $ are positive integers less than $ n $ such that $ a + b = n + 1 $. By induction hypothesis, assume that $ f ( m ) = m + 1 $ for every positive integer $ m $ less than $ n $. This shows that $ f ^ { n - m } ( m ) = n $ for every positive integer $ m $ less than $ n $. Thus, using \eqref{0} we have
$$ f ( n + 1 ) ^ 2 = f ( a + b ) ^ 2 = f ^ { f ( a ) } ( b ) f ^ { f ( b ) } ( a ) = f ^ { a + 1 } ( b ) f ^ { b + 1 } ( a ) \\
= f ^ { n + 2 - b } ( b ) f ^ { n + 2 - a } ( a ) = f ^ 2 \left( f ^ { n - b } ( b ) \right) f ^ 2 \left( f ^ { n - a } ( a ) \right) = \left( f ^ 2 ( n ) \right) ^ 2 \text . $$
Therefore we have $ f ( n + 1 ) = f \big( f ( n ) \big) $, and hence by injectivity of $ f $, $ f ( n ) = n + 1 $, as desired.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2257621",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
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} |
Solving Recurrence Relation with substitution How to solve T(n) = T(n-2) + n using iterative substitution
Base case:
T(0) = 1
T(1) = 1
Solve:
T(n) = T(n-2) + n
Currently I have:
T(n) = T(n-2) + n
= T(n-4) + n - 2 + n = T(n-4) + 2n - 2
= T(n-6) + n - 4 + n - 2 + n = T(n-6) + 3n - 6
= T(n-8) + n - 6 + n - 4 + n -2 + n = T(n-8) + 4n - 12
= T(n-10) + n - 8 + n - 6 + n - 4 + n - 2 + n = T(n-10) + 5n - 20
The pattern I see is:
$$\ T(n-2 \sum_{i=1}^k i) + n \sum_{i=0}^k i - \sum_{i=0}^{k-1} i(i+1) $$
but this may be wrong because I am completely stuck after this
| $$\begin{align}
T(n) &= T(n-2) + n \\
T(n) &= T(n-4) + (n-2) + (n-0) \\
T(n) &= T(n-6) + (n-4) + (n-2) + (n-0) \\
T(n) &= T(n-8) + (n-6) + (n-4) + (n-2) + (n-0) \\
\vdots\\
T(n) &= T(n-2r) + \sum_{k=0}^{r-1}(n-2k)\\
T(n) &= T(n-2r) + \left(n\sum_{k=0}^{r-1}(1)\right)-2\left(\sum_{k=0}^{r-1}k\right)\\
T(n) &= T(n-2r) + nr-2\dfrac{r(r-1)}{2}\\
T(n) &= T(n-2r) + nr - r^2 + r
\\
\end{align}$$
We know that $T(0) = T(1) = 1$, so we need to set $n-2r = 0$ or $1$ in order for $T(n-2r) = 1$ to be reached, which is achieved when $r = \left\lfloor\dfrac{n}{2}\right\rfloor$ where $\left\lfloor x\right\rfloor$ is the floor function of $x$.
Therefore:
$$T(n) = 1 + n\left\lfloor\dfrac{n}{2}\right\rfloor - \left\lfloor\dfrac{n}{2}\right\rfloor^2 + \left\lfloor\dfrac{n}{2}\right\rfloor$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2259243",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Given $x,y,z >0$ and $xy^2z^3 = 108 $, what is the minimum value of $x+y+z$? Given $x,y,z >0$ and $xy^2z^3 = 108 $, what is the minimum value of $x+y+z$ ?
This is a homework problem, so if someone could just give me an outline/hint of the method used to solve this it would be much appreciated.
Thanks
| By AM-GM Inequality,$$x+y+z = x+\frac{y}{2}+\frac{y}{2}+\frac{z}{3}+\frac{z}{3}+\frac{z}{3}\geq6\left(\frac{xy^2z^3}{2^2\cdot3^3}\right)^\frac{1}{6}=6\left(\frac{108}{2^2\cdot3^3}\right)^\frac{1}{6}=6$$
Equality holds when $x=\frac{y}{2}=\frac{z}{3}$, so plugging this into $xy^2z^3=108$ we know the equality occurs when $x=1, y=2,z=3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2261563",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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} |
How many twin primes are of the form $2^n-1$ and $2^n+1$? The first pair is $(3,5)$ for $n=2$. Is there any other pair beside this?
| No, there are no more. A number of the form $2^n-1$ can only be prime if $n$ is prime (although there are prime $n$ which doesn't work, like $n = 11$, so $2^{11}-1$ is not prime), while a number of the form $2^n+1$ can only be prime if $n$ is a power of $2$ (again, there are power-of-two's which do not work, like $32$, so $2^{2^{5}} + 1$ is not prime). These two notions coincide only for $n = 2$.
Proofs:
Let $n$ be a composite number, say $n = pq$ for some $p, q\geq 2$. Then we have
$$
2^n-1 = 2^{pq}-1 = (2^p-1)(2^{(q-1)p} + 2^{(q-2)p} + \cdots + 2^p + 1)
$$
That this is composite follows from $p, q\geq 2$, which makes both of the above factors greater than $1$.
If $n$ is not a power of $2$, then $n = 2^mk$ where $k\geq 3$ is odd and $m\geq 0$. This gives us
$$
2^n + 1 = 2^{2^mk} + 1 = (2^{2^m} + 1)(2^{2^m(k-1)} - 2^{2^m(k-2)} + 2^{2^m(k-3)} - \cdots - 2^{2^m} + 1)
$$
which is necessarily composite, again because $k\geq 3$ so both factors are greater than $1$ ($k$ being odd is exactly what makes the signs work out so that you get $+1$ and not $-1$ at the end).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2262070",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Integral $\int_0^\pi \big( (1+\alpha \cos x) \cos x \big)^n dx $ I have been struggling with the integral
$$ I_n(\alpha) = \frac{1}{\pi} \int_0^\pi \big( (1+\alpha \cos x) \cos x \big)^n dx,$$
where $\alpha$ is real and $n$ is a non-negative integer.
It is relatively easy to get the values for specific $n$;
$$I_0(\alpha) = 1,~~I_1(\alpha) = \alpha/2,~~I_2(\alpha) =\frac{1}{8} \left(3 \alpha ^2+4\right), \ldots$$
But how do I get the expression for general $I_n(\alpha)$?
I have tried building a recursion but did not quite succeed.
I also tried taking the derivatives of 3.661.3 from Gradshteyn-Ryzhik
but did not get anything nice.
Edit I
Attempt using the binomial theorem:
\begin{align}
\frac{1}{\pi} \int_0^{\pi} dx~ \big( 1 + \alpha\cos (x) \big)^n \cos (x)^n
&= \frac{1}{\pi}
\int_0^{\pi} dx~ \cos (x)^n \sum_{m=0}^n{ {n}\choose{m}} \alpha^m \cos(x)^m \\
&= \sum_{m=0}^n{ {n}\choose{m}} \alpha^m \frac{1}{\pi} \int_0^{\pi} dx~ \cos (x)^{n+m} \\
&= \sum_{m=0}^n{ {n}\choose{m}} \alpha^m
\frac{ 2^{n+m} \pi }{(n+m)! \Gamma\left( \frac{1}{2}(1-n-m) \right)^2}
\begin{cases}
1,~~n+m ~ {\rm even}\\
0,~~n+m ~ {\rm odd}
\end{cases}
\end{align}
But I did not quite manage to express the last sum in some nice form...
the best I got is:
\begin{align} I_n(\alpha) &=
\left(1 + (-1)^n\right) \frac{\Gamma (n+1)}
{2^{n+1}\Gamma \left(\frac{n+2}{2}\right)^2}
\,_3F_2\left(\tfrac{1}{2}(1-n),\tfrac{1}{2}(1+n),-\tfrac{n}{2};\tfrac{1}{2},1+\tfrac{n}{2};\alpha ^2\right) \\
&~~~~+
\left(1 + (-1)^{n+1}\right)
\frac{n \Gamma (n+1)}
{2^n(n+1) \Gamma \left(\frac{n+1}{2}\right)^2}
\alpha \,_3F_2\left(\tfrac{1}{2}(1-n),1-\tfrac{n}{2},1+\tfrac{n}{2};\tfrac{3}{2},\tfrac{1}{2}(3+n);\alpha ^2\right)
\end{align}
But it is not very instructive...
P.S. The same can be also be obtained from the solution suggested by orlp.
Edit II
Following a nice suggestion of Igor's I got the result
$$ I_n(\alpha) = \frac{2i}{(2n)!} \lim_{z\to i} \frac{d^{2n}}{dz^{2n}}
\frac{\left(1 +\alpha + (1-\alpha) z^2 \right)^n \left(1 - z^2\right)^n}{(z+i)^{2n+1}}.
$$
However, it is quite tricky now (at least for me) to explicitly compute this derivative.
Any suggestions?
| I am not entirely sure the binomial theorem can be avoided, but also the Weierstrass substitution ($(t = \tan x/2)$ transforms your integral into an integral of a rational function from $0$ to infinity, at which point the residue theorem is your friend.
EDIT
A similar method is the following: first note that your integrand is even, so the integral is half of the integral from $-\pi$ to $\pi.$ Now, make the substitution $z = \exp(i x),$ so that $\cos x = \frac12\left(z + \frac1z\right),$ and so your integral is (half of) the integral over the unit circle:
$$ \begin{multline}-i \int_C ((1+ a (1+z^2)/2z)(1+z^2)/2z)^n \frac{dz}z\\ = \frac{-i}{2^{2n}} \int_C((az^2 + 2 z + a)(1+z^2))^n/ z^{2n+1} dz \\= \frac{-i}{2^{2n}}\int_C (a + 2 z + + 2az^2 + 2 z^3 + az^4)^n/z^{2n+1} d z.\end{multline}$$
So, your goal in life is to find the coefficient of $z^{2 n}$ in $(a + 2 z + + 2az^2 + 2 z^3 + az^4)^n,$ since that will give you the residue at $0.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2263220",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
} |
proving $ \lim_{x \rightarrow 1} \frac{1-\sqrt{x}}{1-x} = \frac{1}{2}$. I'm having trouble with proving $ \lim_{x \rightarrow 1} \frac{1-\sqrt{x}}{1-x} = \frac{1}{2}$.
So far I have:
$\frac{1-\sqrt{x}}{1-x} =\frac{1}{1+\sqrt{x}} $
If $x\in dom (\frac{1}{1+\sqrt{x}}$) and $|x-1| <\delta$ then $ |\frac{1}{1+\sqrt{x}}- \frac{1}{2} |< \epsilon$
So I started writing out $|\frac{1}{1+\sqrt{x}}- \frac{1}{2}|$, and got that it's equal to$ |\frac{1-x}{2(1+2\sqrt{x}+x)}|$. And since we have $|x-1| <\delta$ ,
$ |\frac{1-x}{2(1+2\sqrt{x}+x)}|< |\frac{\delta}{2(1+2\sqrt{x}+x)}|$.
I'm kind of stuck here. Any help on how to continue is much appreciated!
| Hint: for $x>0$:
$|\frac{1-x}{2(1+2\sqrt(x)+x)}| \le |\frac{1-x}{2}|$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2263811",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $\cos A+\cos B=p$ and $\sin A+\sin B=q$, then find $\cos\left( \frac {A+B}{2}\right)$ in terms of $p$ and $q$ If $\cos A+\cos B=p$ and $\sin A+\sin B=q$ then find $\cos \left( \dfrac {A+B}{2}\right)$ in terms of $p$ and $q$.
My Attempt:
$$\cos A+\cos B=p$$
$$2\cos \left( \dfrac {A+B}{2}\right)\cos \left( \dfrac {A-B}{2}\right)=p$$
And,
$$\sin A+ \sin B=q$$
$$2\sin \left( \dfrac {A+B}{2} \right)\cos \left( \dfrac {A-B}{2} \right)=q$$
Now,
$$\tan \left( \dfrac {A+B}{2}\right)=\dfrac {q}{p}$$.
How do I proceed further?
| here is a geometric interpretation of your work: think of the points $(\cos A, \sin A)$ and $(\cos B , \sin B)$ as two points also called $A$ and $B$ on the unit circle. the arc length is measured from the initial point $O = (1,0).$ now,
the mid point of the chord $AB$ has the coordinate $(p/2, q/2).$ we push this point onto the midpoint of the arc $(A+B)/2$ by dividing by its length to get the cordinates $$\left(\frac p{\sqrt{p^2 + q^2}}, \frac q{\sqrt{p^2+q^2}}\right).$$
the $x$-coordinate $\frac p{\sqrt{p^2 + q^2}}$ is $\cos((A+B)/2).$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2265706",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Parameterize Intersection of Surfaces I need to parameterize the intersection of $$4x^2 + y^2 + z^2 = 9\tag{1}$$ and $$z=x^2+y^2\tag{2}$$.
First, I'll solve (2) for $y^2$ and substitute the result into (1):
$$3x^2+z+z^2 = 9 \tag{3}$$
Next, I'll make the substitution $u=\sqrt{3}x$, such that we can complete the square in (3) by adding $1/4$ to each side and arrive at
$$\frac{u^2}{r^2} + \frac{(z+\frac{1}{2})^2}{r^2} = 1$$
where $r^2 = 9 + \frac{1}{4} = \frac{37}{4}$
Now I'll write a parameterization:
$$u = r\cos \phi \implies x(\phi) = \frac{1}{\sqrt{3}}r\cos\phi$$
$$z(\phi) = r\sin \phi -\frac{1}{2}$$
$$y(\phi) = \pm \sqrt{z-x^2} = \pm\left(\sqrt{r\sin \phi - \frac{1}{2} - \left(\frac{1}{\sqrt{3}}r\cos\phi\right)^2}\right)$$
such that we have two branches:
$$\mathbf{r}(\phi)_1 = \big<x(\phi), y(\phi), z(\phi)\big>$$
$$\mathbf{r}(\phi)_2 = \big<x(\phi), -y(\phi), z(\phi)\big>$$
Is this correct?
| It is not necessary to convert to polar coordinates, really. You have
$$z = x^2 + y^2 \tag{1}\label{1}$$
$$4 x^2 + y^2 + z^2 = 9 \tag{2}\label{2}$$
Equation $\eqref{1}$ can be written as
$$x^2 = z - y^2$$
and substituting it into equation $\eqref{2}$ yields
$$4 (z - y^2) + y^2 + z^2 = -3 y^2 + z^2 + 4 z = 9$$
which we can solve for $y$ by writing it as
$$3 y^2 = z^2 + 4 z - 9$$
nothing that
$$y^2 = \frac{1}{3} z^2 + \frac{4}{3} z - 3 \tag{3}\label{3}$$
Solving equation $\eqref{3}$ for $y$ we get
$$y = \pm \sqrt{\frac{1}{3} z^2 + \frac{4}{3} z - 3} \tag{4}\label{4}$$
The $\pm$ above means the intersection is symmetric with respect to the $y$ axis.
If we substitute equation $\eqref{3}$ into equation $\eqref{1}$, we get
$$z = x^2 + \frac{1}{3} z^2 + \frac{4}{3} z - 3$$
which we can trivially solve for $x^2$,
$$x^2 = - \frac{1}{3} z^2 - \frac{1}{3} z + 3$$
and therefore for $x$,
$$x = \pm \sqrt{-\frac{1}{3} z^2 - \frac{1}{3} z + 3} \tag{5}\label{5}$$
Here the $\pm$ means the intersection is also symmetric with respect to the $x$ axis.
By combining $\eqref{4}$ and $\eqref{5}$, we can write the intersection in Cartesian coordinates:
$$\begin{cases}
x = \pm \sqrt{-\frac{1}{3} z^2 - \frac{1}{3} z + 3} \\
y = \pm \sqrt{\frac{1}{3} z^2 + \frac{4}{3} z - 3}
\end{cases} \; \iff \; \begin{cases}
x = \pm \sqrt{\frac{1}{3}}\sqrt{ 9 - z (z + 1) }\\
y = \pm \sqrt{\frac{1}{3}}\sqrt{ z (z + 4) - 9 }
\end{cases}$$
It is a nice saddle-shaped curve. As JeanMarie already mentioned, it is much easier to visualize as a continuous curve in the cylindrical coordinates, though.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove by induction that $2^n + 4^n \leq 5^n$ I'm trying to prove by induction that $2^n + 4^n \leq 5^n$. Through some value plugging I've established that the induction must start from $n = 2$ because $2^2 + 4^2 \leq 5^2 \equiv 20 \leq 25$; for $n = 1$ it doesn't hold since $2 + 4 \geq 5$.
Now I assume that $2^k + 4^k \leq 5^k$ is true and I want to prove that implies $k+1$. Using the inductive hypothesis I multiply both sides by $4$ to get this:
$$4 \cdot 2^{k} + 4\cdot 4^{k} \leq 4 \cdot 5^{k}$$
$$2^{k+2} + 4^{k+1} \leq 4 \cdot 5^{k}$$
I will use again the induction hypothesis, this time I'll multiply both side by $5$ to get:
$$5 \cdot (2^{k} + 4^{k}) \leq 5^{k+1}$$
I can say that $2^{k+2} + 4^{k+1} \leq 5 \cdot (2^{k} + 4^{k})$ and $4 \cdot 5^{k} \leq 5^{k+1}$ so I concatenate them:
$$2^{k+2} + 4^{k+1} \leq 5 \cdot (2^{k} + 4^{k}) \leq 4 \cdot 5^{k} \leq 5^{k+1}$$
However this doesn't feel right. I'm assuming that $5 \cdot (2^{k} + 4^{k}) \leq 4 \cdot 5^{k}$ which there's no way I can be sure about. At this point I'm stuck since the whole reasoning seems wrong.
| You already have established from the induction hypothesis that
$$2^{k+2} + 4^{k+1} \leq 4 \cdot 5^{k}$$
So:
$$ 2^{k+1} + 4^{k+1} < 2^{k+2} + 4^{k+1} \leq 4\cdot5^k < 5 \cdot 5^k =5^{k+1}$$
Which already shows the required inequality for $k+1$.
