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Question about an integral trick This is an excerpt from brilliant.org on integral tricks:
A similar method to the above is to reverse the interval of
integration: to "integrate backwards." For a function $f$ and real
numbers $a < b$,
$\int_a^b f(x) \, dx = \int_a^b f(a+b-x) \, dx.$
Instead of the function being centered at $0$, the function is now
centered at $(\tfrac{a+b}{2})$. Then,
$\int_a^b f(x) \, dx = \frac{1}{2} \int_a^b f(x) + f(a+b-x) \, dx.$
I understand that the trick is to integrate backwards, from $b$ towards $a$, but I don't understand what is meant by that the function is centered at $(\tfrac{a+b}{2})$. What is meant by that and why is it significant? Also, how is the last integral derived?
|
If we let $x=a+b-y$ then when $x=a$, $y=a+b-a=b$ and when $x=b$, $y=a+b-b=a$ and $dx=-dy$, so
$$\int_a^bf(x)dx=\int_b^af(a+b-y)(-dy)=\int_a^bf(a+b-y)dy=\int_a^bf(a+b-x)dx$$
Where running the integral backwards changes its sign and we renamed the variable of integration back to $x$. Then
$$\begin{align}\int_a^bf(x)dx&=\frac12\int_a^bf(x)dx+\frac12\int_a^bf(x)dx\\
&=\frac12\int_a^bf(x)dx+\frac12\int_a^bf(a+b-x)dx\\
&=\frac12\int_a^b\left(f(x)+f(a+b-x)\right)dx\end{align}$$
The function is centered at $\frac{b+a}2$ means it is even about that point: let $g(x)=\frac12\left(f(x)+f(a+b-x)\right)$. Then
$$\begin{align}g\left(\frac{b+a}2-x\right)&=\frac12\left(f\left(\frac{b+a}2-x\right)+f\left(b+a-\frac{b+a}2+x\right)\right)\\
&=\frac12\left(f\left(b+a-\frac{b+a}2-x\right)+f\left(\frac{b+a}2+x\right)\right)\\
&=\frac12\left(f\left(b+a-\left(\frac{b+a}2+x\right)\right)+f\left(\frac{b+a}2+x\right)\right)\\
&=g\left(\frac{b+a}2+x\right)\end{align}$$
Usually when we talk about even functions the point of reflection is implicitly $0$ so $g(-x)=g(x)$.
Here's an integral from the CRC Handbook of Chemistry and Physics and the Putnam exam:
$$\begin{align}\int_0^{\frac{\pi}2}\frac{d\theta}{1+\tan^{\sqrt2}\theta}&=\frac12\int_0^{\frac{\pi}2}\left(\frac{1}{1+\tan^{\sqrt2}\theta}+\frac{1}{1+\tan^{\sqrt2}\left(\frac{\pi}2-\theta\right)}\right)d\theta\\
&=\frac12\int_0^{\frac{\pi}2}\left(\frac{1}{1+\tan^{\sqrt2}\theta}+\frac{1}{1+\cot^{\sqrt2}\left(\theta\right)}\right)d\theta\\
&=\frac12\int_0^{\frac{\pi}2}\left(\frac{1}{1+\tan^{\sqrt2}\theta}+\frac{\tan^{\sqrt2}\theta}{\tan^{\sqrt2}\theta+1}\right)d\theta\\
&=\frac12\int_0^{\frac{\pi}2}1d\theta=\frac{\pi}4\end{align}$$
Cool how folding about $\theta=\frac{\pi}4$ knocked down what seemed like a very daunting integral.
|
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|
Is function $f(x) = \frac{x^{2}-2}{x-\sqrt{2}}$ continuous for all $x$?
Is function $f(x) = \frac{x^{2}-2}{x-\sqrt{2}}$ continuous for all $x$?
I understand that the function
$$ g(x) = \frac{x^{2}-1}{x-1} $$
has one discontinuity for $x=1$.
[Added later] In general my question is for the function
$$
\frac{x^{2}-a}{x-\sqrt a},
$$
Can we say that there is a discontinuity at point $x = a$? If $a$ is an irrational number, then how to define its point of discontinuity?
|
$\require{cancel}$
Indeed, $$f(x) = \frac{x^{2}-2}{x-\sqrt{2}}$$ has a discontinuity at $x=\sqrt 2$ (the point at which the domominator is 0). But it is removable.
$$f(x) = \frac{x^2 - 2}{x-\sqrt 2} =\frac{x^2 - (\sqrt 2)^2}{x-\sqrt 2}= \frac{(x+\sqrt 2)(\cancel{x-\sqrt 2})}{\cancel{x-\sqrt 2}} = {x+\sqrt 2}$$ when $x \neq \sqrt 2$.
All that remains is to define $f(\sqrt 2) = 2\sqrt 2$. Now, we have defined the function so that $f(x)$ is continuous!
$$f(x) = \begin{cases} x+\sqrt 2 & x \neq \sqrt 2\\ \\2\sqrt 2 & x = \sqrt 2\end{cases}$$
In general, suppose $c\in \mathbb R$. Then, the function defined by $$f(x) = \frac{x^2−c^2}{x−c}$$ is continuous everywhere in $\mathbb R$, except at the point $x=c$.
