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Differentiation using Chain Rule Find $\frac{dy}{dx}$ if $y=7+5^{x^2+2x-1}$.
So far I have done $\frac{dy}{dx}=(5^{x^2+2x-1})'$. Now, the RHS can be found by $(e^{\ln 5\cdot (x^2+2x-1)})'=e^{\ln 5\cdot (x^2+2x-1)}(x^2+2x-1)'\ln 5=5^{x^2+2x-1}(2x+2)\ln 5$.
However, my textbook says the answer is $(2x^3+6x^2+2x-2)(5^{x^2+2x-1})$. Where did I go wrong? Thanks.
| First note that
$$\frac{d}{dx}a^{f(x)}=a^{f(x)}(\ln a)\frac{d}{dx}f(x)$$
So now we have
$$\frac{d}{dx}\left[7+5^{x^2+2x-1}\right]$$
$$=\frac{d}{dx}[7]+\frac{d}{dx}\left[5^{x^2+2x-1}\right]$$
$$=0+5^{x^2+2x-1}(\ln 5)\frac{d}{dx}\left[x^2+2x-1\right]$$
$$=5^{x^2+2x-1}(\ln 5)\left(2x+2\right)$$
$$=2\cdot 5^{x^2+2x-1}(x+1)\ln 5$$
| {
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"timestamp": "2023-03-29T00:00:00",
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$AB=AC$, $BD$ is the angle bisector of $\angle B$ , find $\angle A$ Let $ABC$ be an triangle, $AB=AC$. $BD$ is the angle bisector of $\angle B$, $BD$ intersect $AC$ at point $D$, and $AD=BC+BD$.
show that: $\measuredangle BAC=20^\circ$
Well, If $\measuredangle BAC=20^\circ$, I can prove that $AD=BC+BD$.
But, if $AD=BC+BD$, How to show $\measuredangle BAC=20^\circ$?
Thanks a lot
| By Stewart's theorem we have:
$$ BD^2 = \frac{ab}{(a+b)^2}\left((a+b)^2-c^2\right)=\frac{a^2 b}{(a+b)^2}(a+2b) $$
and by the bisector theorem we have $AD=\frac{b^2}{a+b}$, so $BD^2=(AD-a)^2$ leads to:
$$ a^2 b(a+2b)= (a^2+ab-b^2)^2 $$
so by assuming $b=1$, $a=2\sin\frac{\widehat{BAC}}{2}$ is a root of:
$$ p(x) = x^4+x^3-3x^2-2x+1 = (x+1)(x^3-3x+1). $$
In order to prove $\widehat{BAC}=20^\circ$, we just need to prove that $\sin\frac{\pi}{18}=\cos\frac{4\pi}{9}$ is a root of:
$$ q(x) = 8x^3-6x+1 = 2\cdot T_3(x)+1, $$
that is trivial by using Chebyshev polynomials of the first kind.
| {
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How can $f(x,y)= x^4+x^3y+x^2y^2+xy^3+y^4$ be factorized into a product of two polynomials? Let $x,y$ be 2 coprime integers. I assume the following polynomial:$$f(x,y)= x^4+x^3y+x^2y^2+xy^3+y^4$$ is not irreducible. So there must be at least 2 other polynomials of degree $\leq 4$ such that: $$f(x,y)=g(x,y)h(x,y)$$
How can one find $g(x,y)$ and $h(x,y)$?
| Here's an efficient route, in the spirit of the remark in André Nicolas' answer:
On one hand, the product of $f(x, y)$ with $(x - y)$ telescopes:
$$(x - y) (x^4 + x^3 y + x^2 y^2 + x y^3 + y^4) = x^5 - y^5.$$
On the other,
$$x^5 - y^5 = \prod_{k = 0}^4 (x - \zeta^k y),$$
where $\zeta := e^{2 \pi i / 5},$ and so over $\Bbb C$ (in fact, over $\Bbb Q[\zeta]$), $f$ factors as
$$f(x, y) = \prod_{k = 1}^4 (x - \zeta^k y).$$
(Note that the lower limit in this product is not the same as in the earlier one.)
The irreducible factors of $f$ over $\Bbb R$ are thus the (quadratic) products of the conjugate factors, namely,
\begin{align}(x - \zeta^2 y) (x - \bar{\zeta}^2 y) = (x - \zeta^2 y) (x - \zeta^3 y) = x^2 - 2 (\Re (\zeta^2)) xy + y^2 &= x^2 + \phi xy + y^2 \\ (x - \zeta y) (x - \bar{\zeta} y) = (x - \zeta y) (x - \zeta^4 y) = x^2 - 2 (\Re \zeta) xy + y^2 &= x^2 + \tilde{\phi} xy + y^2 . \end{align}
Here, $\phi := \frac{1}{2}(1 + \sqrt{5})$ is the Golden Ratio and $\tilde{\phi} := \frac{1}{2}(1 - \sqrt{5})$ is its conjugate (in $\Bbb Q[\sqrt{5}]$). In particular, since some of the irreducible coefficients of the factors of $f$ over $\Bbb R$ are irrational, $f$ is irreducible over $\Bbb Q$.
| {
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The value of $a$ for which $f(x)=x^3+3(a-7)x^2+3(a^2-9)x-1$ have a positive point of maximum lies in the interval $(a_1,a_2)\cup(a_3,a_4)$
The value of $a$ for which $f(x)=x^3+3(a-7)x^2+3(a^2-9)x-1$ have a
positive point of maximum lies in the interval
$(a_1,a_2)\cup(a_3,a_4)$.find the value of $a_2+11a_3+70a_4$
I differentiated the equation $f(x)=x^3+3(a-7)x^2+3(a^2-9)x-1$ and put it equal to zero, to get $f'(x)=3x^2+6(a-7)x+3(a^2-9)=0$. Now what to do to get the desired interval. What is the significance of positive point of maximum in this question.
Please help. Thanks in advance.
| HINT:
The solution to the equation
$$f'(x)=3x^2+6(a-7)x+3(a^2-9)=0$$
is
$$\frac{-6(a-7)\pm \sqrt{(6a-42)^2-4\cdot 3 \cdot 3(a^2-9)}}{2\cdot 3}$$
If there exists a point of maximum, then you have
$$(6a-42)^2-4\cdot 3 \cdot 3(a^2-9)\geq 0$$
Moreover, since this point is positive, you have
$$\sqrt{(6a-42)^2-4\cdot 3 \cdot 3(a^2-9)}\geq 6(a-7)$$
Solving those equations should give you the desired intervals.
| {
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Calculus $\int_0^{+\infty}\frac{\sin^2x}x\mathrm dx$
Calculus $$\int_0^{+\infty}\frac{\sin^2x}x\mathrm dx$$
I have just approached to improper integrals, and it may be rather complex to me.
| For each positive integer $n$,
$$\int_{n\pi}^{(n+1)\pi} \frac{\sin^2 x}{x}\, dx = \int_0^\pi \frac{\sin^2(x + n\pi)}{x + n\pi}\, dx = \int_0^\pi \frac{\sin^2 x}{x + n\pi}\, dx \ge \frac{C}{n},$$
where $C = \int_0^\pi \frac{\sin^2 x}{x + \pi}\, dx$. Given $T > 0$, let $N$ be a positive integer such that $\sum\limits_{n = 1}^N \frac{C}{n} > T$ (such an $N$ can be chosen since the series $\sum\limits_{n = 1}^\infty \frac{C}{n}$ diverges). Then
$$\int_0^{(N+1)\pi} \frac{\sin^2 x}{x}\, dx \ge \int_{\pi}^{(N+1)\pi} \frac{\sin^2 x}{x}\, dx = \sum_{n = 1}^N \int_{n\pi}^{(n+1)\pi} \frac{\sin^2 x}{x}\, dx \ge \sum_{n = 1}^N \frac{C}{n} > T.$$
Since $T$ was arbitrary, the improper integral $\int_0^\infty \frac{\sin^2 x}{x}\, dx$ diverges.
Note: Using the bound $\frac{1}{x + \pi} \ge \frac{1}{2\pi}$ on $[0,\pi]$, we have $\int_0^\pi \frac{\sin^2 x}{x + \pi}\, dx \ge \frac{1}{2\pi} \int_0^\pi \sin^2 x\, dx = \frac{1}{4}$. So we may take $C = \frac{1}{4}$ if we wish.
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluation of $\int\frac{2a\sin x+b\sin 2x}{(b+a\cos x)^3}dx$ Evaluation of $\displaystyle\int\frac{2a\sin x+b\sin 2x}{(b+a\cos x)^3}dx$
$\bf{My\; Try::}$Let $$\displaystyle I = \int\frac{2a\sin x+b\sin 2x}{(b+a\cos x)^3}dx = \int\left(\frac{a+b\cos x}{b+a\cos x}\right)\cdot \frac{2\sin x}{(b+a\cos x)^2}dx$$
Now Let $$\displaystyle \left(\frac{a+b\cos x}{b+a\cos x}\right) = t\;,$$ Then $$\displaystyle \frac{\left[a^2-b^2\right]\sin x}{(b+a\cos x)^2}dx = dt\Rightarrow \frac{\sin x}{(b+a\cos x)^2}dx = \frac{1}{(a^2-b^2)}dt$$
So Integral $$\displaystyle I = \frac{2}{(a^2-b^2)}\int tdt = \frac{1}{(a^2-b^2)}\cdot \left(\frac{a+b\cos x}{b+a\cos x}\right)^2+\mathcal{C}$$
My Question is can we solve it any other Substution, If yes then plz explain here
Thanks
| $\int\limits_{0}^{\pi/2}\frac{1+2\cos x}{(2+\cos x)^2}dx$
Using Same process..
Let $$\displaystyle I = \int\frac{2a\sin x+b\sin 2x}{(b+a\cos x)^3}dx\;,$$ Now Divide both $\bf{N_{r}}$ and $\bf{D_{r}}$ by $\sin^3 x\;,$ We get
$$\displaystyle I = \int\frac{2a\csc^2 x+2b\cot x\cdot \csc x}{(b\csc x+a\cot x)^3}dx$$
Now Let $(b\csc x+a\cot x) =t\;,$ Then $\left(b\csc x\cdot \cot^2 x+a\csc^2 x\right)dx = -dt$
So Integral $$\displaystyle I = -2\int\frac{1}{t^3}dt = \frac{1}{t^2}+\mathcal{C} = \frac{1}{(b\csc x+a\cot x)^2}+\mathcal{C}$$
| {
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cos(4v) + cos(v) = 0 I am given the following equation:
$$\cos 4v + \cos v = 0$$
My attempt:
$$\cos4v = -\cos v$$
$$\cos4v = \cos(\pi \pm v)$$
$$4v = \pm \pi \pm v + 2\pi n$$
$$4v_1 = \pi + v_1 + 2\pi n$$
$$4v_2 = -\pi - v_2 + 2\pi n$$
$$4v_3 = +\pi - v_3 + 2\pi n$$
$$4v_4 = -\pi + v_4 + 2\pi n$$
$$v_1 = \frac{\pi}{3} + \frac{2\pi n}{3}$$
$$v_2 = -\frac{\pi}{5} + \frac{2\pi n}{5}$$
$$v_3 = \frac{\pi}{5} + \frac{2\pi n}{5}$$
$$v_4 = -\frac{\pi}{3} + \frac{2\pi n}{3}$$
However, the answer is simply the positive solutions i.e:
$$v_1 = \frac{\pi}{3} + \frac{2\pi n}{3}$$
$$v_3 = \frac{\pi}{5} + \frac{2\pi n}{5}$$
Why? What am I missing?
| HINT: use that $$\cos(x)-\cos(y)=-2\sin\left(\frac{x-y}{2}\right)\sin\left(\frac{x+y}{2}\right)$$
sorry i misreaded the post use that
$$\cos(x)+\cos(y)=2 \cos \left(\frac{x}{2}-\frac{y}{2}\right) \cos
\left(\frac{x}{2}+\frac{y}{2}\right)$$
| {
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How to calculate $\lim_{x\to0}\frac{1}{x}\left(\sqrt[3]{\frac{1-\sqrt{1-x}}{\sqrt{1+x}-1}}-1\right)$ I've been studying limits on Rudin, Principles of Mathematical Analysis for a while, but the author doesn't exactly explain how to calculate limits...so, can you give me a hint on how to solve this? $$\lim_{x\to0}\frac{1}{x}\left(\sqrt[3]{\frac{1-\sqrt{1-x}}{\sqrt{1+x}-1}}-1\right)$$
| Here's a more elementary approach. Since
\begin{eqnarray}
\frac{1-\sqrt{1-x}}{\sqrt{1+x}-1}&=&
\frac{(1-\sqrt{1-x})\color{blue}{(1+\sqrt{1-x})}\color{green}{(\sqrt{1+x}+1)}}{(\sqrt{1+x}-1)\color{green}{(\sqrt{1+x}+1)}\color{blue}{(1+\sqrt{1-x})}}=\frac{(1-1+x)(\sqrt{1+x}+1)}{(1+x-1)(1+\sqrt{1-x})}\\
&=&\frac{x(\sqrt{1+x}+1)}{x(1+\sqrt{1-x})}=\frac{\sqrt{1+x}+1}{1+\sqrt{1-x}},
\end{eqnarray}
using the identity
$$
\sqrt[3]{a}-1=\frac{a-1}{\sqrt[3]{a^2}+\sqrt[3]{a}+1},
$$
with
$$
a=a(x)=\frac{\sqrt{1+x}+1}{1+\sqrt{1-x}}
$$
we have:
\begin{eqnarray}
\frac{1}{x}\left(\sqrt[3]{\frac{1-\sqrt{1-x}}{\sqrt{1+x}-1}}-1\right)&=&\frac{1}{x}\left(\sqrt[3]{\frac{\sqrt{1+x}+1}{1+\sqrt{1-x}}}-1\right)=\frac{a-1}{x(1+\sqrt[3]{a}+\sqrt[3]{a^2})}\\
&=&\frac{\sqrt{1+x}-\sqrt{1-x}}{x(1+\sqrt{1-x})(1+\sqrt[3]{a}+\sqrt[3]{a^2})}\\
&=&\frac{(\sqrt{1+x}-\sqrt{1-x})\color{blue}{(\sqrt{1+x}+\sqrt{1-x})}}{x(1+\sqrt{1-x})\color{blue}{(\sqrt{1+x}+\sqrt{1-x})}(1+\sqrt[3]{a}+\sqrt[3]{a^2})}\\
&=&\frac{1+x-(1-x)}{x(1+\sqrt{1-x})\color{blue}{(\sqrt{1+x}+\sqrt{1-x})}(1+\sqrt[3]{a}+\sqrt[3]{a^2})}\\
&=&\frac{2x}{x(1+\sqrt{1-x})\color{blue}{(\sqrt{1+x}+\sqrt{1-x})}(1+\sqrt[3]{a}+\sqrt[3]{a^2})}\\
&=&\frac{2}{(1+\sqrt{1-x})\color{blue}{(\sqrt{1+x}+\sqrt{1-x})}(1+\sqrt[3]{a}+\sqrt[3]{a^2})}.
\end{eqnarray}
Taking the limit and using the fact that
$$
\lim_{x\to 0}a(x)=1,
$$
we get:
$$
\lim_{x\to0}\frac{1}{x}\left(\sqrt[3]{\frac{1-\sqrt{1-x}}{\sqrt{1+x}-1}}-1\right)=\frac{2}{2\cdot2\cdot3}=\frac16.
$$
| {
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Integrating $\frac{x^3}{(81-x^2)^2}$ I've been trying to figure out this integral for an hour or so now, but keep failing. I can't figure out where I go wrong:
$$I = \int \frac{x^3}{(81-x^2)^2} dx$$
Let $x = 9sin\theta \implies dx = 9 \cos \theta d \theta$
$$I =\int \frac{9^3 \sin^3 \theta}{9^4(1-\sin^2 \theta)^2} 9 \cos \theta d\theta$$
$$I = \int \tan^3 \theta d\theta$$
$$I = \int \tan \theta \sec^2 \theta d \theta - \int \tan \theta d\theta$$
$$I = \frac{1}{2}tan^2 \theta + \ln| \cos \theta | + C$$
Using $cos \theta = \frac{\sqrt{81-x^2}}{9}$, $\tan \theta = \frac{x}{\sqrt{81-x^2}}$
$$I = \frac{1}{2}(\frac{x^2}{81-x^2}) + \frac{1}{2} \ln |\frac{81-x^2}{81}|+C$$
Somewhere I went terribly wrong, but I just can't figure out where. I've done this problem around 5 times now.
| Alternatively write $81-x^2=y$
$$\dfrac{x^3}{(81-x^2)^2}=\dfrac{81-(81-x^2)}{(81-x^2)^2}\cdot x $$
| {
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If $\gcd(a,b) = p$, prime number then, what we can tell about the gcd of $(a^2,b^3)$?
If $\gcd(a,b) = p$, prime number, then what we can tell about $\gcd(a^2,b^3)$?
No idea how to solve.
| Raise the equation
$$
\frac apx+\frac bpy=1
$$
to the fourth power
$$
\frac{a^4}{p^4}x^4+4\frac{a^3}{p^3}x^3\frac bpy+6\frac{a^2}{p^2}x^2\frac{b^2}{p^2}y^2+4\frac apx\frac{b^3}{p^3}y^3+\frac{b^4}{p^4}y^4=1
$$
which is
$$
a^2p\left(\frac{a^2}{p^2}x^4+4\frac{a}{p}x^3\frac bpy+6x^2\frac{b^2}{p^2}y^2\right)+b^3\left(4\frac apxy^3+\frac{b}{p}y^4\right)=p^3
$$
Thus, $(a^2,b^3)\mid p^3$. Since $p\mid a$ and $p\mid b$, we have that $p^2\mid(a^2,b^3)$. Therefore,
$$
(a^2,b^3)\in\left\{p^2,p^3\right\}
$$
| {
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Find the sum of the n terms of the series $2\cdot2^0+3\cdot2^1+4\cdot2^2+\dots$ Find the sum of the n terms of the series:
$2\cdot2^0+3\cdot2^1+4\cdot2^2+\dots$
I don't know how to proceed. Please explain the process and comment on technique to solve questions of similar type.
Source: Barnard and Child Higher Algebra.
Thanks in Advance!
| The formula for the sum of a geometric sequence gives
$$
\sum_{k=0}^n2^kx^{k+2}=\frac{x^2-2^{n+1}x^{n+3}}{1-2x}\tag{1}
$$
Differentiating $(1)$ yields
$$
\sum_{k=0}^n(k+2)2^kx^{k+1}=\frac{2x-(n+3)2^{n+1}x^{n+2}}{1-2x}+\frac{2x^2-2^{n+2}x^{n+3}}{(1-2x)^2}\tag{2}
$$
Plugging in $x=1$ leads to
$$
\sum_{k=0}^n(k+2)2^k=(n+1)2^{n+1}\tag{3}
$$
| {
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$24$ divides $a^2+23$
Prove that $24$ divides $a^2 + 23$ if $a$ is not divisible by $2$ or $3$.
Well $a^2=8k+1$
for some $k$.
So, $$a^2+23 = (8k+1)+23=8(k+3)$$
So, $8$ divides $a^2+23$.
Now can $3$ divide $a^2+ 23$ so that in the end $24$ divides $a^2+23$?
| $a^2+23=(a-1)(a+1)+24$
Since $a$ is odd; $a-1$ and $a+1$ are even.
Moreover, since $4$ does not divide $a$, 4 divides either $a-1$ or $a+1$.
Since $3$ does not divide $a$, $3$ divides either $a-1$ or $a+1$.
Hence $2\times 4\times 3$ divides $(a-1)(a+1)$.
| {
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What is the appropriate method to find the value of $1$ - $1\over 7$ + $1\over 13$ - ... upto infinite terms? What is the appropriate method to find the value of $1$ - $1\over 7$ + $1\over 13$ - ... upto infinite terms?
(The denominators increase by 6 in consecutive terms)
I approximated it by integrating $\frac{1}{1+x^6}$ putting x=1...is there a better and nicer method ? :)But what should I take as limits of the integration?
| For first, we have:
$$\begin{eqnarray*}\sum_{n\geq 0}\frac{(-1)^n}{6n+1}&=&\sum_{n\geq 0}(-1)^n\int_{0}^{1}x^{6n}\,dx=\int_{0}^{1}\frac{dx}{1+x^6}\tag{1}\end{eqnarray*}$$
Now we may compute the last integral through partial fraction decomposition.
If $\xi_i$, $1\leq i\leq 6$, is a root of $1+x^6$, we have:
$$\text{Res}\left(\frac{1}{x^6+1},x=\xi_i\right)=\frac{1}{6\xi_i^5}=-\frac{\xi_i}{6}\tag{2}$$
hence:
$$ \int_{0}^{1}\frac{dx}{1+x^6}=-\frac{1}{6}\sum_{i=1}^{6}\int_{0}^{1}\frac{\xi_i}{x-\xi_i}\,dx=-\frac{1}{6}\sum_{i=1}^{6}\xi_i \log\left(1-\frac{1}{\xi_i}\right)\tag{3} $$
and:
$$\begin{eqnarray*} \int_{0}^{1}\frac{dx}{1+x^6}&=&-\frac{1}{6}\sum_{j=0}^{5}e^{\frac{\pi i}{6}(2j+1)}\log\left(1-e^{-\frac{\pi i}{6}(2j+1)}\right)\\&=&-\frac{1}{6}\sum_{j=0}^{5}e^{\frac{\pi i}{6}(2j+1)}\left(-\frac{\pi i}{12}(2j+1)+\log\left(2i\sin\frac{\pi(2j+1)}{12}\right)\right)\\&=&\frac{\pi}{6}-\frac{1}{6}\sum_{j=0}^{5}e^{\frac{\pi i}{6}(2j+1)}\left(\frac{\pi}{2}i+\log\sin\frac{\pi(2j+1)}{12}\right)\\&=&\color{red}{\frac{\pi+\sqrt{3}\log(2+\sqrt{3})}{6}}.\tag{4}\end{eqnarray*}$$
Another possible approach is the following: we have
$$ \sum_{n\geq 0}\frac{(-1)^n}{6n+1}=\sum_{n\geq 0}\left(\frac{1}{12n+1}-\frac{1}{12n+7}\right)=\frac{\psi\left(\frac{7}{12}\right)-\psi\left(\frac{1}{12}\right)}{12}\tag{5}$$
then the result follows from combining the reflection formula:
$$ \psi(z)-\psi(1-z)=-\pi\cot(\pi z)\tag{6}$$
with the duplication formula:
$$ \psi(z)+\psi\left(z+\frac{1}{2}\right)=-2\log 2+2\,\psi(2z)\tag{7}$$
and the triplication formula:
$$ 3\,\psi(3z)=(3\log 3)z+\psi(z)+\psi\left(z+\frac{1}{3}\right)+\psi\left(z+\frac{2}{3}\right)\tag{8}$$
for the digamma function.
| {
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Evaluate $\sum_{n=1}^\infty \frac{1}{n(n+2)(n+4)}$? How can I evaluate this?
$$\sum_{n=1}^\infty \frac{1}{n(n+2)(n+4)} = \frac{1}{1\cdot3\cdot5}+\frac{1}{2\cdot4\cdot6}+\frac{1}{3\cdot5\cdot7}+ \frac{1}{4\cdot6\cdot8}+\cdots$$
I have tried:
$$\frac{1}{1\cdot3\cdot5}+\frac{1}{3\cdot5\cdot7}+\frac{1}{2\cdot4\cdot6}+ \frac{1}{4\cdot6\cdot8}+\cdots
= \frac{1}{3\cdot5}\left(1+\frac{1}{7}\right)+\frac{1}{4\cdot6}\left(\frac{1}{2}+\frac{1}{8}\right)+\cdots$$
and so on... Been stuck for a while. Result should be $\dfrac{11}{96}$
| HINT:
$$\displaystyle\dfrac4{n(n+2)(n+4)}=\dfrac{n+4-n}{n(n+2)(n+4)}=\dfrac1{n(n+2)}-\dfrac1{(n+2)(n+4)}$$
$$=g(n)-g(n+2)$$
where $g(m)=\dfrac1{m(m+2)}$
See Telescoping series
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1433583",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 3
} |
Find all pairs of prime numbers $p , q$ for which: $p^2 | q^3 + 1$ and $q^2 | p^6 − 1$.
Find all pairs of prime numbers $p , q$ for which:
$$p^2 \mid q^3 + 1 \tag{A}$$
and
$$q^2 \mid p^6 − 1 \tag{B}$$
The question is from the Bulgaria National Olympiad 2014.
I'm looking for any solution I may have missed, and generally any alternative method that might reduce the case work (could combine cases 2.1 and 3.1, I suppose).
I will split the work into three cases:
*
*$p=q\ge2$.
*$p>q\ge2$.
*$q>p\ge2$.
Case 1
It is clear that neither (A) nor (B) are met.
Case 2
First consider two subcases:
*
*$p>q$ and $q\in\{2,3\}$.
*$p>q\ge5$.
Case 2.1
$$\begin{align}
q=2 &\implies p^2\mid9 &\implies p^2=9 &\implies p=3 \\
q=3 &\implies p^2\mid28 &\implies p^2=4 &\implies\text{ no solution} \\
\end{align}$$
So $\boxed{(p,q)=(3,2)}$ is the only solution for this case.