It's simpler than you think...
| {
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"url": "https://math.stackexchange.com/questions/2266291",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find $b$ in $2\sin(\frac{x}{2}+b)$ given the graph Following is the graph of the function. $2\sin \left(\dfrac{x}{2}+b\right)$
I tried solving the equation using the point $\left (\dfrac{\pi}{2},2\right)$:
$$2 = 2\sin\left(\dfrac{x}{2}+b\right) \Leftrightarrow \\
1 = \sin\left(\dfrac{x}{2}+b\right)\Leftrightarrow \\
\frac{x}{2}+b = k\pi + (-1)^k\arcsin(1) \Leftrightarrow \\
b = 2k\pi$$
But my book $b = \dfrac{\pi}{4}$. How do I solve this?
| $\sin\left( \dfrac x 2 + b \right) = 1$ when $\dfrac x 2 + b = \dfrac \pi 2,$ and that implies $x=\pi-2b.$
Since it appears that the point where the curve reaches its maximum is half-way between $0$ and $\pi/2,$ we have $\dfrac\pi 2 = \pi - 2b,$ and from that we get $b=\dfrac \pi 4.$
A bit of an optical illusion seems to afflict this graph: The positioning of the $\text{“}0\text{''}$ a bit to the right of the zero point, presumably to avoid putting it right on top of that vertical axis, makes it look as if that maximum point may be closer to $0$ than to $\pi/2.$ But closer inspection dispells that.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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What's the limit of $\sqrt{2 + \sqrt{2-\sqrt{2+\sqrt{2-\sqrt{2-\sqrt{2 + ...}}}}}} $? Let's look at the continued radical
$ R = \sqrt{2 + \sqrt{2-\sqrt{2+\sqrt{2-\sqrt{2-\sqrt{2 + ...}}}}}} $
whose signs are defined as $ (+, -, +, -, -, + ,-, -, -,...)$, similar to the sequence $101001000100001...$, where
$1 = +$
$0 = - $
This radical seems to converge to a constant approximately equal to $1.567883...$.
The question is: Is it possible to find this limit $R$ in closed form?
Remark: In the article "On the periodic continued radicals of 2 and generalization for Vieta’s product", it is proved that a periodic sequence of signs composed of nested square roots of two converges to $2\sin(q\pi)$ for some rational number $q$. I have tried with non periodic sequences of plus and minus, and they also converge to numbers between $0$ and $2$. if this radical has a closed form, It can be the sine of an irrational multiple of $\pi$, since both are transcendental numbers.
| Let $R$ be the iterated square root in question
One easily check that any finite expression of the form
$$\sqrt{2\pm\sqrt{2\pm\sqrt{2\pm\cdots}}}$$
is between $0$ and $2$. So $0\le R\le 2$.
Suppose that $a=2\cos t$. Then
$$\sqrt{2+a}=\sqrt{2(1+\cos t)}=\sqrt{4\cos^2(a/2)}=2\cos\frac a2$$
and
$$\sqrt{2-a}=\sqrt{2(1-\cos t)}=\sqrt{4\sin^2(a/2)}=2\sin\frac a2
=2\cos\left(\frac\pi2-\frac a2\right).$$
Therefore
$$\begin{align}
\cos^{-1}\frac R2&=\frac12\cos^{-1}\frac12\sqrt{2-\sqrt {2+\sqrt{2
-\sqrt{2-\sqrt{2-\cdots}}}}}\\
&=\frac{\pi}4-\frac14\cos^{-1}
\frac12\sqrt {2+\sqrt{2
-\sqrt{2-\sqrt{2-\cdots}}}}\\
&=\frac{\pi}4-\frac18\cos^{-1}
\frac12\sqrt{2
-\sqrt{2-\sqrt{2-\cdots}}}\\
&=\frac{\pi}4-\frac{\pi}{16}+\frac1{16}\cos^{-1}
\frac12
\sqrt{2-\sqrt{2-\cdots}}\\
\end{align}
$$
etc. So we can get a series for $\cos^{-1}(R/2)$. I think it might be
something like
$$\frac\pi2\sum_{n=0}^\infty\frac{(-1)^n}{2^{n(n+1)/2}}$$
which is some sort of theta function.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2270730",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "39",
"answer_count": 3,
"answer_id": 1
} |
System of Linear Congruences Find all $x$ such that
\begin{align}
x&\equiv 1 \pmod {12}\\
x&\equiv 4 \pmod {21}\\
x&\equiv 18 \pmod {35}
\end{align}
Im not quite sure if this system of linear congruence is solvable. Since
$\gcd(12,21) =3$, $\gcd (12,35)=1$ and $\gcd(21,35) = 7$, and the CRT states that "If(m1, m2) = 1, then the system has its complete solution a single resident class (mod m1.....mr).
| \begin{align}
x&\equiv 1\phantom {8} \pmod {{\color{blue}{12}}}\tag {1}\\
x&\equiv 4\phantom {8} \pmod {{\color{teal}{21}}}\tag {2}\\
x&\equiv 18 \pmod {{\color{blue}{35}}}\tag{3}
\end{align}
Since $\gcd(12,35)=1 $ you can first apply the Chinese Remainder Theorem to find the solution to the two simultaneous congruences $(1)$ and $(3) $, which is of the form of
$$x\equiv x_0\pmod{12\cdot 35} \equiv x_0\pmod { 420}.\tag {4}$$
To compute the smallest positive value $x_{\min} $ of $ x_0 $, using the Chinese Remainder Theorem proceed as follows:
*
*since $12$ and $35$ are relatively prime, there is an $r$ and an $s$ such that $r\cdot 12 + s\cdot 35 = 1$. Indeed, $3\cdot 12 + (-1)\cdot 35 = 1$, so $r = 3$ and $s = -1$;
*in general, if $\gcd(m,n)=1 $, then the set of all solutions of the system
\begin{align}
x& \equiv a \pmod {m}\\
x& \equiv b \pmod {\; n}
\end{align}
is given by the condition $x\equiv a\cdot s\cdot n + b\cdot r\cdot m \pmod {m\cdot n}$. Hence
$$x \equiv 1\cdot (-1)\cdot 35 + 18\cdot 3\cdot 12 \equiv 613\equiv 613-420\equiv 193\pmod {420};\tag {5}$$
Finally solve the following two simultaneous congruences
\begin{align}
x\equiv &193\pmod{420} \tag {5}\\
x\equiv &4\phantom{93}\pmod{\phantom { 4}{\color {teal}{21}}};\tag {2}
\end{align}
Since $420\equiv 0\pmod{\color{teal}{21}}$ and $193\equiv 4\pmod{\color {teal}{21}}$, the solution of the given system is
$$x\equiv 193\pmod{420}. \tag {6}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2272593",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Find the arc length of a path I'm trying to solve the arc length for the following. $c(t) = (2, 6t^2, 4t^3)$ from $0\le t\le 1.$
I've checked on WolframAlpha and I get the answer $8\sqrt2 - 4$ but when I work it out I get $16\sqrt2$ where am I going wrong?
\begin{align*}
L&= \int_0^1 \sqrt{(12t)^2 + (12t^2)^2}\, dt\\
&= \int_0^1 \sqrt{144t^2 + 144t^4}\, dt\\
&= \int_0^1 12t\sqrt{1 + t^2}\,dt
\end{align*}
Then I do
$$ 8t\sqrt{(1+t^2)^3} = 16\sqrt2.$$
| From $$\int_{0}^{1}12t\sqrt{1+t^2}dt$$
Use $u=1+t^2\implies \frac{1}{2}du=tdt$ to get $$\int_{0}^{1}12t\sqrt{1+t^2}dt=\int12\sqrt{u}\cdot\frac{1}{2}du=6u^{3/2}\cdot\frac{2}{3}=4(1+t^2)^{3/2}\bigg\vert_{0}^{1}=4\cdot2^{3/2}-4\cdot1$$ $$=4\cdot2\cdot2^{1/2}-4\cdot1=8\sqrt{2}-4$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2273950",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Find parameter of inequality Find the parameter $m \in \mathbb R$ so that $$x^2 - 2xy + 2y^2 + 2x - 6y + m \ge 0, \forall x, y \in \mathbb R$$
I tried forming some squares but it seems it doesn't help me
$$(x-y)^2 + y^2 + 2x -6y + m \ge 0$$
| $y^2+x^2+1+2x-2y-2xy=(y-x-1)^2$ and $y^2-4y+4=(y-2)^2$. Total of these, we yields
$$x^2 - 2xy + 2y^2 + 2x - 6y + m = (y-x-1)^2 + (y-2)^2 + m - 5 \geq 0$$
Hence, we find that $ m\geq 5$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2275145",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solve for xy in terms of a and b $$x^2 + xy + y^2 = a$$
$$x + y = b$$
$xy = ?$
I tried this and did this:
$xy = a - x^2 - y^2$
$xy = a - (x^2 + y^2)$
$xy = a - (x + y)(x - y)$
$xy = a - b(x - y)$
At this point I can't think of anything to do to represent the $x - y$ part in terms of $a$ and $b$.
Any help would be appreciated.
| We are given \begin{align}x^2+xy+y^2&=a\tag{1}\\
x+y&=b\tag{2}\end{align}
We can square the equation $(2)$ to give us \begin{align}b^2&=(x+y)^2\\
&=x^2+2xy+y^2\end{align}
Therefore, we now have \begin{align}a&=x^2+xy+y^2\tag{1}\\
b^2&=x^2+2xy+y^2\tag{3}\end{align}
We can rearrange equation $(3)$ and substitute equation $(1)$ into it to give us the answer: \begin{align}b^2&=(x^2+xy+y^2)+xy\\
b^2&=a+xy\\
xy&=b^2-a\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2278301",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
} |
Functions and Sequences Problem
The function $F(k)$ is defined for positive integers as $F(1) = 1$,
$F(2) = 1$, $F(3) = -1$ and $F(2k) = F(k)$, $F(2k + 1) = F(k)$ for $k \geq
2$. Then $$F(1) + F(2) + \dotsb + F(63)$$ equals
$\begin{array}{lr}
(\text{A}) & 1 \\
(\text{B}) & -1 \\
(\text{C}) & -32 \\
(\text{D}) & 32 \\
\end{array}$
My approach: $F(4)=1$, $F(5)=1$, $F(6)=-1$, $F(7)=-1$ (i.e., all values of $F$ are either $1$ or $-1$). I tried to find a pattern for which $F(x)$ repeats after a certain integer but tried till $F(30)$ and cannot find a solution. Where am I going wrong?
| With induction it can be shown that:$$F(2^k)+\cdots+F(2^{k+1}-1)=0$$
for $k=1,2,\dots$
This because $F(2^1)+F(2^2-1)=F(2)+F(3)=1+(-1)=0$ and:$$F(2^{k+1})+F(2^{k+1}+1)+\cdots+F(2^{k+2}-2)+F(2^{k+2}-1)=2F(2^k)+\cdots+2F(2^{k+1}-1)$$
Then $$F(1)+\cdots+F(63)=F(1)+\sum_{k=1}^5\left[F(2^k)+\cdots+F(2^{k+1}-1)\right]=F(1)=1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2278951",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 0
} |
Identify the plane defined by $|z-2i| = 2|z+3|$ I tried:
$$|z-2i| = 2|z+3| \Leftrightarrow \\
|x+yi-2i|=2|x+yi+2|\Leftrightarrow \\
\sqrt{x^2+(y-2)^2}=\sqrt{4((x+2)^2+y^2)} \Leftrightarrow \\
\sqrt{x^2+y^2-4y+4} = \sqrt{4x^2+24x+36+4y^2} \Leftrightarrow \\
x^2+y^2-4y+4 = 4x^2+24x+36+4y^2 \Leftrightarrow \\
y^2-4y-4y^2=4x^2+24x+36+x^2 \Leftrightarrow \\
-3y^2-4y=5x^2+24x+26 \Leftrightarrow \\
???$$
What do I do next?
| You can square without adding spurious solutions, because $|w|\ge0$ by definition. So $|z-2i|^2=4|z+3|^2$ and, using $|w|^2=w\bar{w}$,
$$
(z-2i)(\bar{z}+2i)=4(z+3)(\bar{z}+3)
$$
This simplifies to
$$
3z\bar{z}+12(z+\bar{z})-2i(z-\bar{z})+32=0
$$
Now recall that, for $z=x+yi$, $z\bar{z}=x^2+y^2$, $z+\bar{z}=2x$ and $z-\bar{z}=2yi$, to get
$$
3(x^2+y^2)+24x+4y+32=0
$$
that can also be written as
$$
x^2+y^2+8x+\frac{4}{3}y+\frac{32}{3}=0
$$
or, completing the squares,
$$
(x+4)^2+\left(y+\frac{2}{3}\right)^{\!2}=\frac{52}{9}
$$
which represents a circle.
Your computation is also good up to
$$
x^2+y^2-4y+4 = 4x^2+24x+36+4y^2
$$
Then you do wrong simplifications: taking everything to the right hand side you get
$$
3x^2+3y^2+24x+4y+32=0
$$
exactly the same I got.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2279971",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Can this ratio be an integer? Is it true that the number
$$\frac{b^3+1}{ab-1}$$
is never an integer for positive integers $a$ and $b$ such that $a\geq 6$ and $a\geq b$?
| Here is my approach:
By definition you can say that $b^2+{2 \over b} \ge a \ge b$ because $(b^2+{2 \over b})b-1 = b^3+1$ . But because ${2 \over b}$ is only an integer for $b=1$ or $b=2$ your interval for $a$ is effectively $[b, b^2]$ . This means you can write $a=b\cdot b^x$ with $0 \le x \le 1$ or $a=b \sqrt[y]{b}$ with $y = {1 \over x}$ . If you combine this with the original term you get ${b^3+1 \over b^2 \sqrt[y]{b}-1}=n, \quad n \in N \quad \Rightarrow b^3+(1+n) = n b^2 \sqrt[y]{b}$ . This concludes that if you can show that for some $b $ if $n\sqrt[y]{b} \in N$ , $b^2\sqrt[y]{b} \in N$ and $nb^2\sqrt[y]{b} \in N$ lead to a contradiction then for this $b$ ${b^3+1 \over ab-1}$ is not an integer for every $a$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2281075",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Help solving the PDE $u_x^2 + u_y^2 = u$ I'm trying to solve this Cauchy problem:
$$ u_x^2 + u_y^2 = u \\
u(x,0) = x^2+1$$
Here's what I've tried so far:
Letting $p(r,s) = u_x$, $q(r,s) = u_y$ and $z(r,s) = u$, we have:
$$ F = p^2 + q^2 - z = 0 $$
and thus:
$$ \begin{align} \frac{dx}{ds} &= F_p = 2p \\
\frac{dy}{ds} &= F_q = 2q \\
\frac{dz}{ds} &= pF_p + qF_q = 2p^2 + 2q^2 = 2z\\
\frac{dp}{ds} &= -F_x -pF_z = -(0) -p(-1) = p \\
\frac{dq}{ds} &= -F_y -qF_z = -(0) -q(-1) = q \\ \end{align}$$
Also, $\Gamma(r,0) = r^2 +1 $, and
$$ \begin{align} (r^2+1)' &= \phi_1 \cdot (r)' + \phi_2 \cdot (0)' \\
\implies 2r &= \phi_1 \end{align}$$
so that:
$$ \begin{align} (2r)^2 + \phi_2^2 &= r^2+1 \\
\implies 4r^2 + \phi_2^2 &= r^2+1 \\
\implies \phi_2^2 &= -3r^2 + 1 \\
\implies \phi_2 &= \pm \sqrt{ -3r^2 +1 } \end{align} $$
Which leads to:
$$ \begin{align} p &= e^s 2r \\
q &= e^s \sqrt{ -3r^2 +1 } \\
\implies x &= 4e^s r - 3r \\
\text{and } y & = 2e^s \cdot ( \pm \sqrt{ -3r^2 +1 } ) \mp \sqrt{ -3r^2 +1 }\end{align} $$
And I'm pretty sure by now it's gone way off the rails, because I can't solve for for $s$ and $r$ in terms of $x$ and $y$, but I can't figure out where I went wrong.
Could anybody please shed some light?
| Obviously, the main difficulty comes from the boundary condition $u(x,0)=x^2+1$ because it involves to solve a polynomial equation leading to huge formula. Even if the solving is theoretically possible, one have to merely accept a result on implicit form.
To make more clear where the difficulty arises, we will avoid the profusion of symbols introduced into the usual method of characteristics, but in following the same approach in fact.
$$u_x^2+u_y^2=u \quad;\quad u(x,0)=x^2+1$$
First change of function :$\quad u=v^2\quad\begin{cases}u_x=2vv_x \\ u_y=2vv_y\end{cases}\quad\to\quad v_x^2+v_y^2=\frac{1}{4}\quad;\quad v(x,0)=\sqrt{x^2+1}$
$$v_y=\sqrt{\frac{1}{4}-v_x^2} \quad\to\quad v_{xy}=\frac{v_xv_{xx}}{\sqrt{\frac{1}{4}-v_x^2}}$$
Second change of function : $\quad w=v_x\quad\to\quad w_{y}=\frac{ww_{x}}{\sqrt{\frac{1}{4}-w^2}}$
$$\sqrt{\frac{1}{4}-w^2}\:w_y-2w\:w_x=0 \quad;\quad w(x,0)=v_x(x,0)=\frac{x}{\sqrt{x^2+1}}$$
This is a first order PDE. The set of characteristic ODEs is : $\quad \frac{dy}{\sqrt{\frac{1}{4}-w^2}}=\frac{dx}{-2w}=\frac{dw}{0}$
First family of characteristic curves, coming from $dw=0 \quad\to\quad w=c_1$
Second family of characteristic curves, from $\frac{dy}{\sqrt{\frac{1}{4}-c_1^2}}=\frac{dx}{-2c_1} \quad\to\quad 2c_1y+\sqrt{\frac{1}{4}-c_1^2}x=c_2$
The general solution can be presented on various forms :$\begin{cases}
\Phi\left(w\:,\: 2wy+\sqrt{\frac{1}{4}-w^2}\:x\right)=0\\
w=f\left(2wy+\sqrt{\frac{1}{4}-w^2}\:x\right)\\
2wy+\sqrt{\frac{1}{4}-w^2}\:x=F(w)\end{cases}$
where $\Phi$ , $f$ , $F$ are any differentiable functions. Any one of these functions has to be determined according to the boundary condition.
$$w(x,0)=\frac{x}{\sqrt{x^2+1}}\quad\to\quad 2\frac{x}{\sqrt{x^2+1}}0+\sqrt{\frac{1}{4}-\left(\frac{x}{\sqrt{x^2+1}}\right)^2}\:x=F\left(\frac{x}{\sqrt{x^2+1}}\right)$$
Let $t=\frac{x}{\sqrt{x^2+1}} \quad\to\quad x=\frac{t}{\sqrt{1-t^2}} \quad\to\quad
\sqrt{\frac{1}{4}-t^2}\:\frac{t}{\sqrt{1-t^2}} =F\left(t\right)$
Now, the function $F$ is determined. We put it into the above general solution :
$$2wy+\sqrt{\frac{1}{4}-w^2}\:x=\sqrt{\frac{1}{4}-w^2}\:\frac{w}{\sqrt{1-w^2}}$$
$$\frac{4}{\sqrt{1-4w^2}}\:y+\frac{x}{w}=\frac{1}{\sqrt{1-w^2}}$$
Solving this equation for $w$ leads to $\quad w(x,y)$
In fact, this is a four degree polynomial equation. One can solve it analytically, but this involves huge formulas. That is the hitch.