Such a discontinuity is removable by defining $f(x) = x+c$, for all $x\neq c$, and otherwise, $f(c)=2c$
|
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|
How do you find the roots of the negative polynomial $(-x^3 - x + 1)$ to further use for partial fractions? I am trying to factor the polynomial $p(x)=-x^3-x+1$ to use in partial fractions, but I am getting stumped on how to break this up into roots. Any tips will be helpful! My teacher did something similar incorporating alpha and beta, and the quadratic formula, but I didn’t follow.
|
Cubic equations in the form $x^3+px+q=0$ can be solved by the substitution $x=rf(\theta)$ so
$$r^3\left[f(\theta)\right]^3+rf(\theta)p+q=0$$
Then multiply by $4/r^3$ to get
$$4\left[f(\theta)\right]^3+\frac{4p}{r^2}f=\frac{-4q}{r^3}$$
Choose the sign of $r=\pm\sqrt{\frac{4|p|}3}$ such that $qr\le0$. Then
$$4\left[f(\theta)\right]^3+3\frac{p}{|p|}f(\theta)=\left|\frac{4q}{r^3}\right|$$
Match with one of
$$\begin{align}\cos(3\theta)&=4\cos^3\theta-3\cos\theta\\
\cosh(3\theta)&=4\cosh^3\theta-3\cosh\theta\\
\sinh(3\theta)&=4\sinh^3\theta+3\sinh\theta\end{align}$$
In our case, $p=1$, $q=-1$, $r=\frac2{\sqrt3}$ and
$$\sinh(3\theta)=4\sinh^3\theta+3\sinh\theta=\frac{3\sqrt3}2$$
So
$$\theta=\frac13\sinh^{-1}\left(\frac{3\sqrt3}2\right)$$
And our solution is
$$a=\frac2{\sqrt3}\sinh\left(\frac13\sinh^{-1}\left(\frac{3\sqrt3}2\right)\right)$$
Then we can do partial fractions:
$$\begin{align}\frac1{x^3+x-1}&=\frac1{\left(x^2+ax+a^2+1\right)(x-a)}\\
&=\frac1{31}\left[\frac{-\left(6a^2+9a+4\right)x-\left(18a^2-4a+12\right)}{x^2+ax+a^2+1}+\frac{6a^2+9a+4}{x-a}\right]\end{align}$$
|
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|
Possible remainders of a perfect square when divided by $3,5,6$. A perfect square will have in its prime factorization all the primes having an even power, so that the square root will simply divide each by $2$.
Let the two constituents of a perfect square number ($p$) be : $mm$, i.e. same positive integer is repeated twice.
There can be two types of primes: odd, even, as say $p= 3^4.2^2, 2^4, 3^4$ .
(i) The even prime has the form of $2^{2k}=4^k$, for some positive integer $k$.
(ii) The odd prime has the form of $(2n+1)^{2k}$, for some positive integer $k,n$.
Need find: (i) the remainders by congruence arithmetic approach for these two forms of perfect number constituents separately, & (ii)find product of these termsunder modulo $3,5,6$.
Edit Let us take the first few positive integers' squares $\gt1$, and based on each prime's residue class, will find their prime factorization.:
$2^2 => \equiv 1 \pmod 3, \equiv -1 \pmod 5, \equiv -2 \pmod 6$
$3^2 => \equiv 0 \pmod 3, \equiv -1 \pmod 5, \equiv 3 \pmod 6$
$4^2 = 2^4 => \equiv 1^2 \pmod 3, \equiv (-1)^2 \pmod 5, \equiv (-2)^2 \pmod 6 => \equiv 1 \pmod 3, \equiv 1 \pmod 5, \equiv 4 \pmod 6$
$5^2 => \equiv 1 \pmod 3, \equiv 0 \pmod 5, \equiv 1 \pmod 6$
$6^2=(2.3)^2=2^2.3^2 $ $=> (\equiv 1 \pmod 3, \equiv -1 \pmod 5, \equiv -2 \pmod 6)\cdot(\equiv 0 \pmod 3, \equiv -1 \pmod 5, \equiv 3 \pmod 6)$ $=> (\equiv (1.0) \pmod 3, \equiv (-1.-1) \pmod 5, \equiv (-2.3) \pmod 6)$ $=> \equiv 0 \pmod 3, \equiv 1 \pmod 5, \equiv 0\pmod 6$
$7^2=> \equiv 1 \pmod 3, \equiv 3 \pmod 5, \equiv 1 \pmod 6$
$8^2=2^6 => (\equiv 1 \pmod 3, \equiv -1 \pmod 5, \equiv -2 \pmod 6)^3$ $=> \equiv 1^3 \pmod 3, \equiv (-1)^3 \pmod 5, \equiv (-2)^3 \pmod 6$=> $\equiv 1 \pmod 3, \equiv -1 \pmod 5, \equiv -2 \pmod 6$
$9^2=3^4 => (\equiv 0^2 \pmod 3, \equiv (-1)^2 \pmod 5, \equiv 3^2 \pmod 6)$ $=>\equiv 0 \pmod 3, \equiv 1 \pmod 5, \equiv 3 \pmod 6$
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If $p$ is an odd prime $>3$, then
*
*$p\equiv \pm 1\mod 3$, so $p^2\equiv 1\mod 3$;
*$p\equiv \pm 1,\pm 2\mod 5$, so $p^2\equiv \pm 1\mod 5$;
*$p\equiv \pm 1\mod 6$, so $p^2\equiv 1\mod 6$.