Case 2.2
(A) factorises as $p^2 \mid (q+1)(q^2-q+1)$. Now
$$q^2-q+1=(q+1)(q-2)+3 \implies \gcd(q+1,q^2-q+1)=
\begin{cases} 3,\quad\text{if }3\mid q+1\\1,\quad\text{otherwise}\end{cases}$$
Since $p>5$ is prime, we must have either $p^2\mid q+1$ or $p^2\mid q^2-q+1$. But this is impossible because $p>q\ge5 \implies p^2>q+1\text{ and }p^2>q^2>q^2-q+1$. So there are no solutions here.
Case 3
Consider two subcases:
*
*$q>p$ and $p\in\{2,3\}$.
*$q>p\ge5$.
Case 3.1
$$\begin{align}
p=2 &\implies q^2\mid63 &\implies q^2=9 &\implies q=3 \\
p=3 &\implies q^2\mid728 &\implies q^2=4 &\implies\text{ no solution} \\
\end{align}$$
So $\boxed{(p,q)=(2,3)}$ is the only solution for this case.
Case 3.2
(B) factorises as $q^2 \mid (p^3+1)(p^3-1)$. Now
$$p^3+1=(p^3-1)+2 \implies \gcd(p^3+1,p^3-1)=
\begin{cases} 2,\quad\text{if }p\text{ is odd}\\1,\quad\text{otherwise}\end{cases}$$
Since $q>5$ is prime, we must have either $q^2\mid p^3+1$ or $q^2\mid p^3-1$. If $q^2\mid p^3+1$, the the same arguments as in case 2.2 can be applied to show that $q^2\mid p+1$ or $q^2\mid p^2-p+1$ neither of which is possible when $q>p$.
So the only remaining possibility is $q^2\mid p^3-1$. This factorises as $q^2\mid (p-1)(p^2+p+1)$. Now
$$p^2+p+1=(p-1)(p+2)+3 \implies \gcd(p-1,p^2+p+1)=
\begin{cases} 3,\quad\text{if }3\mid p-1\\1,\quad\text{otherwise}\end{cases}$$
Since $q>5$ is prime, we must have either $q^2\mid p-1$ or $q^2\mid p^2+p+1$. But this is impossible because $q>p\ge5 \implies q^2>p-1\text{ and }q^2\ge (p+1)^2=p^2+2p+1>p^2+p+1$. So there are no solutions here either.
| Here is a shorter solution. If $p=2$, then $q^2\mid p^6-1=63$ implies that $q=3$. If $p=3$, then $q^2\mid 3^6-1=728$ leads to $q=2$. If $q=2$, then $p^2\mid q^3+1=9$ means $p=3$. If $q=3$, then $p^2\mid q^3+1=28$ gives $p=2$. That is, solutions with $p\leq 3$ or $q\leq 3$ are $(p,q)=(2,3)$ and $(p,q)=(3,2)$. From now on, suppose that $p,q>3$.
Observe that $p-1$, $p+1$, $p^2-p+1$, and $p^2+p+1$ are pairwise coprime integers. Thus, $$q^2\mid p^6-1=(p-1)(p+1)(p^2-p+1)(p^2+p+1)$$ implies that $q^2$ divides one of the four numbers $p-1$, $p+1$, $p^2-p+1$, and $p^2+p+1$. First, we assume that $q^2$ divides either $p-1$ or $p+1$. Then,
$$q^2\leq p+1\text{ so that }q<p\,.$$
Now, from $p^2\mid q^3+1=(q-1)(q^2-q+1)$ and $q-1$ is relatively prime to $q^2-q+1$, then $$p^2\mid q-1\text{ or }p^2\mid q^2-q+1\,,\tag{*}$$ but this is impossible as $q<p$. Hence, $q^2$ must divide either $p^2-p+1$ or $p^2+p+1$.
Now, we have $$q^2\leq p^2+p+1<(p+1)^2\,,\text{ whence }q\leq p\,$$
Again, from (*), we obtain
$$p^2\leq q^2-q+1 <q^2\leq p^2\,,$$
which is a contradiction. Hence, there are no other solutions than the two listed in the first paragraph.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1435718",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Proving $(x+y)^n = \sum\limits_{k=0}^n \binom{n}{k} x^k y^{n-k}$ I'm reading Serge Lang's 'Analysis I', and there's a problem I cannot figure out how to prove:
Problem: Prove by induction that $$(x+y)^n = \sum_{k=0}^n \begin{pmatrix} n \\ k \end{pmatrix} x^k y^{n-k} . $$
Attempt at proof : I established the base case, which is easily verified. Now I want to prove the inductive step. So assume the assertion holds for $n \geq 1$, and we want to show it also holds for $n + 1$. So we have to proof that: $$(x+y)^{n+1} = \sum_{k=0}^{n+1} \begin{pmatrix} n+1 \\ k \end{pmatrix} x^k y^{n+1-k} . $$
I started with the the LHS and applied the induction hypothesis:
\begin{align*} (x+y)^{n+1} &= (x+y)^n (x+y) \\ & = \sum_{k=0}^n \begin{pmatrix} n \\ k \end{pmatrix} x^k y^{n-k} (x+y) \\ &= \sum_{k=0}^n \begin{pmatrix} n \\ k \end{pmatrix} x^{k+1} y^{n-k} + \sum_{k=0}^n \begin{pmatrix} n \\ k \end{pmatrix} x^k y^{n-k+1} \\ &= \sum_{k=0}^n \begin{pmatrix} n \\ k \end{pmatrix} \bigg[ x^{k+1} y^{n-k} + x^k y^{n-k+1} \bigg] \end{align*} Now I don't know how to proceed. Any help would be appreciated!
| The best way to think of the inductive step is to multiply the sum by $(x+y)$ and think of Pascal's Triangle. Each new coefficient is the sum of the two coefficients from the previous exponent combinations which differed from its exponent by $1$. So adding them together produces, e.g. $\binom{n}{k} + \binom{n}{k+1} = \binom{n+1}{k+1}$, when you work it out. You end up with a solution to your problem and a visualization of how it got there.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1436306",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Finding $z$ such that $\Re(\frac{z}{-3+2i}) = \frac{10}{13},\;|3-2\overline{z}| = |2z+3i|$ I wish to find $z$ such that $\displaystyle \Re(\frac{z}{-3+2i}) = \frac{10}{13},\;|3-2\overline{z}| = |2z+3i|$.
We may simplify $\displaystyle \frac{z}{-3+2i} = \frac{x+iy}{-3+2i} \frac{-3-2i}{-3-2i} = \frac{-3x+2y + i(-2x+3y)}{13}$.
Taking the real part of the fraction above gives us $\displaystyle \frac{-3x+2y}{13}$
Then, $-3x+2y = 10$
So now we move onto the other equation.
Recall that $|z|=z\overline{z}$ and $z=x+iy$. Then, $|3-2(x-iy)| = |2(x+iy)+3i|$, $|3-2x+2iy| = |2x+2iy+3i|$,
Then, $$(3-2x+2iy)(3-2x-2iy) = (2x+i(3+2y))(2x-i(3+2y))$$
which simplifies to $$4x^2-12x+4y^2+9 = 4x^2 - i(2y+3)^2$$.
But this is still messy and does not simplify further. Can anyone see where I messed up?
| Given $$|3-2\bar{z}| = |2z+3i|\Rightarrow |(2\bar{z}-3)| = |2z+3i|\;,$$ Now put $$z=x+iy$$
We get $$\displaystyle |(2x-3)+i(2y)| = |2x+i(2y+3)|\Rightarrow (2x-3)^2+4y^2=(2x)^2+(2y+3)^2$$
So we get $$\displaystyle 9-12x = 9+12y\Rightarrow x+y=0.....(1)$$
and First part $$\displaystyle \bf{\Re\left(\frac{z}{-3+2i}\right)} = \frac{10}{13},$$ You have got $$-3x+2y=10..............(2)$$
so we get $$\displaystyle x=-2,y=2$$, So we get $$z=x+iy=-2+2i$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1436776",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Calculating $\int_{-a}^{a} \frac{x\cdot dy}{(x^2+y^2)^{3/2}}$ requires unusual substitution? Can someone help me understand how to solve this integral? The official solution says to substitute $y=x\cdot \tan(u)$ and $dy=x\cdot \sec^2(u)du$, but I don't understand how I should know that myself.
Here is the integral:
$$\int_{-a}^{a} \frac{x\cdot dy}{(x^2+y^2)^{3/2}}$$
A step-through solution would be appreciated. I haven't done integrals in a while so it would be helpful if you can explain any particularly complicated steps.
Thank you!!!
| Let $$\displaystyle I = \int \frac{x}{(x^2+y^2)^{\frac{3}{2}}}dy\;,$$ Here $x$ is Constant.
Here Denominator is in the form of $(x^2+y^2)$
So we will put $y=x\tan \phi,$ Then $\displaystyle dy = d(x\tan \phi)=x\frac{d}{d\phi}\left[\tan \phi\right]d\phi$
so we get $\displaystyle dy = x\sec^2 \phi d\phi$
So Integral $$\displaystyle I = \int\frac{x^2\cdot \sec^2 \phi}{x^3\sec^3 \phi}d\phi\;,$$ Here we put $(1+\tan^2 \phi) = \sec^2 \phi$$
So we get $$\displaystyle \frac{1}{x}\int \cos \phi d\phi = \frac{\sin \phi}{x}+\mathcal{C}$$
Now above we take $\displaystyle y=x\tan \phi\Rightarrow \tan \phi = \frac{y}{x}$
So Using Right angle $\triangle\;,$ We get $\displaystyle \sin \phi = \frac{y}{\sqrt{x^2+y^2}}$
So Integral $$\displaystyle I = \frac{y}{x\sqrt{x^2+y^2}}+\mathcal{C}$$
So $$\displaystyle \int_{-a}^{a}\frac{y}{(x^2+y^2)^{\frac{3}{2}}} = \left[\frac{y}{x\sqrt{x^2+y^2}}\right]_{-a}^{a} $$
$$\displaystyle = \left[\frac{a}{x\sqrt{x^2+a^2}}\right]-\left[-\frac{a}{x\sqrt{x^2+a^2}}\right] = \left[\frac{2a}{x\sqrt{x^2+a^2}}\right]$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1441935",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Minimalpolynomial of a Matrix without calculating the characterstic polynomial Is there a way calculating the minimal polynomial of a matrix without using the characteristic polynomial?
For example following matrix $$M = \left(
\begin{array}{cccc}
1 & 0 & -2 & 0 \\
-2 & -1 & 4 & 2 \\
0 & 0 & -1 & 0 \\
0 & 0 & 2 & 1 \\
\end{array}
\right).$$
Calculating the characteristic polynomial I get $\chi_M = (x-1)^2(x+1)^2$ with eigenvalues $-1$ and $1$ with algebraic multiplicity 2. Using the characteristic polynomial I have to check all possibilities for the minimal polynomial $\mu_M$ which are $$(x-1), (x-1)^2, (x+1), (x+1)^2, (x-1)(x+1), (x-1)^2(x+1), (x-1)(x+1)^2, (x-1)^2(x+1)^2.$$
Is there a way to see from the form of the matrix, what the minimal polynomial could be? Since $M$ could be partitioned as follows:
$$M = \left(
\begin{array}{cccc}
1 & 0 & -2 & 0 \\
-2 & -1 & 4 & 2 \\
0 & 0 & -1 & 0 \\
0 & 0 & 2 & 1 \\
\end{array}
\right)= \left(
\begin{array}{cc}
A & B \\
0 & C \\
\end{array}
\right),$$ with $A =\left(
\begin{array}{cc}
1 & 0 \\
-2 & -1 \\
\end{array}
\right), B = \left(
\begin{array}{cc}
-2 & 0 \\
4 & 2 \\
\end{array}
\right)$ and $C = \left(
\begin{array}{cc}
-1 & 0 \\
2 & 1 \\
\end{array}
\right)$.
| In fact, if we take
$$
A = \pmatrix{1&-2\\0&-1}, \quad
B = \pmatrix{1&0\\-2&-1}
$$
Then we have
$$
M = \pmatrix{a_{11}B & a_{12}B\\a_{21}B & a_{22}B} = A \otimes B
$$
Where $\otimes$ denotes the Kronecker product. Since both $A$ and $B$ have eigenvalues $\{-1,1\}$, $A \otimes B$ will have the spectrum $\{-1,-1,1,1\}$.
Note also that the Kronecker product of diagonalizable matrices is diagonalizable, so $M$ must have the minimal polynomial $q_M(x) = (x-1)(x+1)$.
In general, the exponent of a linear term in the minimal polynomial of $A \otimes B$ cannot exceed any of the exponents in the minimal polynomials of $A$ or $B$.
In fact, given the Jordan forms of $A$ and $B$, we can deduce the Jordan form of $A \otimes B$.
Another approach: the characteristic polynomial of a matrix of the form
$$
M = \pmatrix{A&B\\0&C}
$$
with $A,C$ square will be $\chi_M(x) = \chi_A(x)\chi_C(x)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1442430",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Indefinite integral of $\int\sqrt{x^2-1}dx$ How to calculate $\int\sqrt{{x^2}-1}dx$?
I tried to use 1st and 3rd Euler substitution and got the answer, but I suppose there's an trigonometrical replacing that gives the answer analytically.
| Integration without substitution
Notice, there is another method without any substitution, using integration by parts, let
$$I=\int\sqrt{x^2-1}\ dx$$$$=\int \sqrt{x^2-1}\cdot 1dx$$
$$I=\sqrt{x^2-1}\int 1\ dx-\int \left(\frac{d}{dx}\left(\sqrt{x^2-1}\right)\cdot \int1 dx\right)dx$$
$$I=\sqrt{x^2-1}(x)-\int \frac{x(2x)}{2\sqrt{x^2-1}}dx$$
$$I=x\sqrt{x^2-1}-\int \frac{(x^2-1)+1}{\sqrt{x^2-1}}dx$$
$$I=x\sqrt{x^2-1}-\int\sqrt{x^2-1} dx+\int \frac{1}{\sqrt{x^2-1}}dx$$
$$I=x\sqrt{x^2-1}-I+\ln|x+\sqrt{x^2-1}|+C$$
$$2I=x\sqrt{x^2-1}+\ln|x+\sqrt{x^2-1}|+C$$
$$I=\color{red}{\frac{1}{2}\left(x\sqrt{x^2-1}+\ln|x+\sqrt{x^2-1}|\right)+C}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1445147",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Compute $\lim_{n\to +\infty}n\left(\tan\left(\frac{\pi}{3}+\frac{1}{n} \right)-\sqrt{3}\right)$ without using L' Hôpital
Compute $$\lim_{n\to +\infty}n\left(\tan\left(\frac{\pi}{3}+\frac{1}{n} \right)-\sqrt{3}\right)$$ without using L'Hospital's rule.
By using L'Hospital's rule and
$$\tan'( \Diamond )=( \Diamond )'(1+\tan^{2}( \Diamond ))$$
I mean by $\Diamond $ a function
so I got
\begin{align}
\lim_{n\to +\infty}n\left(\tan\left(\dfrac{\pi}{3}+\dfrac{1}{n} \right)-\sqrt{3}\right)
&=\lim_{n\to +\infty}\dfrac{\left(\tan\left(\dfrac{\pi}{3}+\dfrac{1}{n} \right)-\sqrt{3}\right)}{\dfrac{1}{n}}\\
&=1+\tan^{2}\left(\dfrac{\pi}{3}\right)=1+\sqrt{3}^{2}=1+3=4
\end{align}
I'm interested in more ways of computing limit for this sequence.
| \begin{align}
\lim_{n\to\infty}n\left(\tan\left(\frac{\pi}{3}+\frac 1n\right)-\tan\frac{\pi}{3} \right)&=\lim_{n\to\infty}n\left(\tan(\frac{\pi}{3}+\frac 1n-\frac{\pi}{3})\Big(1+\tan(\frac{\pi}{3}+\frac 1n)\tan\frac{\pi}{3}\Big) \right)\\
&=\lim_{n\to\infty}\frac{n\sin\frac 1n}{\cos \frac 1n}\times \lim_{n\to\infty}\Big(1+\tan(\frac{\pi}{3}+\frac 1n)\tan\frac{\pi}{3}\Big)\\
&=\frac{\lim_{n\to\infty}n\sin\frac 1n}{\lim_{n\to\infty}\cos \frac 1n}\times \Big(1+\lim_{n\to\infty}\tan(\frac{\pi}{3}+\frac 1n)\tan\frac{\pi}{3}\Big)\\
&=\frac{1}{1}\times(1+\sqrt 3\sqrt 3)\\
&=4
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1447837",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 4
} |
Is there another way to solve this quadratic equation? $$\frac { 4 }{ x^{ 2 }-2x+1 } +\frac { 7 }{ x^{ 2 }-2x+4 } =2$$
Steps I took:
$$\frac { 4(x^{ 2 }-2x+4) }{ (x^{ 2 }-2x+4)(x^{ 2 }-2x+1) } +\frac { 7(x^{ 2 }-2x+1) }{ (x^{ 2 }-2x+4)(x^{ 2 }-2x+1) } =\frac { 2(x^{ 2 }-2x+4)(x^{ 2 }-2x+1) }{ (x^{ 2 }-2x+4)(x^{ 2 }-2x+1) } $$
$$4(x^{ 2 }-2x+4)+7(x^{ 2 }-2x+1)=2(x^{ 2 }-2x+4)(x^{ 2 }-2x+1)$$
$$11x^{ 2 }-22x+23=2x^{ 4 }-8x^{ 3 }+18x^{ 2 }-20x+8$$
I can keep going with all the steps I took, but is there a more elegant way to arrive at the solution for this equation? It seems as if I keep going the way I am, I will hit a dead end. No actual solution, please. Hints are much better appreciated.
| To simplify computations a little bit, I would put $y=(x-1)^2$, so that the equation becomes $$\frac{4}{y}+\frac{7}{y+3}=2$$ which you can easily solve to find the values of $y$ and finally the corresponding values of $x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1448260",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "22",
"answer_count": 3,
"answer_id": 0
} |
$f(x)+f(x+y+z)\geq f(x+y)+f(x+z)$ for convex functions? For a convex function $f$, is it always true that
$$f(x)+f(x+y+z)\geq f(x+y)+f(x+z)$$ for all $x,y,z>0$?
I tried to use the definition of convexity: $$f(tx+(1-t)y)\leq tf(x)+(1-t)f(y).$$ but since the definition contains three occurrences of $f$ while the question contains four, I don't see how to plug in values to make them the same.
| $f$ is convex then the graph of $f$ is below all its chords :
$$\forall z \in [x, y] \in \mathbb{R}, f(z) \leq f(x) + \dfrac{f(y) - f(x)}{y - x} (y - x)$$
Let $x, y, z > 0$.
*
* We have $x < x + y < x + y + z$ then :
$$f(x + y) \leq f(x) + \dfrac{f(x + y + z) - f(x)}{(x + y + z) - x} ((x + y) - x) = f(x) + \dfrac{f(x + y + z) - f(x)}{y + z} y$$
* We have $x < x + z < x + y + z$ then :
$$f(x + z) \leq f(x + y + z) + \dfrac{f(x + y + z) - f(x)}{(x + y + z) - x} ((x + z) - (x + y + z)) = f(x + y + z) - \dfrac{f(x + y + z) - f(x)}{y + z} y$$
By summing the two inequalities :
$$f(x + y) + f(x + z) \leq f(x) + f(x + y + z)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1448345",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Proving the given inequality. (Complex Analysis). We're given a function $P_n(x)$ for $-1\leq x\leq1$ as follows :
$$P_n(x) = \int \limits_0^\pi \dfrac{1}{\pi}(x+i\sqrt{1-x^2} \cos\theta)^n \, d\theta$$
for $n=(0,1,2,3,\ldots)$, we need to prove that $|P_n(x)| \leq 1$.
I tried the following :
Let $z=x+i\sqrt{1-x^2}\cos\theta$
I somehow want to prove that $|z|=|x+i\sqrt{1-x^2}\cos\theta|\leq1$, as that would imply that $|z^n|\leq1$, as $|z^n|=|z|^n$.
$$|z| = \sqrt{x^2+(1-x^2)\cos^2 \theta} \Longrightarrow \sqrt{x^2 \sin^2\theta + \cos^2\theta}.$$
Now, $x^2\leq1$ and $\sin^2\theta \leq1 \Longrightarrow x^2\sin^2\theta \leq1$.
Also, $\cos^2\theta \leq1 \Longrightarrow \sqrt{x^2\sin^2\theta+\cos^2\theta} \leq \sqrt{2}$,
But " $\leq1$" condition is required...
That's the point where I am stuck, could anyone help me with this?
| \begin{align}
x^2\sin^2 t+\cos^2 t
&=x^2(1-\cos^2 t+\cos^2 t\\
&=(1-x^2)\cos^2t+x^2\\
\text{so}\\
|x^2\sin^2t+\cos^2t|
&\le|(1-x^2)\cos^2t+x^2|\\
&\le|(1-x^2)\cos^2t|+|x^2|\\
&\le|1-x^2|+|x^2|\\
&\le(1-x^2)+x^2\\
&=1
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1451301",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Calculating $\lim_{x\to-\infty}\left(\sqrt{4x^2-6}-\sqrt{4x^2+x}\right)$ Solving without L'Hopital
$$\lim_{x\to-\infty}\left(\sqrt{4x^2-6}-\sqrt{4x^2+x}\right)$$
That's
$$\lim_{x\to-\infty}\left(\sqrt{4x^2-6}\right)-\lim_{x\to-\infty}\left(\sqrt{4x^2+x}\right)$$
I have been taught to get the highest exponents, so...
$$\sqrt{4x^2}-\sqrt{4x^2}$$
It's the same for both sides, no?
$$2x-2x$$
$$-\infty+\infty$$
Which is wrong. The correct answer is
$$\frac{1}{4}$$
Why? I always just grab the highest exponent ($\sqrt{4x^2}$) and work with it. But this time it didn't go well.
| \begin{align}
\lim_{x\to-\infty}\left(\sqrt{4x^2-6}-\sqrt{4x^2+x}\right)&=\lim_{x\to-\infty}\left(\sqrt{4x^2-6}-\sqrt{4x^2+x}\right)\cdot\frac{\sqrt{4x^2-6}+\sqrt{4x^2+x}}{\sqrt{4x^2-6}+\sqrt{4x^2+x}}\\
&=\lim_{x\to-\infty}\frac{(\sqrt{4x^2-6})^2-(\sqrt{4x^2+x})^2}{\sqrt{4x^2-6}+\sqrt{4x^2+x}}\\
&=\lim_{x\to-\infty}\frac{4x^2-6-(4x^2+x)}{\sqrt{4x^2-6}+\sqrt{4x^2+x}}\\
&=\lim_{x\to-\infty}\frac{(-x-6)/|x|}{\sqrt{4-6/x^2}+\sqrt{4+x/x^2}}\\
&=\lim_{x\to-\infty}\frac{1+\frac{6}{x}}{\sqrt{4-\frac{6}{x^2}}+\sqrt{4+\frac{1}{x}}}\\
&=\frac{1}{\sqrt{4}+\sqrt{4}}\\
&=\frac{1}{4}
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
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Proof of the inequality $e^x\le e^{x^2} + x$ The question is to prove the inequality $e^x\le e^{x^2} + x$. I tried the Taylor expansion like ${e^x} = 1 + x + \frac{{{x^2}}}{{2!}} + \frac{{{x^3}}}{{3!}} + ...$ and $x + {e^{{x^2}}} = 1 + x + \frac{{{x^4}}}{{2!}} + \frac{{{x^6}}}{{3!}} + ...$ but cannot see anything useful out of this. Anyone can provide some help? Thank you.
| Note that for any $x$, we have
$$
(e^{x^2} + x) - e^x = \\
\frac{1}{2!}x^2 - \frac{1}{3!}x^3 + \left(\frac{1}{2!} - \frac{1}{4!}\right)x^4 - \frac 1{5!}x^5 +
\left( \frac{1}{3!} - \frac{1}{6!} \right)x^6 - \cdots \geq \\
\frac{1}{2!}x^2 - \frac{1}{3!}x^3 + \frac {1}{4!}x^4 - \frac 1{5!}x^5 +
\frac{1}{6!}x^6 - \cdots =\\
e^{-x} + x - 1
$$
Consider the function $f(x) = e^{-x} + x - 1$. We note that
$$
f'(x) = 1 - e^{-x}
$$
so that $f$ achieves its minimum at $x = 0$, where we find $f(0) = 0$. So, $f(x) \geq 0$.
Thus, we have
$$
(e^{x^2} + x) - e^x \geq f(x) \geq 0
$$
The conclusion follows.
| {
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"timestamp": "2023-03-29T00:00:00",
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Enumerative Combinatorics problem from Stanley
Let $n$ and $k$ be two positive integers. Show that
\begin{align}
\sum a_1\cdot a_2 \cdot \cdots \cdot a_k={n+k-1\choose 2k-1} ,
\end{align}
where the sum ranges over all compositions $(a_1,a_2,\ldots,a_k)$ of $n$ into $k$ parts.
Reminder: A composition of $n$ into $k$ parts is defined as a $k$-tuple $(a_1,a_2,\ldots,a_k)$ of positive integers such that $a_1 + a_2 + \cdots + a_k = n$.