So, we let $w$ on the implicit form of the above equation and, from it, we consider that $w(x,y)$ is known.
$w=\frac{u_x}{2\sqrt{u}}\quad\to\quad \int \frac{u_x}{2\sqrt{u}}=\sqrt{u}=\int w(x,y)dx$
$$u(x,y)=\left(\int w(x,y)dx \right)^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2281386",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Find the straight line from two plane vectors Two planes have equations $x + 3y - 2z = 4$ and $2x +y +3z = 5$. The planes intersect in the straight line $l$.
Find a vector equation for the line $l$.
How can I do this type of question?
| Let $\vec{l}(a,b,c)$.
Hence, $a+3b-2c=0$ and $2a+b+3c=0$, which gives $b=-\frac{7}{11}a$ and $c=-\frac{5}{11}a$.
Let $a=11$. Hence, $b=-7$ and $c=-5$ and $\vec{l}(11,-7,-5)$.
But for $z=0$ we get $x=\frac{11}{5}$ and $y=\frac{3}{5}$, which gets an answer:
$$\left\{\left(\frac{11}{5}+11t,\frac{3}{5}-7t,-5t\right)|t\in\mathbb R\right\}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2281719",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Bound on $x^TAx$ for spd matrix $A$ in terms of diagonal Let $A\in\mathbb{R}^{n\times n}$ be a symmetric positive definite matrix and $x\in\mathbb{R}^n$ some vector. I want to find a bound of the form
$$x^T Ax \leq c \sum_{i=1}^n a_{ii} x_i^2$$
with a constant $c>0$.
If $\lambda_\text{min}$ and $\lambda_\text{max}$ denote the smallest and largest eigenvalue of $A$ (both are positive due to spd assumption), then one bound that holds is
$$x^T Ax \leq \frac{\lambda_\text{max}}{\lambda_\text{min}} \sum_{i=1}^n a_{ii} x_i^2.$$
Proof: $x^TAx \leq \lambda_\text{max} \Vert x\Vert^2 = \frac{\lambda_\text{max}}{\lambda_\text{min}} \sum_{i=1}^n \lambda_\text{min} x_i^2 \leq \frac{\lambda_\text{max}}{\lambda_\text{min}} \sum_{i=1}^n a_{ii} x_i^2$, where we used that $a_{ii} \geq \lambda_\text{min}$ for all $i$.
However, I think there should be a tighter bound. For example, the inequality holds with $c=1$ if $A$ is diagonal. Any ideas?
| You can substitute $y_i = \sqrt{a_{ii}}\;x_i$. Then you are looking for the smallest $c$ with
$$
y^T
\begin{pmatrix} \frac{1}{\sqrt{a_{11}}} & & & 0 \\
& \frac{1}{\sqrt{a_{22}}} & & \\
& & \ddots & \\
0 & & & \frac{1}{\sqrt{a_{nn}}}
\end{pmatrix}
A
\begin{pmatrix} \frac{1}{\sqrt{a_{11}}} & & & 0 \\
& \frac{1}{\sqrt{a_{22}}} & & \\
& & \ddots & \\
0 & & & \frac{1}{\sqrt{a_{nn}}}
\end{pmatrix}
y
\leq c \| y \|_2^2 \;\;\forall y\in\mathbb{R}^n
$$
which is the largest eigenvalue of
$$
\begin{pmatrix} \frac{1}{\sqrt{a_{11}}} & & & 0 \\
& \frac{1}{\sqrt{a_{22}}} & & \\
& & \ddots & \\
0 & & & \frac{1}{\sqrt{a_{nn}}}
\end{pmatrix}
A
\begin{pmatrix} \frac{1}{\sqrt{a_{11}}} & & & 0 \\
& \frac{1}{\sqrt{a_{22}}} & & \\
& & \ddots & \\
0 & & & \frac{1}{\sqrt{a_{nn}}}
\end{pmatrix}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2281959",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
find oblique asymptotes : $f(x) =\sqrt[q]{\frac{a_mx^m+a_{m-1}x^{m-1}+\cdots+a_0}{b_nx^n+b_{n-1}x^{n-1}+\cdots+b_0}}$ let : $$f(x) =\sqrt[q]{\frac{a_mx^m+a_{m-1}x^{m-1}+\cdots+a_0}{b_nx^n+b_{n-1}x^{n-1}+\cdots+b_0}}$$
shuch that : $$m-n=q$$
then : fine oblique asymptotes :
i know that : $y=mx+h$ is oblique asymptotes such that :
$$m=\lim_{x\to \infty} \frac{f(x)}{x}$$
and :
$$h=\lim_{x\to \infty}(f(x)-mx)$$
But I can not answer
please help .
| For the slope, you can do the same calculation you mentioned:
$$\begin{align}
\displaystyle\lim_{x\to\infty}\frac{f(x)}{x} &= \displaystyle\lim_{x\to\infty} \sqrt[q]{\frac{a_mx^m+a_{m-1}x^{m-1} + \cdots+a_0}{b_nx^n+b_{n-1}x^{n-1}+\cdots+b_0}}\cdot\frac{1}{x}\\
&= \displaystyle\lim_{x\to\infty} \sqrt[q]{\frac{a_mx^m+a_{m-1}x^{m-1}+\cdots+a_0}{b_nx^n+b_{n-1}x^{n-1} + \cdots+b_0}}\cdot\sqrt[q]{\frac{1}{x^q}}\\
&= \displaystyle\lim_{x\to\infty} \sqrt[q]{\frac{a_mx^m+a_{m-1}x^{m-1}+\cdots+a_0}{b_nx^m+b_{n-1}x^{m-1} + \cdots+b_0x^q}}\\
&= \sqrt[q]{\displaystyle\lim_{x\to\infty}\frac{a_mx^m+a_{m-1}x^{m-1}+\cdots+a_0}{b_nx^m+b_{n-1}x^{m-1} + \cdots+b_0x^q}}\\
&= \sqrt[q]{\frac{a_m}{b_n}}
\end{align}$$
Similarly:
$$h = \displaystyle\lim_{x\to\infty}f(x) - (\text{slope})x = \displaystyle\lim_{x\to\infty} \left( \sqrt[q]{\frac{a_mx^m+a_{m-1}x^{m-1} + \cdots+a_0}{b_nx^n+b_{n-1}x^{n-1}+\cdots+b_0}} - \sqrt[q]{\frac{a_m}{b_n}}x \right)$$
This limit is trickier. Do you know any methods for evaluating it?
Let's try this:
$$\begin{align}
\sqrt[q]{\frac{a_mx^m+a_{m-1}x^{m-1} + \cdots+a_0}{b_nx^n+b_{n-1}x^{n-1}+\cdots+b_0}} &= \left(\sqrt[q]{\frac{a_m}{b_n}}\right)x \cdot \sqrt[q]{\frac{x^n + \frac{a_{m-1}}{a_m}x^{n-1} + \cdots + \frac{a_0}{a_m}x^{-q}}{x^n + \frac{b_{n-1}}{b_n}x^{n-1} + \cdots + \frac{b_0}{b_n}}}\\
&= \left(\sqrt[q]{\frac{a_m}{b_n}}\right)x \cdot \left(1 + \left(\frac{a_{m-1}}{a_m} - \frac{b_{n-1}}{b_n}\right)x^{-1} + \cdots\right)^{1/q}
\end{align}$$
where everything in the dots at the end involves powers $x^{-2}$ or smaller. Now, close to $x=1$, we know that $(1+h)^{1/q}$ is well-approximated by $1+\frac{h}{q}$. Thus, we replace the complicated factor on the right with an approximation:
$$\left(\sqrt[q]{\frac{a_m}{b_n}}\right)x \cdot \left(1 + \frac{1}{q}\left(\frac{a_{m-1}}{a_m} - \frac{b_{n-1}}{b_n}\right)x^{-1} + \cdots\right)$$
Multiplying this out, we get that
$$f(x) = \left(\sqrt[q]{\frac{a_m}{b_n}}\right)x + \sqrt[q]{\frac{a_m}{b_n}}\cdot\frac{1}{q}\left(\frac{a_{m-1}}{a_m} - \frac{b_{n-1}}{b_n}\right) + O(x^{-1})$$
Based on this, and encouraged by numeric experiments, I'm pretty sure that we can say
$$h=\sqrt[q]{\frac{a_m}{b_n}}\cdot\frac{1}{q}\left(\frac{a_{m-1}}{a_m} - \frac{b_{n-1}}{b_n}\right).$$
I'd love to see a simpler way to get there. I got this idea from one of the answers to: How to find the oblique asymptote of root of a function?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2283767",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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partially differentiate $\frac{x-y}{\sqrt{x^2-y^2}}$ How do you partially differentiate $\frac{x-y}{\sqrt{x^2-y^2}}$, with respect to $y$ then $x$?
I have done it and got the answer $$\frac{y^2+yx}{\sqrt[3]{x^2-y^2}}$$
However, the answer given in my paper gives it with a $-y^2$ on the top with all the rest the same. Any help?
| We should get the following steps:
\begin{align}\frac{\partial}{\partial y}\left(\frac{x-y}{\sqrt{x^2-y^2}}\right)&=\frac{-x}{(x+y)\sqrt{x^2-y^2}}\\
\frac{\partial}{\partial x}\left(\frac{-x}{(x+y)\sqrt{x^2-y^2}}\right)&=\frac{x^2-xy+y^2}{(x-y)(x+y)^2\sqrt{x^2-y^2}}\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2285032",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Counting sets which can be partition in two subsets satisfying certain criterion I have a question that involves some ability in combinatorics that I don't have. Any hint could be extremely appreciated.
Take $n>1$ with $n\in \mathbb{N}$.
Let $\mathcal{G}$ be the set of all possible binary matrices of dimension $n\times (n-1)$. $\mathcal{G}$ has cardinality $|\mathcal{G}|=2^{n(n-1)}$.
For example, if $n=2$, then $\mathcal{G}:=\{\begin{pmatrix}
1\\
1
\end{pmatrix},\begin{pmatrix}
1\\
0
\end{pmatrix}, \begin{pmatrix}
0\\
1
\end{pmatrix} , \begin{pmatrix}
0\\
0
\end{pmatrix}\}$ with $|\mathcal{G}|=4$.
Question: For $M=2,...,|\mathcal{G}|-1$, I want to count all possible sets $C\subset \mathcal{G}$ with cardinality $M$ such that:
there exists a non-empty set $D\subset C$ such that for every pair of matrices, one taken from $D$ and the other taken from $C-D$, the two matrices in the pair considered differ for at least two rows.
Is this counting possible? If not, is there any known upper bound or lower bound?
I have looked at this question whose comments link to bounds for binary constant weight codes, but it does not seem to help.
[$C-D$ denotes the complement of $D$ in $C$].
For example: when $n=2$ the result is $2$. Indeed, for $M=2$, we can have
$$
C=\{\begin{pmatrix}
1\\
1
\end{pmatrix}, \begin{pmatrix}
0\\
0
\end{pmatrix}\}
$$
where $D=\{\overbrace{\begin{pmatrix}
1\\
1
\end{pmatrix}}^{d_1}\}$, $C-D=\{\overbrace{\begin{pmatrix}
0\\
0
\end{pmatrix}}^{c_1}\}$, row $1$ of $d_1$ $\neq $ row $1$ of $c_1$, and row $2$ of $d_1$ $\neq $ row $2$ of $c_1$
and
$$
C=\{\begin{pmatrix}
0\\
1
\end{pmatrix}, \begin{pmatrix}
1\\
0
\end{pmatrix}\}
$$
where $D=\{\overbrace{\begin{pmatrix}
0\\
1
\end{pmatrix}}^{d_1}\}$, $C-D=\{\overbrace{\begin{pmatrix}
1\\
0
\end{pmatrix}}^{c_1}\}$, row $1$ of $d_1$ $\neq $ row $1$ of $c_1$, and row $2$ of $d_1$ $\neq $ row $2$ of $c_1$
For $M=3$, there is no $C$ satisfying the considered criterion.
Remark: Notice that the order of the rows matter.
| This is not an answer but an input regarding the maximum of $M$.
Claim:
For $M \gt 2^{n(n-1)}-n(2^{n-1}-1)$ there is no $C$ satisfying the criterion.
Argument:
When pairing matrices, at least two rows must differ. This means the only pairing which isn't allowed is the pairing of matrices which differ in exactly one row. For a given matrix, how many such disallowed pair-mates are there?
If a row has $n-1$ binary digits there are $2^{n-1}$ possible permutations of that row. For a given row and a given matrix, there are thus $2^{n-1}-1$ matrices which differ from the given matrix in only that row. With $n$ rows there are thus $n(2^{n-1}-1)$ matrices which differ from the given matrix in one row.
An example with $n=3$. The given matrix:
$$
\begin{pmatrix}
0 & 0 \\
0 & 0 \\
0 & 0 \\
\end{pmatrix}
$$
has the following $n(2^{n-1}-1)$ disallowed pair-mates:
$$
\begin{pmatrix}
0 & 1 \\
0 & 0 \\
0 & 0 \\
\end{pmatrix}\begin{pmatrix}
1 & 0 \\
0 & 0 \\
0 & 0 \\
\end{pmatrix}\begin{pmatrix}
1 & 1 \\
0 & 0 \\
0 & 0 \\
\end{pmatrix}
$$
$$
\begin{pmatrix}
0 & 0 \\
0 & 1 \\
0 & 0 \\
\end{pmatrix}\begin{pmatrix}
0 & 0 \\
1 & 0 \\
0 & 0 \\
\end{pmatrix}\begin{pmatrix}
0 & 0 \\
1 & 1 \\
0 & 0 \\
\end{pmatrix}
$$
$$
\begin{pmatrix}
0 & 0 \\
0 & 0 \\
0 & 1 \\
\end{pmatrix}\begin{pmatrix}
0 & 0 \\
0 & 0 \\
1 & 0 \\
\end{pmatrix}\begin{pmatrix}
0 & 0 \\
0 & 0 \\
1 & 1 \\
\end{pmatrix}
$$
This means a given matrix can only be paired with $2^{n(n-1)}-n(2^{n-1}-1)-1$ other matrices (the $-1$ at the end is because a given matrix cannot be paired with itself). A subset $C$, where $D$ consists only of the given matrix, can therefore at most have $|C-D| = 2^{n(n-1)}-n(2^{n-1}-1)-1$ other elements and can therefore at most have cardinality $M = 2^{n(n-1)}-n(2^{n-1}-1)$.
If the sole element of $D$ is the given matrix, we know from above that we cannot add extra matrices in $C-D$, but perhaps we can add extra matrices in $D$, thus increasing $M$?
If we add a disallowed matrix to $D$ (say the upper left matrix in the example above) we see that this new member of $D$ doesn't share $(n-1)(2^{n-1}-1)$ of the disallowed with the given matrix. Since the total number of matrices that a matrix can be paired with remains constant for a given $n$, the number of matrices that both the given matrix and the new member can be paired with, is reduced by $(n-1)(2^{n-1}-1)$. So one extra member of $D$, but a greater reduction in $|C-D|$.
If we add an allowed matrix to $D$, the reduction of $|C-D|$ would be even greater.
Let me know what you think.
| {
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"url": "https://math.stackexchange.com/questions/2286399",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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Help in evaluating integral using elementary methods I have the following integral:
$\int\limits_{0}^{\pi/2}\frac{x\sin x \cos x}{a^2 \cos^2x+ b^2\sin^2x} \ dx$, for $a,b \geq 0$, both not zero.
I have tried several substitutions without any success at all. How can this be tackled?
Thanks.
| First we deal with the case $b= a$, then:
$$I(a,a)=\int\limits_{0}^{\pi/2}\frac{x\sin x \cos x}{a^2 \cos^2x+ a^2\sin^2x} \ dx=\frac{1}{a^2} \int\limits_{0}^{\pi/2} x\sin x \cos x dx$$
Now we use integration by parts with:
$$u=x,\qquad dv=\sin x \cos x dx$$
$$v=\int \sin x \cos x dx=\frac{1}{4} \int \sin 2x d2x=-\frac{\cos 2x}{4} $$
$$uv \bigg|_0^{\pi/2}=\frac{\pi}{8}$$
$$I(a,a)=\frac{\pi}{8a^2}+\frac{1}{4a^2} \int\limits_{0}^{\pi/2} \cos 2x ~dx=\frac{\pi}{8a^2}+\frac{1}{8a^2} \int\limits_{0}^{\pi} \cos 2x ~d2x=\frac{\pi}{8a^2}$$
Now we assume $a \neq b$. We use the same method of integration by parts:
$$u=x,\qquad dv=\frac{\sin x \cos x}{a^2 \cos^2x+ b^2\sin^2x} dx$$
Suprisingly, $v$ has a simple form, which I'm not going to derive here, as it can be checked by direct differentiation:
$$v=-\frac{\log \left(a^2+b^2+(a^2-b^2) \cos 2x \right)}{2(a^2-b^2)}$$
$$uv \bigg|_0^{\pi/2}=-\frac{\pi \log \left(2 b^2 \right)}{4(a^2-b^2)} $$
So:
$$I(a,b)=-\frac{\pi \log \left(2 b^2 \right)}{4(a^2-b^2)}+\frac{1}{4(a^2-b^2)}\int\limits_{0}^{\pi}\log \left(a^2+b^2+(a^2-b^2) \cos u \right) du=$$
$$=\frac{\pi \left(\log \left(a^2+b^2\right)- \log \left(2 b^2 \right) \right)}{4(a^2-b^2)}+\frac{1}{4(a^2-b^2)}\int\limits_{0}^{\pi}\log \left(1+p \cos u \right) du$$
Where $p=\frac{a^2-b^2}{a^2+b^2}$ and thus $|p|<1$.