|
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|
Is $577^3$ the smallest cube that is expressible as the sum of $3$ positive cubes in $6$ different ways? A Taxi Cab Number is a number $\text{Ta}(n)$ that can be written as the sum of two cubes in $n$ different ways. More formally, they are known as Hardy-Ramanujan Numbers or, as Ramanujan had called them, Magic Numbers.
$$\begin{align} 577^3 &= 356^3 + 385^3 + 448^3 \\ &= 1^3 + 426^3 + 486^3 \\ &= 172^3 + 318^3 + 537^3 \\ &= 41^3 + 244^3 + 562^3 \\ &= 153^3 + 174^3 + 568^3 \\ &= 90^3 + 201^3 + 568^3 \end{align}$$ Is this the smallest cube that is expressible as the sum of $3$ positive cubes in $6$ different ways? In symbols, does $$577^3 = \text{taxicab}(3, 3, 6)\,?\tag*{$\bigg(\begin{align} \verb|S|&\verb|uch that the LHS| \\ &\verb|must be a cube.|\end{align}\bigg)$}$$ I have been trying to find numbers similar to Taxi Cab Numbers. The best example of a taxicab number is $1729$ because it is the smallest number that can be expressed as the sum of two cubes in two different ways, i.e. $$1729 = 1^3 + 12^3 = 9^3 + 10^3.$$ I developed a more general case of finding numbers $a_n^{ \ \ 2n - 1}$ that can be written as the sum of $(n+1)$ cubes in $2(2n-1)$ different ways, trying to find a potential pattern in the values of $a_n$ for which $a_n \in\mathbb{Z^+}$.
Given that $n = 1$, we obtain that $a_1 = 1729$. Given $n = 2$, we apparently obtain $a_2 = 577$. If this is true, the only pattern I can find is that $$a_n^{\ \ 2n - 1} - 1 = \left\{\sum_{k=1}^n b_k^{\ \ 3} : b_k\in\mathbb{Z^+}\right\}.$$ Also, could somebody find the value of $a_3$ and $a_4$ and if they want, $a_5$ (though I think the smallest number that is the sum of $9$ cubes in $18$ different ways might be pretty large).
Thank you in advance.
P.S. I was not able to find any other appropriate tags apart from > (number-theory) <
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No.
$$\begin{align}216^3 &=24^3+ 144^3+ 192^3\\ &=26^3 +102^3+208^3\\ &=30^3 +164^3+ 178^3\\ &=48^3 +76^3+ 212^3\\ &= 102^3 +117^3 +195^3\\ &=108^3 +144^3+ 180^3\end{align}$$
Find more here.
|
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|
If $A$, $B$ and $C$ are the angles of a triangle then find the value of $\Delta$ I'll state the question from my book below:
If $A$, $B$ and $C$ are the angles of a triangle, then find the determinant value of
$$\Delta = \begin{vmatrix}\sin^2A & \cot A & 1 \\ \sin^2B & \cot B & 1 \\ \sin^2C & \cot C & 1\end{vmatrix}.$$
Here's how I tried solving the problem:
$\Delta = \begin{vmatrix}\sin^2A & \cot A & 1 \\ \sin^2B & \cot B & 1 \\ \sin^2C & \cot C & 1\end{vmatrix}$
$R_2 \to R_2 - R_1$
$R_3 \to R_3 -R_1$
$= \begin{vmatrix}\sin^2A & \cot A & 1 \\ \sin^2B-\sin^2A & \cot B-\cot A & 0 \\ \sin^2C-\sin^2A & \cot C-\cot A & 0\end{vmatrix}$
Expanding the determinant along $C_3$
\begin{align}
&= (\sin^2B-\sin^2A)(\cot C-\cot A)-(\cot B-\cot A)(\sin^2C-\sin^2A) \\
&= \sin(B+A) \sin(B-A) \left[\frac {\cos C}{\sin C} - \frac {\cos A}{\sin A}\right] - \left[\frac {\cos B}{\sin B} - \frac {\cos A}{\sin A}\right]\sin(C+A) \sin(C-A) \\
&= \frac {\sin(B+A) \sin(B-A) \sin(A-C)} {\cos A \cos C} - \frac {\sin(A-B) \sin(C+A) \sin(C-A)} {\cos A \cos C} \\
&= \frac {\sin(B-A) \sin (A-C)} {\cos A} \left[\frac {\sin(A+B)} {\cos C} - \frac {\sin(A+C)} {\cos B}\right] \\
&= \frac {\sin(B-A) \sin (A-C)} {\cos A} \left[\frac {\sin C} {\cos C} - \frac {\sin B} {\cos B}\right] \\
&= \frac {\sin(B-A) \sin (A-C) \sin (C-B)} {\cos A \cos B \cos C}
\end{align}
I tried solving further but the expression just got complicated. I don't even know if the work I've done above is helpful. My textbook gives the answer as $0$. I don't have any clue about getting the answer. Any help would be appreciated.