This is Exercise 1.29 in Stanley's Enumerative Combinatorics volume 1 (2nd edition). I have no idea how to do it.
| It suffices to prove they have the same generating function:
\begin{align*}
\sum\limits_{n} \sum a_1\cdots a_k x^n &= \sum\limits_{1\leq a_1,\ldots,a_k} a_1\cdots a_k x^{a_1 + \ldots + a_k} \\
&= \sum\limits_{1\leq a_1,\ldots,a_k} (a_1 x^{a_1}) \cdots (a_k x^{a_k}) \\
&= \left(\sum\limits_{1\leq a_1} a_1 x^{a_1}\right)\cdots\left(\sum\limits_{1\leq a_k} a_k x^{a_k}\right) \\
&= \left( \frac{x}{(1 - x)^2} \right)^k \\
&= x^k \left( \frac{1}{(1 - x)^{2k}}\right) \\
&= x^k \sum\limits_m \binom{m + 2k - 1}{m}x^m \\
&= x^k \sum\limits_n \binom{n - k + 2k - 1}{n - k}x^{n-k} \\
&= \sum\limits_{n} \binom{n + k - 1}{2k - 1}x^n.
\end{align*}
| {
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"source": "stackexchange",
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How to find subspace $W_1\cap W_2$? Where $W_1$ and $W_2$ are subspaces of $\mathbb{R}^5$. Consider $S_1=\{(2,0,3,1,1),(1,0,2,1,1),(2,0,3,1,3)\}$ and $S_2=\{(2,1,1,0,1),(3,2,3,2,3),(1,1,1,1,1)\}$. Suppose $W_1$ and $W_2$ are subspaces of $\mathbb{R}^5$ generated by $S_1$ and $S_2$ respectively. Then what will be $W_1\cap W_2$. I am not able to understand how to start.
| Well, there must be an easier method, but here is a solution. Let $\vec z \in W_1 \cap W_2$. Then it holds:
$$a_1\cdot \begin{bmatrix} 2\\0\\3\\1\\1 \end{bmatrix}+a_2\cdot \begin{bmatrix}1\\0\\2\\1\\1 \end{bmatrix}+a_3 \cdot \begin{bmatrix} 2\\0\\3\\1\\3\end{bmatrix}-\lambda_1\cdot \begin{bmatrix}2\\1\\1\\0\\1\end{bmatrix}-\lambda_2\cdot\begin{bmatrix}3\\2\\3\\2\\3 \end{bmatrix}-\lambda_3\cdot \begin{bmatrix}1\\1\\1\\1\\1\end{bmatrix}=\begin{bmatrix}0\\0\\0\\0\\0\end{bmatrix},$$for some $a_1,a_2,a_3,\lambda_1,\lambda_2,\lambda_3 \in \mathbb R$, which we have to define.
Equivalently, it holds:
$$\begin{bmatrix} 2&1&2&2&-3&-1\\0&0&0&-1&-2&-1\\3&2&3&-1&-3&-1\\1&1&1&0&-2&-1\\1&1&3&-1&-3&-1\end{bmatrix}\cdot\begin{bmatrix}a_1\\a_2\\a_3\\\lambda_1\\\lambda_2\\\lambda_3\end{bmatrix}=\begin{bmatrix} 0\\0\\0\\0\\0\end{bmatrix}.$$
We can find the reduced row echelon form of the coefficient matrix, which is (after some work):
$$R=\text{rref }A = \begin{bmatrix}
1 & 0 & 0 & 0 & 2.5 & 1.5\\
0 & 1 & 0 & 0 & -5 & -3\\
0 & 0 & 1 & 0 & 0.5 & 0.5\\
0 & 0 & 0 & 1 & 2 & 1\\
0 & 0 & 0 & 0 & 0 &0
\end{bmatrix}.$$
Due to the fact that the system is homogeneous, we can solve the system $$R\cdot\begin{bmatrix}a_1\\a_2\\a_3\\\lambda_1\\\lambda_2\\\lambda_3\end{bmatrix}=\mathbf{0}.$$
Now, we can easily find that $\lambda_3 = y \in \mathbb R, \, \lambda_2 = x \in \mathbb R$ and $\lambda_1 =-2x-y$.
Thus every $\vec z \in W_1\cap W_2$ can be written in the form
$$(-2x-y)*\begin{bmatrix}2\\1\\1\\0\\1\end{bmatrix} + x*\begin{bmatrix}3\\2\\3\\2\\3\end{bmatrix}+ y*\begin{bmatrix}1\\1\\1\\1\\1\end{bmatrix} = x*\begin{bmatrix}-1\\0\\1\\2\\1\end{bmatrix}+y*\begin{bmatrix}-1\\0\\0\\1\\0\end{bmatrix},$$ where $x,y \in \mathbb R$.
Thus, $$W_1\cap W_2 = \langle(-1,0,1,2,1),\, (-1,0,0,1,0)\rangle.$$
Also, from the rref we can find $a_1 = -2.5x-1.5y$, $a_2 = 5x +3y$, $a_3= -0.5x-0.5y$.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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Find four integers summing to zero, with sum of cubes 24 I'm stuck on the following problem from Terence Tao's "Solving Mathematical Problems"
Find all integers $a,b,c,d$ such that $a+b+c+d=0$ and $a^3+b^3+c^3+d^3=24$. (Hint: it is not hard to guess some solutions ti these equations, but to show that you have all of them, substitute the first equation into the second and factorize.)
My attempt:
After said substitution, I now want to solve
$$-3 a^2 b-3 a^2 c-3 a b^2-6 a b c-3 a c^2-3 b^2 c-3 b c^2-24=0 $$
by factoring the LHS. There is some evident symmetry, so I tried going with $c=Aa+Bb+C$ for one of the factors without success. Moreover, Mathematica can't seem to factorize this polynomial as well.
I'd appreciate any help on this problem, thanks!
| Divide by $-3$ of your expression, we have
$$a^2 b+ a^2 c+ a b^2+2 a b c+ a c^2+ b^2 c+ b c^2+8=0 $$
Note that
$$\begin{split}(a+b)(b+c)(c+a) &= (ab+ac + b^2 +bc)(c+a)\\
&= abc + ac^2 + b^2c + bc^2 + a^2b + a^2 c+ ab^2 + abc \\
&= -8
\end{split}$$
As $a, b, c$ are integers, there are not many choices. You need then to solve
$$\begin{split} a + b &= \alpha \\ b+c &= \beta \\ a+c &= \gamma \end{split}$$
$$\Rightarrow a = \frac{\alpha -\beta +\gamma}{2},\ \ b = \frac{\alpha + \beta - \gamma}{2},\ \ c = \frac{-\alpha + \beta + \gamma}{2}. $$
where $\alpha, \beta, \gamma$ are any integers so that $\alpha\beta\gamma = -8$. Now you need only to check which choices of $\alpha, \beta, \gamma$ would make $a, b, c$ integers.
(To come up with that factorization, one observe that the expression is of homogeneuos degree three (except the constant term) and are highly symmetric in $a, b, c)$. Thus $(a+b)(b+c)(c+a)$ is more or less the first thing you want to try.)
| {
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How to solve $\arctan{\sqrt{\frac{10-x}{x}} + \arccos\sqrt{\frac{10-x}{10}}}= \frac\pi{2}$ Can you tell me if it is possible to solve $$\arctan{\sqrt{\frac{10-x}{x}} + \arccos\sqrt{\frac{10-x}{10}}}= \frac\pi{2}\ ?$$
wolfram doesn't say it is not computable, but says there are no real solutions, whereas we know the solution is $x=1$
Any hints on how to proceed?
Thanks
| Let $y=\sqrt{\frac{10-x}{x}}$, let $z=\sqrt{\frac{10-x}{10}}$. Take the cosine of both sides:
$$\cos\left(\arctan y+\arccos z\right)=0.$$
Apply the angle addition formula.
$$\cos(\arctan y)z-\sin(\arctan y)\sin(\arccos z)=0.$$
From the definition of trigonometric ratios, we know that $\cos(\arctan y)=\frac{1}{\sqrt{y^2+1}}$, $\sin(\arctan y)=\frac{y}{\sqrt{y^2+1}}$, and $\sin(\arccos z)=\sqrt{1-z^2}$. This leaves us with
$$\frac{z}{\sqrt{y^2+1}}-\frac{y}{\sqrt{y^2+1}}\sqrt{1-z^2}=0$$
$$z-y\sqrt{1-z^2}=0$$
$$\sqrt{\frac{10-x}{10}}-\sqrt{\frac{10-x}{x}}\sqrt{1-\frac{10-x}{10}}=0$$
$$\sqrt{\frac{10-x}{10}}-\sqrt{\frac{10-x}{x}}\sqrt{\frac{x}{10}}=0$$
$$0=0.$$
This equation boils down to a tautology, so every value of $x$ is a solution.
| {
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"timestamp": "2023-03-29T00:00:00",
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Coloring the windmill A windmill has $5$ wings and each of these is symmetrically connected to the axis and consists of two parts.
If on the wings of the windmill $4$ parts are colored black, $3$ parts are colored red, and $3$ parts are colored orange, in how many ways can one then color the windmill? (In different ways when one looks at the windmill from the front and the wings rotate.)
Use Pólya's theorem and the fact that the group $C_n$ that consists of rotations of a regular $n$-polygon has the cycle index
$$\frac 1n\sum_{d\,|\, n} \varphi(d)t_d^{\frac nd},$$
where $\varphi(d)=\left |\{j\,:\, 0\leq j\leq d-1,\; \textrm{gcd}(j,d)=1\}\right |$, but observe that in this case the number of orbits is double that of a regular $n$-polygon because each wing is built from two parts.
Here $\sum_{ d\,|\,n}$ means that we take the sum over all integers $d$ such that $1\leq d\leq n$ and $n$ is divisible by $d$ and the term $t_d^{\frac nd}$ is due to the fact that in the rotation group there is an element so that the cyclic group generated by it has $\frac nd$ orbits with length $d$.
Pólya's theorem says that if in the cycle index the variable $t_j$ is replaced by $b^j+r^j+o^j$ then the coefficient of the term $b^4\cdot r^3\cdot o^3$ is the number one is looking for!
My attempt:
I set $n = 4 + 3 + 3 = 10$ and determined all divisors $d$ of $n$ to be $1,2,5,10$. Now $\varphi (1) = 1,\,\varphi (2) = 1,\,\varphi (5) = 4,\,\varphi (10) = 4$. Plugging into equation i got the cycle index $$\frac{1}{{10}}\left( {t_1^{10} + t_2^5 + 4t_5^2 + 4{t_{10}}} \right)$$
Calculating for $b^4\cdot r^3\cdot o^3$
$$\frac{1}{{10}}\left( {{{(b + r + o)}^{10}} + {{({b^2} + {r^2} + {o^2})}^5} + 4{{({b^5} + {r^5} + {o^5})}^2} + 4({b^{10}} + {r^{10}} + {o^{10}})} \right)$$
I get the coefficient for $b^4\cdot r^3\cdot o^3$ $420$, which is wrong. Now, could someone explain my mistake?
| Let me typeset some of the content from the comments. We have five panels attached to the center and five is prime, so the cycle index is really simple here namely $$Z(C_5) = \frac{1}{5} (a_1^5 + 4 a_5).$$ Taking into account that the outer and inner elements of the panels move in sync but never enter into the same orbit we get $$Z(W) = \frac{1}{5} (a_1^{10} + 4 a_5^2).$$
Now we are looking for $$[B^4 R^3 O^3] Z(W)(B+R+O).$$
We get for $$Z(W)(B+R+O) =
1/5\, \left( B+R+O \right) ^{10}+4/5\, \left( {B}^{5}+{O}^{5}+{R}^{5}
\right) ^{2}$$
and $$[B^4 R^3 O^3] Z(W)(B+R+O) = 840.$$
Note that the second term produces multiples of five in the exponents and hence does not contribute to the coefficient, which is
$$\frac{1}{5} {10\choose 4,3,3} = 840.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1469164",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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A golden ratio series from a comic book The eighth installment of the Filipino comic series Kikomachine Komix features a peculiar series for the golden ratio in its cover:
That is,
$$\phi=\frac{13}{8}+\sum_{n=0}^\infty \frac{(-1)^{n+1}(2n+1)!}{(n+2)!n!4^{2n+3}}$$
How might this be proven?
| Like Calculating $1+\frac13+\frac{1\cdot3}{3\cdot6}+\frac{1\cdot3\cdot5}{3\cdot6\cdot9}+\frac{1\cdot3\cdot5\cdot7}{3\cdot6\cdot9\cdot12}+\dots? $,
$$T(n)=\frac{(-1)^{n+1}(2n+1)!}{(n+2)!n!4^{2n+3}}=\dfrac1{64}\cdot\dfrac{1\cdot3\cdot5\cdots(2n+1)2^n}{(n+2)!16^n}$$
As the denominator has $(n+2)!,$
$$T(n)=-\dfrac1{8^{n+2}}\cdot\dfrac{-1\cdot1\cdot3\cdot5\cdots(2n+1)}{(n+2)!} $$ $$=\dfrac1{4^{n+2}}\cdot\dfrac{\dfrac12\left(\dfrac12-1\right)\left(\dfrac12-2\right)\cdots\left(\dfrac12-(n+1)\right)}{(n+2)!}$$
Using Generalized Binomial Expansion, $$(1+x)^m=1+mx+\frac{m(m-1)}{2!}x^2+\frac{m(m-1)(m-2)}{3!}x^3+\cdots$$ given the converge holds
$$\sum_{n=0}^\infty T(n)=\left(1+\dfrac14\right)^{1/2}-1-T(-1)=\dfrac{\sqrt5}2-1-\dfrac14\dfrac12$$
See also: A Series For the Golden Ratio
| {
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How to prove this inequality Let $x,y,z>0$,and such $xyz=1$,show that
$$\dfrac{1+x^2y^2}{1+x}+\dfrac{1+y^2z^2}{1+y}+\dfrac{1+z^2x^2}{1+z}\ge 3$$
I tried using $AM-GM$ inequality, but i'm not sure how it would work ,Thanks in advance.
$$\sum\dfrac{1+x^2y^2}{1+x}\ge3\sqrt[3]{\prod\dfrac{1+x^2y^2}{1+x}}$$
| Let me try. $$LHS \geq 3\sqrt[3]{\prod \frac{1+x^2y^2}{1+x}} \geq 3 \sqrt[3]{\prod \frac{(1+xy)^2}{2(1+x)}} = 3 \sqrt[3]{\prod \frac{(xy)^2(1+z)^2}{2(1+x)}} = 3 \sqrt[3]{\frac{(xyz)^4}{8}\prod (1+x)} = \frac{3}{2}\sqrt[3]{(1+x)(1+y)(1+z)} \geq \frac{3}{2}\sqrt[3]{8\sqrt{xyz}} = 3.$$
| {
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Express $x^3+3x^2-2x+1$ in the form $ax^3+b(x-1)^2+cx+d$.
Express $x^3+3x^2-2x+1$ in the form $ax^3+b(x-1)^2+cx+d$.
I have no idea what the question is asking. Any help would be appreciated.
| $$x^3+3x^2-2x+1=x^3+3(x^2-2x+1)+6x-3-2x+1$$
$$x^3+3(x-1)^2+4x-2$$
| {
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Show by induction: $(1+\frac{1}{n})^{n}Show by induction that for all natural numbers n>3
$(1+\frac{1}{n})^n<n$
Let $(1+\frac{1}{n})^n<n$ be true !
We show that $(1+\frac{1}{n+1})^{n+1})<n+1$
$(1+\frac{1}{n})^n(1+\frac{1}{n})<n(1+\frac{1}{n})$
$(1+\frac{1}{n})^{n+1}<n+1$
So now we just need to show that
$(1+\frac{1}{n+1})^{n+1}<(1+\frac{1}{n})^{n+1}$
So here , If I show that :
$(1+\frac{1}{n+1})<(1+\frac{1}{n})$
Does that mean im done with exercise ?
| You have to prove for $n=4$ in order that mathematical induction work! To show the desired implication, note that if $n \geq 4$ is such that
$$
\bigg( 1 + \frac{1}{n} \bigg)^{n} < n,
$$
then
$$
\bigg( 1 + \frac{1}{n+1} \bigg)^{n+1} = \bigg( 1 + \frac{1}{n+1} \bigg)^{n} \bigg( 1 + \frac{1}{n+1} \bigg) < \bigg( 1 + \frac{1}{n} \bigg)^{n}\bigg(1 + \frac{1}{n+1} \bigg) < n + \frac{n}{n+1} < n+1.
$$
Note also that this implication makes no use of the hypothesis that $n \geq 4$, so if you do not check the initial proposition then a wrong conclusion could be drawn!
| {
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How to solve these recurrence relations by using generating function First of all, I want to make an apology for my English for I'm not an English native speaker.
I'm reading Discrete Mathematics and its Applications recent days, and I am stumped by these three problems.
*
*$a_k=a_{k-1}+2a_{k-2}+2^k, a_0=4, a_1=12$
*$a_k=4a_{k-1}-4a_{k-2}+k^2, a_0=2, a_1=5$
*$a_k=2a_{k-1}-3a_{k-2}+4^k+6, a_0=20, a_1=60$
Anyone has any idea? Your replies will be highly appreciated.
Problem solved. Thanks to you all. And here are my solutions.
Problem 1.
$$\begin{cases}a_k=a_{k-1}+2a_{k-2}+2^k\\ a_0=4\\ a_1=12\end{cases}$$
Let $f(x)$ denote the generating function for the sequence $a_k$, then we get $f(x)=\sum\limits_{k\ge0}a_kx^k$
Take the first equation, then multiply each term by $x^k$
$$a_kx^k=a_{k-1}x^k+2a_{k-2}+2^kx^k$$
And sum each term from 2 since it's a 2-order recurrence relation.
$$\begin{align*}\sum\limits_{k\ge2}a_kx^k&=\sum\limits_{k\ge2}a_{k-1}x^k + \sum\limits_{k\ge2}2a_{k-2}x^k+\sum\limits_{k\ge2}2^kx^k\\&=\sum\limits_{k\ge2}a_{k-1}x^k + 2\sum\limits_{k\ge2}a_{k-2}x^k+\sum\limits_{k\ge2}2^kx^k\end{align*}$$
The idea is to manipulate each term so that you can write them as expressions in terms of the generating function $f(x)$ and known series representations.
$\begin{align*}\sum\limits_{k\ge2}a_kx^k&=\left(\sum\limits_{k\ge2}a_kx^k+a_1x+a_0\right)-a_1x-a_0\\&=\sum\limits_{k\ge0}a_kx^k-a_1x-a_0\\&=f(x)-4x-12\end{align*}$
$\begin{align*}\sum\limits_{k\ge2}a_{k-1}x^k&=x\sum\limits_{k\ge2}a_{k-1}x^{k-1}\\&=x\sum\limits_{k\ge1}a_kx^k\\&=x\left(\sum\limits_{k\ge0}a_kx^k-a_0\right)\\&=x(f(x)-4)\end{align*}$
$\begin{align*}\sum\limits_{k\ge2}a_{k-2}x^k&=x^2\sum\limits_{k\ge2}a_{k-2}x^{k-2}\\&=x\sum\limits_{k\ge0}a_kx^k\\&=x^2f(x)\end{align*}$
$\begin{align*}\sum\limits_{k\ge2}2^kx^k&=\sum\limits_{k\ge0}2^kx^k-2x-1\\&=\frac{1}{1-2x}-2x-1\end{align*}$
Right now, we have
$$f(x)-4x-12=x(f(x)-4)+2x^2f(x)+\frac{1}{1-2x}-2x-1$$
Rewrite the equation to solve $f(x)$, and we have
$$f(x)=\frac{2}{3}\frac{1}{(1-2x)^2}+\frac{38}{9}\frac{1}{1-2x}-\frac{8}{9}\frac{1}{1+x}$$
Finally,
$$\begin{align*}a_k&=\frac{2}{3}C(2+k-1, 2-1)2^k+\frac{38}{9}·2^k-\frac{8}{9}(-1)^k\\&=\frac{2}{3}(1+k)2^k+\frac{38}{9}·2^k-\frac{8}{9}(-1)^k\\&=\frac{44}{9}·2^k+\frac{2}{3}k·\frac{k}{3}-\frac{8}{9}(-1)^k\end{align*}$$
Problem 2.
Most steps are the same as they were in problem 1, so I'll only write the steps of conjugating $\sum\limits_{k\ge2}k^2x^k$
$$\begin{align*}\sum\limits_{k\ge2}k^2x^k&=\sum\limits_{k\ge0}k^2x^k-x\\&=x\sum\limits_{k\ge0}k^2x^{k-1}-x\\&=x\sum\limits_{k\ge0}k·kx^{k-1}-x\\&=x\left(\sum\limits_{k\ge0}k·x^k\right)'-x\\&=x\left(x\sum\limits_{k\ge0}k·x^{k-1}\right)'-x\\&=x\left(x\left(\sum\limits_{k\ge0}x^k\right)'\right)'-x\\&=x\left(x\left(\frac{1}{1-x}\right)'\right)'-x\\&=x\left(\frac{x}{(1-x)^2}\right)'-x\\&=\frac{x(1+x)}{(1-x)^3}-x\end{align*}$$
My notes, using generating functions to solve recurrence relations
| Here's a (partial) solution for the second recurrence relation. There's no particular reason for me having chosen this particular recurrence among the other two; this is just used as an example to give you an idea of how to solve the others. I intentionally left out most of the details to let this serve more as a reference for you to check your solution at different checkpoints rather than as an answer key.
$$\begin{cases}a_k=4a_{k-1}-4a_{k-2}+k^2\\ a_0=2\\ a_1=5\end{cases}$$
Let $f(x)$ denote the generating function for the sequence $a_k$; that is, $f(x)=\sum\limits_{k\ge0}a_kx^k$. Taking the first equation, multiply each term by $x^k$ and sum each term over all positive $k\ge2$. (Why?)
$$\color{red}{\sum_{k\ge2}a_kx^k}=4\color{blue}{\sum_{k\ge2}a_{k-1}x^k}-4\color{green}{\sum_{k\ge2}a_{k-2}x^k}+\color{orange}{\sum_{k\ge2}k^2x^k}$$
The idea is to manipulate each term so that you can write them as expressions in terms of the generating function $f(x)$ and known series representations. For instance, consider the LHS:
$$\begin{align*}\sum_{k\ge2}a_kx^k&=\sum_{k\ge2}a_kx^k+a_1x+a_0-a_1x-a_0\\&=\sum_{k\ge0}a_kx^k-a_1x-a_0\\&=f(x)-5x-2\end{align*}$$
Doing the same throughout, you end up with a rewritten equation that can be solved for $f(x)$.
$$\begin{align*}\color{red}{f(x)-5x-2}&=4\color{blue}{x(f(x)-2)}-4\color{green}{x^2f(x)}+\color{orange}{\frac{x(1+x)}{(1-x)^3}-x}\\[1ex]
f(x)&=\frac{2}{(1-x)^3}+\frac{5}{(1-x)^2}+\frac{13}{1-x}+\frac{6}{(1-2x)^2}-\frac{24}{1-2x}
\end{align*}$$
From here, you need to determine an appropriate series representation for $f(x)$. All the information you need happens to be contained in the geometric series,
$$\sum_{k\ge0}x^k=\frac{1}{1-x}\quad\text{for }|x|<1$$
and its derivatives.
You should arrive at the final solution
$$a_k=20-9(2^{1+k})+8k+3k(2^{1+k})+k^2$$
| {
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"timestamp": "2023-03-29T00:00:00",
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How to prove that $x+y \ge 2$ for $xy = 1$ and both positive? For every $x,y\in\mathbb{R}$ such that $x,y>0$ and that satisfy the condition xy=1, the following will hold:
$$x+y\ge 2$$
Steps I took and my thoughts on this:
$$xy=1\Rightarrow \frac { xy }{ x } =\frac { 1 }{ x } \Rightarrow y=\frac { 1 }{ x } \quad and\quad xy=1\Rightarrow \frac { xy }{ y } =\frac { 1 }{ y } \Rightarrow x=\frac { 1 }{ y } $$
$$x\cdot \frac { 1 }{ x } =1\quad and\quad y\cdot \frac { 1 }{ y } =1$$
$$x+\frac { 1 }{ x } \ge 2\quad and\quad \frac { 1 }{ y } +y\ge 2$$
$$x(x+\frac { 1 }{ x } )\ge (2)x\quad and\quad y(\frac { 1 }{ y } +y)\ge (2)y$$
$$x^{ 2 }+1\ge 2x\quad and\quad y^2+1\ge 2y$$
$$x^{ 2 }-2x+1\ge 0\quad and\quad y^{ 2 }-2y+1\ge 0$$
$$(x-1)^{ 2 }\ge 0\quad and\quad (y-1)^{ 2 }\ge 0$$
$$x\in\mathbb{R} \quad and \quad y\in\mathbb{R} $$
I'm not sure where I am going wrong, but I don't know how to organize my thoughts about this in order to prove it with a formal mathematical proof. I'd like to be guided in the right direction. Hints are much appreciated.
| $$
x+\dfrac{1}{x}=\dfrac{x^2+1}{x}=\dfrac{(x-1)^2+2x}{x}=\dfrac{(x-1)^2}{x}+2
$$
so, for $x>0$:
$$
x+\dfrac{1}{x}-2>0
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Let $M$ be a point on the straight line $L\equiv x+y=0$ such that $|AM-BM|$ is minimum,then the area of triangle AMB Consider two points $A(1,2)$ and $B(3,-1)$.Let $M$ be a point on the straight line $L\equiv x+y=0$ such that $|AM-BM|$ is minimum,then the area of triangle AMB equals
$(A)\frac{13}{4}\hspace{1cm}(B)\frac{13}{2}\hspace{1cm}(C)\frac{13}{6}\hspace{1cm}(D)\frac{13}{8}$
I asked a similar problem What is the coordinate of a point $P$ on the line $2x-y+5=0$ such that $|PA-PB|$ is maximum where $A=(4,-2)$ and $B=(2,-4)$ and i was answered and i understood the concept and now i am applying the same thing here(the method suggested by Juantheron),but i am not getting the desired answer.