Going further, we can expand the logarithm as Taylor series, then express the integrals as Beta function, and then find a closed form for the resulting series. The latter I have done with the help of Wolfram Alpha.
This way turns out to be quite tedious. Maybe it's better to expand the original integral as a series in $p$ right from the start. However, I will write the final answer I obtained (which can be confirmed by numerical experiments):
$$I(a,b)=\int\limits_{0}^{\pi/2}\frac{x\sin x \cos x}{a^2 \cos^2x+ b^2\sin^2x} \ dx= \frac{\pi}{2} \frac{\log (a+b)-\log b- \log 2}{a^2-b^2}$$
$$a \neq b$$
Remark: while the methods here are not completely elementary, neither is this integral. The best way to evaluate integrals like this is contour integration.
| {
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Limit with 2 variables: $\frac{x^2y}{x^4 +y^2}$, $\frac{e^{xy^3}-1}{x^2 +y^4}$ and $(x^2 +y^2)^{xy}$ at $(0,0)$ I am new to this topic so I appreciate any help on this.
a)$$\lim_{(x,y) \rightarrow (0,0)} f(x,y) = \frac{x^2y}{x^4 +y^2}$$
For $x=0 \lor y=0 : f(x,y) \longrightarrow 0$ but for $y = x^2$ the limit is $\frac{1}{2}$ so the limit does not exist.
b)
$$\lim_{(x,y) \rightarrow (0,0)} f(x,y) = \frac{e^{xy^3}-1}{x^2 +y^4}$$
I don't have any ideas here
c)
$$\lim_{(x,y) \rightarrow (0,0)} f(x,y) = (x^2 +y^2)^{xy}$$
Let $x \geq y$ then $|x^2 +y^2|^{xy} \leq |2y^2|^{y^2} \longrightarrow 1 \quad $for $y \longrightarrow 0$ but for $x = \frac{1}{y} \Longrightarrow f(x,y) = (\frac{1}{y^2} + y^2)^1 \longrightarrow \infty \quad$ for $y \longrightarrow 0 $ so the limit does not exist.
| You've made a mistake in c): $(1/y,y)$ does not approach $(0,0)$ as $y\to 0.$
To do it, we can apply $\ln$ to get $xy\ln (x^2+y^2).$ Recall $|xy|\le (x^2+y^2)/2.$ Thus
$$\tag 1 |xy \ln (x^2+y^2)| \le \frac{(x^2+y^2)\ln (x^2+y^2)}{2}.$$
Now $u\ln u \to 0$ as $u\to 0^+.$ Thus the limit of $(1)$ as $(x,y) \to (0,0)$ is $0.$ This implies the orginal limit is $1.$
| {
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Confusion about proving $\frac{1}{1\cdot2} + \frac{1}{2\cdot3} + \frac{1}{3\cdot4} + \dotsb + \frac{1}{n(n+1)} = \frac{n}{n+1}$ by induction I know what the answer to this question is, but I am not sure how the answer was reached and I would really like to understand it! I am omitting the base case because it is not relevant for my question.
Inductive hypothesis:
$$\frac{1}{1\cdot2} + \frac{1}{2\cdot3} + \frac{1}{3\cdot4} + \dotsb + \frac{1}{n(n+1)} = \frac{n}{n+1}$$ is true when $n = k$ and $k > 1$
Therefore: $$\frac{1}{1\cdot2} + \frac{1}{2\cdot3} + \frac{1}{3\cdot4} + \dotsb + \frac{1}{k(k+1)} = \frac{k}{k+1}$$
Inductive step:
Prove that $$\frac{1}{1\cdot2} + \frac{1}{2\cdot3} + \frac{1}{3\cdot4} + \dotsb + \frac{1}{k(k+1)} = \frac{k+1}{k+1+1} = \frac{k+1}{k+2}$$
$$\frac{1}{1\cdot2} + \frac{1}{2\cdot3} + \frac{1}{3\cdot4} + \dotsb + \frac{1}{k(k+1)} = \left[\frac{1}{1\cdot2} + \frac{1}{2\cdot3} + \frac{1}{3\cdot4} + \dotsb + \frac{1}{k(k+1)}\right] + \frac{1}{(k+1)(k+2)}$$
$$\frac{1}{1\cdot2} + \frac{1}{2\cdot3} + \frac{1}{3\cdot4} + \dotsb + \frac{1}{k(k+1)} = \frac{k}{k+1} + \frac{1}{(k+1)(k+2)}$$
$$\frac{1}{1\cdot2} + \frac{1}{2\cdot3} + \frac{1}{3\cdot4} + \dotsb + \frac{1}{k(k+1)} = \frac{k+1}{k+2}$$
What I am confused about is where the $\frac{1}{(k+1)(k+2)}$ comes from in the first line of the inductive step. Can someone please explain this in a little more detail? The source of the answer explains it as "break last term from sum", but I am unclear on what that means.
| "What I am confused about is where the 1/(k+1)*(k+2) comes from in the first line of the inductive step."
It comes because you are trying to evaluate for $n = k+1$
You want to prove the statement $\frac 1{1*2} + .....+ \frac 1{(n-1)(n)} + \frac 1{n*(n+1)} = \frac n{n+1}$.
For $n = k$ if you replace $n$ with $k$ you are assuming $\frac 1{1*2} + .....+ \frac 1{(k-1)(k)} + \frac 1{k*(k+1)} = \frac k{k+1}$.
If you replace $n$ with $k + 1$ you get the statement:
$\frac 1{1*2} + .....+ \frac 1{(k-1)(k)} + \frac 1{k*(k+1)}+ \frac 1{(k+1)(k+1)} = \frac {k+1}{k+2}$.
And that is the statement you wish to prove.
We are assuming $\frac 1{1*2} + .....+ \frac 1{(k-1)(k)} + \frac 1{k*(k+1)} = \frac k{k+1}$
So $\frac 1{1*2} + .....+ \frac 1{(k-1)(k)} + \frac 1{k*(k+1)}+ \frac 1{(k+1)(k+1)} = \frac k{k+1} + \frac 1{(k+1)(k+1)}$
You just need to prove that $\frac k{k+1} + \frac 1{(k+1)(k+1)}= \frac {k+1}{k+2}$.
.....
Anyway, you certainly did not transcribe to proof correctly.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to take $\int_0^{+\infty} \frac{x^2+1}{x^4+1}dx$? The integral:
$$\int_0^{+\infty} \frac{x^2+1}{x^4+1}dx$$
If num were greater than denum I would just devide it normally with long division, but it is not, how should I handle it then?
| By symmetry,
$$2\int_0^{+\infty}\frac{x^2+1}{x^4+1}\ dx=\int_{-\infty}^{+\infty}\frac{x^2+1}{x^4+1}\ dx$$
By taking a semi-circle contour in the upper half of the complex plane and applying Cauchy's residue formula, we find that
$$\begin{align}\int_{-\infty}^{+\infty}\frac{x^2+1}{x^4+1}\ dx&=2\pi i\left[\lim_{a\to e^{\pi i/4}}\frac{(a^2+1)(a-e^{\pi i/4})}{a^4+1}+\lim_{b\to e^{3\pi i/4}}\frac{(b^2+1)(b-e^{3\pi i/4})}{b^4+1}\right]\\&=2\pi i\left[\frac{1+i}{4e^{3\pi i/4}}+\frac{1-i}{4e^{9\pi i/4}}\right]\\&=\frac{\pi i}{2\sqrt2}\left[(1-i)^2-(1+i)^2\right]\\&=\sqrt2\pi\end{align}$$
Thus,
$$\int_0^{+\infty}\frac{x^2+1}{x^4+1}\ dx=\frac{\sqrt2\pi}2=\frac\pi{\sqrt2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2292541",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Find the last digit of $43^{43^{43}}$ Find the last digit of $43^{43^{43}}$. My attempt:
$$(40+3)^3 \equiv 3\cdot 40\cdot 9+27 \equiv 1080+27\equiv 07\pmod{100}$$
Am I right or is there any other way to find the answer?
| There will come some number $n $ where the last digit of $43^n $ is $1$.( $43^{n}\equiv 1 \mod 10$) Then any $43 ^{kn+m}=(43^n)^k43^m\equiv 1^k43^m \mod 10$ which will probably be easier to solve.
So our first step is to find that $n $.
Since $43=40+3$ then $43^n=40^n+a_140^{n-1}*3+......+a_{n-1}40*3^{n-1}+3^n $ and all but the last term is a multiple of $10$. So $43^n $ will have the same last digit as $3^n $.
Now there is something called Euler's theorem that will tell me exactly what the $n$ so that $3^n\equiv 1 \mod 10$. But I'll assume you don't know it.
Instead we'll try it till we get it.
$3^2=9\equiv -1\mod 10$ so $3^4\equiv (-1)^2\equiv 1 \mod 10$. So $n=4$ was the number we wanted: $43^{4k+m}\equiv 3^{4k+m}\equiv (3^4)^k*3^m\equiv 1^k*3^m\equiv 3^m\mod 10$.
So the last digit of $43^{43} $ is the same last digit of $3^3=27$ which is $7$.
But that does't help us solve $43^{43^{43}} $.
We need to now how $43^{43}=4k+m $ is so the last digit of $43^{43^{43}}=43^ {4k+m}\equiv 3^m\mod 10$.
So that's the next thing we need to figure out. What is $m$ if $43^{43}=4k+m $. (Note:we don't care what $k $ is.)
$43= 40+3$ so $(40+3)^s $ will be $40^s+......+3^s $. Just like before, only the last term is not a multiple of $4$ so $43^s\equiv 3^s \mod 4$.
Like above $3^2=9\equiv 1\mod 4$ so $3^{2k+i}\equiv (3^2)^k 3^i\mod 4$.
So $43^{43}\equiv 3^{43}\equiv (3^2)^{21}*3\equiv 3\mod 4$.
So $43^{43}=4k+3$ for some $k $.
So $43^{43^{43}}\equiv 3^{4k+3}\equiv 3^3\equiv 27 \equiv 7\mod 10$.
So the last digit is $7$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2294490",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Without the aid of a calculator, compute $\sin\frac{5\pi}{12}+\sin\frac{\pi}{12}$
Without the aid of a calculator, compute $\sin\frac{5\pi}{12}+\sin\frac{\pi}{12}$
The only method in doing this that I know was to just plug it in into the calculator, but I really had no clue on how to do this without the aid of a calculator,
I was thinking more along the lines of using the Taylor polynomial series but that really didn't work out for me though.
| One more method:
\begin{align*}
\left(\sin\frac{5\pi}{12}+\sin\frac{\pi}{12}\right)^2&=\left(\cos\frac{\pi}{12}+\sin\frac{\pi}{12}\right)^2\\
&=\cos^2\frac{\pi}{12}+\sin^2\frac{\pi}{12}+2\sin\frac{\pi}{12}\cos\frac{\pi}{12}\\
&=1+\sin\frac{\pi}{6}\\
&=\frac{3}{2}
\end{align*}
As $\displaystyle \sin\frac{5\pi}{12}>0$ and $\displaystyle \sin\frac{\pi}{12}>0$,
$$\sin\frac{5\pi}{12}+\sin\frac{\pi}{12}=\sqrt{\displaystyle \frac{3}{2}}=\frac{\sqrt{6}}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2295696",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 3
} |
Solve: $(i+i^2+i^3+...+i^{102})^2$
Solve $$(i+i^2+i^3+...+i^{102})^2$$
I have tried 2 approaches:
*
*to look at power of $2 (=-1)$ and power of $4 (=1)$ and there are $102$ elements which is even number, but because there are $4$ "types" $\{i,-1,-i,1\}$ it did not work it out
*to look at
$$(i+i^2+i^3+...+i^{102})^2=\left(\sum_{k=1}^{102}i^k\right)^2$$ but $$\left(\sum_{k=1}^{102}i^k\right)^2\neq \sum_{k=1}^{102}i^{2k}$$ so it was not useful.
Any suggestion how to tackle this?
| $$a + a^2 + a^3 + \dots a^n = a\frac{1-a^{n}}{1-a}$$
Therefore,
\begin{align*}
(i+i^2 + \cdots + i^{102}) &= i \frac{1-i^{102}}{1-i}\\
& = i \frac{1-i^{100}i^2}{1-i} \\
&= i \frac{1+(i^{4})^{25}}{1-i}\\
&= \frac{2i}{1-i}\\
&= -1 + i
\end{align*}
Squaring yields $-2i$.
| {
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Find all complex roots for $z^6 + (1 + i)z^3 + i = 0$ Find all complex roots of $z^6 + (1 + i)z^3 + i = 0$.
I've tried by first settings $w = z^3$, so then I get $$w^2 + (1 + i)w + i = 0$$
Now I rewrite it as $$(w+(\frac{1+i}{2}))^2 = -i + (\frac{1+i}{2})^2$$
Set $w+(\frac{1+i}{2}) = a+bi$
$$(a+bi)^2 = -i + 2i=i$$
Which implies that
$$a^2-b^2=0$$
$$a^2+b^2=\sqrt{0^2+1^2} = 1$$
$$2ab=1$$
So I get $w=\frac{1+i}{2}\pm(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}i)$. Not sure where to go from here. I get answers for $w$, but how do I get the answers for $z$?
| It's $$(z^3+i)(z^3+1)=0.$$
We have two cases.
*
*$z^3+i=0$ or $z^3-i^3=0$ or
$$(z-i)(z^2+iz-1)=0,$$ which gives
$$\left\{i,\pm\frac{\sqrt3}{2}-\frac{1}{2}i\right\}$$
*$z^3+1=0$ or
$$(z+1)(z^2-z+1)=0,$$ which gives
$$\left\{-1,\frac{1}{2}\pm\frac{\sqrt3}{2}i\right\}$$
Done!
| {
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"url": "https://math.stackexchange.com/questions/2300151",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find the power series representation of the function I have to find $a_n$ so, my first attempt was to do partial fraction, but with no real solution. Any recommendation will be appreciate $$\frac{1}{(2x-3)(x^2-x+1)}=\sum_{n=0}^\infty a_n (x-1)^n$$
| Let us replace $x$ with $z+1$ for the sake of simplicity. We have to find the sequence $\{a_n\}_{n\geq 0}$ given by:
$$ f(z)=\frac{1}{(2z-1)(z^2+z+1)}=\sum_{n\geq 0}a_n z^n \tag{1} $$
but $f(z)$ is a meromorphic function with simple poles at $z=\frac{1}{2},\,z=\omega,\,z=\omega^2$, where $\omega=e^{\frac{2\pi i}{3}}$.
By computing the residues of $f(z)$ at such points we get a partial fraction decomposition for $f(z)$:
$$ f(z) = \frac{2}{7}\cdot \frac{1}{x-\frac{1}{2}}-\frac{1}{7}\cdot\frac{3-x-2x^2}{1-x^3} \tag{2}$$
and by performing expansions as geometric series it follows that:
$$ a_n = -\frac{4}{7} 2^n-\frac{1}{7}\cdot\left\{\begin{array}{rcl}3 & \text{if} & n\equiv 0\pmod{3}\\ -1 & \text{if} & n\equiv 1\pmod{3}\\ -2 & \text{if} & n\equiv 2\pmod{3}\end{array}\right.\tag{3}$$
that can be stated as:
$$ a_n \text{ is the closest integer to } -\frac{2^{n+2}}{7}. \tag{4}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2300409",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove $(1-\frac{8}{\pi^2})x^2\cot^2(x)+\frac{8}{\pi^2}x\cot(x)+\frac{4}{\pi^2}x^2-1\ge0$ for $x\in[0,\pi/2]$
How to prove the following inequality:
$$\left(1-\frac{8}{\pi^2}\right)x^2\cot^2(x)+\frac{8}{\pi^2}x\cot(x)+\frac{4}{\pi^2}x^2-1\ge0,\quad x\in[0,\pi/2]?$$
I evaluated the LHS of the above inequality numerically. I found that equality holds at $x=0$ and $x=\pi/2$, and strict inequality holds elsewhere. The LHS is not a monotonic function; as $x$ increases it first increases and then decreases. I derived the derivative of the LHS and the results are quite messy. Is there any easier ways of proving the inequality?
| A partial answer - I am going to show that the given inequality holds for every $x\in(0,1]$.
Since the cotangent is the logarithmic derivative of the sine function we have
$$ \cot x = \frac{1}{x} + \sum_{n\geq 1}\left(\frac{1}{x-n\pi}+\frac{1}{x+n\pi}\right) \tag{1} $$
and
$$ \frac{1-\pi z\cot(\pi z)}{2} = \sum_{n\geq 1}\zeta(2n) z^{2n}.\tag{2} $$
This gives the Taylor series of $z^2\cot(z)^2$ too, and allow us to state that the given LHS is
$$ \left(-\frac{2}{3}+\frac{20}{3 \pi ^2}\right) x^2+\left(\frac{1}{15}-\frac{32}{45 \pi ^2}\right) x^4+\sum_{k\geq 3}c_k x^{2k},\qquad c_k\geq 0\tag{3} $$
for any $x\in\left(0,\frac{\pi}{2}\right)$, so the given inequality holds over the interval $\left(0,1\right]$ for sure.
That should be most of the job.
The given inequality is indeed equivalent to
$$\forall x\in\left(0,\frac{\pi}{2}\right),\quad x\cot(x)\geq \frac{-4+\sqrt{\left(\pi ^2-4\right)^2-4 \left(\pi ^2-8\right) x^2}}{\pi ^2-8} $$
where the RHS is a remarkably good approximation of the LHS.