|
In the standard notation we obtain:
$$\Delta=\sum_{cyc}\sin^2\alpha(\cot\beta-\cot\gamma)=\sum_{cyc}\frac{4S^2}{b^2c^2}\left(\frac{\frac{a^2+c^2-b^2}{2ac}}{\frac{2S}{ac}}-\frac{\frac{a^2+b^2-c^2}{2ab}}{\frac{2S}{ab}}\right)=$$
$$=S\sum_{cyc}\frac{a^2+c^2-b^2-a^2-b^2+c^2}{b^2c^2}=2S\sum_{cyc}\frac{c^2-b^2}{b^2c^2}=2S\sum_{cyc}\left(\frac{1}{b^2}-\frac{1}{c^2}\right)=0.$$
|
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|
what is the best way to solve $\int\frac{\cos^2x\,\mathrm dx}{\sin^2x+4\sin x\cos x}$? I have the integral:
$$\int\frac{\cos^2x\,\mathrm dx}{\sin^2x+4\sin x\cos x}$$
of course there is $u = \tan\frac{x}{2}$
But I want to avoid it if possible, also neither $\cos x = u$ nor $\sin x = u$ made the life easier :(
|
I must say Lab bhattachargee shew a method that is good, here is my method which may also be used. First we split the numerator using $\sin^2 x + \cos ^2 x = 1$.
$$\begin{align}
I &=\int\frac{ dx}{\sin^2x+4\sin x\cos x}-\int\frac{\sin^2x dx}{\sin^2x+4\sin x\cos x}\\
&= \int \frac{\csc^2 x}{1+4 \cot x}dx - \int \frac{\sin x}{\sin x + 4 \cos x}dx
\\
&= \frac{-1}{4} \ln|1+4\cot x|- \frac{1}{17}\left(\int \frac{\sin x + 4 \cos x}{\sin x + 4 \cos x} dx -4\int\frac{\cos x - 4 \sin x}{\sin x + 4 \cos x} dx \right)
\end{align}$$
You can see why we split the second integral into two parts, in one numerator contains the denominator $(\sin x + 4 \cos x)$ and in other numerator is its derivative $(\cos x - 4 \sin x)$. So we had to solve the equation :
$$\sin x = a(\sin x + 4 \cos x)+b(\cos x + 4 \sin x)$$
to find the coefficients $a,b$ by comparing coefficients of $\sin x$ and $\cos x$.
The final antiderivative is simple to write:
$$\frac{-1}{4} \ln|1+4\cot x|- \frac{1}{17}\left(x -4 \ln|\sin x + 4 \cos x| \right)$$
|
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|
Prove that $24|n^2-1$, if $(n,6)=1$ $n^2-1 = (n-1)(n+1)$
Then $24|(n-1)(n+1)$
$(n,6)=1$: $\exists a,b\in\mathbb{Z}$ that $n = 6\cdot a+b$
Investigate the residues, which arise when dividing the number n by two and three:
$\frac{6\cdot a+b}{3} = \frac{6\cdot a}{3}+\frac{b}{3} = 2\cdot a+\frac{b}{3}$
$\frac{6\cdot a+b}{2} = 3\cdot a+\frac{b}{2}$
Is this correct? How to prove it using resiudes?
Thank You.
|
If $\gcd(n,6)=1$, then there is some integer $k$ such that $n=6k\pm 1$. We compute $$n^2-1=(6k\pm 1)^2-1 = 36k^2\pm 12k = 12k (3k\pm 1)$$
Now, if $k$ is even, then $24|12k$ already. If instead $k$ is odd, then $3k\pm 1$ is even, so again $24|12k(3k\pm 1)$.
|
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|
Prove $f(x) = \frac{1}{1+x^2}$ is uniformly continuous on $\mathbb{R}$ Prove $f(x) =\frac{1}{1+x^2}$ is uniformly continuous on $\mathbb{R}$.
My attempt...
Proof
$$\left| f(x) - f(y) \right| = \left| \frac{1}{1+x^2} - \frac{1}{1+y^2}\right| = \frac{\left|x+y\right|}{\left(1+x^2\right)\left(1+y^2\right)}\left|x-y\right| $$
By the triangle inequality
$$\frac{\left|x+y\right|}{\left(1+x^2\right)\left(1+y^2\right)}\left|x-y\right| \leq \left(\frac{\left|x\right|}{\left(1+x^2\right)\left(1+y^2\right)} +\frac{\left|y\right|}{\left(1+x^2\right)\left(1+y^2\right)} \right) \left|x-y\right| \tag{$\star$}$$
Note that for all $x\in \mathbb{R}$
$$\left|x\right| < 1 +x^2 \implies \left|x\right| < \left(1 +x^2\right)\left(1+y^2\right)$$
Therefore
$$\frac{\left|x\right|}{\left(1 +x^2\right)\left(1+y^2\right)} \leq 1$$
Applying this fact to $(\star)$ we see that
$$\left(\frac{\left|x\right|}{\left(1+x^2\right)\left(1+y^2\right)} +\frac{\left|y\right|}{\left(1+x^2\right)\left(1+y^2\right)} \right) \left|x-y\right| \leq \left(1 + 1\right)\left|x-y\right| \leq 2\left|x - y \right|$$
Therefore $f$ is a Lipschitz function, which implies $f$ is uniformly continuous on $\mathbb{R}$.