$A(1,2),B(3,-1)$,let angle $AMB=\theta,M$ lies on the line $x+y=0$.Using the cosine law and the concept $\cos\theta\geq-1,$ we get $(AM+BM)^2\geq AB^2\Rightarrow|AM+BM|\geq AB$,but this is not helpful as this gives the minimum value of $|AM+BM|$.
Then i used the cosine law and the fact that $\cos\theta\leq 1$,we get $(AM-BM)^2\leq AB^2\Rightarrow|AM-BM|\leq AB$ but this is also not helpful as it gives the maximum value of $|AM-BM|$.
What should i do to find $M$ so that $|AM-BM|$ is minimum.Then i could find the area of triangle $AMB$.Please help me.
|
It can be clearly observed that $|PA - PB| \geq 0$. So, the least value is zero. But when is this achieved?
We know that the locus of points which are equidistant from two points is the perpendicular bisector of the line segment joining that two points. So, Let $P$ be any point on the perpendicular bisector of $AB$. So, from the definition of perpendicular bisector, we get that $PA = PB$, which implies that $PA - PB = 0$. So, the required point i.e. the point that satisfies the given condition lies on the perpendicular bisector $AB$. But the equation of $\perp$ bisector of $AB$ is
$$\left( y-\dfrac{1}{2} \right) = \dfrac{2}{3} \left(x-2 \right) \Leftrightarrow 4x-6y-5=0 $$
But this point lies on $L \equiv x+y=0$. So, the required point is actually the point of intersection of the lines $x+y=0$ and $ 4x-6y-5=0$. Therefore, the required point is $ \left( \dfrac{1}{2} , -\dfrac{1}{2} \right)$. So, the area of the triangle formed this point, $A$ and $B$ is equal to
$$\dfrac{1}{2} \left| \dfrac{1}{2} \left( 2+1 \right) + 1 \left(-1+\dfrac{1}{2}\right) +3 \left(-\dfrac{1}{2} - 2 \right) \right| = \dfrac{13}{4} $$
| {
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"timestamp": "2023-03-29T00:00:00",
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Integral $\int \sqrt{x+\sqrt{x^2+2}}dx$ $$\int \sqrt{x+\sqrt{x^2+2}}\ dx$$
I tried various solving methods but I am not coming forward. I unformed the term also to ${x+\sqrt{x^2+2} \over \sqrt{x+\sqrt{x^2+2}}}$ and even multiplied with ${\sqrt{x-\sqrt{x^2+2}} \over \sqrt{x-\sqrt{x^2+2}}}$.
Trigonometric and hyperbolic substitution didn't help either.
| HINT: set $t=\sqrt{x+\sqrt{x^2+2}}$ then we get $x=\frac{t^4-2}{2t^2}$ and we have $$dx=\frac{t^4+2}{t^3}dt$$
| {
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Let $k$ be a natural number . Then $3k+1$ , $4k+1$ and $6k+1$ cannot all be square numbers.
Let $k$ be a natural number. Then $3k+1$ , $4k+1$ and $6k+1$ cannot all be square numbers.
I tried to prove this by supposing one of them is a square number and by substituting the corresponding $k$ value. But I failed to prove it.
If we ignore one term, we can make the remaining terms squares. For example, $3k+1$ and $4k+1$ are both squares if $k=56$ (they are $13^2$ and $15^2$); $3k+1$ and $6k+1$ are both squares if $k=8$ (they are $5^2$ and $7^2$); $4k+1$ and $6k+1$ are both squares if $k=20$ (they are $9^2$ and $11^2$).
| Edit. There must be an error in the following answer, I cannot locate it yet, will leave it as is.
This is a reduction of the problem to another problem that I could not solve.
Are there positive integers $t,i,j$ with $j>i$ such that:
$\displaystyle \frac{t(t+1)}2=\frac{2i(j-i)j(j+i)}3$ ?
I do not know if the above problem is any easier or more elementary.
The reduction to this problem is elementary (using Pythagorean triples), as follows.
Let $x^2 = 3k+1$, $y^2 = 4k+1$, $z^2 = 6k+1$ (as in @WillJagy 's answer: I do not know if the other answers already might contain something that is equivalent to my attempt).
Then $(x^2)^2=(3k+1)^2=(3k)^2+6k+1=(3k)^2+z^2$, hence $x^2=m^2+n^2$, $3k=2m n$, $z=m^2-n^2$ (for some integers $m>n>0$), where I used that $z$ is odd, $k$ is even.
$1=x^2-3k=m^2+n^2-2m n=(m-n)^2$,
hence $m=n+1$, $x^2=2n^2+2n+1$, $z=2n+1$, $3k=2(n+1)n=2n^2+2n$.
Since $x^2=m^2+n^2$, we have $x=j^2+i^2$, and
either $m=2ij, n= j^2-i^2$, or $m= j^2-i^2, n=2ij$ for some integers $j>i>0$
(the case $i=0$ resulting only in the solution $(x,y,z)=(1,1,1)$).
Either way, $mn=(n+1)n=2ij(j^2-i^2)=2i(j-i)j(j+i)$.
$\displaystyle z^2-y^2=2k=\frac{4(n+1)n}3$, hence (using that $z=2n+1$)
$\displaystyle y^2=4n^2+4n+1-\frac{4(n+1)n}3 =4(n+1)n(1-\frac13)+1 =$
$\displaystyle\ =\frac{8(n+1)n}3+1 =\frac{16i(j-i)j(j+i)}3+1$. Thus
$\displaystyle y^2-1=(y-1)(y+1)=\frac{16i(j-i)j(j+i)}3$.
Using that $y$ is odd and letting $y=2t+1$, we obtain
$\displaystyle 2t(2t+2)=4t(t+1)=\frac{16i(j-i)j(j+i)}3$, hence
$\displaystyle \frac{t(t+1)}2=\frac{2i(j-i)j(j+i)}3$.
Obviously the latter has no solutions with $j>i>0$ (but I do not see how to show this elementary, perhaps it is difficult, perhaps I am overlooking something) since if it did have solutions then it would produce a solution to the original problem. It is easily seen that both sides of the latter equation are integers, indeed $\displaystyle\frac{t(t+1)}2$ is a triangular number, while if $3\not|\ i$ then $3|(j-i)j(j+i)$.
I posted the above equation as a separate question.
| {
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Calculate the limit of a function of two variables How to calculate this limit?
$$\lim_{(x,y) \rightarrow (0,0)}\frac{x^2+y^2}{\sqrt{x^2+y^2+9}-3}$$
| You have for $(x,y) \neq (0,0)$:
\begin{align}
\frac{x^2+y^2}{\sqrt{x^2+y^2+9}-3}
=\frac{(x^2+y^2)}{\sqrt{x^2+y^2+9}-3} \cdot \frac{\sqrt{x^2+y^2+9}+3}{\sqrt{x^2+y^2+9}+3}
=\sqrt{x^2+y^2+9}+3
\end{align}
Hence:
\begin{align}
\lim_{(x,y) \rightarrow (0,0)}\frac{x^2+y^2}{\sqrt{x^2+y^2+9}-3}
=\lim_{(x,y) \rightarrow (0,0)}\sqrt{x^2+y^2+9}+3=\color{blue}{6}
\end{align}
| {
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Definition of limit with$ f(x)=|x^3|$ Using the definition of the limit I tried to find the derivative of $f(x)=|x^3|$. I came up with: $$f'(x)=\frac{3x^5}{|x^3|}$$
Question: Why is the derivative (according to this answer) not defined for $x=0$ (division by zero) while $f'(0)$ actually exists, since: $$f'(0)=\lim_{x\to 0}\frac{|x^3|-|0^3|}{x-0}=0$$
Elaboration using the definition of the limit:
$$f'(x)=\lim_{h\to 0}\frac{|(x+h)^3|-|x^3|}{h}$$
$$=\lim_{h\to 0}\frac{\sqrt{(x+h)^6}-\sqrt{x^6}}{h}$$
$$=\lim_{h\to 0}\frac{(x+h)^6-x^6}{h(\sqrt{(x+h)^6}+\sqrt{x^6})}$$
$$=\lim_{h\to 0}\frac{h^5 + 6 h^4 x + 15 h^3 x^2 + 20 h^2 x^3 + 15 h x^4 + 6 x^5}{(\sqrt{(x+h)^6}+\sqrt{x^6})}$$
$$=\frac{6x^5}{2\sqrt{x^6}}=\frac{3x^5}{|x^3|}$$
Which is not defined for $x=0$.
| Your problem was saying that
$$=\lim_{h\to 0}\frac{h^5 + 6 h^4 x + 15 h^3 x^2 + 20 h^2 x^3 + 15 h x^4 + 6 x^5}{(\sqrt{(x+h)^6}+\sqrt{x^6})}=\lim_{h\to 0}\frac{6x^5}{2\sqrt{x^6}}$$
which is not true. If you set $h$ to $0$, then you must set all values of $h$ to $0$, and drop the limit all together. You cannot just set some of them to $0$. For exapmle, you cannot say
$$\lim_{h\to 0} \frac hh = \lim_{h\to 0} \frac0h = 0$$
because the first equality is not true.
What you get in your limit calculation is
$$=\lim_{h\to 0}\frac{h^5 + 6 h^4 x + 15 h^3 x^2 + 20 h^2 x^3 + 15 h x^4 + 6 x^5}{(\sqrt{(x+h)^6}+\sqrt{x^6})}=\\=\lim_{h\to 0}\frac{6x^5}{\sqrt{(x+h)^6}+\sqrt{x^6}}+\frac{h(h^4 + 6 h^3 x + 15 h^2 x^2 + 20 h^1 x^3 + 15 x^4)}{\sqrt{(x+h)^6}+\sqrt{x^6}}$$
Which is equal to the sum of limits of both sumands, if both limits exist.
Now, you can easily see that
$$\lim_{h\to0} \frac{6x^5}{\sqrt{(x+h)^6}+\sqrt{x^6}} = 3\mathrm{sign}(x)x^2$$ no matter what the value of $x$ is. If $x=0$, the limit becomes $$\lim_{h\to0} \frac{0}{\sqrt{h^6}}$$ and is equal to $0=3\cdot 0^2.$
For the second limit, you can also quickly see that it is $0$ no matter what the value of $x$ is. If $x\neq 0$, then the result is obvious, if $x=0$, then the limit is $$\lim_{h\to0}\frac{h^5}{\sqrt{h^6}}=\lim_{h\to 0} h^2\cdot\mathrm{sign}(h)=0$$
| {
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Evaluate $\int \frac{dx}{3\sin(x)+2\cos(x)+3}$ $$\int \frac{1}{3\sin(x)+2\cos(x)+3}\ \text{d}x$$
This is one of the question which appeared in my exam today and I tried solving this but just couldn't solve it.
I tried converting sine and cosine in terms of $\tan\left(\frac{x}{2}\right)$ but didn't help very well..
Does anyone have idea how to evaluate this integral?
| Notice, $$\int \frac{1}{3\sin x+2\cos x+3}\ dx$$
$$=\int \frac{1}{3\frac{2\tan\frac{x}{2}}{1+\tan^2\frac{x}{2}}+2\frac{1-\tan^2\frac{x}{2}}{1+\tan^2\frac{x}{2}}+3}\ dx$$
$$=\int \frac{1+\tan^2\frac{x}{2}}{\tan^2\frac{x}{2}+6\tan\frac{x}{2}+5}\ dx$$
$$=\int \frac{\sec^2\frac{x}{2}}{\left(\tan\frac{x}{2}+3\right)^2-4}\ dx$$
Let $\tan\frac{x}{2}=t\implies \frac{1}{2}\sec^2\frac{x}{2}\ dx=dt$
$$=\int \frac{2\ dt}{t^2-4}$$ $$=\frac{2}{4}\ln\left|\frac{t-2}{t+2}\right|+C$$
$$=\color{blue}{\frac{1}{2}\ln\left|\frac{\tan\frac{x}{2}-2}{\tan\frac{x}{2}+2}\right|+C}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Is my proof for this claim correct? Prove that for every $x,y\in\mathbb{R} $ such that $x,y>0$ and that fulfill $x\cdot y=1$, the following holds:
$$x+y=2 \quad iff \quad x=y=1$$
Steps I took:
$$\frac { xy }{ x } =\frac { 1 }{ x } \Rightarrow y=\frac { 1 }{ x } $$
$$1+\frac { 1 }{ 1 } =2$$
Is this good enough? It feels like something is missing.
Constructing my proof using input from answers below:
Assume: $x,y\in\mathbb{R}$ such that $x,y>0$ and $xy=1$
1) $x+y=2\Rightarrow x=y=1$
$y=2-x\Rightarrow x+2-x=2$; therefore, $2-x=1$
2) $x=y=1\Rightarrow x+y=2$
$x+(2-x)=2$
$x(2-x)=1\Rightarrow 2x-x^{ 2 }=1\Rightarrow x^2-2x+1=0$
$(x-1)^2=0\Rightarrow x=1$
$y=2-x\Rightarrow y=1$
| The approaches described in the answers already posted are the most natural ones. But the following may be of interest. By expanding, we can show that for any $x$ and $y$ we have
$$(x-y)^2+4xy=(x+y)^2.\tag{1}$$
If $xy=1$ and $x+y=2$, then substituting in (1) we get $(x-y)^2=0$, and therefore $x=y$. Now from $x+y=2$ we find that $x=y=1$.
So if our two given equations hold, then $x=y=1$. And it is easy to verify that $x=y=1$ does indeed satisfy the equations.
| {
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Sum of smallest roots
If 2 and -3 are the roots of a biquadratic equation, then the sum of the two smallest
roots of this equation is: $\{-1, -3, -5, \text{cannot be determined} \}$ those are the options.
The answer key says C = $-5$.
But,
$P(x) = (x-2)(x+3)(f(x))$ where $f(x)$ has degree two.
It is possible to have: $P(x) = (x-2)(x+3)(x+2)(x-3)$ which gives $-5$ but there are other options such as:
$P(x) = (x-2)(x+3)(x+1999919199919912919199191)(x+1919359237597373792397)$
What is it then?
| Biquartic functions have the form
$$
Q(z)=a_4z^4+a_2z^2+a_0
$$
Each solution $z_\pm(\neq 0)$ of $Q(z)=0$ splits in two: $\pm\sqrt{z_\pm}$.
So when $2$ and $-3$ are root, $-2$ and $3$ are too, so the sum of the smallest is $-5$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Integration of Arc Length I am asked to calculate the length of $\mathbf{x}(t)=(7,t,t^2)$ on the interval $[1,3]$.
Here is my work:
$$
\mathbf{x}'(t)=(0,1,2t)
$$$$
\int_1^3\|\mathbf{x}'(t)\|dt=\int_1^3\sqrt{4t^2+1}dt
$$
How do I calculate $$\int_1^3\sqrt{4t^2+1}dt$$
| Notice, let $$I=\int_{1}^{3}\sqrt{4t^2+1}\ dt=2\int_{1}^{3} \sqrt{t^2+\frac{1}{4}}\ dt$$
using integration by parts
one, we can find $$\color{blue}{\int \sqrt{x^2+a^2}\ dx=\frac{1}{2}\left(x\sqrt{x^2+a^2}+a^2\ln\left|x+\sqrt{x^2+a^2}\right|\right)+C}$$
Hence, applying proper limits, we get $$2\int_{1}^{3} \sqrt{t^2+\left(\frac{1}{2}\right)^2}\ dt=\frac{2}{2}\left[t\sqrt{t^2+\left(\frac{1}{2}\right)^2}+\frac{1}{4}\ln\left|t+\sqrt{t^2+\left(\frac{1}{2}\right)^2}\right|\right]_{1}^{3}$$
$$=3\sqrt{\frac{37}{4}}+\frac{1}{4}\ln\left|3+\sqrt{\frac{37}{4}}\right|-\sqrt{\frac{5}{4}}-\frac{1}{4}\ln\left|1+\sqrt{\frac{5}{4}}\right|$$
$$=\color{blue}{\frac{3\sqrt{37}-\sqrt{5}}{2}+\frac{1}{4}\ln\left(\frac{6+\sqrt{37}}{2+\sqrt{5}}\right)}$$
Proof:
let $$I=\int \sqrt{x^2+a^2}\ dx=\int \underbrace{\sqrt{x^2+a^2}}_{I}\cdot \underbrace{1}_{II}\ dx$$ using integration by parts $$I=x\sqrt{x^2+a^2}-\int \frac{2x}{2\sqrt{x^2+a^2}}x\ dx$$
$$I=x\sqrt{x^2+a^2}-\int \frac{(x^2+a^2)-a^2}{\sqrt{x^2+a^2}}\ dx$$
$$I=x\sqrt{x^2+a^2}-\int \sqrt{x^2+a^2}\ dx+a^2\int \frac{1}{\sqrt{x^2+a^2}}\ dx$$
$$I=x\sqrt{x^2+a^2}-I+a^2\int \frac{1}{\sqrt{x^2+a^2}}\ dx$$$$\implies I=\frac{1}{2}\left(x\sqrt{x^2+a^2}+a^2\int \frac{1}{\sqrt{x^2+a^2}}\ dx\right)\tag 1$$
Now, let $x=a\tan\theta\implies dx=a\sec^2\theta\ d\theta$ hence,
$$\color{red}{\int \frac{1}{\sqrt{x^2+a^2}}\ dx}=\int\frac{a\sec^2\theta\ d\theta}{\sqrt{a^2\tan^2\theta+a^2}}=\int\frac{a\sec^2\theta\ d\theta}{a\sec\theta}=\int \sec\theta\ d\theta$$$$=\ln\left|\sec\theta+\tan\theta\right|+c=\ln\left|\tan\theta+\sqrt{\tan^2\theta+1}\right|+c=\ln\left|\frac{x}{a}+\sqrt{\left(\frac{x}{a}\right)^2+1}\right|+c$$ $$=\ln\left|x+\sqrt{x^2+a^2}\right|+\ln|a|+c=\color{red}{\ln\left|x+\sqrt{x^2+a^2}\right|+c_1}$$
now, setting the value of integral: $\int\frac{1}{\sqrt{x^2+a^2}}\ dx$ in (1), we get $$I=\frac{1}{2}\left(x\sqrt{x^2+a^2}+a^2\ln\left|x+\sqrt{x^2+a^2}\right|\right)+C$$
hence, we get
$$\bbox[5pt, border:2.5pt solid #FF0000]{\color{blue}{\int\sqrt{x^2+a^2}\ dx=\frac{1}{2}\left(x\sqrt{x^2+a^2}+a^2\ln\left|x+\sqrt{x^2+a^2}\right|\right)+C}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1493383",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
A Diophantine Equation
Finding the number of $(a, b, c)$, where $a, b, c$ are positive
integers, that
$$
\frac{a^2+b^2-c^2}{ab}+\frac{c^2+b^2-a^2}{cb}+\frac{a^2+c^2-b^2}{ac}=2+\frac{15}{abc}
$$
I factored it in the following form
$(a+b+c)(2ab+2ac+2bc-a^2-b^2-c^2)=8abc+15$. But I can't make any progress from here.
Also I tried working with case when $a, b, c$ are sides of a triangle to get,
$\cos A+\cos C+\cos B=1+\frac{15}{2abc} < \frac{3}{2}$.
But that gives a lower bound on $abc$ which is not very helpful either.
| Hint
$$\Longleftrightarrow c(a^2+b^2-c^2)+a(b^2+c^2-a^2)+b(a^2+c^2-b^2)-2abc=15$$
and note that
$$c(a^2+b^2-c^2)+a(b^2+c^2-a^2)+b(a^2+c^2-b^2)-2abc=(c-a+b)(a-b+a)(c-a-b)$$
so
$$(c-a+b)(c-b+a)(c-a-b)=15=1\cdot 3\cdot 5$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1493935",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
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} |
Find the approximation for the interpolation of $f(x)$ by a polynomial of second degree
Assume that $f(x)$ has a minimum in the interval $x_{n-1}\leq x\leq x_{n+1}$ where $x_k=x_0+kh$, $k$ being an integer. Show that the interpolation of $f(x)$ by a polynomial of second degree yields the approximation $$f_n-\frac{1}{8}\left[\frac{(f_{n+1} - f_{n-1})^2}{f_{n+1} -2f_{n}+f_{n-1}}\right],$$ $f_k=f(x_k)$.
I know that the error term for second degree polynomial approximation
$$error \leq \frac{M_{3}}{(3)!}\max_{x\in[x_{k-1},x_{k+1}]}|(x-x_{k-1})(x-x_{k})(x-x_{k+1})|$$
where $$M_3=\max_{\xi\in[x_{k-1},x_{k+1}]}|f^3(\xi)|$$
Please help me to solve the problem.
| The function is approximated by a second order polynomial. Let:
$$
f(x) = a(x-x_n)^2+b(x-x_n)+c
$$
Specify for the points $x_k$ with $k=n-1,n,n+1$ ,
with $x_{n+1}-x_n=h$ and $x_{n-1}-x_n=-h$ :
$$
\begin{array}{l}
f_{n-1} = a h^2 - b h + c \\
f_{n} = c \\
f_{n+1} = a h^2 + b h + c
\end{array}
$$
Subtract the first equation from the last one:
$$
f_{n+1} - f_{n-1} = 2 b h \quad \Longrightarrow \quad b = \frac{f_{n+1} - f_{n-1}}{2h}
$$
Substitute this and $\,c\,$ into the last equation:
$$
f_{n+1} = a h^2 + \frac{f_{n+1} - f_{n-1}}{2h} h + f_{n}
\quad \Longrightarrow \quad a = \frac{f_{n+1} -2 f_n + f_{n-1}}{2h^2}
$$
The minimum of $f(x)$ - find it e.g. by differentiation - is obtained for $x-x_n = -b/(2a)$ .
And the minimum is:
$$
f(x_n-b/(2a)) = a\left[-\frac{b}{2a}\right]^2 + b\left[-\frac{b}{2a}\right] + c = c-\frac{b^2}{4a}
$$
Now substitute the values found for $\{a,b,c\}$ and you're done.
EDIT. Combine the condition $x_{n-1}\leq x\leq x_{n+1}$ with the place of the minimum:
$$
x-x_n = -b/(2a) = -\frac{(f_{n+1}-f_{n-1})/(2h)}{2(f_{n+1} -2 f_n + f_{n-1})/(2h^2)}
\quad \Longrightarrow \\
x_{n-1} \leq x_n-\frac{f_{n+1}-f_{n-1}}{f_{n+1} -2 f_n + f_{n-1}} \frac{1}{2} h \leq x_{n+1}
\quad \Longrightarrow \\
-h \leq \frac{f_{n+1}-f_{n-1}}{f_{n+1} -2 f_n + f_{n-1}} \frac{1}{2} h \leq +h
\quad \Longrightarrow \\
\left| f_{n+1}-f_{n-1} \right| \leq 2\left|f_{n+1} -2 f_n + f_{n-1}\right|
$$
Unless $f$ is a constant, this ensures that the denominator is nonzero in:
$$
f_n-\frac{1}{8}\left[\frac{(f_{n+1} - f_{n-1})^2}{f_{n+1} -2f_{n}+f_{n-1}}\right]
$$
Also note that still there can be a maximum instead of a minimum, all depending on the sign of the denominator (which is $\sim$ the discretization of the second order derivative). From the question as it is formulated therefore can be concluded that $f_{n+1} -2f_{n}+f_{n-1}$ is positive, or:
$$ f_n < \frac{f_{n+1}+f_{n-1}}{2} $$
So $f_n$ is smaller than the mean of its neighbors. If $f_n$ is itself the minimum, then its neighbors are equal: $f_{n-1}=f_{n+1}$ . If not, then the minimum is $< f_n$ .
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1499082",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How to find the maximum value of $3^x + 5^x - 9^x + 15^x - 25^x$ as $x$ varies over the reals?
How to find the maximum value of $3^x + 5^x - 9^x + 15^x - 25^x$ as x
varies over the reals ?
Suggestions please!
| Almost Same as grant_chat.
Let $a=3^x$ and $b=5^x\;,$ Then expression convert into $f(a,b) = a+b-a^2+ab-b^2$
So $$f(a,b) = -\frac{1}{2}[2a^2+2b^2-2ab-2a-2b]$$
So $$f(a,b) = -\frac{1}{2}\left[(a^2-2a+1)+(b^2-2b+1)+(a^2+b^2-2ab)-2\right]$$
So $$f(a,b) = -\frac{1}{2}\left[(a-1)^2+(b-1)^2+(a-b)^2-2\right]$$
So $$f(a,b) = \frac{1}{2}\left[2-(a-1)^2-(b-1)^2-(a-b)^2\right]\leq 1$$
And equality hold when
$$(a-1)^2=0\Rightarrow a=1\Rightarrow 3^x=1=3^0$$ and $$(b-1)^2=0\Rightarrow b=1\Rightarrow 5^x=1=5^0$$ and $$(a-b)^2=0\Rightarrow a=b\Rightarrow 3^x=5^x$$
So from above we get $x=0$ for which $f(x)=3^x+5^x-4^x+15^x-9^x.$ is maximum.