The (generalized) Shafer-Fink inequality should completely settle the question, but I have to perform some numerical experiments. In a Shafer-Fink-suitable form, the previous line can be written as
$$ \forall x>0,\quad \frac{\arctan x}{x}\geq \frac{-4+\sqrt{\left(\pi^2-4\right)^2+4 \pi ^2 x^2}}{\pi^2-8+4 x^2}\tag{4} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2304793",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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} |
fermat's last theorem by elementary methods I was wondering about correctness of the following theorem/proof.
Theorem:
Suppose $p$ is prime in:
$ \quad \quad p^k | x - y$
$ \quad \quad p^{k + 1} \not | x - y$
$ \quad \quad gcd(x,y,z) = 1$
$ \quad \quad n > 2$
$ \quad \quad x^n = y^n + z^n$
Then:
$ \quad \quad p | n$
Proof:
Let us consider the following identity:
$\quad \quad (ap^k + b)^n = b^n + rp^k \pmod {p^{2k}} \iff r = nab^{n - 1} \pmod {p^k}$
which we can prove by induction on $n$.
When we solve for $a,b$ in:
$ \quad \quad ap^k + b = x$,
$ \quad \quad b = y$,
we get:
$ \quad \quad a = (x - y)/p^k \implies gcd(a,p) = 1$ because $x - y \not \equiv 0 \pmod {p^{k + 1}}$,
also note:
$ \quad \quad gcd(b,p) = 1$ otherwise we would have $gcd(x,y,z) \gt 1$;
We now have:
$ \quad \quad p^k$ a prime-power;
$ \quad \quad gcd(a,p) = gcd(b,p) = 1$;
Now assume $z$ exists in:
$ \quad \quad x^n \equiv y^n + z^n \pmod {p^{2k}} \iff (ap^k + b)^n \equiv b^n + rp^k \pmod {p^{2k}}$
$ \quad \quad \implies z^n \equiv rp^k \pmod {p^{2k}}$
We must consider the following three cases:
*
*$n \gt k \implies p | z \implies z = sp$ for some $s$
$ \quad \quad \implies z^n \equiv (sp)^n \equiv rp^k \pmod {p^{2k}}$
$ \quad \quad \implies s^np^{n - k} \equiv r \pmod {p^k}$
$ \quad \quad $ so clearly $p | r$
*$n = k \implies p | z \implies z = sp$ for some $s$
$ \quad \quad \implies z^n \equiv (sp)^n \equiv rp^n \pmod {p^{2n}}$
$ \quad \quad \implies s^n \equiv r \equiv nab^{n - 1} \pmod {p^n}$
$ \quad \quad $ Now let:
$ \quad \quad \quad \quad na \equiv b \pmod {p^n}$
$ \quad \quad \quad \quad \implies r \equiv nab^{n - 1} \equiv b^n \pmod {p^n}$
$ \quad \quad $ So we get:
$ \quad \quad \quad \quad (ap^n + b)^n \equiv b^n + (bp)^n \pmod {p^{2n}}$
$ \quad \quad \quad \quad \implies gcd(x,y,z) > 1$, which is impossible
*$n < k \implies p^t | z \implies z = sp^t$ for some $s$ and with $t = ceil(k/n)$
$ \quad \quad \implies z^n \equiv (sp^t)^n \equiv rp^k \pmod {p^{2k}}$
$ \quad \quad \implies s^np^{tn - k} \equiv r \pmod {p^n}$
$ \quad \quad$ and since $tn \gt k$ we have $p | r$
So we have shown $p | r \implies n \equiv 0 \pmod p$ always holds
Note that we cannot say anything about $n = 2$, because we always have:
$ \quad \quad (ap + b)^2 \equiv b^2 + z^2 = b^2 \pmod {p^2}$
$ \quad \quad \iff r \equiv 0 \pmod {p}$
no matter what $a,b$ will be
This proves our theorem.
| Let $n$ be a positive integer, $n > 1$, and suppose $b,c,d$ are positive integers with $\gcd(b,c,d)=1$ such that $d^n = b^n + c^n$.
Next, suppose $p$ is a prime such that $d=ap+b$.
\begin{align*}
\text{Then}\;\;&(ap + b)^n = b^n + c^n\\[4pt]
\implies\;&p\mid c^n\\[4pt]
\implies\;&p\mid c\\[4pt]
\end{align*}
So you can't choose $p$ arbitrarily, since $p$ must be a factor of $c$.
What you proved is that if $p$ is a prime factor of $c$, then $p\mid a$ or $p\mid n$.
Stated in full, what you proved is equivalent to this:
\begin{align*}
&\text{If:}\\[4pt]
&\;\;{\small\bullet}\;\;\text{$n$ is a positive integer with $n > 1$}\\[4pt]
&\;\;{\small\bullet}\;\;\text{$b,c,d$ are positive integers with $\gcd(b,c,d)=1$}\\[-1pt]
&{\phantom{\;\;{\small{\bullet}}\;\;}}\text{such that $d^n = b^n + c^n$}\\[4pt]
&\;\;{\small\bullet}\;\;\text{$p$ is a prime factor of $c$}\\[8pt]
&\text{Then:}\\[4pt]
&\;\;{\small\bullet}\;\;\text{$p\mid n\;$ or $\;p^2 \mid (d-b)$}\\[4pt]
\end{align*}
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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every integer can be represented in the form $x^2+y^2-z^2$ Every integer can be represented in the form $x^2+y^2-z^2$ and show that $6$ actually requires all three terms.
I put
*
*$z=y+1$
*$x=n^2+3$
*$y=3n^2+4+(n^4-n)/2$
what does it mean that $6$ actually requires all three terms?
| It means that if $x,y,z$ are such that
$$x^2+y^2-z^2=6$$
then $x,y,z$ are all non-zero.
You can prove this by looking at the equation mod $4$.
You know that a square is always $0$ or $1$ mod $4$, and $6\equiv 2\pmod 4$.
So if $x^2+y^2-z^2=6$, then $z$ must be $0$ mod $4$, otherwise you are not able to get to $2$ mod $4$.
But $6$ can't be a sum of two squares mod $4$ because $3\mid 6$ (look there).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2308391",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How multiply Blocked Matrices? I am having a hard time understanding how to multiply blocked matrices with rectangle matrices and blocking into non-square matrices. Can someone please explain me how that works?
| I hope it can help you
Example:
$A=
\begin{bmatrix}
2 & 3 & 2\\
4 & 1 & 3\\
3 & 2 & 1\\
\end{bmatrix}
$$\to$
$ \left[
\begin{array}{cc|c}
2&3&2\\
4&1&3 \\
\hline
3 &2 &1
\end{array}
\right] $
$ B=
\begin{bmatrix}
1 & 3 \\
2 & 4 \\
2 & 1 \\
\end{bmatrix}
$$\to$
$ \left[
\begin{array}{}
1&3\\
2&4 \\
\hline
2 &1
\end{array}
\right] $
$AB=
\begin{bmatrix}
A_{11} & A_{12} \\
A_{21} & A_{22} \\
\end{bmatrix}
\times\begin{bmatrix}
B_{1} \\
B_{2} \\
\end{bmatrix}
=\begin{bmatrix}
A_{11}B_{1} + A_{12}B_{2} \\
A_{21}B_{1} + A_{22}B_{2} \\
\end{bmatrix}
$
$A_{11}B_{1}=
\begin{bmatrix}
2 & 3 \\
4 & 1 \\
\end{bmatrix}
\times\begin{bmatrix}
1 & 3 \\
2 & 4 \\
\end{bmatrix}
=\begin{bmatrix}
8 & 18 \\
6 & 16 \\
\end{bmatrix}
$
$A_{12}B_{2}=
\begin{bmatrix}
2 \\
3 \\
\end{bmatrix}
\times\begin{bmatrix}
2 & 1 \\
\end{bmatrix}
=\begin{bmatrix}
4 & 2 \\
6 & 3 \\
\end{bmatrix}
$
$A_{21}B_{1}=
\begin{bmatrix}
3 & 2 \\
\end{bmatrix}
\times\begin{bmatrix}
1 & 3 \\
2 & 4 \\
\end{bmatrix}
=\begin{bmatrix}
7 & 17 \\
\end{bmatrix}
$
$A_{22}B_{2}=
\begin{bmatrix}
1 \\
\end{bmatrix}
\times\begin{bmatrix}
2 & 1 \\
\end{bmatrix}
=\begin{bmatrix}
2 & 1 \\
\end{bmatrix}
$
$A_{11}B_{1}+A_{12}B_{2}=
\begin{bmatrix}
8 & 18 \\
6 & 16 \\
\end{bmatrix}
+\begin{bmatrix}
4 & 2 \\
6 & 3 \\
\end{bmatrix}
=\begin{bmatrix}
12 & 20 \\
12 & 19 \\
\end{bmatrix}
$
$A_{21}B_{1}+A_{22}B_{2}=
\begin{bmatrix}
7 & 17 \\
\end{bmatrix}
+\begin{bmatrix}
2 & 1 \\
\end{bmatrix}
=\begin{bmatrix}
9 & 18 \\
\end{bmatrix}
$
$$AB=
\begin{bmatrix}
12 & 20 \\
12 & 19 \\
\hline
9 & 18 \\
\end{bmatrix}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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How to prove that $\sqrt{-3-2i}+\sqrt{-3+2i} = \sqrt{2(\sqrt{13}-3)}$? Is there a trick to show that
$$\sqrt{-3-2i}+\sqrt{-3+2i} = \sqrt{2(\sqrt{13}-3)}$$
is true ?
| Start by squaring the LHS and RHS
\begin{align*}
(\sqrt{-3-2i} + \sqrt{-3+2i} )^2 &= \sqrt{2(\sqrt{13}-3)}^2 \\
(-3-2i) + 2\sqrt{(-3-2i)(-3+2i)} + (-3 + 2i) &= 2(\sqrt{13}-3) \\
-6 + \sqrt{4 (9 + 4)} &= 2(\sqrt{13}-3) \\
-6 + 2\sqrt{13} &= 2(\sqrt{13}-3) \\
2(\sqrt{13}-3) &= 2(\sqrt{13}-3)
\end{align*}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluate series $\sum\limits_{n=1}^{\infty}\frac{x^{2^{n-1}}}{1-x^{2^n}}$
Determine the value of
$$\sum_{n=1}^{\infty}\frac{x^{2^{n-1}}}{1-x^{2^n}}$$
or $$\frac{x}{1-x^2}+\frac{x^2}{1-x^4}+\frac{x^4}{1-x^8}+\cdots$$
for $x\in\mathbb{R}$.
The answer is $\dfrac{x}{1-x}$ for $x\in(0,1)$. To prove this, notice
$$\frac{x}{1-x^2}=x+x^3+x^5+\cdots$$
$$\frac{x^2}{1-x^4}=x^2+x^6+x^{10}+\cdots$$
$$\cdots$$
Add them all and get the answer. Unfortunately, I havn't got a direct method to calculate it. Appreciate for your help!
| As soon as $|x|<1$ we have
$$ \frac{x^{2^{n-1}}}{1-x^{2^n}} = x^{2^{n-1}}+x^{3\cdot 2^{n-1}}+x^{5\cdot 2^{n-1}}+\ldots \tag{1}$$
hence:
$$ \sum_{n\geq 1}\frac{x^{2^{n-1}}}{1-x^{2^n}} = \sum_{m\geq 0}\sum_{h\geq 0} x^{(2h+1) 2^m} = \sum_{n\geq 1} x^n = \frac{x}{1-x}\tag{2} $$
since every positive integer can be represented in a unique way as the product between a power of two and an odd integer.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "7",
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Prove that $\frac{(\sin 20^\circ + \cos 20^\circ)^2}{\cos 40^\circ} = \cot 25^\circ$ So I'm trying to come up with an answer to this question for hours now. I don't know what I'm doing wrong and none of the calculators on the internet couldn't help so I figured I should ask people.
What have I done so far:
$\frac{(\sin 20^\circ + \cos 20^\circ)^2}{\cos 40^\circ} = \frac{(\frac{2\sin(45^\circ+20^\circ)}{\sqrt{2}})^2}{\cos 40^\circ} = \frac{\sin^2 65^\circ}{\cos 40^\circ} = \frac{1 - \cos 130^\circ}{\cos 40^\circ} = \frac{1 + \cos 50^\circ}{\cos 40^\circ} = \frac{2\cos^2 25^\circ}{\cos 40^\circ}$ ... etc.
I can't seem to figure out where to go from here so I'm just stuck.
I also tried the classic approach:
$\frac{(\sin 20^\circ + \cos 20^\circ)^2}{\cos 40^\circ} = \frac{1 + \sin 40^\circ}{\cos 40^\circ} = \frac{1}{\cos 40^\circ} + {\tan 40^\circ} = \sec 40^\circ + \tan 40^\circ$
But how can I prove that
$\sec 40^\circ + \tan 40^\circ = \cot 25^\circ$ ?
What am I doing wrong? Any hints or solutions would be great. Thanks in advance.
| Generalization:
$$\dfrac{(\cos x+\sin x)^2}{\cos2x}=\dfrac{(\cos x+\sin x)^2}{\cos^2x-\sin^2x}=\dfrac{\cos x+\sin x}{\cos x-\sin x}$$ provided $\cos x+\sin x\ne0\iff\tan x\ne-1$
Now $$\dfrac{\cos x+\sin x}{\cos x-\sin x}=\dfrac{1+\tan x}{1-\tan x}=\tan\left(45^\circ+x\right)$$
Can you identify $x$ here?
Alternatively, $$\dfrac{(\cos x+\sin x)^2}{\cos2x}=\dfrac{1+\sin2x}{\cos2x}=\dfrac{1+\dfrac{2\tan x}{1+\tan^2x}}{\dfrac{1-\tan^2x}{1+\tan^2x}}=\cdots=\dfrac{1+\tan x}{1-\tan x}=\cdots$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Which way is correct to represent this differentiation? $$3x^2+x^3$$
Option 1
$$\frac{d}{dx}(3x^2+x^3)=6x+3x^2$$
Option 2
$$\frac{d(3x^2+x^3)}{dx}=6x+3x^2$$
My student wrote the Option 2 style, not sure whether I can consider is ok or not.
| Both of the given representations are equivalent.
For:
$$y(x)=3x^2+x^3$$
$$\frac{dy}{dx} \equiv \frac{d(3x^2+x^3)}{dx} \equiv \frac{d}{dx}(3x^2+x^3) \equiv \frac{d(3x^2)}{dx}+\frac{d(x^3)}{dx} \equiv \frac{d}{dx}(3x^2)+\frac{d}{dx}(x^3)
=6x+3x^2$$
| {
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Confusion about infinitely many bounded prime gaps? I was working at something and I had this idea: Let $f(x) = \frac{1}{x}$ for all $x > 0$. Let $p_n$ denote the $n$'th prime number. Then by mean value theorem $$ f(p_{n+1}) - f(p_n) = f'(c_n) (p_{n+1} - p_n)$$ where $c_n \in (p_n,p_{n+1})$. Hence $$ \frac{1}{p_{n+1}} = -\frac{1}{c_n^2} (p_{n+1} - p_n) + \frac{1}{p_n}$$ Suppose that $\exists N \in \mathbb{N}$ such that $\forall n > N$ $p_{n+1} - p_n > D$ where $D$ is a positive constant. Then $-\frac{p_{n+1} - p_n}{c_n^2} < -\frac{D}{p_{n+1}^2}$ hence $$ \frac{1}{p_{n+1}} < -D\sum_{i=N+1}^{n+1}\frac{1}{p_i^2} + \frac{1}{p_N} = -D \sum_{i=1}^{n+1} \frac{1}{p_i^2} + D\sum_{i=1}^{N} \frac{1}{p_i^2} + \frac{1}{p_N}$$ Now $Z_p(2,n+1) = \sum_{i=1}^{n+1} \frac{1}{p_i^2} $ is bounded (converges to prime zeta function of 2). Therefore
$$ \frac{1}{p_{n+1}} < -D \cdot Z_p(2,n+1) + D\cdot C_2 + \frac{1}{p_N}$$ where $C_2 = \sum_{i=1}^{N} \frac{1}{p_i^2}$. Obviously $C_2 < Z_p(2,n+1)$. We take $D$ such that $D \cdot \left( C_2 - Z_p(2,n+1)\right) + \frac{1}{p_N} < 0$ that is $$ D > \frac{1}{p_N \cdot \left( Z_p(2,n+1) - C_2\right)} > 0$$ to obtain a contradiction ! Therefore there are infinitely many $k \in \mathbb{N}$ such that $p_{n_k+1} - p_{n_k} < D$ ?!
As mentioned in the comments the question is whether $\frac{1}{p_N \cdot \left( Z_p(2,n+1) - C_2\right)}$ is bounded above for $N > 0$ ?
Edit:
The answer is NO!
| You conclude at some point:
Therefore $$\frac{1}{p_{n+1}}<−D\cdot Z_p(2,n+1)+D\cdot C_2+\frac{1}{p_N}$$
which basically is the same as;
$$\color{red}{0<}\frac{1}{p_{n+1}}<\color{red}{D\cdot \left(C_2- Z_p(2,n+1)\right)+\frac{1}{p_N}}$$
and the next thing you do is:
We take $D$ such that $D\cdot \left(C_2−Z_p(2,n+1)\right)+\frac{1}{p_N}\color{red}{<0}$ ...
Do you see the problem?
| {
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Manipulating an algebraic equation This came in a competition, I went back home and tried it but was unsuccessful.
If $$\frac{a^3+3ab^2}{3a^2b+b^3} = \frac{x^3+3xy^2}{3x^2y+y^3}$$ then -
$$1) ~ax =by
~~2)~ xy = ab
~~3)~ ay = bx
~~4) ~ax = b$$
It is clear that if the following given condition holds for the variables then there would be a case in which $a = x$ and $b = y$. Hence then options should also satisfy this and hence we can eliminate options a and d. But I was not able to solve the question further. I tried adding and subtracting 1 from both sides but found nothing helpful.