Please comment on validity and/or readability, thank you.
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Big Hint: Use this and the fact that $\vert f'(x)\vert <1$.
|
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|
Show that $f(x)=x^{5/3}-kx^{4/3}+k^2x$ is increasing for $k\neq0$ So to show the function is increasing/decreasing we differentiate and show it is more than zero/less than zero:
We have
$$f(x)=x^{5/3}-kx^{4/3}+k^2x$$
Hence,
$$f'(x)=\frac{5}{3}x^{2/3}-\frac{4k}{3}x^{1/3}+k^2$$
But how do I show
$$\frac{5}{3}x^{\frac{2}{3}}-\frac{4k}{3}x^{\frac{1}{3}}+k^2>0$$
|
Let
$$y = x^{\frac{1}{3}}$$
So,
$$\frac{5}{3}y^{2} - \frac{4k}{3}y + k^{2} > 0 \Longleftrightarrow 5y^{2} - 4ky + 3k^{2} > 0$$
and
$$(-4k)^{2} - 4.5.3k^{2} = 16k^{2} - 60k^{2} < 0.$$
Therefore, $\frac{5}{3}y^{2} - \frac{4k}{3}y + k^{2} > 0$. Complete the solution.
|
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|
Prove $\lim\limits_{n\to∞}{\sum\limits_{x=0}^n\binom nx(1+{\rm e}^{-(x+1)})^{n+1}\over\sum\limits_{x=0}^n\binom nx(1+{\rm e}^{-x})^{n + 1}}=\frac 13$ I am trying to find limit of the following function:
$$
\lim_{n\rightarrow \infty}\frac{\sum\limits_{x = 0}^{n}\binom{n}{x}\left[1 + \mathrm{e}^{-(x+1)}\right]^{n + 1}}{\sum\limits_{x=0}^{n} \binom{n}{x}\left[1 + \mathrm{e}^{-x}\right]^{n + 1}}.
$$
When I wrote a Python code for this, I saw that it converges to $1/3$.
I am not sure how to approach this. Could somebody give a pointer about how to go about it?
EDIT: I want just some positive lower bound on this. So even if limit cannot be evaluated exactly, it is okay if I get some lower bound on this.
|
$\def\e{\mathrm{e}}$For $n \in \mathbb{N}_+$, denote$$
S_n = \sum_{k = 0}^n \binom{n}{k} (1 + \e^{-(k + 1)})^{n + 1},\ T_n = \sum_{k = 0}^n \binom{n}{k} (1 + \e^{-k})^{n + 1}.
$$
First,\begin{align*}
S_n &= \sum_{k = 0}^n \binom{n}{k} (1 + \e^{-(k + 1)})^{n + 1} = \sum_{k = 0}^n \binom{n}{k} \sum_{j = 0}^{n + 1} \binom{n + 1}{j} \e^{-(k + 1)j}\\
&= \sum_{k = 0}^n \sum_{j = 0}^{n + 1} \binom{n}{k} \binom{n + 1}{j} \e^{-(k + 1)j} = \sum_{k = 0}^n \binom{n}{k} + \sum_{k = 0}^n \sum_{j = 1}^{n + 1} \binom{n}{k} \binom{n + 1}{j} \e^{-(k + 1)j}\\
&= 2^n + \sum_{k = 0}^n \sum_{j = 0}^n \binom{n}{k} \binom{n + 1}{j + 1} \e^{-(k + 1)(j + 1)}, \tag{1}
\end{align*}
thus $S_n \geqslant 2^n$. Note that$$
(k + 1)(j + 1) \geqslant k + j + 1 \Longleftrightarrow kj \geqslant 0,
$$
thus from (1) there is\begin{align*}
S_n &\leqslant 2^n + \sum_{k = 0}^n \sum_{j = 0}^n \binom{n}{k} \binom{n + 1}{j + 1} \e^{-(k + j + 1)}\\
&= 2^n + \e^{-1} \sum_{k = 0}^n \sum_{j = 0}^n \binom{n}{k} \e^{-k} \binom{n + 1}{j + 1} \e^{-j}\\
&= 2^n + \e^{-1} \left( \sum_{k = 0}^n \binom{n}{k} \e^{-k} \right)\left( \sum_{j = 0}^n \binom{n + 1}{j + 1} \e^{-j} \right)\\
&\leqslant 2^n + \e^{-1} (1 + \e^{-1})^n (1 + \e^{-1})^{n + 1}\\
&= 2^n + \e^{-1} (1 + \e^{-1})^{2n + 1}.