And $\max{f(x)}=1$ at $x=0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1504006",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
Find the indefinite integral $\int {dx \over {(1+x^2) \sqrt{1-x^2}}} $ Find the indefinite integral
$$\int {dx \over {(1+x^2) \sqrt{1-x^2}}} $$
Is there a smart substitution or algebric trick that I'm missing? Because integration by parts hasn't helped..
| Substitute $u = \frac{x\sqrt2}{\sqrt{1-x^2}}, \frac{dx}{du} =\frac{(1-x^2)^{3/2}}{\sqrt2} $
$\begin{align*}
\int {\frac1 {(1+x^2) \sqrt{1-x^2}}dx} &= \frac1{\sqrt2}\int \frac{1-x^2}{1+x^2} du \\
&= \frac1{\sqrt2}\int \frac1 {\frac{1+x^2}{1-x^2}} du \\
&= \frac1{\sqrt2}\int \frac1 {1+\frac{2x^2}{1-x^2}} du \\
&= \frac1{\sqrt2}\int \frac1 {1+ (\frac{x\sqrt2}{\sqrt{1-x^2}})^2} du \\
&= \frac1{\sqrt2}\int \frac1 {1+ u^2} du \\
&= \frac1{\sqrt2} \tan^{-1} \left(\frac{x\sqrt2}{\sqrt{1-x^2}} \right) + C
\end{align*}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1505148",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
Existence of a function in $C([0,1], \mathbb{R}), ||.||_{\infty}$ s.t. $2f(x) -f(x)^2 + f(x^2)=g(x)$ Let $g \in E= C([0,1], \mathbb{R}), ||.||_{\infty}$ s.t. $\|g\|_\infty< \frac{1}{4}$. I need to show that there exists $f$ in the same set such that
$$2f(x) - f(x)^2 + f(x^2)=g(x).$$
Can anyone please provide me with a hint? Should I maybe look into $\Gamma :E \to E$, $\Gamma (f)= 2f(x)-f(x)^2+ f(x^2)$?
| Let $X = \{ f\in E: \|f\|_\infty \le C\}$, where $C<\frac 12$ is close to $\frac 12$ such that
$$\left(C-\frac 12\right)^2 \le \frac 14 - \| g\|_\infty.$$
Consider the mapping
$$L : X \to X, \ \ Lf (x) = \frac{1}{2} (g(x) - f(x^2) + f(x)^2).$$
First we check that $L$ really maps $X$ to $X$. Note that if $f\in X$,
$$\begin{split} \|Lf\|_\infty &\le \frac{1}{2}( \| g\|_\infty + C + C^2) = \frac{1}{2} (C^2 - C + \frac 14 - \frac 14 + \|g\|_\infty + 2C)\\
&= \frac{1}{2}( (C-\frac 12)^2 - (\frac 14 - \|g\|_\infty) + 2C)\\
&\le C
\end{split}$$
Thus $Lf \in X$. Now we check that $L$ is a contractoin:
$$\begin{split}
\|Lf_1 - Lf_2\|_\infty &= \left\| \frac{1}{2} \left(f_2(x^2) - f_1(x^2) + f_1(x)^2 - f_2(x)^2 \right)\right\|_\infty\\
&\le \frac 12 \|f_1 - f_2\|_\infty + \frac 12 \left(C+C \right)\|f_1 - f_2\|_\infty \\
&= \frac{1}{2} (1+ 2C) \|f_1 - f_2\|_\infty.
\end{split}$$
Since
$$\frac{1}{2} (1+ 2C) < \frac{1}{2}(1+ 1) =1$$
as $C<\frac 12$, $L$ is a contraction. As $X$ is a complete metric space, there is $f\in X$ so that $Lf = f$, by the contraction mapping theorem. But this is the same as
$$2 f(x) - f(x)^2 + f(x^2) = g(x).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1506193",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
roots of the quadratic equation $(a^4+b^4)x^2+4abcdx+(c^4+d^4)=0$ are real
If $a,b,c,d\in \mathbb{R}$ and roots of the quadratic equation $(a^4+b^4)x^2+4abcdx+(c^4+d^4)=0$
are real.Then prove that roots are equal.
$\bf{My\; Try::}$ Given $(a^4+b^4)x^2+4abcdx+(c^4+d^4)=0\;$ Then we can write it as
$$\left[(a^2x)^2+c^4-2a^2c^2x+(b^2x)^2+d^4-2b^2d^2x+2a^2c^2x+2b^2d^2x+4abcdx\right]=0$$
So $$(a^2x-c^2)^2+(b^2x-d^2)^2+2x(ac+bd)^2=0$$
Now I did not understand How can I proceed further.
Although I have a knowledge of $\bf{Discriminant\; Method.}$
So plz explain me above method which i am trying above.
Thanks.
| We have
$$\begin{align}&(a^4+b^4)(c^4+d^4)\\&\ge(a^2c^2+b^2d^2)^2\\&\ge \left(2\sqrt{(a^2c^2)\times (b^2d^2)}\right)^2\\&=4a^2b^2c^2d^2\end{align}$$
Hence, the discriminant
$$(4abcd)^2-4(a^4+b^4)(c^4+d^4)$$
is non-positive.
Thus, the discriminant has to be $0$, which implies that the real roots are equal.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1506409",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
} |
Prove n-th derivative function Let $f(x)=(3x+5)^{1/2}$. Obtain and prove a formula for the $n$-th derivative $f^{(n)}$
I may need to find some derivatives of the function, and prove by induction, but I do not know how to do these.
F’(x) = 3/2(3x+5)^(-1/2)
F''(x) = -9/4(3x+5)^(-3/2)
F'''(x) = 81/8(3x+5)^(-5/2)
F''''(x) = -1215/16(3x+5)^(-7/2)
Pattern:
F^(n)(x)=(1/2−(n−1))(1/2−(n−2))...(1/2−(n−n))3^n(3x+5)(1/2−n)
These repeated steps ...(1/2−(n−n)) continue until the number(n - number) reaches n value. E.g. if n is 1 then F'(x)=(1/2-(1-1))(3x+5)^(-1/2)
| Let
$f(x)
=(ax+b)^c
$.
What is
$f^{(n)}(x)
$?
Each derivative
decreases the exponent
by 1.
So,
let
$f^{(n)}(x)
=g(n)(ax+b)^{c-n}
$.
Then
$f^{(n+1)}(x)
=g(n)(c-n)a(ax+b)^{c-n-1}
$,
so that
$g(n+1)
=g(n)(c-n)a
$,
or
$g(n+1)/g(n)
=(c-n)a
$.
Since
$g(0) = 1$,
$\begin{array}\\
g(n)
&=\prod_{k=0}^{n-1} g(k+1)/g(k)\\
&=\prod_{k=0}^{n-1} (c-k)a\\
&=a^{n}\prod_{k=0}^{n-1} (c-k)\\
&=a^{n}\Gamma(c+1)/\Gamma(c-n+1)\\
&=a^{n}(c)!/(c-n)!
\qquad\text{if } c \text{ is an integer}\\
\end{array}
$
If
$c = u/v$,
$\begin{array}\\
g(n)
&=a^{n}\prod_{k=0}^{n-1} (c-k)\\
&=a^{n}\prod_{k=0}^{n-1} (u/v-k)\\
&=a^{n}\prod_{k=0}^{n-1} (u-kv)/v\\
&=(a/v)^{n}\prod_{k=0}^{n-1} (u-kv)\\
\end{array}
$
If $c=\frac12$,
so $u=1, v=2$,
$\begin{array}\\
g(n)
&=\frac{a^n}{2^n}\prod_{k=0}^{n-1} (1-2k)\\
&=\frac{(-1)^{n}a^n}{2^n}\prod_{k=0}^{n-1} (2k-1)\\
&=\frac{(-1)^{n+1}a^n}{2^n}\prod_{k=1}^{n-1} (2k-1)\\
&=\frac{(-1)^{n+1}a^n}{2^n}\prod_{k=1}^{n-1} (2k-1)\frac{2k}{2k}\\
&=\frac{(-1)^{n+1}a^n}{2^n}\frac{\prod_{k=1}^{n-1} (2k-1)(2k)}{\prod_{k=1}^{n-1}(2k)}\\
&=\frac{(-1)^{n+1}a^n}{2^n}\frac{(2n-2)!}{2^{n-1}(n-1)!}\\
&=\frac{(-1)^{n+1}a^n}{2^{2n-1}}\frac{(2n-2)!}{(n-1)!}\\
\end{array}
$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1508523",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Inverse function of $f(x) = \frac{x+5}{x-2}$ Find the inverse of the function
$f(x) = \frac{x+5}{x-2}$
Here's what I have so far:
$y = \frac{x+5}{x-2}$
$x = \frac{y+5}{y-2}$
$(x)(y-2) = (y+5)$
but this seems to be a dead end.
How should I approach this?
Thank you!
| I'm just continuing from where you stopped
$(x)(y-2) = (y+5)$
$xy-2x = y+5$
$xy-y=2x+5$
$y(x-1)=2x+5$
$y=\frac{2x+5}{x-1}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1511766",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
How to compute this gross limit.
How do I compute this limit?
$$
\lim_{n \to \infty}
\frac{\left(1 + \frac{1}{n} + \frac{1}{n^2}\right)^n -
\left(1 + \frac{1}{n} - \frac{1}{n^2}\right)^n
}{
2 \left(1 + \frac{1}{n} + \frac{1}{n^2}\right)^n -
\left(1 + \frac{1}{n} - \frac{1}{n^2 + 1}\right)^n -
\left(1 + \frac{1}{n} - \frac{1}{n^2 (n^2 +1)}\right)^n
}
$$
I think I got the correct limit by using fast converging limits to $e$.
In particular I used truncated Taylor series for the sqrt and 4th root.
Or squares and bisquares.
Example
$(1+1/2n)^{2n}$
Becomes
$(1 + 1/n + 1/4n^2)^n.$
In combination with l'hopital it gives me the answer.
But I guess that is not a very good (fast) method.
| Hint: First show $(1+1/n +o(1/n))^n \to e.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1512063",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 0
} |
Show that $\frac{ -1 }{ 2 } \le \frac{ x }{ 1+x^2 } \le \frac{ 1 }{ 2 }$ So I'm trying to show that:
$\dfrac{ -1 }{ 2 } \le \dfrac{ x }{ 1+x^2 } \le \dfrac{ 1 }{ 2 }$
for every value of x.
I know I have to use mean value theorem so I tried to show it with cases. First I tried showing that $\dfrac{-1}{2} \le \dfrac{x}{1+x^2}$ and then $\dfrac{1}{2} \ge \dfrac{x}{1+x^2}$ using MVT. Is that correct? Can someone guide me?
Thanks!
| WLOG $x=\tan y$
$$\implies\dfrac{2x}{1+x^2}=\sin2y$$
Alternatively,
$$\dfrac x{1+x^2}=\dfrac1{x+\dfrac1x}$$
Now if $x>0,\dfrac{x+\dfrac1x}2\ge\sqrt{x\cdot\dfrac1x}=1$
What if $x<0?$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1515563",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
proving $ \sqrt 2 + \sqrt 3 $ is irrational I need to proof that $\sqrt{3} + \sqrt{2}$ is irrational, without using the fact that an irrational number plus a rational number equals irrational. also, i can't use the rational root theorem. that's why i posted a new question..
thanks for help!
| Assume to get contradiction $\sqrt{3} + \sqrt{2}=\frac{p}{q}$ where $p,q\in \mathbb{N}$. $\sqrt{3} + \sqrt{2}=\frac{p}{q} \Rightarrow (\sqrt{3} + \sqrt{2})^2=(\frac{p}{q})^2 \Rightarrow 3+2\sqrt{6}+2=\frac{p^2}{q^2} \Rightarrow \sqrt{6}=\frac{1}{2}(\frac{p^2}{q^2}-5)$. So now we have $\sqrt{3}+\sqrt{2}\not \in \mathbb{Q} \iff \sqrt{6} \not \in \mathbb{Q}$. Let us show $\sqrt{6} \not \in \mathbb{Q}$. Assume $\sqrt{6}=\frac{r}{t}$ where $r,t\in \mathbb{N}$ s.t. $gcd(r,t)=1$. $\sqrt{6}=\frac{r}{t} \Rightarrow 6=\frac{r^2}{t^2} \Rightarrow r^2=6t^2 \Rightarrow 2\mid r \Rightarrow r=2m \Rightarrow 6t^2=(2m)^2=4m^2$. So we get $4 \mid 6t^2$ but we know $6=2\cdot 3$ which implies $2^2 \not \mid 6$ thus we must have $2\mid t^2 \Rightarrow 2\mid t \Rightarrow t=2n$. But now we have $gcd(r,t)=gcd(2m,2n)\geq 2 > 1 \Rightarrow \Leftarrow$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1516482",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Proving that the limit of a function doesn't exist using negation of epsilon delta $\lim \limits_{x \to \frac{1}{2}}\frac{1}{4x-2}$
I want to use the negation, $\exists \epsilon>0$ such that $ \forall \delta>0$ , $\lvert\frac{1}{4x-2}-L \rvert \ge \epsilon$, $\forall x$ with $0<\lvert x-\frac{1}{2} \rvert <\delta$
So can I say that because $\lvert \frac{1}{4x-2}\rvert =\lvert \frac{1}{4(x-\frac{1}{2})} \rvert = \frac{1}{4}\lvert \frac{1}{x-\frac{1}{2}} \rvert \ge \epsilon $
Then $\frac{1}{4} \ge \epsilon \lvert x-\frac{1}{2} \rvert$
Can I then let $\epsilon =\frac{1}{4 \delta}$
| Let $\epsilon = 1$, then for any $\delta > 0$, you can find an $x$ such that: $|x-\dfrac{1}{2}| < \delta$, and $\dfrac{1}{\left|4x-2\right|} \geq 1\iff \left|x-\dfrac{1}{2}\right| \leq \dfrac{1}{4}$. You can take $x$ such that $|x-\dfrac{1}{2}| < \min{{\dfrac{1}{4}, \delta}}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1516969",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
} |
Simplifying using the rule of logs How would you simplify: $\left[\frac{16}{5}\ln(x + 2) - \frac{1}{5}\ln(x - 3) - \ln x \right]_4^6$ and put it in the form $\ln\frac{m}{n}$. Stating the values of m and n.
Note:
$a\ln b = \ln b^a$
$\ln a - \ln b = \ln \frac{a}{b}$
I just cant get the answer, my answer is always a decimal.
| Notice, $$\left[\frac{16}{5}\ln(x+2)-\frac{1}{5}\ln(x-3)-\ln (x)\right]_{4}^{6}$$
$$=\frac{1}{5}\left[\ln(x+2)^{16}-\ln(x-3)-\ln (x^5)\right]_{4}^{6}$$
$$=\frac{1}{5}\left[\ln\left(\frac{(x+2)^{16}}{x^5(x-3)}\right)\right]_{4}^{6}$$
$$=\frac{1}{5}\left[\ln\left(\frac{(6+2)^{16}}{6^5(6-3)}\right)-\ln\left(\frac{(4+2)^{16}}{4^5(4-3)}\right)\right]$$
$$=\frac{1}{5}\left[\ln\left(\frac{8^{16}}{3\cdot 6^5}\right)-\ln\left(\frac{6^{16}}{4^5}\right)\right]$$
$$=\frac{1}{5}\ln\left(\frac{8^{16}\cdot 4^{5}}{3\cdot 6^{5}\cdot 6^{16}}\right)$$
$$=\ln\left(\frac{2^{48}\cdot 2^{10}}{3\cdot 3^{21}\cdot 2^{21}}\right)^{1/5}$$
$$=\ln\left(\frac{2^{37}}{3^{22}}\right)^{1/5}$$
$$=\ln\left(\frac{2^{37/5}}{3^{22/5}}\right)$$
now, comparing with $\ln\left(\frac{m}{n}\right)$, one should get
$$m=2^{37/5}, \ \ \ \ n=2^{22/5} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1518732",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the inverse of $x^3+x$ mod $x^4+x+1$ in $F_2[x]$ I did a huge amount of algebra and I think I ended up with the wrong answer. I got $(-x^4+3x^3-x^2-2x)$ which is obviously not the right answer.
So I started by saying I need $a(x)$ and $b(x)$ such that
$a(x^4+x+1) + b(x^3 +x)=1$ and $b(x)$ will be the inverse. Then I tried to use the extended Euclidiean algorithm which took forever. I got:
$$x^4+x+1 = (x^3+x)(x) + (-x^2+x+1)$$
$$x^3+x = (-x^2+x+1)(-x-1) + (3x+1)$$
Then since $3x \equiv x \pmod 2$,
$$(-x^2+x+1) = (x+1)(-x+2) + (-1)$$
Since -1 = 1, I stopped there and got the gcd as 1. However when I did the extended version it did not seem to work. Where am I going wrong ?
| The Euclidean Algorithm can be done using matrices:
$$\left[\begin{array}{cc|c}1&0&x^4 + x + 1\\0&1&x^3+x\end{array}\right]$$
Now take row 1 and add/subtract $x$ times row 2 from it:
$$\left[\begin{array}{cc|c}1&x&x^2 + x + 1\\0&1&x^3+x\end{array}\right]$$
Now take row 2 and add/subtract $x+1$ times row 1 from it:
$$\left[\begin{array}{cc|c}1&x&x^2 + x + 1\\x+1&x^2+x+1&x+1\end{array}\right]$$
Now take row 1 and add/subtract $x$ times row 2 from it:
$$\left[\begin{array}{cc|c}x^2+x+1&x^3+x^2&1\\x+1&x^2+x+1&x+1\end{array}\right]$$
Now read off row 1 as an equation, giving the Bezout coefficients:
$$(x^2+x+1)(x^4+x+1)+(x^3+x^2)(x^3+x)=1,$$
which implies that $(x^3+x)^{-1}\equiv(x^3+x^2)\pmod{x^4+x+1}$.
(Note that this also works for the gcd of two or more numbers; just make sure you're subtracting an integer times a row.)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1522068",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Express $\frac{1-x}{(x-1)^2+y^2}-\bigg(\frac{y}{(x-1)^2+y^2}\bigg)i$ in the form of $f(z)$ I need to express $f(z)$ from the form $\color{blue}{u+vi}$ to the form $\color{blue}z$ for example if: $g(z)=\frac{1}{x+yi}$ so $ =g(z)=\frac{1}{z}$
$$f(z)=\underbrace {\frac{1-x}{(x-1)^2+y^2}}_{=u(x,y)}+\underbrace {\bigg(-\frac{y}{(x-1)^2+y^2}\bigg)}_{=v(x,y)}i$$
My try:
$$f(z)=\frac{1-x-yi}{(x-1)^2+y^2}$$
$$=\frac{1-\overbrace{x-yi}^{=\bar z}}{\underbrace{x^2+y^2}_{=|z|^2}-2x+1^2}$$
$$-\frac{1-\bar z}{|z|^2-2x+1}$$
I'm stuck here
| HINT: $$f(z)=\frac{1-x-yi}{(x-1)^2+y^2}$$
$$=\frac{-[(x-1)+iy]}{[(x-1)+iy][(x-1)-iy]}$$
$$=\frac{-(z-1)}{(z-1)(\bar z-1)}$$
I hope the rest can be understood.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1522760",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Show that, under the mapping $f(z)=\frac{1}{z+2}$, all points on the circle $|z|=2$ get mapped onto $\{z\in \mathbb{C}: \Re(z)=\frac{1}{4}\}$.
Show that, under the mapping $f(z)=\frac{1}{z+2}$, all points on the circle $|z|=2$ get mapped onto the set $\{z\in \mathbb{C}: \Re(z)=\frac{1}{4}\}$.
My attempt:
Let $z= x+iy$ then we have \begin{align}f(z) &= \frac{1}{z+2}\\ &=\frac{\bar{z}-2}{(z+2)(\bar{z}-2)} \\ &= \frac{(x-2)+i(-y)}{(x+2)^2 +y^2} \\ &= \frac{(x-2)}{(x+2)^2+y^2} +i(-y)\end{align}
Now let \begin{align}u(x,y)= \frac{(x-2)}{(x+2)^2 +y^2}~~ \text{and}~~v(x,y) = -y\end{align}
then we have $$f(z)= u(x,y) + iv(x,y)$$
Now the part that I am stuck with, is showing that the circle of radius 2 centered at the origin gets mapped onto the vertical line through $\frac{1}{4}$ on the real axis.
What I've tried is \begin{align}x^2 + y^2 &= 4\end{align}
Then, since we're interested in the real part, we now only look at \begin{align}u(x,y) &= \frac{(x-2)}{(x+2)^2 + y^2} \\ &= \frac{(x-2)}{x^2 + 4x +4 +y^2} \\ &= \frac{(x-2)}{4x + 8} \\ &= \frac{(x-2)}{4(x+2)}\end{align}
Now this is where I see I have a problem - I cannot cancel those two out, did I make a mistake somewhere?
| I think you have an algebra error in your third line:
$$(z+2)(\bar{z}-2)=x^2+y^2-4iy-4\neq(x+2)^2+y^2.$$
As suggested, try using polar coordinates: the circle of radius 2 can be written $2e^{it}$ for $-\pi<t<\pi$. Plug this in to $f$ and take the real part:
$$Re\left(f(2e^{it})\right)=\frac{1}{2}Re\left(\frac{1}{e^{it}+1}\right)=\cdots=\frac{1}{4}.$$
You should be able to get that the whole line is filled in by considering $Im(f(2e^{it}))$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Functional equation $f(x^3) - f(x^3 - 2) = (f(x))^2 + 12$ $$f(x^3) - f(x^3 - 2) = (f(x))^2 + 12$$
Given the functional equation above, I am trying to find the value of $f(3)$.
I do not remember the exact statement of the problem precisely, so I am not sure whether an initial value of the form $f(a) = p$ was provided.
I started off by finding the degree of $f(x)$ on both sides, which must be equal. If the degree of $f(x)$ is $n$, then the right side clearly has degree $2n$, while the left side seems to have a degree of the form $3(n-1)$.
Equating this, I got that $f(x)$ is a polynomial of degree $3$.
Now I am stuck — I thought of setting $f(x) = a x^3 + b x^2 + cx + d$ but I don't believe that will help much.
EDIT: Ok I did that, and substituted back into the functional equation to get:
$$a x^9-a \left(x^3-2\right)^3+b x^6-b \left(x^3-2\right)^2+c x^3-c \left(x^3-2\right)=\left(a x^3+b x^2+c x+d\right)^2+12$$
Ew. I guess I could expand and equate coefficients on both sides (and indeed, churching through that with Mathematica gives $6x^3 - 6$), but seeing as this is an AIME-esque problem, there should probably be an easier way. Is there one?
| Note that the left side depends only on $x^3$, so the right side should also. Thus we look for a function of the form $f(x) = g(x^3)$. As you say, $f$ should have degree $3$, so $g$ should have degree $1$. If $f(x) = a x^3 + b$, then taking $x = 0$ we get $b - (-8a + b) = b^2 + 12$, i.e.
$8 a = b^2 + 12$. Substitute $a= (b^2+12)/8$ in to the equation,
expand, and subtract right from left: I get
$$ \left(-\frac{b^4}{64} + \frac{3}{8} b^2 + \frac{27}{4}\right) x^6
+ \left(-\frac{b^3}{4} - \frac{3}{2} b^2 - 3 b - 18 \right) x^3 = 0
$$
so both the coefficients of $x^6$ and $x^3$ must be $0$. Factoring the first, which is a quadratic in $b^2$:
$$ -\frac{b^4}{64} + \frac{3}{8} b^2 + \frac{27}{4} = - \frac{1}{64} (b^2 - 36)(b^2 + 12) = 0 $$
so if we're looking for real solutions, $b = \pm 6$. Of those only $b = -6$ makes the other coefficient $0$. Thus the only polynomial solution
with real coefficients is $b = -6$, $a = 6$: $f(x) = 6 x^3 - 6$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Fermat's infinite descent for finding the squares that sum to a prime Fermat's theorem on sum of two squares states that an odd prime $p = x^2 + y^2 \iff p \equiv 1 \pmod 4$
Applying the descent procedure I can get to $a^2 + b^2 = pc$ where $c \in \mathbb{Z} \gt 1$
I want $c = 1$, so how do I proceed from here? How do I apply the procedure iteratively?