Moreover we are required to show if the correct option is sufficient to show that the first condition is true for the given variables.
| Rewrite $$\frac{a^3+3ab^2}{3a^2b+b^3} = \frac{x^3+3xy^2}{3x^2y+y^3}\tag 1$$ as
$$\frac{ ({a \over b})^3+3{a \over b}}{3({a\over b})^2+1} = \frac{({x \over y})^3+3{x\over y}}{3({x\over y})^2+1} \tag 2$$ and set $t={a\over b},\; s={x\over y}.$ Then
$$0=\frac{t^3+3t}{3t^2+1}-\frac{s^3+3s}{3s^2+1}=\frac{(t-s)\left[(t-s)^2+3(ts-1)^2\right]}{(3t^2+1)(3s^2+1)}.$$ Consequently $t=s,$ which is ${a \over b} = {x \over y},$ or $$ax=by \tag 3.$$
Sufficiency
Assume $ax=by.$ If $b\neq 0 \neq y,$ then clearly $(2)$ holds and $(1)$ follows.
However, $ax=by$ is true if $b=0=y$ as well, but $(1)$ is not defined.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Point out my fallacy, in sequence and series.
The sum of the first $n$--terms of the series $1^2+2\cdot2^2+3^2+2\cdot4^2+\cdots$ is $\dfrac{n(n+1)^2}{2}$, when $n$ is even. When $n$ is odd, the sum is?
I got the correct answer when is replaced $n\rightarrow (n+1)$ to make above valid for odd, but when I tried the different approach then something following had happened.
For $n$ even, last term $=n$ which is even and term before it $=n-1$ which is odd. Clubbing all odds and evens separately as follows:
$\big(1^2+3^2+\cdots +(n-1)^2\big)+2\big(2^2+4^2+\cdots+n^2\big)=\dfrac{n(n+1)^2}{2}\tag{1}$
For $n$ odd, last term $=n$ which is odd and term before it $=n-1$ which is even. Clubbing all odds and evens separately as follows:
$\big(1^2+3^2+\cdots +n^2\big)+2\big(2^2+4^2+\cdots+(n-1)^2\big)\tag*{}$
$=\big(1^2+3^2+\cdots +(n-1)^2\big)+2\big(2^2+4^2+\cdots+n^2\big)-n^2+(n-1)\tag*{}$
From equation $(1)$
$=\dfrac{n(n+1)^2}{2}-n^2+(n-1)\tag*{}$
And answer given is: $\dfrac{n^2(n+1)}{2}$
please help.
| For $n$ odd case: $$\big(1^2+3^2+\cdots +n^2\big)+2\big(2^2+4^2+\cdots+(n-1)^2\big) \\= \big(1^2+3^2+\cdots +n^2\big)+2\big(2^2+4^2+\cdots+(n+1)^2\big) - 2(n+1)^2 \\= \frac{(n+1)(n+2)^2}{2} - 2(n+1)^2 \\= \frac{n^2(n+1)}{2}.$$
Let $T1$ denote number of terms contributed by odd indexes (for example $1^2, 3^2$ etc) and $T2$ denote number of terms contributed by even indexes (for example $2.2^2, 2.4^2$ etc). Note that, for $n$ odd case, $T1 = T2 + 1$ and for $n$ even case, $T1 = T2$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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$a_1+a_2+…+a_n=2008$ where all $a_i$ are positive integers. If $A_k = a_1 a_2 … a_k$ , what is the largest possible value of $A_1+A_2+…+A_n$? The sum of a set of positive integers = $2008$.
$a_1+a_2+…+a_n=2008$ where all $a_i$ are positive integers.
If $A_k=a_1 a_2 … a_k$ , what is the largest possible value of $A_1+A_2+…+A_n$ ?
| All of the $a_i$ must come in weakly decreasing order.
If $a_1+\dots+a_n=2008$ with $a_1$ an even number bigger than $2$, then consider $$(a_1/2)+2+a_2+\dots+a_n+1+\dots+1=2008.$$
If $a_1+\dots+a_n=2008$ with $a_1$ odd and at least $5$, say $a_1=2k+1$, consider
$$
(a_1-2)+2+a_2\dots+a_n=2008
$$
(Clearly $a_1=1$ cannot be maximal).
Once you have figured out $a_1$, apply the same argument to $a_2$, $a_3$, etc.
(edit: at each step above, replace $j$ trailing copies of $1$ by the integer $j$; that is, instead of using $2006+1+1$, use $2006+2$. Doing this does not change the result).
This shows that each $a_i$ must be $3$ or $2$. Simplifying the terms $A_1+\dots+A_n$ is now just adding terms of a 2 geometric series. Finally, if there are $k$ $3s$ then there are $\frac{2008-3k}{2}$ $2s$; thus we want to maximize
$$
\frac{3(3^k-1)}{2}+\frac{3(3^k-1)}{2}\cdot2\cdot\left(2^\frac{2008-3k}{2}-1\right)
$$
From Wolframalpha this gives $668$ copies of $3$ and $2$ copies of $2$ for an answer of roughly $3\times 10^{319}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2316434",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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How to find eigenvector of large matrix without substitution directly. \begin{bmatrix}a&b&b&0&0&0\\b&a&b&0&0&0\\b&b&a&0&0&0\\0&0&0&a&b&b\\0&0&0&b&a&b\\0&0&0&b&b&a\end{bmatrix}
I have a large matrix as above which I already know that
\begin{bmatrix}a&b&b\\b&a&b\\b&b&a\end{bmatrix} have three eigenvectors equal to \begin{bmatrix}-1\\0\\1\end{bmatrix},\begin{bmatrix}-1\\1\\0\end{bmatrix} and \begin{bmatrix}1\\1\\1\end{bmatrix}
Is there any way I can find eigenvector for a large matrix above by using the submatrix that I already solved
Thank you
ps. If you found this question is a duplicate of another, kindly put the link for me please.
| In general, if $A$ and $B$ are diagonalizable as $A=MD_AM^{-1}$ and $B=ND_BN^{-1}$ ( where the columns of $M$ are the eigenvectors of $A$ and the columns of $N$ are the eigenvectors of $B$) than we have:
$$
C=\begin{pmatrix}A&0\\0&B
\end{pmatrix}=\begin{pmatrix}MD_AM^{-1}&0\\0&ND_BN^{-1}
\end{pmatrix}=
$$
$$
=\begin{pmatrix}M&0\\0&N
\end{pmatrix}
\begin{pmatrix}D_A&0\\0&D_B
\end{pmatrix}
\begin{pmatrix}M^{-1}&0\\0&N^{-1}
\end{pmatrix}=\begin{pmatrix}M&0\\0&N
\end{pmatrix}
\begin{pmatrix}D_A&0\\0&D_B
\end{pmatrix}
\begin{pmatrix}M&0\\0&N
\end{pmatrix}^{-1}
$$
So $C$ is diagonalizable as
$$
C=PDP^{-1}
$$
and the columns of $P=\begin{pmatrix}M&0\\0&N
\end{pmatrix}$ are the eigenvectors of $C$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2318474",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find closed formula and limit for $a_1 =1$, $2a_{n+1}a_n = 4a_n + 3a_{n+1}$ Tui a sequence $(a_n)$ defined for all natural numbers given by
$$a_1 =1, 2a_{n+1}a_n = 4a_n + 3a_{n+1}, \forall n \geq 1$$
Find the closed formula for the sequence and hence find the limit.
Here, what I have done:
$$2a_{n+1}a_n = 4a_n + 3a_{n+1}
\implies a_{n+1} = \frac{4a_n} {2a_n - 3}
\implies a_{n+1} = \frac{\frac{4a_n} {a_n} } {\frac{2a_n}{a_n} - \frac{3} {a_n} }
\implies a_{n+1} = \frac{4} {2 - \frac{3} {a_n} }
\implies \frac{1 } {a_{n+1}} = \frac{2 - \frac{3} {a_n} } {4}
\implies \frac{1 } {a_{n+1}} =\frac{1 } {2 } - \frac{3} {4a_n}$$
Then go to where????
| Enforcing $u_n=1/a_n$ gives
$$2a_{n+1}a_n = 4a_n + 3a_{n+1}\Longleftrightarrow 4u_{n+1}+ 3u_n =2\Longleftrightarrow u_{n+1}=- \frac34u_n +\frac12$$
Now set $x_n = u_n-\frac27$ then you can easily check that we have $x_{n}= -\frac34x_{n-1}$ which is a geometric sequence
$$x_{n}= -\frac34x_{n-1}\implies x_n =\left(-\frac{3}{4}\right)^{n-1}x_1 \\\implies \color{red}{\frac{1}{a_n}= u_n =\frac27+\left(-\frac{3}{4}\right)^{n-1}(u_1 -\frac27)\to \frac27}$$
Alternatively, a more general formula is obtained using this How to find the closed form of $u_{n+1}=a_nu_n+b_n~~~\text{where $a_1$, $a_n$ and $b_n$ are given.}$ you get that
$$ u_n = \left(\sum_{p=1}^{n-1}\frac{ \frac12}{\prod_{k=1}^{p}- \frac34}+u_1\right)\prod_{k=1}^{n-1}- \frac34 = \left(-\frac{3}{4}\right)^{n-1}\left(1+\frac12\sum_{p=1}^{n-1}\left(-\frac{4}{3}\right)^{p-1} \right)\\=\color{red}{\frac{1}{a_n}= u_n =\frac27+\frac57\left(-\frac{3}{4}\right)^{n-1} \to \frac27}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find Jordan's normal form It is first time I try normally to figure out how it works.
So there's the matrix:
$$A=\begin{pmatrix} 0 & 6 & 4 & 0 \\ 0 & 4 & 0 & 2 \\ 0 & -1 & 2 & -1 \\1 & -3 & -2 & 2 \end{pmatrix}$$
1.) First of all we have to find eignevalue:
$$\mathrm{det}|A-\lambda{I}|= \begin{vmatrix} 0 - \lambda & 6 & 4 & 0 \\ 0 & 4-\lambda & 0 & 2 \\ 0 & -1 & 2-\lambda & -1 \\ 1 & -3 & -2 & 2-\lambda \end{vmatrix} = (\lambda-2)^4$$
So eigenvalue $\lambda = 2$
2.) Substitute $\lambda$ value into the matrix, got:
$$A_{\varphi}=\begin{pmatrix} -2 & 6 & 4 & 0 \\ 0 & 2 & 0 & 2 \\ 0 & -1 & 0 & -1 \\ 1 & -3 & -2 & 0 \end{pmatrix}$$
3.) As far as I understand I have to find rank of $A_{\varphi}$. Therefore $\mathrm{rank}A_{\varphi}=2$
Because $$A_{\varphi}=\begin{pmatrix} 0 & -1 & 0 & -1 \\ 1 & -3 & -2 & 0
\end{pmatrix}$$
I do not know how to procced from here...
| You can easily, though annoyingly carefully, check that $\;(A-2I)^3=0\;$ and that means the minimal polynomial of your matrix is $\;(x-2)^3\;$ , which means its Jordan form must have at least (and, in this case, obviously also at most) a Jordan block of size three, so
$$J_A=\begin{pmatrix}
2&1&0&0\\
0&2&1&0\\
0&0&2&0\\
0&0&0&2\end{pmatrix}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2320749",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Prove or disprove that where $a, b \in Z$, the number a+b is even if and only if a-b is even.
Prove or disprove that where $a, b \in Z$, the number a+b is even if and only if a-b is even.
Can some help me improve on this or check it please?
If $a + b$ is even, then there is some number $k \in Z$ such that $a + b = 2k$.
If $a - b$ is odd, then there is some number $m \in Z$ such that $a - b = 2m +1$.
$a + b = a - b + 2b = 2m + 1 + 2b = 2(m + b) +1$. This is an odd number.
Now try for $a - b$ is even.
If $a - b$ is even, then there is a number $m \in Z$ such that $a - b = 2m$.
$a + b = a - b + 2b = 2b + 2m = 2(b + m)$. This is an even number.
Therefore, $a + b$ is even if and only if $a - b$ is even.
| We have $a+b = a-b +2b$ and so $a+b$ and $a-b$ differ by an even number. Therefore, they have the same parity.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2323716",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find $(x + 1)^{−3}$ in $\mathbb{Z}_2[x]/(x^5 + x^4 + 1)$ In a previous question, I established that $x^5 + x^4 + 1$ is irreducible in $\mathbb{Z}_2[x]/(x^5 + x^4 + 1)$, so it will not be divisible by $x+1$. Hence $x^5+x^4+1$ will be coprime to $(x+1)^3 = x^3+x^2+x+1$
So far I have used the Euclidean algorithm to get:
$x^5+x^4+1 = (x^3+x^2+x+1)(x^2+1) + x$
$x^3+x^2+x+1 = (x^2+1)(x+1) + 0$
However, I know I have done something wrong - because the Euclidean algorithm should show that the gcd of $x^5+x^4+1$ and $x^3+x^2+x+1$ is $1$, not $(x^2+1)$.
| Using the algorithm detailed in this answer with polynomials (and writing top to bottom instead of left to right),we get
$$
\begin{array}{c|c|c|c}
x^5+x^4+1&0&1&\\
x^3+x^2+x+1&1&0&\\
x&x^2+1&1&x^2+1\\
1&\color{#C00}{x^4+x^3+x}&x^2+x+1&x^2+x+1\\
0&x^5+x^4+1&x^3+x^2+x+1&x
\end{array}
$$
which says that in $\mathbb{Z}_2$,
$$
(x^3+x^2+x+1)^{-1}=x^4+x^3+x\pmod{x^5+x^4+1}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2324443",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$8\sin x\cos x-\sqrt{6} \sin x- \sqrt{6} \cos x+1=0$, Solve for $x$
Solve for $x:0\leq x \leq \frac{\pi}{2}$
$$8\sin x\cos x-\sqrt{6} \sin x- \sqrt{6} \cos x+1=0$$
My attempt,
I changed it into $$1+4 \sin 2x-2\sqrt{3} \sin (x+\frac{\pi}{4})=0$$
| Hint: Let $t=\sin x+ \cos x$ and use $1=\sin^2 x+\cos^2 x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2326355",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Proving a relation related to quadratic equation Question:If $α$ and $β$ be the roots of $ax^2+2bx+c=0$ and $α+δ$, $β+δ$ be those of $Ax^2+2Bx+C=0$, prove that, $\frac{b^2-ac}{a^2}=\frac{B^2-AC}{A^2}$.
My Attempt: Finding the sum of roots and product of roots for both the equations we get,
$α+β=\frac{-2b}{a}$
$αβ=\frac{c}{a}$
$α+δ+β+δ=\frac{-2B}{A}$
⇒ $α+β+2δ =\frac{-2B}{A}$
$(α+δ)(β+δ)=\frac{C}{A}$
⇒ $αβ+αδ+βδ+δ^2=\frac{C}{A}$
⇒$\frac{c}{a}+αδ+βδ+δ^2=\frac{C}{A}$ ⇒ $αδ+βδ+δ^2=\frac{Ca-cA}{Aa}$
$(α+β)^2=\frac{4b^2}{a^2}$
⇒ $α^2+β^2+2αβ=\frac{4b^2}{a^2}$
$α^2+β^2+\frac{2c}{a}=\frac{4b^2}{a^2}$
⇒ $α^2+β^2=\frac{4b^2-2ac}{a^2}$ -(1)
$(α+β+2δ)^2 =\frac{4B^2}{A^2}$
⇒ $α^2+β^2+(2δ)^2+2(αβ+2βδ+2αδ)=\frac{4B^2}{A^2}$
⇒$α^2+β^2+4δ^2+2αβ+4βδ+4αδ=\frac{4B^2}{A^2}$
⇒$α^2+β^2+2αβ+4(δ^2+βδ+αδ)=\frac{4B^2}{A^2}$
⇒$α^2+β^2+\frac{2c}{a}+4(\frac{Ca-cA}{Aa})=\frac{4B^2}{A^2}$
⇒$α^2+β^2=\frac{4aB^2-2A^2 c-4Aac+4cA^2}{A^2a}$ -(2)
From (1) and (2) we get,
$\frac{4b^2-2ac}{a^2}=\frac{4aB^2-2A^2 c-4Aac+4cA^2}{A^2a}$
My problem: I tried simplifying it further but could not reach the required result. A continuation of my method would be more appreciated compared to other methods.
| Alternative hint: we can assume WLOG that $\,a=A=1\,$, since both the roots and the equality to be proved are homogeneous in the respective coefficients.
Then, if $\alpha, \beta$ are the roots of $x^2+2bx+c=0$, the polynomial with roots $\alpha+\delta, \beta+\delta$ is:
$$(x-\delta)^2+2b (x-\delta)+c=0 \;\;\iff\;\; x^2 + 2(b-\delta)x+\delta^2-2b\delta+c=0\,$$
Identifying coefficients gives $B=b-\delta$ and $C=\delta^2-2b\delta+c\,$, then:
$$\require{cancel}
B^2 - C=(b^2-\bcancel{2 b\delta}+\cancel{\delta^2})-(\cancel{\delta^2}-\bcancel{2b\delta}+c) = b^2 - c
$$
[ EDIT ] To answer OP's edit:
A continuation of my method would be more appreciated compared to other methods.
There is an error/typo in formula (2). Once corrected (in red below):
$$\require{cancel}
\frac{4b^2-2ac}{a^\bcancel{2}}=\frac{4aB^2-2A^2 c-4Aa\color{red}{C}+4cA^2}{A^2 \bcancel{a}}
$$
$$
4b^2A^2-\bcancel{2acA^2}=4a^2B^2-\bcancel{2acA^2}-4a^2AC+4acA^2
$$
$$
\bcancel{4}A^2(b^2-ac) = \bcancel{4}a^2(B^2-AC)
$$
$$\frac{b^2-ac}{a^2} =\frac{B^2-AC}{A^2}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2327883",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
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Multiplied X in Quadratic Equation - Delta always lower than 0 I'm new to Quadratic Equations and I am following the two formulas to find $x_1$ and $x_2$ and they are:
$\Delta = b^2 - 4ac$
$x = \dfrac{-(b) \pm \sqrt\Delta}{2a}$
but delta always gives a negative number;
for example I have the following equation:
$6x^2 + 11x - 35 = 0$
What I did was:
1. Calculate the Delta ($\Delta$) by using the formula above
$11^2 - 4ac$
$11^2 = 121$ and $4ac = -840 = 4\times6\times(-35)$
$\Delta = 121 - 840 = 719$
and got $-719$ as result
| $6x^2+11x−35=0$
$a =6; b= 11; c= -35$
$b^2 = 121$.