\end{align*}
Therefore,$$
1 \leqslant \frac{S_n}{2^n} \leqslant 1 + \e^{-1} (1 + \e^{-1}) \left( \frac{1}{2} (1 + \e^{-1})^2 \right)^n.
$$
Note that $\dfrac{1}{2} (1 + \e^{-1})^2 < 1$, thus $S_n \sim 2^n$ $(n \to \infty)$.
Next,\begin{align*}
T_n &= \sum_{k = 0}^n \binom{n}{k} (1 + \e^{-k})^{n + 1} = \sum_{k = 0}^n \binom{n}{k} \sum_{j = 0}^{n + 1} \binom{n + 1}{j} \e^{-kj}\\
&= \sum_{k = 0}^n \sum_{j = 0}^{n + 1} \binom{n}{k} \binom{n + 1}{j} \e^{-kj}\\
&= \sum_{k = 0}^n \binom{n}{k} + \sum_{j = 0}^{n + 1} \binom{n + 1}{j} - 1 + \sum_{k = 1}^n \sum_{j = 1}^{n + 1} \binom{n}{k} \binom{n + 1}{j} \e^{-kj}\\
&= 2^n + 2^{n + 1} - 1 + \sum_{k = 1}^n \sum_{j = 1}^{n + 1} \binom{n}{k} \binom{n + 1}{j} \e^{-kj}\\
&= 3 × 2^n - 1 + \sum_{k = 0}^{n - 1} \sum_{j = 0}^n \binom{n}{k + 1} \binom{n + 1}{j + 1} \e^{-(k + 1)(j + 1)}, \tag{2}
\end{align*}
thus $T_n \geqslant 3 × 2^n - 1$. Also, analogously, from (2) there is\begin{align*}
T_n &\leqslant 3 × 2^n + \sum_{k = 0}^{n - 1} \sum_{j = 0}^n \binom{n}{k + 1} \binom{n + 1}{j + 1} \e^{-(k + j + 1)}\\
&= 3 × 2^n + \e \left( \sum_{k = 0}^{n - 1} \binom{n}{k + 1} \e^{-(k + 1)} \right)\left( \sum_{j = 0}^n \binom{n + 1}{j + 1} \e^{-(j + 1)} \right)\\
&\leqslant 3 × 2^n + \e (1 + \e^{-1} )^n (1 + \e^{-1})^{n + 1},
\end{align*}
which analogously implies that $T_n \sim 3 × 2^n$ $(n \to \infty)$.
Therefore,$$
\lim_{n \to \infty} \frac{\displaystyle\sum_{k = 0}^n \binom{n}{k} (1 + \e^{-(k + 1)})^{n + 1}}{\displaystyle\sum_{k = 0}^n \binom{n}{k} (1 + \e^{-k})^{n + 1}} = \lim_{n \to \infty} \frac{S_n}{T_n} = \lim_{n \to \infty} \frac{2^n}{3 × 2^n} = \frac{1}{3}.
$$
|
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"url": "https://math.stackexchange.com/questions/2674981",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
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|
Proving $f(x,y) = |xy| + a(x^2 + y^2)$ is convex for $a \ge 1/2$ I am working on a problem to prove $$
f(x,y) = |xy| + a(x^2 + y^2)
$$
is convex for $a \ge 1/2$.
My approach is to show that the Hessian is PSD for the cases where $x \not= 0, y \not= 0$.
However, this approach breaks down for cases when $x$ or $y$ are zero since the function is not differentiable at these points.
Any suggestions on other approaches that get around this?
|
$\begin{array}\\
f(x,y)
&= |xy| + a(x^2 + y^2)\\
&= |xy| + (a-\tfrac{1}{2})(x^2+y^2)+\tfrac{1}{2}|x|^2+\tfrac{1}{2}|y|^2\\
&= (a-\tfrac{1}{2})(x^2+y^2) + \tfrac{1}{2}(|x|+|y|)^2\\
\end{array}
$
The first term, $(a-\tfrac{1}{2})(x^2+y^2)$, is clearly convex.
The second term is the composition of the (outer) convex increasing nonnegative function $\xi\mapsto \tfrac{1}{2}\max\{0,\xi^2\}$ and the inner function $(x,y)\mapsto |x|+|y|$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2675784",
"timestamp": "2023-03-29T00:00:00",
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|
Factorize $\det\left[\begin{smallmatrix}yz-x^2&zx-y^2&xy-z^2\\zx-y^2&xy-z^2&yz-x^2\\xy-z^2&yz-x^2&zx-y^2\end{smallmatrix}\right]$ using factor theorem Factorize and prove that
$$
\Delta=\begin{vmatrix}
yz-x^2&zx-y^2&xy-z^2\\
zx-y^2&xy-z^2&yz-x^2\\
xy-z^2&yz-x^2&zx-y^2
\end{vmatrix}\\=\frac{1}{4}(x+y+z)^2\Big[(x-y)^2+(y-z)^2+(z-x)^2\Big]^2
$$
using factor theorem.
My Attempt:
$\Delta$ is a homogeneous symmetric polynomial of degree $6$.