Example:
$$
p = 97
$$
$$97 \equiv 1 \pmod 4 \implies \left(\frac{-1}{97}\right) = 1 \implies x^2 \equiv -1 \pmod {97}$$ has a solution
$$x^2 + 1 \equiv 0 \pmod {97}$$
$$x^2 + 1 = 97m$$
We find an $x,m$ that solves the equation.
$$x = 75, m = 58$$
Now, we pick an $a,b$ such that $\frac{-m}{2} \leq a,b \leq \frac{m}{2}$
$$a \equiv x \pmod m = 17$$
$$b \equiv y \pmod m = 1$$
Observations:
*
*$ a^2 + b^2 \equiv x^2 + 1 \equiv 0 (\mod m)$
*$ (a^2 + b^2) = mc$
*$ (x^2 + 1) = mp$
Plugging in $a,b,m$ for 2, we get $c = 5$
By this identity, we know that
$(a^2 + b^2)(c^2 + d^2) = (ac + bd)^2 + (ad - bc)^2$
**$(a^2 + b^2)(x^2 + 1^2) = (ax + b)(a - bx) = m^2pc$
Dividing ** by $m^2$, $pc = (\frac{ax+b}{m})^2 + (\frac{a-bx}{m})^2$
Plugging in $a,b,m,p,c$ we get that $22^2 + (-1)^2 = 97*5$
So we have two squares that add up to 5 times our $p$. How do we turn the 5 into a 1? What is the next step in the descent?
| We need to know that if $q$ is prime and $q\equiv 3\pmod 4$ then $q|(x^2+y^2)\to (q|x\land q|y)$. Now for prime $p\equiv 1\pmod 4$ we have $c p=a^2+b^2$ where $p/2>a>0<b<p/2$. This implies $0<c<p$. We ask for the smallest possible positive $c$. We show that if $c>1$ then $c$ is not the minimum. First,if $q$ is prime and $q|c$, where $q\equiv 3\pmod 4$, then $q|c\to q|(a^2+b^2) \to (q|a\land q|b) \to q^2|c\to (c/q^2)p=(a/q)^2+(b/q)^2$; and $0<c/q^2<c$ so $c$ is not the minimum. Second, suppose that every prime $r$ such that $r<p$ and $r\not \equiv 3\pmod 4$ is the sum of two squares. (This is the Inductive Hypothesis.) Then if the prime $r$ divides $c$ and $r\not \equiv 3\pmod 4$ then $r=(s^2+t^2) $ for some positive integers $s,t,$....Recall that $r$ also divides $a^2+b^2$. Let $s\equiv i t\pmod r$ and $a\equiv j b\pmod r$ where $i^2\equiv j^2\equiv -1\pmod r$ and $j\equiv \pm i\pmod r. $ In the case $j\equiv i\pmod r$ we have, with $c'=c/r,$ $$ p c' r^2=p c r=c'(a^2+b^2)(s^2+t^2)=c'((a s +b t)^2+(a t-b s)^2).$$ Now $(a s +b t)\equiv j i b t+b t\equiv i^2 b t+ b t\equiv 0\pmod r$ and similarly $a t-b s\equiv 0\pmod r$. Therefore for some $u,v$ we have $$0\ne (r u)^2+(r v)^2=(a s+b t)^2+(a t-b s)^2=p c' r^2$$ from which $$0\ne p c'=u^2+v^2$$ Since $0<c'<c$, we see that $c$ is not minimum.....In the case $j\equiv -i\pmod r$ we use the equality $(a^2+b^2)(s^2+t^2)=(a s-b t)^2+(a t+b s)^2.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1525113",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Examine $\lim_{(x,y)\to (0,0)}\frac{ax^2+bxy+cy^2}{x^2+y^2}=0$ The question is what can we conclude about $a,b,c$ if $\displaystyle \lim_{(x,y)\to (0,0)}\frac{ax^2+bxy+cy^2}{x^2+y^2}=0$. Let $f(x,y)$ be the given function.
Firstly, if we switch to polar coordinates we get that $f(x,y)=f(\theta)=a\cos^2\theta+b\cos\theta \sin\theta+c\sin^2\theta$. I can't see how this can give us more information.
We can try this approach: the limit exists and it is equal to $0$, so if we approach $(0,0)$ from any path, the function will take the value $0$. For example, we can examine the paths
$y=x$ then $\displaystyle f(x,x)=\frac{ax^2+bx^2+cx^2}{x^2+x^2}=\frac{a+b+c}{2}$ is $x$ independent so it must be equal to $0\Leftrightarrow a+b+c=0$. I tried some function examples when $a,b,c$ follow that condition in Wolfram Alpha but their limit didn't exist when $(x,y)\to(0,0)$. Therefore $a+b+c=0$ doesn't seem to be a sufficient condition.
another path: $y=0$ then $\displaystyle f(x,0)=\frac{ax^2}{x^2}=a$ it must be equal to $0$.
Similarly the path $x=0$ gives us $\displaystyle f(0,y)=\frac{cy^2}{y^2}=c=0$ and by combining the results we get that $a=b=c=0$.
This conclusion makes me think that I've done a mistake somewhere.
Any hints?
Thank you in advance!
| Consider a linear approach $y=mx$. Since the multivariable limit exists, we find that
$$
\lim_{x \to 0} \frac{ax^2+bmx^2 + cm^2x^2}{x^2+m^2x^2} = \frac{a + bm+cm^2}{1+m^2}
$$
exists and does not depend on $m$. However, obviously this limit will depend on $m$ if we don't impose some conditions on $a,b,c$.
Taking $m=0$ the limit must equal $a$ and taking $m\to \infty$ the limit must equal $c$, so we must have $a=c$. Now the limit reduces to
$$
\frac{a(1+m^2) + bm}{1+m^2} = a + \frac{b m}{1+m^2}
$$
which must equal $a$, so we find $b=0$.
Thus we have shown that in order for the multivariable limit to exist, we must have $a=c$ and $b=0$. Moreover, for any $a$
$$
\lim_{(x,y)\to(0,0)}\frac{ax^2 + 0xy + ay^2}{x^2+y^2} = \lim_{(x,y) \to (0,0)} a = a
$$
which does exist. Thus, we have show that the limit exists if and only if $a=c$ and $b=0$, and that when these conditions hold the limit is $a$.
| {
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"timestamp": "2023-03-29T00:00:00",
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If $x$ is a positive integer such that $x(x+1)(x+2)(x+3)+1=379^2$, find $x$
If $x$ is a positive integer such that $x(x+1)(x+2)(x+3)+1=379^2$, find $x$
This is a 1989 ARML problem. One, ugly way to solve this is:
Approximate this as $x^4=379^2$, so $x\approx \sqrt{379}\approx 19$ and guess and check around there to see that $18$ works.
What's a nicer way?
Hint
Difference of squares
| About general solution of this equation.
$x(x+1)(x+2)(x+3)=a^2-1$
$⇔(x^2+3x)(x^2+3x+2)=(a-1)(a+1) $
obviously
$x^2+3x=a-1 $ so,
$x=\dfrac{-3+\sqrt{5+4a}}{2}$
$a_n=1,5,11,19・・・
=1+\sum2(n+1)
=1+n(n-1)+2(n-1)=n^2+n-1 $
Therefore when $a=n^2+n-1$, this equation has one positive solution $x=n-1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1528187",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Trick for a definite integral
Solve $$I=\displaystyle\int_0^{\pi}x^2(\pi-x)\sin(n\pi x)\, dx$$
How can solve for this case? Integration by parts? I think that I can do this:
$$u=x^2(\pi-x),\quad dv=\sin(n\pi x)\, dx$$
Does it work?
| $$\int_0^{\pi}x^2(\pi-x)\sin(n\pi x)\, dx\\
=x^2(x-\pi)\frac{\cos(n\pi x)}{n\pi}\Bigg|_0^\pi+
\int_0^{\pi}(2x\pi-3x^2)\frac{\cos(n\pi x)}{n\pi}\, dx\\=
\int_0^{\pi}x(2\pi-3x)\cos(n\pi x)\, dx\\=
x(2\pi-3x)\frac{\sin(n\pi x)}{n^2\pi^2}\Bigg|_0^\pi - \int_0^{\pi}(2\pi-6x)\frac{\sin(n\pi x)}{n^2\pi^2}\, dx\\=
-\frac{\sin(n\pi^2)}{n^2} - \int_0^{\pi}(2\pi-6x)\frac{\sin(n\pi x)}{n^2\pi^2}\, dx
\\=
-\frac{\sin(n\pi^2)}{n^2} +
(2\pi-6x)\frac{\cos(n\pi x)}{n^3\pi^3}\Bigg|_0^\pi-
\int_0^{\pi}(-6)\frac{\cos(n\pi x)}{n^3\pi^3}\, dx\\=
-\frac{\sin(n\pi^2)}{n^2} -
\frac{4\cos(n\pi^2)-2}{n^3\pi^2}
+6
\int_0^{\pi}\frac{\cos(n\pi x)}{n^3\pi^3}\, dx
\\=
-\frac{\sin(n\pi^2)}{n^2} -
\frac{4\cos(n\pi^2)-2}{n^3\pi^2}
+6
\frac{\sin(n\pi x)}{n^4\pi^4}\Bigg|_0^\pi\\=
-\frac{\sin(n\pi^2)}{n^2} -
\frac{4\cos(n\pi^2)-2}{n^3\pi^2}
+6
\frac{\sin(n\pi ^2)}{n^4\pi^4}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1537745",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solving the 4th Degree Equation $x^4- 8\sqrt{3}x^2 - 16 = 0$ I'm learning radical simplification and our teacher gave us this equation to solve:
$$x^4-8\sqrt{3}x^2-16=0$$
She told us to consider $y=x^2$ to transform the equation into a quadratic equation, which we can solve. However, when I apply the quadratic formula to the equation:
$$y^2-8\sqrt{3}y-16=0$$
I get:
$$y={8\sqrt{3}\pm\sqrt{\sqrt{192}-64}\over 2}\equiv y=4\sqrt{3}\pm8$$
And after that I get stuck. Our teacher solved another equation in class and she transformed the result of the quadratic formula into the square of a binomial, so that you're able to square root it and get the value of $x$, however, I haven't been able to transform $y=4\sqrt{3}\pm8$ into a square of a binomial. Maybe there's another way around it, but I can't seem to find it.
Any help would be greatly appreciated, thanks in advance! :)
| You got to $x^2 = 4\sqrt 3 \pm 8$
Since $4\sqrt 3 - 8 < 0$, then $x^2 = 4\sqrt 3 + 8$. What can you do if you cant see that $(\sqrt 2 + \sqrt 6)^2 = 4 \sqrt 3 + 8$?
Let $$\text{$x^2 = 8 + 4\sqrt 3 \ $ and $ \ y^2 = 8 - 4\sqrt 3$}$$
We can assume that $x > y > 0$.
Then $x^2y^2 = 8^2 - (4\sqrt 3)^2 = 16$. So $xy = 4$.
$(x+y)^2 = (x^2+y^2) + 2xy = 16 + 8 = 24$. So $x+y = 2\sqrt 6$.
$(x-y)^2 = (x^2+y^2) - 2xy = 16 - 8 = 8$. So $x+y = 2\sqrt 2$.
So
\begin{align}
(x+y)+(x-y) &= 2\sqrt 6 + 2 \sqrt 2 \\
2x &= 2\sqrt 6 + 2 \sqrt 2 \\
x &= \sqrt 6 + \sqrt 2
\end{align}
Since $x$ can also be negative, then $x = \pm(\sqrt 6 + \sqrt 2)$
We compute
$(x + (\sqrt 6 + \sqrt 2))(x - (\sqrt 6 + \sqrt 2)) = x^2 - 8 + 4\sqrt 3$
We find that
$x^4- 8\sqrt{3}x^2 - 16 = (x^2 - 4\sqrt 3 - 8)(x^2 - 4\sqrt 3 + 8)$
It only seems proper to check that
$(\sqrt 6 - \sqrt 2)^2 = 8 - 4\sqrt 3$
So the roots are
$$\{\pm(\sqrt 6 + \sqrt 2), \pm i(\sqrt 6 - \sqrt 2) \}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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what is the remainder when $7^{2015}$ is divided by $25$? The question says that what is the remainder when $7^{2015}$ is divided by $25$. Please help me out, I am out of ideas (I know that I have to find out the last two digits and I will be done... but I only know how to find the units digit.) and I need your help.
| I imagine you found the last digit by looking for a pattern when you do powers of 7 like this:
$7^0=1$
$7^1=7$
$7^2=49\equiv9\pmod{10}$
$7^3\equiv9\times7\equiv63\equiv3\pmod{10}$
$7^4\equiv3\times7\equiv21\equiv1\pmod{10}$
So every 4 powers brings you back to 1. So:
$$7^{2015}=7^{4\times503+3}\equiv7^3\equiv3\pmod{10}$$
You can do the same by looking at the remainder each step when you divide by 25.
$7^0=1$
$7^1=7$
$7^2=49\equiv-1\pmod{25}$
$7^3=-1\times7\equiv-7\pmod{25}$
$7^4\equiv-7\times7\equiv-49\equiv1\pmod{25}$
So it follows the same pattern so it has similar working:
$$7^{2015}=7^{4\times503+3}\equiv7^3\equiv-7\equiv18\pmod{25}$$
| {
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"url": "https://math.stackexchange.com/questions/1540989",
"timestamp": "2023-03-29T00:00:00",
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Evaluate $\sum_{k=0}^n (-1)^k{n \choose k}{m-k \choose r}$ using generating functions. I am trying to evaluate this with generating functions under the assumption that $m \geq n$. From there I need to come up with a proof of the identity obtained using principle of inclusion exclusion.
I tried to set this up as a binomial convolution but can't seem to make it work.
| Here is the closed form with complex analysis.
Suppose we are interested in
$$\sum_{k=0}^n (-1)^k {n\choose k}
{m-k\choose r}$$
where $m\ge n$ and $r\ge n.$ We will first solve this by complex
variables and that will tell us what the generating functions are.
Introduce
$${m-k\choose r} =
\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{r+1}} (1+z)^{m-k} \; dz.$$
We get for the sum
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{r+1}} (1+z)^{m}
\sum_{k=0}^n (-1)^k {n\choose k} \frac{1}{(1+z)^k} \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{r+1}} (1+z)^{m}
\left(1-\frac{1}{1+z}\right)^n
\; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{r+1}} (1+z)^{m-n}
z^n \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{r-n+1}} (1+z)^{m-n}
\; dz
\\ = {m-n\choose r-n}.$$
Now that we have this we can reverse-engineer the solution using
generating functions.
We thus get
$${n\choose k} (-1)^k =
[z^k]\sum_{k=0}^n (-1)^k {n\choose k} z^k = [z^k] (1-z)^n$$
and $${m-k\choose r} = {m-k-r + r \choose r}
= [z^{m-k-r}] \frac{1}{(1-z)^{r+1}}
\\ = [z^{n+m-k-r}] \frac{z^n}{(1-z)^{r+1}}
\\ = [z^{n-k}] \frac{z^{n-m+r}}{(1-z)^{r+1}}.$$
Therefore the desired OGFs are
$$(1-z)^n \quad\text{and}\quad \frac{z^{n-m+r}}{(1-z)^{r+1}}.$$
Doing the multiplication we get
$$\frac{z^{n-m+r}}{(1-z)^{r-n+1}}.$$
Extract the coefficient on $[z^n]$ we have
$$[z^n] \frac{z^{n-m+r}}{(1-z)^{r-n+1}}
= [z^{n-(n-m+r)}] \frac{1}{(1-z)^{r-n+1}}
\\ = [z^{m-r}] \frac{1}{(1-z)^{r-n+1}} =
{m-r+r-n\choose r-n} = {m-n\choose r-n}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1543186",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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} |
Calculating $\sum_{n=1}^\infty {\frac{1}{2^nn(3n-1)}}$ I'd appreciate any help, I know it has something to do with the geometric series but I still can't figure out how. I thought about integration but couldn't find a way to do it.
$$\sum_{n=1}^\infty \frac{1}{2^nn(3n-1)}$$
Thanks.
| $$\log\frac{1}{1-x}=\sum_{n=1}^\infty \frac{x^n}{n}$$
$$\log\frac{1}{1-x^3}=\sum_{n=1}^\infty \frac{x^{3n}}{n}$$
$$\frac{1}{x^2}\log\frac{1}{1-x^3}=\sum_{n=1}^\infty \frac{x^{3n-2}}{n}$$
$$\int_{0}^{\frac{1}{\sqrt[3]{2}}}\frac{1}{x^2}\log\frac{1}{1-x^3}dx=\sum_{n=1}^\infty \int_{0}^{\frac{1}{\sqrt[3]{2}}}\frac{x^{3n-2}}{n}dx$$
$$\int_{0}^{\frac{1}{\sqrt[3]{2}}}\frac{1}{x^2}\log\frac{1}{1-x^3}dx=\sum_{n=1}^\infty \frac{\sqrt[3]{2}}{n2^n(3n-1)}$$
so
$$\sum_{n=1}^\infty \frac{1}{n2^n(3n-1)}=\frac{1}{\sqrt[3]{2}}\int_{0}^{\frac{1}{\sqrt[3]{2}}}\frac{1}{x^2}\log\frac{1}{1-x^3}dx$$
$$=-\log 2+\frac{1}{2\sqrt[3]{2}}\log(\frac{1}{\sqrt[3]{4}}+\frac{1}{\sqrt[3]{2}}+1)-\frac{1}{\sqrt[3]{2}}\log(1-\frac{1}{\sqrt[3]{2}})-\frac{\sqrt{3}}{\sqrt[3]{2}}\tan^{-1}(\frac{\sqrt[3]{4}+1}{\sqrt{3}})+\frac{\pi}{2\sqrt[3]{2}\sqrt{3}}$$
$$=0.282319554....$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1543454",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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On the convergence of a trigonometric series If the following is duplicated, please let me know.
Suppose that $a_n = \frac{\sin\frac 1n\sin\frac 1{n+1}}{\cos\frac 1{n(n+1)}}$. I wonder if the series $\sum a_n$ is convergent? And if so, then what is its sum? Any suggestion would be helpful.
| Recall that the sine function satisfies the inequalities
$$x\cos x\le \sin x\le x \tag 1$$
for $0\le x\le \pi/2$. From $(1)$, it is easy to show that fpr $0\le x\le \pi/2$, the cosine function satisfies the inequalities
$$\sqrt{1-x^2}\le \cos x\le 1 \tag 2$$
Then, from $(1)$ and $(2)$, we can write
$$0\le\sin\left(\frac1n\right)\sin\left(\frac{1}{n+1}\right)\le \frac{1}{n(n+1)}\tag 3$$
and
$$0\le \frac{1}{\cos \left(\frac{1}{n(n+1)}\right)}\le \frac{1}{\sqrt{1-\left(\frac{1}{n(n+1)}\right)^2}} \tag 4$$
For $n\ge 1$, it is easy to arrive at the inequality
$$\frac{1}{\sqrt{1-\left(\frac{1}{n(n+1)}\right)^2}} \le 1+\frac1n \tag 5$$
Putting $(3)$, $(4)$, and $(5)$ together yields
$$0\le \frac{\sin\left(\frac1n\right)\sin\left(\frac{1}{n+1}\right)}{\cos \left(\frac{1}{n(n+1)}\right)}\le \frac{1}{n(n+1)\sqrt{1-\left(\frac{1}{n(n+1)}\right)^2}}\le \frac1{n^2}$$
Then, given that $\sum_{n=1}^\infty \frac1{n^2}=\frac{\pi^2}{6}$, we have
$$0\le \sum_{n=1}^\infty \frac{\sin\left(\frac1n\right)\sin\left(\frac{1}{n+1}\right)}{\cos \left(\frac{1}{n(n+1)}\right)}\le \frac{\pi^2}{6}$$
and we have not only proven convergence, but provided an upper bound.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Volume between cone and sphere - First octant Find the volume between $z=\sqrt{x^2+y^2}$ and the sphere $x^2+y^2+z^2=1$ that lies in the first octant using cylindrical coordinates.
So I found the intersection and got $r=\frac{\sqrt2}{2}$.
I know theta has to be between $0$ and $\frac{\pi}{2}$ but not sure about z
| We can do this as a single variable integration, along the $z$-axis. We integrate just the cone from $z=0$ to $z=\sqrt{2}/2$ and then just the sphere from $z=\sqrt{2}/2$ to $z=1$, because in those ranges the region is simply the part of the cone and the part of the sphere, respectively. We finally divide by $4$ because we are only interested in the first octant (which is $1$ of the $4$ octants with positive $z$ coordinate).
$$V = \frac{1}{4}\left(\displaystyle\int_{0}^{\sqrt{2}/2} \pi z^2 \,\mathrm{d}z+\displaystyle\int_{\sqrt{2}/2}^1 \pi\left(1-z^2\right)\,\mathrm{d}z\right)$$
So the problem is now a straightforward single-variable integral.
$$V=\frac{\pi}{4}\left(\displaystyle\int_{0}^{\sqrt{2}/2} z^2 \,\mathrm{d}z+\displaystyle\int_{\sqrt{2}/2}^1 \left(1-z^2\right)\,\mathrm{d}z\right)$$
$$V=\frac{\pi}{4}\left(1-\frac{\sqrt{2}}{2}+ \displaystyle\int_{0}^{\sqrt{2}/2} z^2 \,\mathrm{d}z - \displaystyle\int_{\sqrt{2}/2}^1 z^2\,\mathrm{d}z\right)$$
$$V=\frac{\pi}{4}\left(1-\frac{\sqrt{2}}{2}+ \frac{1}{3}\left(\frac{\sqrt{2}}{2}\right)^3 - \frac{1}{3}+\frac{1}{3}\left(\frac{\sqrt{2}}{2}\right)^3\right)$$
$$V=\frac{\pi}{4}\left(\frac{2}{3}-\frac{\sqrt{2}}{2}+ \frac{2}{3}\left(\frac{2\sqrt{2}}{8}\right) \right)$$
$$V=\frac{\pi}{4}\left(\frac{2}{3}-\frac{\sqrt{2}}{3} \right)$$
$$V= \boxed{\frac{2\pi-\pi\sqrt{2}}{12}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1547997",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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How does one get from $p+3=3k+2$ then $2^{p+3} \equiv 4 \pmod 7$, to $5 \cdot 2^{p+3} -31 \equiv 3 \pmod7$ How does one get from $p+3=3k+2$ then $2^{p+3}\equiv4 \pmod 7$, to $5 \cdot 2^{p+3} -31 \equiv 3 \pmod7$.
I am just starting with modular arithmetic so any help would be greatly appreaciated.
Credit to user236182 for original answer, just wanted to know how he got to it
| First look at the powers of $2$ modulo $7$. In particular note that $2^3\equiv1\pmod7$. Hence $2^{3k+a}\equiv 2^{3k}2^a\equiv2^a\pmod7$.
This gives us the first part of the question:
$$p+3=3k+2\hspace{5mm}\implies\hspace{5mm} 2^{p+3}\equiv2^{3k+2}\equiv2^2\equiv4\pmod7$$
The second part is pretty straightforward too:
$$5\cdot4-31\equiv-11\equiv3\pmod 7$$
hence
$$5\cdot2^{p+3}-31\equiv3\pmod7$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1550027",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove $\lim_{x\to 7/4^+}\tfrac{3x}{4x-7}=\infty$ by definition $$\lim_{x\to \frac{7}{4}^+}\frac{3x}{4x-7}=\infty $$
I want to prove that for every $M>0$ exists $\delta$ for which $ 0<x-\frac{7}{4}<\delta $ such that $ f(x)>M $
What I tried:
$$f(x)=\frac{3x}{4x-7}>M\iff\frac{1}{M}>\frac{4x-7}{3x} =* $$
Then I took some $\delta_1 = \frac{1}{4}$ and then:
$$0<x-\frac{7}{4}<\frac{1}{4}\iff \frac{7}{4}<x<2\iff7<4x<8\iff0<4x-7<1 $$
Final proof:
We take arbitrary $M>0 $.
Then we take some $ \delta_{1}=\frac{1}{4}$. Then
$$ 0<x-\cfrac{7}{4}<\cfrac{1}{4}\iff \cfrac{7}{4}<x<2\leftrightarrow\cfrac{21}{4}<3x<6 $$
So $$ f(x)=\cfrac{3x}{4x-7}>\cfrac{5}{4x-7}=\cfrac{5}{4\left(x-\cfrac{7}{4}\right)}=\cfrac{5}{4}\cdot\cfrac{1}{x-\cfrac{7}{4}}>M\iff \cfrac{1}{x-\cfrac{7}{4}}>\cfrac{4M}{5}\iff x-\cfrac{7}{4}<\frac{5}{4M}$$
Then $ \delta_{2}=\frac{5}{4M} $
$$ \delta=\min\{\delta_{1},\delta_{2}\}=\min\left\{\cfrac{5}{4M},\cfrac{1}{8}\right\} $$
| The part that makes the limit go to infinity is the denominator, the numerator will be larger than 3*7/4 > 5.
For x>7/4 we have $$f(x) > \frac{5}{4x-7} = \frac{5}{4} \cdot \frac{1}{x-7/4} > M$$
This holds if
$$ 0 < x-7/4 < \frac{5}{4M} = \delta$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1551764",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
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Is $\int_{-\pi}^{\pi}\int_{-\pi}^{\pi}\int_{-\pi}^{\pi}\frac{1}{1 - \cos x\cos y\cos z}\, dx\, dy\, dz$ finite? Is $$\int_{-\pi}^{\pi}\int_{-\pi}^{\pi}\int_{-\pi}^{\pi}\frac{1}{1 - \cos x\cos y\cos z}\, dx\, dy\, dz$$ finite? Mathematica is throwing me an error, but I suspect the integral converges, since the Taylor expansion of $\cos x\cos y\cos z$ should have a leading term of $1$ followed by a $x^{2} + y^{2} + z^{2}$ and $\frac{1}{x^{2} + y^{2} + z^{2}}$ is integrable in 3 dimensions if I switch to polar.
| While it is a bit surprising, the integral should converge. Around each of the corners, $(0, 0, 0)$, $(\pi, \pi, 0)$, $\dots$, etc, we have (very loose bounds)
$$
\begin{aligned}
\cos x\cos y\cos z
&\le |\cos x||\cos y||\cos z| \\
&\le \left( \frac{|\cos x| + |\cos y| + |\cos z|}{3} \right)^3
\\
&\le \left( 1 - \frac{ x^2 + y^2 + z^2 } {9} \right)^3
\quad \mathrm{for}\; |x|, |y|, |z| \in (0, \pi/2)
\\
&\le 1 - \frac{ x^2 + y^2 + z^2 } { 12 }.