$4ac = 4*6*(-35) =-840$
$b^2 - 4ac = 121 - (-840) = 121 + 841 = 961=31$.
So $\Delta = 961=31^2 > 0$ and $\sqrt{961} = 31$
$x = \frac {-11 \pm 31}{2*6}$.
$x = \frac {-42}{12}$ or $x = \frac {20}{12}$
$x = -\frac 72$ or $x = \frac 53$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2328997",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Infinitely many positive integers $n$ such that $n^2+1 \mid 1 \cdot 2 \cdot 5 \cdot 10 \cdots ((n-1)^2+1)$
Prove that there exist infinitely many positive integers $n$ such that $$n^2+1 \mid 1 \cdot 2 \cdot 5 \cdot 10 \cdots ((n-1)^2+1).$$
We need to prove that there are infinitely many $n$ such that $\prod_{m=0}^{n-1}(m^2+1)$ is divisible by $n^2+1$. Some examples I found for when it is divisible are $n = 3,7,8,13,17,18,21$, but I didn't see how to find infinitely many.
| I conjecture that $n=2^{2k+1}$ is always a solution. We have that $$2^{4k+2}+1=(2^{2k+1}+1)^2-2^{2k+2}=(2^{2k+1}-2^{k+1}+1)(2^{2k+1}+2^{k+1}+1)$$
Now we can pick $m=2^{k+1}+1$ then $$m^2+1=2^{2k+2}+2^{k+2}+2=2(2^{2k+1}+2^{k+1}+1)$$ and $m=2^{k+1}-1$ then $$m^2+1=2^{2k+2}-2^{k+2}+2=2(2^{2k+1}-2^{k+1}+1)$$
And both $m$'s are obviously part of the RHS product (if $k>1$) the conjecture is true hence infinitely many solutions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2331734",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 2,
"answer_id": 1
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$\arctan(\tan(-\frac{3\cdot \pi}{5}))+\operatorname{arccot}(\cot(-\frac{3\cdot \pi}{5}))=?$ We know that $$\DeclareMathOperator{\arccot}{arccot}\arctan(-x)=-\arctan(x)$$ $$\arccot(-x)=\pi-\arccot(x)$$ $$\cot(-x)=-\cot(x)$$ $$\tan(-x)=-\tan(x)$$So I have figured out that $$\arctan(\tan(-\frac{3\cdot \pi}{5}))=\arctan(-\tan(\frac{3\cdot \pi}{5}))=-\arctan(\tan(\frac{3\cdot \pi}{5}))=-\frac{3\cdot \pi}{5},$$ $$\arccot(\cot(-\frac{3\cdot \pi}{5}))=\arccot(-\cot(-\frac{3\cdot \pi}{5}))=\pi-\arccot(\cot(\frac{3\cdot \pi}{5}))=\pi-\frac{3\cdot \pi}{5}.$$Coming from these, I have solved the problem as follows:$$\arctan(\tan(-\frac{3\cdot \pi}{5}))+\arccot(\cot(-\frac{3\cdot \pi}{5}))=-\frac{3\cdot \pi}{5}+\pi-\frac{3\cdot \pi}{5}=-\frac{6\cdot \pi}{5}
+\pi=-\frac{\pi}{5}.$$But the book says it is incorrect. Could somebody point out where I have made a mistake?
| For all real $x$ we have $-\frac{\pi}{2}<\arctan{x}<\frac{\pi}{2}$ and $0<\operatorname{arccot}{x}<\pi$.$\DeclareMathOperator{\arccot}{arccot}$
$$\arctan\tan\left(-\frac{3 \pi}{5}\right)+\arccot\cot\left(-\frac{3\pi}{5}\right)=$$
$$=\arctan\tan\frac{2 \pi}{5}+\arccot\cot\frac{2\pi}{5}=\frac{4\pi}{5}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2332090",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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A simple trigonometric inequality How to sole this inequalities:
$$2\cos\left(\frac{x}{2}\right) \leq -\cos(x).$$
I use the equality $$\cos\left(\frac{x}{2}\right)=\pm\sqrt{\frac{1+\cos x}{2}}$$
to find the solution for the equality
$$2\cos\left(\frac{x}{2}\right) = -\cos(x).$$
I get a second degree equation $t^2-2t-2=0$ where $t=\cos x$. I think that the solution of the equality are: $$x=\arccos(1-\sqrt3)+2k\pi \\ x=2\pi -\arccos(1-\sqrt3)+2k\pi.$$
How can I find the final solution of the inequality?
| \begin{align}
2\cos\frac{x}{2}&\le-\cos x\\
\cos x+2\cos\frac{x}{2}&\le0\\
2\cos^2\frac{x}{2}-1+2\cos\frac{x}{2}&\le0\\
\left(2\cos\frac{x}{2}+1\right)^2&\le3\\
\frac{-1-\sqrt{3}}{2}\le\cos\frac{x}{2}&\le\frac{-1+\sqrt{3}}{2}
\end{align}
So, $\displaystyle x\in\bigcup_{n\in\mathbb{Z}}\left[4n\pi+2\cos^{-1}\left(\frac{-1+\sqrt{3}}{2}\right),4(n+1)\pi-2\cos^{-1}\left(\frac{-1+\sqrt{3}}{2}\right)\right]$.
Note 1:
Consider the equation $\displaystyle 2\cos\frac{x}{2}=-\cos x$. If we prefer to write it as an equation in $\cos x$, we have to square both sides.
\begin{align}
4\cos^2\frac{x}{2}&=\cos^2x\\
2\cos x+2&=\cos^2x\\
\cos^2x-2\cos x-2&=0
\end{align}
The roots are $2n\pi\pm\arccos(1-\sqrt{3})$.
But as squaring the equation $\displaystyle 2\cos\frac{x}{2}=\cos x$ will also yield $\displaystyle 4\cos^2\frac{x}{2}=\cos^2x$, The general solution $x=2n\pi\pm\arccos(1-\sqrt{3})$ includes solutions to both the equations $\displaystyle 2\cos\frac{x}{2}=-\cos x$ and $\displaystyle 2\cos\frac{x}{2}=\cos x$.
Note 2:
For the inequality, as $\displaystyle 2\cos\frac{x}{2}+\cos x$ has a period $4\pi$, we can consider those $x\in[0,4\pi]$ only.
(a) When $x\in[0,\frac{\pi}{2}]$, $\cos\frac{x}{2}>0$ and $-\cos x\le0$. The inequality has no solution.
(b) When $x\in[\frac{\pi}{2},\pi]$, $\cos\frac{x}{2}\ge0$ and $-\cos x\ge0$. So, $2\cos\frac{x}{2}\le-\cos x$ can be written as
\begin{align}
4\cos^2\frac{x}{2}&\le\cos^2x\\
2\cos x+2&\le\cos^2x\\
\cos^2x-2\cos x-2&\ge0\\
\cos x&\le 1-\sqrt{3}\\
\arccos(1-\sqrt{3})&\le x\le \pi
\end{align}
(c) When $x\in[\pi,\frac{3\pi}{2}]$, $\cos\frac{x}{2}\le0$ and $-\cos x\ge0$. The inequality always holds.
(d) When $x\in[\frac{3\pi}{2},\frac{5\pi}{2}]$, $\cos\frac{x}{2}\le0$ and $-\cos x\le0$. The inequality always holds. So, $2\cos\frac{x}{2}\le-\cos x$ can be written as
\begin{align}
-2\cos\frac{x}{2}&\ge\cos x\\
4\cos^2\frac{x}{2}&\ge\cos^2x\\
2\cos x+2&\ge\cos^2x\\
\cos^2x-2\cos x-2&\le0\\
1-\sqrt{3}\le\cos x&\le 1+\sqrt{3}\\
\end{align}
which is always true if $x\in[\frac{3\pi}{2},2\pi]$.
(f) When $x\in[\frac{5\pi}{2},3\pi]$, $\cos\frac{x}{2}\le0$ and $-\cos x\ge0$. The inequality always holds.
(g) When $x\in[3\pi,\frac{7\pi}{2}]$, $\cos\frac{x}{2}\ge0$ and $-\cos x\ge0$. So, $2\cos\frac{x}{2}\le-\cos x$ can be written as
\begin{align}
4\cos^2\frac{x}{2}&\le\cos^2x\\
2\cos x+2&\le\cos^2x\\
\cos^2x-2\cos x-2&\ge0\\
\cos x&\le 1-\sqrt{3}\\
3\pi\le x &\le 4\pi-\arccos(1-\sqrt{3})
\end{align}
(h) When $x\in[\frac{7\pi}{2},4\pi]$, $\cos\frac{x}{2}>0$ and $-\cos x\le0$. The inequality has no solution.
Combining the results in all cases, $\displaystyle \arccos(1-\sqrt{3})\le x\le 4\pi-\arccos(1-\sqrt{3})$.
The general solution is $\displaystyle x\in\bigcup_{n\in\mathbb{Z}}\left[4n\pi+\arccos(1-\sqrt{3}),4(n+1)\pi-\arccos(1-\sqrt{3})\right]$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Arithmetic-Geometric mean sequences limit If $a,b$ are positive quantities such that $(a<b)$ and if
\begin{align}
a_1 &= \frac{a+b}2 & b_1 & = \sqrt{a_1 b} \\
a_2 &= \frac{a_1+b_1}2 & b_2 &= \sqrt{a_2 b_1} \\
&\phantom{36pt} \vdots & & \\
a_n &= \frac{a_{n-1}+b_{n-1}}2 & b_n & = \sqrt{a_n b_{n-1}} \\
&\phantom{36pt} \vdots & &
\end{align} then show that $\displaystyle\lim_\limits{{n\to \infty}} b_n=\frac{\sqrt{b^2-a^2}}{\arccos(\frac{a}{b})}$
| If we assume that $b=1$ and $a=\cos\theta$ with $\theta\in\left(0,\frac{\pi}{2}\right)$ we get:
$$ a_1 = \frac{1+\cos\theta}{2} = \cos^2\frac{\theta}{2},\qquad b_1=\cos\frac{\theta}{2}$$
$$ a_2 = \cos\frac{\theta}{2}\left(\frac{1+\cos\frac{\theta}{2}}{2}\right),\qquad b_2=\cos\frac{\theta}{2}\cos\frac{\theta}{4}$$
so by induction it follows that
$$ b_n = \prod_{k=1}^{n}\cos\frac{\theta}{2^k} = \frac{\sin\theta}{2^n\sin\frac{\theta}{2^n}} $$
and $\lim_{n\to +\infty}b_n = \frac{\sin\theta}{\theta}. $ The claim easily follows by rescaling, since $\theta=\arccos\frac{a}{b}$.
| {
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Somebody help me please. I have a difficult inequality. Let $ab+bc+ca=1$. Prove that $2 \ge \sqrt{1+a^2} + \sqrt{1+b^2}+\sqrt{1+c^2}-a-b-c \geq \sqrt3 $.
| The right inequality.
We need to prove that
$$ \sum\limits_{\text{cyc}}\sqrt {\left(a+c\right)\left(a+b\right)}\geq a+b+c + \sqrt3\cdot\sqrt {ab+ac+bc}$$ or
$$2\left(a+b+c - \sqrt {3(ab+ac+bc)}\right) - \sum_{cyc}\left(\sqrt {a+c} - \sqrt {b+c}\right)^{2}\geq0.$$
But $$ 2\left(a+b+c- \sqrt {3(ab+ac+bc)}\right) - \sum_{cyc}\left(\sqrt {a+c} - \sqrt {b+c}\right)^{2} =$$
$$ = \sum_{cyc}(a+b)^2\left(\frac {1}{a+b+c+ \sqrt {3(ab+ac+bc)}}- \frac {1}{\left(\sqrt {a+c} + \sqrt {b+c}\right)^2}\right)=$$
$$ = \sum_{cyc}\frac {(a-b)^2\left(c + 2\sqrt {(a+c)(b+c)} - \sqrt {3(ab+ac+bc)}\right)}{\left(a+b+c+ \sqrt {3(ab+ac+bc)}\right)\left(\sqrt {a+c} + \sqrt {b+c}\right)^2}=$$
$$ = \sum_{cyc}\frac {(a-b)^2\left(c + \frac {4c^{2} + ab+ac+bc}{2\sqrt {(a+c)(b+c )} + \sqrt {3(ab+ac+bc)}}\right)}{\left(a+b+c + \sqrt {3(ab+ac+bc)}\right)\left(\sqrt {a+c} + \sqrt {b+c}\right)^2}\geq0.$$
The left inequality.
Let $a=\tan\frac{\alpha}{2}$, $b=\tan\frac{\beta}{2}$ and $c=\tan\frac{\gamma}{2}.$
Thus, $\alpha+\beta+\gamma=180^{\circ}$.
Let $\alpha\geq\beta\geq\gamma$.
Thus, $\beta$ and $\gamma$ are acute angles and we need to prove that $$2+\sum\limits_{cyc}f(\alpha)\geq0,$$ where $$f(x)=\tan\frac{x}{2}-\frac{1}{\cos\frac{x}{2}}.$$
But $$f''(x)=\frac{\tan\frac{x}{4}-1}{4\cos^2\frac{x}{4}\left(\tan\frac{x}{4}+1\right)^3}<0$$
for $0<x<\frac{\pi}{2}$, which says that $f$ is a concave function on $\left(0,\frac{\pi}{2}\right)$.
Thus, since $\left(\beta+\gamma,0^{\circ}\right)\succ(\beta,\gamma)$, by Karamata we obtain:
$$2+\sum_{cyc}f(\alpha)>2+f(\alpha)+f(\beta+\gamma)+f\left(0^{\circ}\right)=$$
$$=2+f(\alpha)+f(180^{\circ}-\alpha)+f\left(0^{\circ}\right)=1+\tan\frac{\alpha}{2}-\frac{1}{\cos\frac{\alpha}{2}}+\cot\frac{\alpha}{2}-\frac{1}{\sin\frac{\alpha}{2}}=$$
$$=1+\frac{1}{\sin\frac{\alpha}{2}\cos\frac{\alpha}{2}}-\frac{1}{\cos\frac{\alpha}{2}}-\frac{1}{\sin\frac{\alpha}{2}}=\frac{\left(1-\sin\frac{\alpha}{2}\right)\left(1-\cos\frac{\alpha}{2}\right)}{\sin\frac{\alpha}{2}\cos\alpha\frac{\alpha}{2}}>0.$$
Done!
| {
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"question_score": "2",
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Directional derivatives of a function with partial derivatives The prompt is to 1. find the directional derivatives of the function $f$ at a point $P(1, -1,2)$ in the direction towards origin and we are given $$f(1, -1, 2) = 5$$ $$f_x(1, -1, 2) = 1$$ $$f_y(1, -1, 2) = 3$$ $$f_z(1, -1, 2) = -1$$
*Find the plane tangent to the surface $f(x, y, z) = 5$ at point $P = (1, -1, 2)$
*Suppose we are told that there is a direction $u$ in which the directional derivative of $f$ is $6$, is this possible? If yes, find such a direction $u$, else explain why not.
*The equation $f(x, y, z) = 5$ determines $z(x, y)$, evaluate $z_x(1, -1, 2)$
Here's what I tried doing,
For part 1,
$$\nabla f = (1, 3, -1)$$
$$\vec v = (0, 0, 0) - (x, y, z) = (-x, -y, -z)$$
Unit vector along $\vec v$
$$\vec v = \frac{(-x, -y, -z)}{\sqrt{x^2 + y^2 + z^2}}$$
Directional vector at point $P(1, -1, 2)$
$$\nabla f.\vec v = (1, 3, -1).\frac{(-x, -y, -z)}{\sqrt{x^2 + y^2 + z^2}}$$
$$\nabla f.\vec v = \frac{(-x)}{\sqrt{x^2 + y^2 + z^2}} +\frac{(-3y)}{\sqrt{x^2 + y^2 + z^2}} + \frac{(z)}{\sqrt{x^2 + y^2 + z^2}}$$
For part 2,
The plane tangent is given by
$$\nabla f_{(1, -1, 2)} = (x - 1) + 3(y + 1) -(z - 2) = 0$$
$$x +3y-z= -4$$
I'm not sure how to go on about part 3 and 4 also I'm not sure if part 1 and 2 are correct.
| The unit vector at $P(1, -1,2)$ in the direction of the origin is $\mathbf{u}=\left(-\frac{1}{\sqrt{6}},\frac{1}{\sqrt{6}},-\frac{2}{\sqrt{6}}\right)$ and the gradient is $\nabla f = (1, 3, -1)$ therefore the directional derivative at $P$ in the direction $\mathbf{u}$ is
$$ D_{\mathbf{u}}f(1,-1,2)=\nabla f\cdot\mathbf{u}=(1, 3, -1)\cdot\left(-\frac{1}{\sqrt{6}},\frac{1}{\sqrt{6}},-\frac{2}{\sqrt{6}}\right)=\frac{4}{\sqrt{6}}=\frac{2\sqrt{3}}{3} $$
Your equation for the tangent plane is correct.
The largest possible value for the directional derivative of $f$ at $P$ is in the direction of the gradient, which would be
$$ \mathbf{v}=\left(\dfrac{1}{\sqrt{11}},\dfrac{3}{\sqrt{11}},-\dfrac{1}{\sqrt{11}}\right) $$
giving
$$ D_{v}=\nabla f\cdot\mathbf{v}=\sqrt{11}<6 $$
so this is not possible.
| {
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"question_score": "2",
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sum of integer parts of roots
If $\alpha,\beta,\gamma$ are real roots of the equation $\displaystyle \frac{1}{x}+\frac{1}{x-1}+\frac{1}{x-2} = \frac{11}{54}x^2.$ Then $\lfloor \alpha \rfloor + \lfloor \beta \rfloor +\lfloor \gamma \rfloor $
$\bf{Attempt}$ From equation $$\frac{x^2-3x+2+x^2-2x+x^2-x}{x(x-1)(x-2)} = \frac{11}{54}x^2$$
So $$\frac{3x^2-6x+2}{x^3-3x^2+2x} = \frac{11}{54}x^2$$
$$162x^2-324x+108 = 11x^5-33x^4+22x^3$$
$$11x^5-33x^4+22x^3-162x^2+324x-108 =0$$
Could some help me how to solve it, thanks
| If $x\ge 3$, then the LHS is $\le \frac13+\frac12+1=\frac{11}{6}$ and the RHS is $\ge \frac{11}6$ with equality iff $x=3$. Thus we have found our first real root and shown that it is the largest: $x=3$.