When $(x-y)^2+(y-z)^2+(z-x)^2=0$, i.e. $x=y=z$
$$
\Delta=\begin{vmatrix}
0&0&0\\
0&0&0\\
0&0&0\\
\end{vmatrix}=0
$$
Thus, $(x-y)^2+(y-z)^2+(z-x)^2$ is a factor.
How do I extract the other $(x-y)^2+(y-z)^2+(z-x)^2$ from $\Delta$ $\color{red}{?}$
Does this have anything to do with all rows (or columns) being zero when $(x-y)^2+(y-z)^2+(z-x)^2=0$ $\color{red}{?}$
If I can extract that then i think I know how to proceed. The remaining factor must be a homogeneous quadratic symmetric polynomial, i.e. $p(x,y,z)=a(x^2+y^2+z^2)+b(xy+yz+zx)$
$$
\Delta(x,y,z)=\Big[(x-y)^2+(y-z)^2+(z-x)^2\Big]^2.a(x^2+y^2+z^2)+b(xy+yz+zx)
$$
$$
\Delta(1,0,0)=\begin{vmatrix}
-1&0&0\\
0&0&-1\\
0&-1&0\\
\end{vmatrix}=1=4.a\implies a=\frac{1}{4}
$$
$$
\Delta(1,1,0)=\begin{vmatrix}
-1&-1&1\\
-1&1&-1\\
1&-1&-1\\
\end{vmatrix}=\begin{vmatrix}
0&0&1\\
-2&0&-1\\
0&-2&-1\\
\end{vmatrix}\\
=\begin{vmatrix}
-2&0\\
0&-2\\
\end{vmatrix}=4=4.(2a+b)=4(1/2+b)=2+4b\\
\implies b=\frac{1}{2}
$$
$$
\Delta(x,y,z)=\Big[(x-y)^2+(y-z)^2+(z-x)^2\Big]^2.\frac{1}{4}(x^2+y^2+z^2)+\frac{1}{2}(xy+yz+zx)\\
=\frac{1}{4}\Big[(x-y)^2+(y-z)^2+(z-x)^2\Big]^2.(x^2+y^2+z^2+2xy+2yz+2zx)\\
=\frac{1}{4}\Big[(x-y)^2+(y-z)^2+(z-x)^2\Big]^2(x+y+z)^2
$$
Note:
I am trying to factorize the determinant using factor theorem given the fact that the determinant is a homogeneous symmetric polynomial of degree 6.
|
Note that
$$
M:=
\begin{bmatrix}
yz-x^2&zx-y^2&xy-z^2\\
zx-y^2&xy-z^2&yz-x^2\\
xy-z^2&yz-x^2&zx-y^2
\end{bmatrix}
$$
is the matrix of $2\times 2$ cofactors of the matrix:
$$
N:=
\begin{bmatrix}
x & y & z\\
y & z & x\\
z& x & y \\
\end{bmatrix}.
$$
Then as usual $MN=(\det N) I$, so that $\det M \det N =(\det N)^3$. As $\det N\not=0$ (as a polynomial) we have that $\Delta =\det M= (\det N)^2$ is the perfect square of a polynomial of degree $3$ -- which is what was asked.
But from this we can see everything: there is clearly a factor $(x+y+z)$ in $\det N$ and a factor $\frac{1}{2}( (x-y)^2 +(y-z)^2 +(z-x)^2))$ in $\det M$. Establishing the value of the constant is trivial.
|
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "4",
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|
Find all real solutions for $x$ in $2(2^x−1)x^2+(2^{x^2}−2)x=2^{x+1}−2$.
Find all real solutions for $x$ in $2(2^x−1)x^2+(2^{x^2}−2)x=2^{x+1}−2$.
I started off by dividing $2(2^x- 1) x^2 + (2^{x^2}-2)x = 2^{x+1} -2$ by $2$, and I got $(2^x-1)x^2 + (2^{x^2-1}-1)x = 2^x -1$. I tried dividing by $2^x-1$ on both sides which would give me a simple quadratic to solve for $x$, but I don't know how to simplify $\frac{(2^{x^2}-1)}{(2^x-1)}$. Am I missing a simple trick here, or am I on a completely wrong path to solving this?
|
$\begin{align*} 2\left (2^{x}-1 \right )x^{2}+\left(2^{x^{2}}-2\right)x&=2^{x}.2-2\\ 2\left(2^{x}-1\right)x^{2}-\left(2^{x^{2}}-2\right)x&=2(2^{x}-1) \end{align*}$
Let $2^{x}-1=a$ and $2^{x^{2}}-2=b$ and equation becomes:
$2ax^{2} + bx=2a$ or $2ax^{2}+bx-2a=0$. It's a quadratic equation. The quadratic equation has real solution if and only if $b^{2}-4ac>0$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2677239",
"timestamp": "2023-03-29T00:00:00",
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|
What is the area of the triangle ABC?
$ABC$ is an equaliteral triangle.
Suppose $DB=4$, $DA=4\sqrt{3}$ and $DC=8$.
Find the area of the triangle $ABC$.
|
Reflect $D$ in $AB$ to obtain $P$. Reflect $D$ in $BC$ to obtain $Q$. Reflect $D$ in $CA$ ro obtain $R$. Then $[APBQCR]=2[\triangle ABC]$.