\end{aligned}
$$
So
$$
\begin{aligned}
\frac{1}{1-\cos x\cos y\cos z}
&\le \frac{12}{ x^2 + y^2 + z^2 } = \frac{12}{r^2}.
\end{aligned}
$$
$$
\begin{aligned}
\int_0^{\pi/2}\int_0^{\pi/2}\int_0^{\pi/2}\frac{1}{1-\cos x\cos y\cos z}
\,dx \, dy \, dz
&\le \frac{\pi}{2} \int_0^{\pi/2} \frac{12}{r^2} r^2 \, dr = 3\pi^2.
\end{aligned}
$$
Together there are 64 such integrals, and the total integral is finite (actually 32 of them are not singular, but this is just an estimate).
For Mathematica, try this
NIntegrate[1/(1 - Cos[x] Cos[y] Cos[z]), {x, -Pi, Pi}, {y, -Pi, Pi}, {z, -Pi, Pi},
Method -> "AdaptiveMonteCarlo", PrecisionGoal -> 3]
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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What is coefficient of $x^k$ in $ n! (x/1! + x^2/2! + x^3/3! + ... )^n$? Given
$n![\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+...\frac{x^n}{n!} ]^n$, how do I find coefficient of $x^k$ in it ?
How to find coefficient in case of above series having infinite terms i.e. $n! [\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+... ]^n$ ?
| So lets consider the infinite case. First note that $x+\frac{x}{2!}+\frac{x}{3!}+...=e^x-1$, and so
\begin{align*}
n!\left(x+\frac{x}{2!}+\frac{x}{3!}+...\right)^n=n!(e^x-1)^n.
\end{align*}
Using the Binomial Theorem, we have that
\begin{align*}
n!(e^x-1)^n & = n!\sum_{i=0}^n \binom{n}{i}e^{x(n-i)}(-1)^{i}\\
& = n!\left(e^{nx}-ne^{x(n-1)}+\binom{n}{2}e^{x(n-2)}-...\pm 1 \right)\\
& = n!\left(\sum_{r=0}^\infty n^r \frac{x^r}{r!}-n\sum_{r=0}^\infty (n-1)^r \frac{x^r}{r!}+\binom{n}{2}\sum_{r=0}^\infty (n-2)^r \frac{x^r}{r!}-...\pm 1\right).
\end{align*}
So then it follows that the coefficient for $x^k/k!$ is
\begin{equation}
n!\left(n^k-n(n-1)^k+\binom{n}{2}(n-2)^k-...\pm \binom{n}{n-1}\right)=n!\sum_{l=0}^{n-1}(-1)^l\binom{n}{l}(n-l)^k.
\end{equation}
So then the coefficient for the $x^k$ would be
\begin{equation}
\frac{n!}{k!}\sum_{l=0}^{n-1}(-1)^l\binom{n}{l}(n-l)^k.
\end{equation}
Note: the finite case follows a similar idea.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "5",
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Zero divided by zero must be equal to zero What is wrong with the following argument (if you don't involve ring theory)?
Proposition 1: $\frac{0}{0} = 0$
Proof: Suppose that $\frac{0}{0}$ is not equal to $0$
$\frac{0}{0}$ is not equal to $0 \Rightarrow \frac{0}{0} = x$ , some $x$ not equal to $0$ $\Rightarrow$ $2(\frac{0}{0}) = 2x$ $\Rightarrow$ $\frac{2\cdot 0}{0} = 2x$ $\Rightarrow$ $\frac{0}{0} = 2x$ $\Rightarrow$ $x = 2x$ $\Rightarrow$ $ x = 0$ $\Rightarrow$[because $x$ is not equal to $0$]$\Rightarrow$ contradiction
Therefore, it is not the case that $\frac{0}{0}$ is not equal to $0$
Therefore, $\frac{0}{0} = 0$.
Q.E.D.
Update (2015-12-01) after your answers:
Proposition 2: $\frac{0}{0}$ is not a real number
Proof [Update (2015-12-07): Part 1 of this argument is not valid, as pointed out in the comments below]:
Suppose that $\frac{0}{0}= x$, where $x$ is a real number.
Then, either $x = 0$ or $x$ is not equal to $0$.
1) Suppose $x = 0$, that is $\frac{0}{0} = 0$
Then, $1 = 0 + 1 = \frac{0}{0} + \frac{1}{1} = \frac{0 \cdot 1}{0 \cdot 1} + \frac{1 \cdot 0}{1 \cdot 0} = \frac{0 \cdot 1 + 1 \cdot 0}{0 \cdot 1} = \frac{0 + 0}{0} = \frac{0}{0} = 0 $
Contradiction
Therefore, it is not the case that $x = 0$.
2) Suppose that $x$ is not equal to $0$.
$x = \frac{0}{0} \Rightarrow 2x = 2 \cdot \frac{0}{0} = \frac{2 \cdot 0}{0} = \frac{0}{0} = x \Rightarrow x = 0 \Rightarrow$ contradiction
Therefore, it is not the case that $x$ is a real number that is not equal to $0$.
Therefore, $\frac{0}{0}$ is not a real number.
Q.E.D.
Update (2015-12-02)
If you accept the (almost) usual definition, that for all real numbers $a$, $b$ and $c$, we have $\frac{a}{b}=c$ iff $ a=cb $, then I think the following should be enough to exclude $\frac{0}{0}$ from the real numbers.
Proposition 3: $\frac{0}{0}$ is not a real number
Proof: Suppose that $\frac{0}{0} = x$, where $x$ is a real number.
$\frac{0}{0}=x \Leftrightarrow x \cdot 0 = 0 = (x + 1) \cdot 0 \Leftrightarrow \frac{0}{0}=x+1$
$ \therefore x = x + 1 \Leftrightarrow 0 = 1 \Leftrightarrow \bot$
Q.E.D.
Update (2015-12-07):
How about the following improvement of Proposition 1 (it should be combined with a new definition of division and fraction, accounting for the $\frac{0}{0}$-case)?
Proposition 4: Suppose $\frac{0}{0}$ is defined, so that $\frac{0}{0} \in \mathbb{R}$, and that the rule $a \cdot \frac{b}{c} = \frac{a \cdot b}{c}$ holds for all real numbers $a$, $b$ and $c$.
Then, $\frac{0}{0} = 0$
Proof: Suppose that $\frac{0}{0}=x$, where $x \ne 0$.
$x = \frac{0}{0} \Rightarrow 2x = 2 \cdot \frac{0}{0} = \frac{2 \cdot 0}{0} = \frac{0}{0} = x \Rightarrow x = 0 \Rightarrow \bot$
$\therefore \frac{0}{0}=0$
Q.E.D.
Suggested definition of division of real numbers:
If $b \ne 0$, then
$\frac{a}{b}=c$ iff $a=bc$
If $a=0$ and $b=0$, then
$\frac{a}{b}=0$
If $a \ne 0$ and $b=0$, then $\frac{a}{b}$ is undefined.
A somewhat more minimalistic version:
Proposition 5. If $\frac{0}{0}$ is defined, so that $\frac{0}{0} \in \mathbb{R}$, then $\frac{0}{0}=0$.
Proof: Suppose $\frac{0}{0} \in \mathbb{R}$ and that $\frac{0}{0}=a \ne 0$.
$a = \frac{0}{0} = \frac{2 \cdot 0}{0} = 2a \Rightarrow a = 0 \Rightarrow \bot$
$\therefore \frac{0}{0}=0$
Q.E.D.
| If $\frac00 = 0$ then $0\times0=0$ would work as a proof. The problem is that also $3\times0=0$ and $7\times0$ and so on would all work.
So $\frac00=0,\space\frac00=1,\space\frac00=3,\space\frac00=$ just about anything.
There are too many possible results, we want such operations to return defined results (one and only one result), otherwise the operation itself is not actually defined. In fact, dividing any number by $0$ is 'undefined', 'illegal'.
When dividing a non-zero integer by zero, we have the same problem. only, in that case, we simply have no-results.
ie: $\frac60 =$ number which multiplied by $0$ equals $6$.
We can say $\frac60=\infty$ only to cut short on a theorem, stating that the limit of $\frac mn$ with m integer and $n$ going to $0$ is infinite. It is obviously a notation-artifact. Otherwise we should say that $\infty\times0$ equals $6$, and $7$, and $8$, and $9$...
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "52",
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Calculate $\Delta J$ for a functional $J(y)=\int_{0}^{1} (x^2-y^2+(y')^2)dx$
$y(x)=x, h(x)=x^2$
I need to calculate $\Delta J$ and I am given this from the answer key:
$\Delta J = J(y + \epsilon h)-J(y) = J(x + \epsilon x^2) - J(x)$
I just need to verify that this comes out to:
$\Delta J = (\frac{17}{15}\epsilon ^2+\frac{3}{2}\epsilon +1) -(1)= \frac{17}{15}\epsilon ^2+\frac{3}{2}\epsilon$ ?
The reason I ask is that our book never gives an example of this, so I want to verify that this is correct.
| It's just a matter of substition: For a given function $h$ and value $\epsilon$, $\Delta J = J(y + \epsilon h) - J(y)$ by definition. So, given that $y(x) = x, h(x) = x^2$, we see that $y'(x) = 1, h'(x) = 2x$. So:
$$\begin{align}\Delta J &= J(y + \epsilon h)-J(y) \\
&= \int_{0}^{1} (x^2-(y + \epsilon h)^2+(y'+\epsilon h')^2)dx - \int_{0}^{1} (x^2-y^2+(y')^2)dx\\
&= \int_{0}^{1} (x^2-(x + \epsilon x^2)^2+(1+2\epsilon x)^2)dx - \int_{0}^{1} (x^2-x^2+(1)^2)dx\\
&= \int_{0}^{1} -\epsilon^2 x^4 - 2\epsilon x^3 + 1+4\epsilon x + 4\epsilon^2 x^2)dx - \int_{0}^{1} 1\ dx\\
&= -\frac{\epsilon^2}{5} - \frac{2\epsilon}{4} + \frac{4\epsilon}{2} + \frac{4\epsilon^2}{3}\\
&=\frac{17}{15}\epsilon ^2+\frac{3}{2}\epsilon\end{align}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Question on partial derivative If $u= \log(\tan x + \tan y)$, prove that
$$\sin(2x) \, \frac{du}{dx} + \sin(2y) \, \frac{du}{dy} = 2$$.
| For $u = \ln(\tan x + \tan y)$ the derivatives with respect to $x$ and $y$ are
\begin{align}
\frac{du}{dx} &= \frac{sec^{2}(x)}{\tan x + \tan y} \\
\frac{du}{dy} &= \frac{sec^{2}(y)}{\tan x + \tan y}.
\end{align}
Now,
\begin{align}
\sin(2x) \, \frac{du}{dx} + \sin(2y) \, \frac{du}{dy} &= 2 \, \frac{\tan x + \tan y}{\tan x + \tan y} = 2.
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
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is $\frac{a-b}{a+b} - \frac{c-d}{c+d}$ equal to $ \frac{a-c}{a+c}$? I am using values out of a cross-correlation analysis and my intuition tells me that the equation in the title is true, however I would like to prove it.
$$ \frac{a-b}{a+b} - \frac{c-d}{c+d} = \frac{a-c}{a+c}$$
Unfortunately my math skills are horrible and I could not find this equation on-line (which now makes me think that my intuition is faulty).
Could you at least point out what mental strategy you would follow to try to solve it?
Thanks for any help
PS= any suggestion to improve my way of asking questions is also welcome.
all the best
| First, we can assume that:
$$
(a+b)\neq 0\qquad c+d\neq 0\qquad a+c\neq 0
$$
Rewrite the left-hand side:
$$
\frac{(a-b)(c+d)-(a+b)(c-d)}{(a+b)(c+d)}=\frac{ac+ad-bc-bd-ac+ad-bc+bd}{(a+b)(c+d)}=
$$
$$
=2\frac{ad-bc}{(a+b)(c+d)}
$$
So we need the following to hold:
$$
2\frac{ad-bc}{(a+b)(c+d)}=\frac{a-c}{a+c}
$$
Which is generally not true, and leads to the equation:
$$
a^2 (c-d)+a \left(3 b c+b d-c^2-3 c d\right)+b c^2-b c d=0
$$
If this is true for eg. all $a$, then all the coefficients should be $0$:
$$
c-d=3 b c+b d-c^2-3 c d=bc(c-d)=0
$$
From $c=d$ we have $3 b c+b d-c^2-3 c d=4 (b - d) d$, so either $d=0$ or $b=d$.
| {
"language": "en",
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"question_score": "2",
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"answer_id": 1
} |
Define $a$ and $b$ rational numbers so they satisfy equation Define $a, b \in \mathbb{Q}$ so that
$$\frac{a}{\sqrt{7 + 4\sqrt{3}}} + \frac{b}{\sqrt{7 - 4\sqrt{3}}} = \sqrt{4 + 2\sqrt{3}}$$
Using $\sqrt{a \pm \sqrt{b}} = \sqrt{\frac{a + \sqrt{a^2 - b}}{2}} \pm \sqrt{\frac{a - \sqrt{a^2 - b}}{2}}$ I got $\frac{a}{2 + \sqrt{3}} + \frac{b}{2 - \sqrt{3}} = \sqrt{3}+1$ which results in $2(a+b)+\sqrt{3}(b-a) = \sqrt{3}+1$, I'm not sure how to proceed from that.
| You can move from your last equation by setting the following system of equations:
$$(b-a)\sqrt3 = \sqrt3$$ $$2(a+b) = 1$$
Since $a$ and $b$ are rational, the coefficient of $\sqrt3$ must agree on each side of the equation $2(a+b)+\sqrt{3}(b-a) = \sqrt{3}+1$ . The rational parts on each side must also agree.
| {
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"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Proof using the definition of the limit. Prove that $\lim_{n\to\infty}$ $\frac{n^3}{3^n}$ $=0$ using the definition of the limit.
Here is what I have so far:
$\left\lvert{\frac{n^3}{3^n} -0}\right\rvert$ = $\left\lvert{\frac{n^3}{(1+2)^n} }\right\rvert$$<\left\lvert{\frac{n^3}{1+\binom{n}{1}2+\binom{n}{2}2^2+\binom{n}{3}2^3+\binom{n}{4}2^4+\cdots}}\right\rvert$
However, I cannot think of anyway to reduce the above equation so it is less of any $\epsilon$ greater than zero.
| I think you did something closely related to the following.
The fifth term in the binomial expansion of $(1+2)^n$ is $\binom{n}{4}2^4$. So for $n\gt 3$ we have
$(1+2)^n \gt 16\cdot\frac{n(n-1)(n-2)(n-3)}{4!}$.
Note that if $n\gt 5$ then $n-1\gt n/2$ and $n-2\gt n/2$ and $n-3\ge n/2$. It follows that
$$16\cdot\frac{n(n-1)(n-2)(n-3)}{4!}\gt 16\cdot\frac{(n)(n/2)(n/2)(n/2)}{4!}=\frac{n^4}{12}.$$
Thus if $n\ge 5$, then
$$\frac{n^3}{3^n}\lt \frac{12}{n}.$$
Now I think you can finish.
Remark: It may have been a strategic mistake to "multiply out." The expression $16\cdot\frac{n(n-1)(n-2)(n-3)}{4!}$ is easy to think about, while a multiplied out version is less easy.
| {
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} |
Prime numbers of form $a^n-1$. Let $a$ and $n$ be integers greater than $1$. Suppose that $a^n - 1$ is prime. Show that $a=2$ and $n$ is prime.
| Since $a^n-1=(a-1)(a^{n-1}+\dots+1)$ if $a>2$ then $a^n-1$ is divisible with $a-1$. $a^n-1$is prime, so $a$ should be 2.
If $n=p\times q$ then $2^n-1= (2^p)^q-1=(2^p-1)((2^p)^{q-1}+\dots+1)$ and $2^n-1$ will divisible with $2^p-1$. $2^n-1$ is prime so $n$ wouldn't be written as $p\times q$, so $n$ is prime.
| {
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Finding the nth derivative of $y=\frac4{(6x+8)^3}$ Find the $n$-th derivative of $y=\frac{4}{(6x+8)^3}$
\begin{align}
y' ={} & 4(-3)(6) \frac{1}{(6x+8)^4}
\\
y''={} & 4(-3)(-4)(6)^2 \frac{1}{(6x+8)^5}
\\
y'''={} & 4(-3)(-4)(-5)(6)^3\frac{1}{(6x+8)^6}
\end{align}
I recognise the pattern but can't interpret that into a formula.
| The general term for nth derivative would be $(-1)^n.4.6^n.(3)..(3+(n-1)\frac{1}{(6x+8)^{n+3}}$. 'n' indicates the nth derivative.
| {
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"answer_count": 4,
"answer_id": 2
} |
Derivative of $10^x\cdot\log_{10}(x)$
Derive $10^x\cdot\log_{10}(x)$
$$10^x\cdot \ln(10)\cdot \log_{10}(x)+\frac{1}{x\cdot \ln(10)}\cdot 10^x$$
But WolframAlpha gives another solution. Where am I wrong?
| Your answer is correct, but you can simplify it further using the change of base formula. Where $\log x$ is in base $10$ and $\ln x$ is the natural log, you have
$$10^x\cdot \ln10\cdot \log x+\frac{1}{x\cdot \ln10}\cdot 10^x$$
$$=10^x\cdot \left(\ln10\cdot \color{green}{\log x}+\frac{1}{x\cdot \ln 10}\right)$$
$$=10^x\cdot \left(\ln10\cdot \color{green}{\frac{\ln x}{\ln 10}}+\frac{1}{x\cdot \ln 10}\right)$$
$$=10^x\cdot \left(\ln x+\frac{1}{x\cdot \ln 10}\right)$$
| {
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"question_score": "3",
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Unitary transformation I have a matrix in following form
$$A=\begin{bmatrix}
1&0&0&0&0&0&0&0\\
0&-1&0&0&0&0&0&0\\
0&0&1&0&0&0&0&0\\
0&0&0&-1&0&0&0&0\\
0&0&0&0&1&0&0&0\\
0&0&0&0&0&-1&0&0\\
0&0&0&0&0&0&1&0\\
0&0&0&0&0&0&0&-1\\
\end{bmatrix}$$
I need some help to reduce the matrix in following form using unitary transformation
$$C=\begin{bmatrix}
1&0&0&0&0&0&0&0\\
0&1&0&0&0&0&0&0\\
0&0&1&0&0&0&0&0\\
0&0&0&1&0&0&0&0\\
0&0&0&0&-1&0&0&0\\
0&0&0&0&0&-1&0&0\\
0&0&0&0&0&0&-1&0\\
0&0&0&0&0&0&0&-1\\
\end{bmatrix}$$
I have used the unitary matrix transformation$UAU^*$ method but the result always come in following form
$$UAU^*=\begin{bmatrix}
1&0&0&0&0&0&0&0\\
0&1&0&0&0&0&0&0\\
0&0&-1&0&0&0&0&0\\
0&0&0&-1&0&0&0&0\\
0&0&0&0&1&0&0&0\\
0&0&0&0&0&1&0&0\\
0&0&0&0&0&0&-1&0\\
0&0&0&0&0&0&0&-1\\
\end{bmatrix}$$
| Let $e_1 = (1,0,\dots,0)^T, e_2 = (0,1,0,\dots,0)^T$, and so on. One $U$ that works here is the $U$ whose columns are
$$
U = [e_1\quad e_3 \quad e_5 \quad e_7 \quad
e_2 \quad e_4 \quad e_6 \quad e_8]
$$
Notice that this $U$ is a permutation matrix.
| {
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"source": "stackexchange",
"question_score": "1",
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Polar radius of a general ellipsoid Is there a proper parametrization of a general ellipsoid in spherical coordinates?
The regular parametrization is this:
$$x=a\cdot \cos\phi \cos\theta\\y=b\cdot \cos\phi \sin \theta\\z=c\cdot \sin\phi.$$
What one can do is extract the polar radius via:
$$x=r\cdot \cos\phi \cos\theta\\y=r\cdot \cos\phi \sin \theta\\z=r\cdot \sin\phi,$$ plug these equations into the standard ellipsoid equation and solve for $r$. Thus,
$$ r(\phi,\theta)=\frac{a\cdot b\cdot c}{\sqrt{
b^2 c^2 \cos^2(\theta) \cos^2(\phi) +
a^2 (c^2 \cos^2(\phi) \sin^2(\theta) +
b^2 \sin^2(\phi))}} .$$
What I'm looking for, is an expression for the polar radius, just as the one above, but which takes into account the three possible rotations of the ellipsoid (via Euler or Tait-Bryan anles); i.e. a function $r_{\alpha,\beta, \gamma}(\phi,\theta)$ which fulfills $x=r_{\alpha,\beta, \gamma}\cdot \cos\phi \cos\theta,y=r_{\alpha,\beta, \gamma}\cdot \cos\phi \sin \theta,z=r_{\alpha,\beta, \gamma}\cdot \sin\phi$ simultaneously and describes the surface of an ellipsoid.
| Ok, I managed to compute the polar radius for a general ellipsoid. Brace yourself, here it is*:
$$r_{\alpha,\beta,\gamma} = \frac{a\cdot b\cdot c}{\sqrt{(b^2 c^2 Sin(\beta)^2 Sin(\phi)^2 +
a^2 c^2 Cos(\phi)^2 Sec(\gamma)^2 Sin(\theta - \
\alpha)^2 +
a^2 b^2 Cos(\phi)^2 Sin(\theta - \alpha)^2 Sin(\
\gamma)^2 +
Cos(\theta)^2 Cos(\phi)^2 Sec(\alpha)^2 (b^2 c^2 \
Cos(\beta)^2 Cos(\alpha)^4 +
a^2 Sin(\beta)^2 (b^2 Cos(\gamma)^2 +
c^2 Sin(\gamma)^2)) +
Cos(\beta)^2 (a^2 b^2 Cos(\gamma)^2 Sin(\phi)^2 +
c^2 (b^2 Cos(\phi)^2 Sin(\theta)^2 Sin(\alpha)^2 \
+ a^2 Sin(\phi)^2 Sin(\gamma)^2)) -
a^2 b^2 Cos(\phi)^2 Sin(\beta) Sin(\theta - \
\alpha)^2 Sin(2 \gamma) Tan(\alpha) +
a^2 b^2 Cos(\phi)^2 Cos(\gamma)^2 Sin(\beta)^2 \
Sin(\theta - \alpha)^2 Tan(\alpha)^2 +
a^2 c^2 Cos(\phi)^2 Sin(\beta)^2 Sin(\theta - \
\alpha)^2 Sin(\gamma)^2 Tan(\alpha)^2 +
Cos(\beta) (-2 b^2 Cos(\phi) Sin(\phi) (c^2 \
Sin(\theta) Sin(\beta) Sin(\alpha) +
a^2 Cos(\gamma) Sin(\theta - \alpha) Sin(\gamma)) +
a^2 Sin(2 \phi) Sin(\theta - \alpha) (b^2 Cos(\
\gamma)^2 Sin(\beta) Tan(\alpha) +
c^2 Sin(\gamma) (Cos(\gamma) +
Sin(\beta) Sin(\gamma) Tan(\alpha)))) +
Cos(\theta) Cos(\phi) Sec(\alpha) (-2 b^2 c^2 Cos(\
\beta) Cos(\alpha)^2 Sin(\beta) Sin(\
\phi) +
2 b^2 c^2 Cos(\beta)^2 Cos(\phi) Cos(\
\alpha)^2 Sin(\theta) Sin(\alpha) +
a^2 (c^2 Sin(2 \beta) Sin(\phi) Sin(\gamma)^2 +
b^2 Cos(\gamma)^2 (Sin(2 \beta) Sin(\phi) +
2 Cos(\phi) Sin(\beta)^2 Sin(\theta - \
\alpha) Tan(\alpha)) +
Cos(\phi) Sin(\beta) Sin(\theta - \
\alpha) (-b^2 Sin(2 \gamma) +
2 c^2 Sin(\gamma) (Cos(\gamma) +
Sin(\beta) Sin(\gamma) Tan(\alpha))))) \
+ 2 a^2 c^2 Cos(\phi)^2 Sin(\beta) Sin(\theta - \
\alpha)^2 Tan(\alpha) Tan(\gamma) -
2 a^2 c^2 Cos(\phi)^2 Sin(\beta) Sin(\theta - \
\alpha)^2 Sin(\gamma)^2 Tan(\alpha) Tan(\gamma) -
2 a^2 c^2 Cos(\phi)^2 Sin(\theta - \alpha)^2 Tan(\
\gamma)^2 +
a^2 c^2 Cos(\phi)^2 Sin(\theta - \alpha)^2 Sin(\
\gamma)^2 Tan(\gamma)^2)}}$$
$\gamma$, $\beta$ and $\alpha$ rotate the ellipsoid around the x, y and z axis
*Mathematica simplified
| {
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How to invert $f(x) = \frac{\sqrt{3}x}{\sqrt{1+3x^2}}$ Consider the function
$f:(-1,1)\to(-1,1)$,
$f(x) = \frac{\sqrt{3}x}{\sqrt{1+3x^2}}$
To find $f^{-1}$ we have to show that f is injective and surjective.
Let $a,b\in(-1,1)$.
Then $f(a)=f(b)$ is equivalent to $\frac{\sqrt{3}a}{\sqrt{1+3a^2}}$=$\frac{\sqrt{3}b}{\sqrt{1+3b^2}}$.
It implies $(a-b)(a+b)=0$, which is problematic with $a=-b$.