If $x<0$, the LHS is negative, the RHS positive, hence we cannot find a real solution there.
If $2<x<3$, then the LHS is strictly decreasing from $+\infty$ to $\frac{11}6$ and the RHS strictly increasing from $\frac{22}{27}$ to $\frac{11}6$. We conclude that there is no real root with floor $2$.
Similarly, in the interval $(1,2)$ and in the interval $(0,1)$, the LHS goes from $+\infty$ to $-\infty$, whereas the RHS remains bounded. We conclude that in each of these intervals, there is at least one real root.
So far, we have found/shown the existence of one real root with $\lfloor x\rfloor =3$, one with $\lfloor x\rfloor =1$, one with $\lfloor x\rfloor =0$.
If we trust the problem statement that a definite answer can be given at all, we thus see that the answer must be $$4.$$
But for a more satisfying nswer, we have to verify that there are no additional roots in $(0,1)$ and/or $(1,2)$ that would allow us to achieve a different sum with different picks of real roots.
One way you might try for this, is to show that in each of these intervals, the LHS is strictly decreasing.
| {
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Let $a,b,c$ be roots of $x^3+px+r=0$. Find the cubic whose roots are $(a-b)^2$,$ (b-c)^2$ and $(c-a)^2$ Question
Let $a,b,c$ be roots of $x^3+px+r=0$. Then find the cubic whose roots are $(a-b)^2, (b-c)^2$ and $ (c-a)^2$
Attempt
I have tried using Vieta's formulas to compute coefficients of the sought cubic.
For sum of roots we have $$\sum_{cyc}(a-b)^2 = 2\left(\sum_{cyc} a^2-\sum_{cyc}{ab}\right)\\
= 2\left(\sum_{cyc} a\right)^2-6\left(\sum_{cyc} ab\right)\\
= -6p$$
This is coefficient of $x^2$ in the sought cubic.
But now computing coefficient of $x^2$ requires simplifying factors like $(a-b)^2\cdot(b-c)^2$ which becomes very lengthy.
Is there a shorter way around?
Thanks!
| Hint: assume WLOG that the three roots $a,b,c$ are all distinct (the case of multiple roots follows by continuity, or could be handled separately). Then subtracting the equations $a^3+pa+r=0\,$ and $b^3+pb+r=0\,$ gives:
$$\require{cancel}
a^3-b^3 + p(a-b) = 0 \iff \cancel{(a-b)}\big(a^2+ab+b^2+p\big) = 0 \iff (a-b)^2=-p-3ab
$$
The problem then reduces to finding the equation with roots $-p-3ab\,$, $-p-3bc\,$, $-p-3ca\,$.
Let $x=-p-y\,$, then the equation in $y$ will have roots $3ab,3bc,3ca\,$. After calculating the symmetric functions, the equation is found to be $y^3-3py^2-27r^2=0\,$. Substituting back $y=-x-p$ then gives the equation in $x$ as $(x+p)^3+3p(x+p)^2+27r^2=0$.
| {
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"source": "stackexchange",
"question_score": "3",
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BMO2 1992 - Triangle Circumscribed in Circle The circumcircle of triangle $ABC$ has radius $R$ satisfying
$$AB^2+AC^2=BC^2-R^2$$
Prove that the angles in the triangle are uniquely determined, and state the values for the angles.
So far I have used the Extended Law of Sines and the fact that $\alpha+\beta+\gamma=180$ to show that $\sin{\beta}\sin{\gamma}\cos{\alpha}=-\frac{1}{8}$ (where $\alpha$, $\beta$, $\gamma$ are the angles opposite $BC$, $CA$, $AB$), but I cannot seem to get any further.
| $b^2+c^2=a^2-R^2$ leads to
$$ 4\sin^2 B+4\sin^2 C=4\sin^2(B+C)-1 \tag{1}$$
that can be written as
$$ \left(2\cos(B+C)-\cos(B-C)\right)^2 + \sin^2(B-C) = 0.\tag{2} $$
It follows that $B=C$ and $2\cos(2B)-1=0$, from which
$$ A=\frac{2\pi}{3},\quad B=C=\frac{\pi}{6}.\tag{3} $$
| {
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What is the probability of the sum of four dice being 22? Question
Four fair six-sided dice are rolled. The probability that the sum of the results being $22$ is $$\frac{X}{1296}.$$ What is the value of $X$?
My Approach
I simplified it to the equation of the form:
$x_{1}+x_{2}+x_{3}+x_{4}=22, 1\,\,\leq x_{i} \,\,\leq 6,\,\,1\,\,\leq i \,\,\leq 4 $
Solving this equation results in:
$x_{1}+x_{2}+x_{3}+x_{4}=22$
I removed restriction of $x_{i} \geq 1$ first as follows-:
$\Rightarrow x_{1}^{'}+1+x_{2}^{'}+1+x_{3}^{'}+1+x_{4}^{'}+1=22$
$\Rightarrow x_{1}^{'}+x_{2}^{'}+x_{3}^{'}+x_{4}^{'}=18$
$\Rightarrow \binom{18+4-1}{18}=1330$
Now i removed restriction for $x_{i} \leq 6$ , by calculating the number of bad cases and then subtracting it from $1330$:
calculating bad combination i.e $x_{i} \geq 7$
$\Rightarrow x_{1}^{'}+x_{2}^{'}+x_{3}^{'}+x_{4}^{'}=18$
We can distribute $7$ to $2$ of $x_{1}^{'},x_{2}^{'},x_{3}^{'},x_{4}^{'}$ i.e$\binom{4}{2}$
We can distribute $7$ to $1$ of $x_{1}^{'},x_{2}^{'},x_{3}^{'},x_{4}^{'}$ i.e$\binom{4}{1}$ and then among all others .
i.e
$$\binom{4}{1} \binom{14}{11}$$
Therefore, the number of bad combinations equals $$\binom{4}{1} \binom{14}{11} - \binom{4}{2}$$
Therefore, the solution should be:
$$1330-\left( \binom{4}{1} \binom{14}{11} - \binom{4}{2}\right)$$
However, I am getting a negative value. What am I doing wrong?
EDIT
I am asking for my approach, because if the question is for a larger number of dice and if the sum is higher, then predicting the value of dice will not work.
| It is the coefficient of $x^{22}$ in the expansion of the generating function $$(x^6+x^5+x^4+x^3+x^2+x^1)^4=\left(\frac{x(1-x^6)}{1-x}\right)^4$$ which is $$1\cdot{{x}^{24}}+4\cdot {{x}^{23}}+10\cdot {{x}^{22}}+20\cdot {{x}^{21}}+35\cdot {{x}^{20}}+56\cdot {{x}^{19}}+80\cdot {{x}^{18}}\\+104\cdot {{x}^{17}}+125\cdot {{x}^{16}}+140\cdot {{x}^{15}}+146\cdot {{x}^{14}}+140\cdot {{x}^{13}}+125\cdot {{x}^{12}}+104\cdot {{x}^{11}}\\+80\cdot {{x}^{10}}+56\cdot {{x}^{9}}+35\cdot {{x}^{8}}+20\cdot {{x}^{7}}+10\cdot {{x}^{6}}+4\cdot {{x}^{5}}+1\cdot{{x}^{4}}$$
so the answer is $10$.
This generalises easily to more dice or even those with different numbers of faces
| {
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$\text{If } x(x+y+z)=20,y(x+y+z)=30 \text{ and } z(x+y+z)=50 \text{ then the value of } 2(x+y+z) is:$
If $x(x+y+z)=20$, $y(x+y+z)=30$ and $z(x+y+z)=50$ then what is the value of $2(x+y+z)$?
Ans. $20$
I have tried the following:
$$ \frac{20}{x}=\frac{30}{y}=\frac{50}{z}$$
From which I get:
$$x:y:z=2:3:5$$
Now,
$$2(x+y+z)=\frac{20}{x}+\frac{30}{y}$$
$$\implies\frac{20y+30x}{xy}$$
$$\implies\frac{20*\frac{3z}{5}+30*\frac{2z}{5}}{\frac{2z}{5}\frac{3z}{5}}$$
Which gives me
$$\frac{100}{z}$$
Back to the 3rd equation!
| $$x(x+y+z)=20,\\ y(x+y+z)=30\\ z(x+y+z)=50\\ $$ by summing all these three eqatuations$$\left( x+y+z \right) \left( z+y+x \right) =100\\ { \left( x+y+z \right) }^{ 2 }=100\\ x+y+z=\pm 10$$
| {
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Does $\lim_{n\to \infty} \left(\frac{a}{a+bn}\right)^{\frac{1}{n}}$ exist? Let $a, b >0$ and $a \ll b$, $a<1$. Does $\lim_{n\to \infty} \left(\frac{a}{a+bn}\right)^{\frac{1}{n}}$ exist ?
| Using
\begin{align}
\frac{a}{a + b \, n} &= 1 - \frac{1}{1 + \frac{a}{b \, n}} \\
&= 1 - \left( 1 - \frac{a}{b \, n} + \frac{a^{2}}{b^{2} \, n^{2}} + \mathcal{O}\left(\frac{1}{n^{3}}\right) \right) \\
&= \frac{a}{b \, n} - \left(\frac{a}{b \, n}\right)^{2} + \mathcal{O}\left(\left(\frac{a}{b \, n}\right)^{3}\right) \\
&= \frac{a}{b \, n} \, \left( 1 - \frac{a}{b \, n} + \mathcal{O}\left(\left(\frac{a}{b \, n}\right)^{2}\right) \right)
\end{align}
then
\begin{align}
\lim_{n \to \infty} \left( \frac{a}{a + b \, n} \right)^{\frac{1}{n}} &= \lim_{n \to \infty} \, \left(\frac{a}{b \, n}\right)^{\frac{1}{n}} \, \left( 1 - \frac{a}{b \, n} + \mathcal{O}\left(\left(\frac{a}{b \, n}\right)^{2}\right) \right)^{\frac{1}{n}}.
\end{align}
By inverting $n$, ie $n \to \frac{1}{n}$, the limit becomes
\begin{align}
\lim_{n \to \infty} \left( \frac{a}{a + b \, n} \right)^{\frac{1}{n}} &= \lim_{n \to 0} \, \left(\frac{a \, n}{b}\right)^{n} \, \left( 1 - \frac{a \, n}{b} + \mathcal{O}\left(\left(\frac{a \, n}{b}\right)^{2}\right) \right)^{n} = 0^{0} = 1.
\end{align}
| {
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solve trigonometric lim with tan Hi guys I tried to solve the following problem lately and got stucked.
I'd love to get some help and guidance.
consider this:
$$\lim_{x \rightarrow\infty}\tan\left(\frac{\pi x}{2x+1}\right)^{\frac{1}{x}}
$$
I did the following:
1. break to sin and cos :
$$
\lim_{x \rightarrow\infty}\tan\left(\frac{\pi x}{2x+1}\right)^{\frac{1}{x}} = \lim_{x\rightarrow\infty}\frac{\left(\sin\left(\frac{\pi x}{2x+1}\right)\right)^\frac{1}{x}}{\left(\cos\left(\frac{\pi x}{2x+1}\right)\right)^\frac{1}{x}}$$
*use log rules:
$$\lim_{x\rightarrow\infty}\frac{\left(\sin\left(\frac{\pi x}{2x+1}\right)\right)^\frac{1}{x}}{\left(\cos\left(\frac{\pi x}{2x+1}\right)\right)^\frac{1}{x}} = \lim_{x\rightarrow\infty}\frac{e^{\frac{1}{x}\ln\left(\sin\left(\frac{\pi x}{2x+1}\right)\right)}}{e^{\frac{1}{x}\ln\left(\cos\left(\frac{\pi x}{2x+1}\right)\right)}}$$
*Numerator converges to one so I am going to work with the denominator:
$$
\lim_{x \rightarrow \infty}e^{\frac{1}{x}\ln\left(\cos\left(\frac{\pi x}{2x+1}\right)\right)} = e^\left(\lim_{x \rightarrow \infty}{\frac{1}{x}\ln\left(\cos\left(\frac{\pi x}{2x+1}\right)\right)}\right)
$$
$$
\lim_{x \rightarrow \infty}{\frac{1}{x}\ln\left(\cos\left(\frac{\pi x}{2x+1}\right)\right)} = \frac{-\infty}{\infty}$$
*at this point I tried to apply l'hopital but encountered a huge mess, and now I feel that maybe I chose the wrong way. Please help me guys.
| Considering $$A=\tan\left(\frac{\pi x}{2x+1}\right)^{\frac{1}{x}}$$ first $$\tan\left(\frac{\pi x}{2x+1}\right)=\tan\left(\frac{\pi} 2-\frac{\pi }{2(2 x+1)}\right)=\cot \left(\frac{\pi }{2(2 x+1)}\right)$$ So $$\log(A)=\frac 1 x \log\left(\cot \left(\frac{\pi }{2(2 x+1)}\right)\right)$$ Now, using Taylor series starting with $$\cot(\epsilon)=\frac{1}{\epsilon }-\frac{\epsilon }{3}-\frac{\epsilon ^3}{45}+O\left(\epsilon
^4\right)$$ and replacing $\epsilon=\frac{\pi }{2(2 x+1)}$ and continuing the expansion $$\cot \left(\frac{\pi }{2(2 x+1)}\right)=\frac{4 x}{\pi }+\frac{2}{\pi }-\frac{\pi }{12
x}+O\left(\frac{1}{x^2}\right)$$ $$\log\left(\cot \left(\frac{\pi }{2(2 x+1)}\right)\right)=\log \left(\frac{4}{\pi }\right)+\log
\left({x}\right)+\frac{1}{2 x}+O\left(\frac{1}{x^2}\right)$$ $$\log(A)=\frac 1x\log \left(\frac{4}{\pi }\right)+\frac{\log
\left({x}\right)}x+O\left(\frac{1}{x^2}\right)$$ $$A=e^{\log(A)}=1+\frac{\log \left(\frac{4 x}{\pi }\right)}{x}+O\left(\frac{1}{x^2}\right)$$ which shows the limpit and how it is approached.
To check how good (or bad) is the approximation, let us set $x=10^k$ and compute the exact and approximate values
$$\left(
\begin{array}{ccc}
k & \text{approximation} & \text{exact} \\
1 & 1.295772742 & 1.254414957 \\
2 & 1.049713236 & 1.048467347 \\
3 & 1.007175440 & 1.007149320 \\
4 & 1.000945642 & 1.000945190 \\
5 & 1.000117552 & 1.000117545 \\
6 & 1.000014057 & 1.000014057
\end{array}
\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2355738",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
if a,b,c are roots of a cubic equation then for the following question... If $a, b, c$ are roots of $x^3 -3x^2 + 2x +4 = 0$ and
$$y= 1 + \frac{a}{x-a} + \frac{bx}{(x-a)(x-b)} + \frac{cx^2}{(x-a)(x-b)(x-c)}$$
then value of $y$ at $x=2$ is:
| We can reduce this to all be over the denominator $(x-a)(x-b)(x-c)$:
\begin{align}y&=1 + \frac{a}{x-a} + \frac{bx}{(x-a)(x-b)} + \frac{cx^2}{(x-a)(x-b)(x-c)}\\\\
&=\frac{(x-a)(x-b)(x-c)}{(x-a)(x-b)(x-c)}+\frac{a(x-b)(x-c)}{(x-a)(x-b)(x-c)}\\
&\qquad\qquad+\frac{bx(x-c)}{(x-a)(x-b)(x-c)}+\frac{cx^2}{(x-a)(x-b)(x-c)}\\\\
&=\frac{(x-a)(x-b)(x-c)+a(x−b)(x−c)+bx(x−c)+cx^2}{(x-a)(x-b)(x-c)}\\\\
&=\frac{-a b c + a b x + a c x - a x^2 + b c x - b x^2 - c x^2 + x^3 +ab c - ab x - ac x + ax^2 +bx^2-bcx+cx^2}{(x-a)(x-b)(x-c)}\\\\
&=\frac{x^3+(-a-b-c+a+b+c)x^2+(ab+ac+bc-ab-ac-bc)x+(-abc+abc)}{(x-a)(x-b)(x-c)}\\
&=\frac{x^3}{(x-a)(x-b)(x-c)}\end{align}
Now, we know that $(x-a)(x-b)(x-c)=x^3−3x^2+2x+4$, and so we have $$y=\frac{x^3}{x^3-3x^2+2x+4}$$
Therefore, when $x=2$, \begin{align}y&=\frac{2^3}{2^3-3\times 2^2+2\times 2+4}\\
&=\frac{8}{8-12+4+4}\\
&=\frac{8}{4}\\
&=2\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2355843",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
How to calculate $\sum_{r=1}^\infty\frac{8r}{4r^4+1}$?
Calculate the following sum: $$\frac{8(1)}{4(1)^4+1} + \frac{8(2)}{4(2)^4+1} +\cdots+ \frac{8(r)}{4(r)^4+1} +\cdots+ \text{up to infinity}$$
MY TRY:- I took $4$ common from the denominator. and used $a^2+b^2=(a+b)^2-2ab$. It gave me two brackets, whose subtraction was written in numerator. so I did the same thing as we do in the method of partial fraction, and started putting $1,2,3$ and so on. my answer came didn't match with the right answer.
| Hint. The sum is telescopic. Note that $4r^4+1=(2r^2-2r+1)(2r^2+2r+1)$, and
$$\frac{8r}{4r^4+1}=\frac{2}{2r(r-1)+1}-\frac{2}{2(r+1)r+1}.$$
Hence
$$\sum_{r=1}^n\frac{8r}{4r^4+1}=\sum_{r=1}^n\left(\frac{2}{2r(r-1)+1}-\frac{2}{2(r+1)r+1}\right)=\frac{2}{2\cdot 1(1-1)+1}-\frac{2}{2(n+1)n+1}.$$
Can you find the sum of the series now?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2357340",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 2
} |
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.