To show surjectivity we have to show that for every $y$ there is an $x$ such that $f(x)=y,$ with which I'm also stuck!
| Try first solving
$$
y=\frac{\sqrt{3}\,x}{\sqrt{1+3x^2}}
$$
First let's try it for $y>0$, so also $x>0$ (if it exists). Then we get
$$
y^2=\frac{3x^2}{1+3x^2}=1-\frac{1}{1+3x^2}
$$
that is
$$
\frac{1}{1+3x^2}=1-y^2
$$
This is possible only if $0<y<1$ and, in this case, we have
$$
1+3x^2=\frac{1}{1-y^2}
$$
so
$$
3x^2=\frac{1}{1-y^2}-1=\frac{y^2}{1-y^2}
$$
and so
$$
x=\frac{1}{\sqrt{3}}\frac{y}{\sqrt{1-y^2}}
$$
If $y\le0$ you find exactly the same formula under the condition $-1<y\le0$.
So $f$ is indeed invertible (over the range $(-1,1)$).
Note that this does show injectivity. However, if you're requested to go the hard way, here it is.
Suppose
$$
\frac{\sqrt{3}\,a}{\sqrt{1+3a^2}}
=
\frac{\sqrt{3}\,b}{\sqrt{1+3b^2}}
$$
Then either $a\ge 0$ and $b\ge0$ or $a<0$ and $b<0$.
Let's do the first case. Squaring we get
$$
\frac{3a^2}{1+3a^2}=\frac{3b^2}{1+3b^2}
$$
or
$$
1-\frac{1}{1+3a^2}=1-\frac{1}{1+3b^2}
$$
that implies
$$
1+3a^2=1+3b^2
$$
and therefore
$$
a^2=b^2
$$
so $a=b$.
Similarly for the case $a<0$ and $b<0$.
| {
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Counting the max integer in each partition of $p(n)$ Assume $n=5$.
We have $p(n)=$
5
4 + 1
3 + 2
3 + 1 + 1
2 + 2 + 1
2 + 1 + 1 + 1
1 + 1 + 1 + 1 + 1
What I want to get is the number of partitions in which the maximum integer is $m$, for each $1\leq m \leq n$.
For the previous example, #5=1, #4=1, #3=2, #2=2, #1=1.
Can anybody help for the general case where $n$ is unknown?
Thanks.
| Your problem is equate to finding factor of $x^n$ in :
$(1+x+x^2+...+x^n)^n = (\frac{1-x^{n+1}}{1-x})^n$
And subtract $n!-1 $ from that.
Because if you difine $f = (1+x+x^2+...+x^n)$ for every summation that is equal $n$ we have one term in polynomial $f^n$ which consist of $x^n$. for example for $1+1+1+...1 = n $ we have term $x.x.x.....x = x^n $ that's enough to choose $x $ from each parantes of $f^n $ or for $n$ we have $x^n .1.1.1....1 $ that's enough to choose $x^n$ from one pranteses of $f^n$ and choose $1$ from others.
But we count some of the summations several time and the number of what we count it several is $n! -1$
So your answer is : (factor of $x^n$ in $f^n$) $- (n!-1)$
| {
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Computing $ \int_0^{2\pi} \frac{1}{a^2\cos^2 t+b^2 \sin^2 t} dt \;; a,b>0$.
Using Residue Theorem find $\displaystyle \int_0^{2\pi} \frac{1}{a^2\cos^2 t+b^2 \sin^2 t} dt \;; a,b>0$.
My Try:
So, I am going to use the ellipse $\Gamma = \{a\cos t+i b \sin t: 0\leq t\leq 2\pi\}$.
On $\Gamma$, $z=a\cos t+i b \sin t$, so $|z|^2=z\bar{z}=a^2\cos^2 t+b^2 \sin^2 t$.
Now, $dz=-a\sin t+i b \cos t dt$.
Hence, the integral becomes $\displaystyle \int_\Gamma \frac{dz}{z(iab+(\sin t \cos t)(b^2-a^2))}$. I know that $\displaystyle \int_\Gamma \frac{dz}{z}=2\pi i$. Now, how do I get rid of $\sin t \cos t$ part? I am stuck here. Can somebody please explain how?
| Enforcing the change of variables $x= t-π$ leads to
$$I=\int_{0}^{2\pi} \frac{dx}{a^2\sin^2x+b^2\cos^2x}=2\int_{0}^{\pi} \frac{dx}{a^2\sin^2x+b^2\cos^2x}=4\int_{0}^{\pi/2} \frac{dx}{a^2\sin^2x+b^2\cos^2x}$$
and we have \begin{align}\frac{1}{a^2\sin^2x+b^2\cos^2x}&=\frac1{ab}\cdot\frac{b^2\cos^2x}{b^2\cos^2x+a^2\sin^2x}\cdot\frac{a}{b\cos^2x}\\&=\frac1{ab}\cdot\frac1{1+\left(\frac ab \tan x\right)^2}\cdot \frac a{b\cos^2x}\\&=\frac1{ab}\left[\arctan{\left(\frac ab\tan x\right)}\right]'\end{align}
$$I=4\int_{0}^{\pi/2}\frac1{ab}\left[\arctan{\left(\frac ab\tan x\right)}\right]'dx = \frac{2π}{ab} $$
| {
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"url": "https://math.stackexchange.com/questions/1570249",
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"source": "stackexchange",
"question_score": "5",
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Why is $\int_0^L [\cos(\frac{n \pi x}{L} - \frac{m \pi x}{L}) - \cos(\frac{n \pi x}{L} + \frac{m \pi x}{L})]dx = L $? Why is $\int_0^L [\cos(\frac{n \pi x}{L} - \frac{m \pi x}{L}) - \cos(\frac{n \pi x}{L} + \frac{m \pi x}{L})]dx = L$ when $n$ does equal $m$ but neither $n$ nor $m$ equal $0$?
$n$ is an integer, $L$ is a constant.
| Hints
$1.$ Note that
$$\cos(x+y)=\cos(x)\cos(y)-\sin(x)\sin(y) \\
\cos(x-y)=\cos(x)\cos(y)+\sin(x)\sin(y)$$
$2.$ Subtract the above relations
$$\cos(x-y)-\cos(x+y)=2\sin(x)\sin(y)$$
$3.$ Conclude that
$$\cos(\frac{n \pi x}{L} - \frac{m \pi x}{L}) - \cos(\frac{n \pi x}{L} + \frac{m \pi x}{L})=2\sin(\frac{n \pi x}{L})\sin(\frac{m \pi x}{L})$$
$4.$ Prove that
$$\int_{0}^{L}\sin(\frac{n \pi x}{L})\sin(\frac{m \pi x}{L})=\delta_{mn}\int_{0}^{L}\sin^2(\frac{n \pi x}{L})$$
| {
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Find $x,y,z>0$ such that $x+y+z=1$ and $x^2+y^2+z^2$ is minimal How can I find $3$ positive numbers that have a sum of $1$ and the sum of their squares is minimum?
So far I have:
$$x+y+z=1 \qquad \implies \qquad z=1-(x+y)$$
So,
$$f(x,y)=xyz=xy(1-x-y)$$
But I'm stuck from here. Hints?
| By Cauchy-Schwarz, for every $x,y,z\in\Bbb R$ we have
$$x+y+z=\langle(1,1,1),(x,y,z)\rangle \leq \left\|(1,1,1)\right\|_2\cdot \|(x,y,z)\|_2=\sqrt{3}\cdot \sqrt{x^2+y^2+z^2}$$
Now, if $x+y+z=1$, then squaring both sides and dividing by $3$ gives
$$ \frac{1}{3}\leq x^2+y^2+z^2 $$
This lower bound is then attained for $|x|=|y|=|z|=1/3$.
| {
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Prove that $\int_a^bxf(x)dx\geq\frac{b+a}{2}\int_a^bf(x)dx$ Let $f:[a,b]\to\mathbb{R}$ be continuous and increasing, show that
$$\int_a^bxf(x)dx\geq\frac{b+a}{2}\int_a^bf(x)dx$$
I am thinking of using integration by parts. First let
$$F(x)=\int_a^xf(t)dt$$
Then
$$\int_a^bxf(x)dx=bF(b)-\int_a^bF(x)dx=b\int_a^bf(x)dx-\int_a^b\int_a^xf(t)dtdx$$
So far I only have these in mind. Any one have any hints?
| Note that $\int_{a}^{b} \left(x - \frac{a + b}{2}\right)dx = 0$. Since $f$ is continuous on a compact set it is bounded and it achieves minimum and since it is increasing the minimum is achieved at the lowest end of an interval and maximum at the highest end of an interval.
Let $I_{1} = \left[a, \frac{a + b}{2} \right]$ and $I_{2} = \left[\frac{a + b}{2}, b\right]$, and $g(x) = \left(x - \frac{a + b}{2} \right)$.
We have $g(x) \leq 0, f(x) \leq f(\frac{a + b}{2})$ on $I_{1}$. So that $g(x)f(x) \geq f(\frac{a + b}{2}) g(x)$.
Similarly, $g(x) \geq 0$, $f(x) \geq f(\frac{a + b}{2})$ on $I_{2}$. So that $g(x) f(x) \geq f(\frac{a + b}{2})g(x)$ on $I_{2}$.
Now we have:
$\int_{a}^{b}\left(x - \frac{a + b}{2} \right) f(x) dx$
$= \int_{a}^{\frac{a + b}{2}}\left(x - \frac{a + b}{2} \right) f(x) dx + \int_{\frac{a + b}{2}}^{b}\left(x - \frac{a + b}{2} \right) f(x) dx$
$\geq f(\frac{a + b}{2})\left[\int_{I_{1}}g(x)dx + \int_{I_{2}} g(x) dx \right]$
$ = f(\frac{a + b}{2} )\int_{a}^{b} g(x)dx = 0$, which implies the desired inequality.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1576300",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
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Find $\tan C$ in a triangle satisfying the constraint Given a triangle with angles $A,B, C$ and sides $a, b, c$ opposite to their respective angles, how can I find $\tan C$ such that
$$c^2={a^3+b^3+c^3\over a+b+c}$$
I used the law of Cosines on the LHS as well as on the cube terms but I don't know exactly I'm looking for in manipulating the expression. Should I try to get $\sin C$ and $\cos C$ separately, or is there a more elegant way of doing things? I believe there is since this is an old competition problem, I'm just not seeing it.
| $$c^2={a^3+b^3+c^3\over a+b+c} \Rightarrow c^2(a+b+c)=(a^3+b^3+c^3)$$
So $$c^2a+c^2b+c^3=a^3+b^3+c^3$$
$$\Rightarrow c^2a+c^2b=a^3+b^3=(a+b)(a^2-ab+b^2)$$
Since $a,b$ are non zero , $a+b \neq0$.
Thus $$c^2=(a^2-ab+b^2) \Rightarrow ab = a^2+b^2-c^2$$
$$\Rightarrow \frac{1}{2} = \frac{a^2+b^2-c^2}{2ab}$$
Thus $$\cos C = \frac{1}{2} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1576847",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Maclaurin's Series for $\sec(x)$ with help of Maclaurin's series for $\tan(x)$
Is there any way to derive Maclaurin's series for $\sec(x)$ with the help of Maclaurin's series for $\tan(x)$?
As we know, the Maclaurin's Series for $\tan(x)$ is:
$$\tan(x)=x+\frac{x^3}{3}+\frac{2x^5}{15}+....$$
| It is probably not the simplest way to do it but, using $$\frac{d}{dx}\tan(x)=\sec ^2(x)$$ you will need to compute $$\tan(x)=x+\frac{x^3}{3}+\frac{2 x^5}{15}+\frac{17 x^7}{315}+O\left(x^9\right)$$ $$\sec ^2(x)=1+x^2+\frac{2 x^4}{3}+\frac{17 x^6}{45}+O\left(x^8\right)$$ $$\sec(x)=\sqrt{1+x^2+\frac{2 x^4}{3}+\frac{17 x^6}{45}+O\left(x^8\right)}$$ Now, consider $$y=x^2+\frac{2 x^4}{3}+\frac{17 x^6}{45}+\cdots$$ $$\sqrt{1+y}=1+\frac{y}{2}-\frac{y^2}{8}+\frac{y^3}{16}-\frac{5 y^4}{128}+O\left(y^5\right)$$ and replace $y$ and expand.
Doing it with the terms given above, you would arrive to $$1+\frac{x^2}{2}+\frac{5 x^4}{24}+\frac{61 x^6}{720}-\frac{41
x^8}{640}+\cdots$$ while the correct series should be $$1+\frac{x^2}{2}+\frac{5 x^4}{24}+\frac{61 x^6}{720}+\frac{277
x^8}{8064}+\cdots$$
Edit
Another possible way to do it based on $$\int \tan(x)=\log(\sec(x))$$ So $$\log(\sec(x))=\frac{x^2}{2}+\frac{x^4}{12}+\frac{x^6}{45}+\frac{17 x^8}{2520}+O\left(x^{10}\right)$$ $$\sec(x)=\exp\left(\frac{x^2}{2}+\frac{x^4}{12}+\frac{x^6}{45}+\frac{17 x^8}{2520}+\cdots \right)$$ Now, consider $$y=\frac{x^2}{2}+\frac{x^4}{12}+\frac{x^6}{45}+\frac{17 x^8}{2520}+\cdots$$ and use $$\exp(y)=1+y+\frac{y^2}{2}+\frac{y^3}{6}+\frac{y^4}{24}+O\left(y^5\right)$$ Replace $y$ and expand.
Doing it with the terms given above, you would arrive to the correct series up to the $O\left(x^{9}\right)$.
Edit
Another way using the tangent half angle substitution. Let us start with $$\sec(x)=\frac{1+t^2}{1-t^2}=1+2 t^2+2 t^4+2 t^6+2 t^8+O\left(t^9\right)$$ using $$t=\tan(\frac x2)=\frac{x}{2}+\frac{x^3}{24}+\frac{x^5}{240}+\frac{17 x^7}{40320}+O\left(x^9\right)$$ Doing it with the terms given above, you would arrive to the correct series up to the $O\left(x^{9}\right)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1577978",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find $f \circ g$, $g \circ f$, $f \circ f$, $g \circ g$ where $f(x)=-1+|x-2|$ and $g(x)=2-|x|$. If $f(x)=-1+|x-2|;0\leq x\leq4$ and
$g(x)=2-|x|;-1\leq x\leq3$
Then find $f \circ g(x),g \circ f(x),f \circ f(x),g \circ g(x)$
In the above figure the red graph shows $g(x)$ and the blue graph shows $f(x)$.
The domain of $f(x)$ is $0\leq x\leq4$ and the domain of $g(x)$ is $-1\leq x\leq3$.
I found the domain of $f \circ g(x)$,it is the intersection of $-1\leq x\leq3$ and $0\leq g(x)\leq 4$ i.e. the intersection of $-1\leq x\leq3$ and $-2\leq x\leq 2$.So its domain is $-1\leq x\leq 2$.
then found the domain of $g \circ f(x)$,it is the intersection of $0\leq x\leq4$ and $-1\leq f(x)\leq 3$ i.e. the intersection of $0\leq x\leq4$ and $-2\leq x\leq 6$.So its domain is $0\leq x\leq 4$.
then i found the domain of $f \circ f(x)$,it is the intersection of $0\leq x\leq4$ and $0\leq f(x)\leq 4$ i.e. the intersection of $0\leq x\leq4$ and $-3\leq x\leq 1\cup 3\leq x\leq 7$.So its domain is $0\leq x\leq 1\cup 3\leq x\leq 4$.
then i found the domain of $g \circ g(x)$,it is the intersection of $-1\leq x\leq3$ and $-1\leq g(x)\leq 3$ i.e. the intersection of $-1\leq x\leq3$ and $-3\leq x\leq 3$.So its domain is $-1\leq x\leq 3$.
Now i found $f \circ g(x)=\left\{ \begin{array}{lcc}
-1-x & -1 \leq x \leq 0 \\
\\ x-1, &0<x\leq2 \\
\\
\end{array}
\right.$ which i got correct.
I found $g \circ f(x)=\left\{ \begin{array}{lcc}
x+1 & 0 \leq x \leq 2 \\
\\ -x+5, &2\leq x\leq 4 \\
\\
\end{array}
\right.$.But my book says this is wrong and correct $g \circ f(x)$ is
$g \circ f(x)=\left\{ \begin{array}{lcc}
x+1 & 0 \leq x <1 \\
\\3-x&1\leq x\leq 2 \\
\\x-1&2<x\leq 3 \\
\\ -x+5, &3 <x\leq 4 \\
\\
\end{array}
\right.$I dont know where have i gone wrong.
I found $f \circ f(x)=\left\{ \begin{array}{lcc}
x & 0 \leq x \leq 1 \\
\\ x-6, &3\leq x\leq 4 \\
\\
\end{array}
\right.$.But my books says this is wrong.And the correct $f \circ f(x)$ is $f \circ f(x)=\left\{ \begin{array}{lcc}
x & 0 \leq x \leq 1 \\
\\ -x+4, &3\leq x\leq 4 \\
\\
\end{array}
\right.$.I dont know where have i gone wrong.
I found $g \circ g(x)=\left\{ \begin{array}{lcc}
x+4 & -1 \leq x \leq 0 \\
\\ x, &0\leq x\leq 3 \\
\\
\end{array}
\right.$ But my books says this is wrong and the correct gog(x) is
$g \circ g(x)=\left\{ \begin{array}{lcc}
-x & -1 \leq x \leq 0 \\
\\ x, &0< x\leq 2 \\
\\ 4-x,&2<x\leq 3\\
\end{array}
\right.$I dont know where have i gone wrong.
Please help me.Thanks.
| I will look at $g\circ f$ for you only, maybe this would help you correcting the other parts.
In general when doing such exercises you should particular attention to when the functions or their derivatiaves change sign. For $f(x)$ these points on the domain are $x=1,2,3$. Here hence you should look at four different ranges: $0<x<1$, $1<x<2$, $2<x<3$ and $3<x<4$. When feeding $f(x)$, i.e. the range you should then care only about the rule of $g(x)$.
Let start with the $0<x<1$. Here $0<f(x)=1-x<1$ therefore $g(f(x))=2-|1-x|$ or that $g(f(x))=2-(1-x)=1+x$.
Now lets look at $1<x<2$, here we have $-1<f(x)=1-x<0$, therefore $g(f(x))=2-|1-x|$ or that $g(f(x))=2-(x-1)=3-x$.
also for $2<x<3$ we have $-1<f(x)=-3+x<0$ hence $g(f(x))=2-|-3+x|$ or that $g(f(x))=2-(3-x)=-1+x$.
and finally for $3<x<4$ we have $0<f(x)=-3+x<1$ hence $g(f(x))=2-|-3+x|$ or that $g(f(x))=2-(-3+x)=5-x$.
Doing similar steps for the rest of the functions you should be able to obtain the correct results.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1578072",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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distinct roots of the equation $A\sin^3 x+B\cos^3 x+C =0$
The number of distinct roots of the equation $A\sin^3 x+B\cos^3 x+C =0$
no two of which differ by $2\pi$ is, Where $A,B,C\in \mathbb{R}$
$\bf{(a)}\;\;\;\;\;\; 3\;\;\;\;\;\; (b)\;\;\;\;\;\; 4\;\;\;\;\;\; (c)\;\;\;\;\;\; 6\;\;\;\;\;\; (d)\;\;\;\;\;\; infinite$
$\bf{My\; Try::}$ Using $$\sin 3x =3\sin x -4\sin^3 x\Rightarrow 4\sin^3 x=3\sin x-\sin 3x$$ and
$$\cos 3x = 4\cos^3 x-3\cos x\Rightarrow 4\cos^3 x=\cos 3x+3\cos x$$
So we get $$A(4\sin^3 x) +B(4\cos^3 x)+4C=0$$
So $$A(3\sin x-\sin 3x)+B(\cos 3x+3\cos x)+4C=0$$
I did not understand How can I solve that question, Help me
Thanks
| First, observe that if $A=0$, $B=0$, and $C=0$, then there are infinitely many solutions. This, however is boring, so we'll exclude this case.
Using the parameterization $\sin(x)=\frac{2t}{1+t^2}$ and $\cos(x)=\frac{1-t^2}{1+t^2}$ for $\sin$ and $\cos$, (and clearing fractions) we get the equation
$$
8At^3+B(1-t^2)^3+C(1+t^2)^3=0.
$$
By expanding, this becomes
$$
(C-B)t^6+3(B+C)t^4+8At^3+3(C-B)t^2+(B+C)=0.
$$
We can use Descartes' rule of signs to bound the number of roots. There are at most $3$ sign changes since some of the coefficients repeat. Replacing $t$ by negative $t$, shows the same structure, so there are at most $3$ negative roots.
The one case that is missing is when $t=\infty$, in this case, $\sin(x)=0$ and $\cos(x)=-1$, which is a solution only when $B+C=0$. In this case, the polynomial also has a root at $t=0$ and the equation above simplifies to
$$
2Ct^6+8At^3+6Ct^2=0
$$
The LHS has at most $2$ sign changes (observe that for $t>0$ or $t<0$, one case will have no sign variation).
This shows that when $A$, $B$, and $C$ are not all zero, there are at most $6$ solutions (and also shows the dependence on the values of $A$, $B$, and $C$).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1580975",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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If $\small {x+\sqrt { (x+1)(x+2) } +\sqrt { (x+2)(x+3) } +\sqrt { (x+3)(x+1) } = 4}$, solve for $x$. I came across this olympiad algebra problem, asking to solve for $x$:
$x\ +\ \sqrt { (x+1)(x+2) } \ +\ \sqrt { (x+2)(x+3) } +\ \sqrt { (x+3)(x+1) } =\ 4$
Here was my try:
If $$x\ +\ \sqrt { (x+1)(x+2) } \ +\ \sqrt { (x+2)(x+3) } +\ \sqrt { (x+3)(x+1) } =\ 4$$
Then $\quad \sqrt { (x+1)(x+2) } +\sqrt { (x+2)(x+3) } +\sqrt { (x+3)(x+1) } =4-x$.
Further, I tried squaring the equation on both sides, but that doesn't seem to solve my problem. Please help.
Thank you.
| While Kf-Sansoo has given an elegant answer, if the problem asks for any real (and not just rational) solution, then it misses a second one which is a root of a quartic equation, hence normally is not easy to do by hand.
In general, the two solutions to,
$$x+\sqrt{(x+1)(x+2)}+\sqrt{(x+2)(x+3)}+\sqrt{(x+3)(x+1)} = n\tag1$$
for real $n>0$ are,
$$x = \frac{(n^2+4n+5)^2}{4(n+1)(n+2)(n+3)}-2\tag{2a}$$
and the appropriate root of,
$$-23 + 48 n - 22 n^2 + n^4 - 4 (30 - 33 n + 6 n^2 + n^3) x \\+
16 (-11 + 6 n) x^2 + 16 (-6 + n) x^3 - 16 x^4=0\tag{2b}$$
For $n = 4$, we have $x_1 = -311/840 \approx -0.37$. Then $x_2 \approx -5.12357$ as a root of,
$$73 - 232 x + 208 x^2 - 32 x^3 - 16 x^4 = 0$$
with both valid for the positive case of $\sqrt{z}$ as the graph from Walpha below shows,
$\color{green}{Edit:}$ When using Kf-Sansoo's method, we end up with an expression of form,
$$\prod^4 (c_1\sqrt{x+1}\pm c_2\sqrt{x+2}\pm c_3\sqrt{x+3}) = 0$$
Let $n=4$ and we get $x = -311/840$. Simpler, but the price to pay is we lose a second solution. Another method is to form an octic,
$$\prod^8 \big(y-(\pm\sqrt{z_1}\pm \sqrt{z_2}\pm \sqrt{z_3})\big)=0 \tag3$$
After it is formed, substitute into $(3)$ the ff,
$$y = n-x\\z_1=(x+1)(x+2)\\z_2=(x+2)(x+3)\\z_3=(x+3)(x+1)$$
and we get linear/quartic factors given by $(2a), (2b)$. Less simpler, but it yields a second valid solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1582932",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 1
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Integrating $\int \frac{8 dx}{3 \cos 2x + 1}$ to arctan rather than log [Edit: It is becoming increasingly likely that the expected answer containing arctan might be a typo from my book, which is transcribed correctly here, so an answer containing log might be correct after all. Sorry about that and thanks for the answers so far, I've learned a lot!]
I'm trying to integrate:
$$
\int \frac{8 dx}{3 \cos 2x + 1} \\
$$
So I tried substituting $x$ for $z$:
$$
z = \tan \frac{x}{2} \\
\cos x = \frac{1 - z^2}{1 + z^2} \\
dx = \frac{2 dz}{1 + z^2} \\
$$
Then I tried replacing $\cos 2x$ with $2 \cos^2 x - 1$ followed by some work:
$$
\int \frac{4(1 + z^2) dz}{z^4 - 4z^2 + 1} \\
$$
If I try substituting $z$ back to $x$, it seems my answer will contain log rather than the expected arctan:
$$
\frac{x}{3} - \frac{5}{6} \arctan (2 \tan \frac{x}{2}) + C \\
$$
Did I take a wrong turn somewhere?
| If we are going to make a Weierstrass-style substitution, it seems more natural, because of the $\cos 2x$, to let $z=\tan x$. Then $\cos 2x=\frac{1-z^2}{1+z^2}$ and $dx=\frac{1}{1+z^2}\,dz$. Thus our integral becomes, after a little manipulation,
$$\int\frac{4}{2-z^2}\,dz.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1585256",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.