Note that $\triangle APR$, $\triangle BQP$ and $\triangle CRQ$ are $30^\circ$- $120^\circ$-$30^\circ$ isosceles triangles.
$[\triangle APR]=\frac{1}{2}(4\sqrt{3})^2\sin 120^\circ $
$[\triangle BQP]=\frac{1}{2}(4)^2\sin 120^\circ $
$[\triangle CRQ]=\frac{1}{2}(8)^2\sin 120^\circ$
By cosine formula, $PQ=4\sqrt{3}$, $PR=(4\sqrt{3})\sqrt{3}=12$ and $QR=8\sqrt{3}$. It is easy to see that $\triangle PQR$ is right-angled and its area is easy to find.
Summing up, we will have the area of $APBQCR$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Definite integrals question $$\int^{\pi/2}_{0} \frac{1}{\sin^4x + \cos^4 x} \,dx$$
First I solve the indefinite integral $$\int \frac{1}{\sin^4x + \cos^4 x} \,dx=\frac{\sqrt2}{2}\arctan\Big(\frac{\tan2x}{\sqrt2}
\Big)+C$$ by the universal substitution $\tan2x=y$.
My question is: when I do this substitution should I bound x around $(-{\pi/4},{\pi/4})$ because arctan function is defined over $(-{\pi/2},{\pi/2})$?
And if I bound it like this how to solve the definite integral?
|
One way to address the original integral is to say that
$$\begin{align}\int_0^{\pi/2}\frac1{\sin^4x+\cos^4x}dx&=\int_0^{\pi/2}\frac1{\frac14(1-\cos2x)^2+\frac14(1+\cos2x)^2}dx\\
&=2\int_0^{\pi/2}\frac1{1+\cos^22x}dx=\int_0^{\pi}\frac1{1+\cos^2y}dy\\
&=\int_0^{\pi}\frac{dy}{\sin^2y+2\cos^2y}\end{align}$$
Now,
$$\int_0^{\pi}\frac{d\theta}{a^2\sin^2\theta+b^2\cos^2\theta}$$
Has been encountered many times on this forum. It's kind of the grapefruit integral in that just about any method we apply to it hits it out of the park. Here, let's do it by folding:
$$\begin{align}\int_0^{\pi}\frac{d\theta}{a^2\sin^2\theta+b^2\cos^2\theta}&=\frac12\int_0^{\pi}\frac{d\theta}{a^2\sin^2\theta+b^2\cos^2\theta}+\frac12\int_0^{\pi}\frac{d\theta}{a^2\cos^2\theta+b^2\sin^2\theta}\\
&=\frac12\int_0^{\pi}\frac{(a^2+b^2)d\theta}{a^2b^2\cos^4\theta+(a^4+b^4)\sin^2\theta\cos^2\theta+a^2b^2\sin^4\theta}\\
&=\frac12\int_0^{\pi}\frac{(a^2+b^2)d\theta}{\frac14a^2b^2(1+\cos2\theta)^2+\frac14(a^4+b^4)\sin^22\theta+\frac14a^2b^2(1-\cos2\theta)^2}\\
&=\frac12\int_0^{\pi}\frac{(a^2+b^2)d\theta}{\frac12a^2b^2(\cos^22\theta+\sin^22\theta)+\frac12a^2b^2\cos^22\theta+\frac14(a^4+b^4)\sin^22\theta}\\
&=\frac12\int_0^{\pi}\frac{2d\theta}{\frac12(a^2+b^2)\sin^22\theta+\frac{2a^2b^2}{a^2+b^2}\cos^22\theta}\\
&=\frac12\int_0^{2\pi}\frac{d\phi}{\frac12(a^2+b^2)\sin^2\phi+\frac{2a^2b^2}{a^2+b^2}\cos^2\phi}\\
&=\int_0^{\pi}\frac{d\phi}{\frac12(a^2+b^2)\sin^2\phi+\frac{2a^2b^2}{a^2+b^2}\cos^2\phi}\end{align}$$
So folding has transformed our original $F(a^2,b^2)$ to $F\left(\frac12(a^2+b^2),\frac2{\frac1{a^2}+\frac1{b^2}}\right)$. Since the limit of the arithmetic-harmonic mean is the geometric mean, it follows that
$$F(a^2,b^2)=F(ab,ab)=\int_0^{\pi}\frac{d\theta}{ab}=\frac{\pi}{ab}$$
In the instant case, $a=1$, $b=\sqrt2$, so
$$\int_0^{\pi/2}\frac1{\sin^4x+\cos^4x}dx=\frac{\pi}{\sqrt2}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
How to prove that $ \frac{(99)!!}{(100)!!} < \frac{1}{10}$ How to prove that $ \dfrac{(99)!!}{(100)!!}=\dfrac{1\cdot3\cdot5\cdot7\cdot9 \cdots99}{2\cdot4\cdot6\cdot8\cdot10\cdots100} < \dfrac{1}{10}$
Any hint to prove it?
|
In general the ratio
$$ r_n = \frac{(2n-1)!!}{(2n)!!}$$
is related to the central binomial coefficient for which many results are known.
In particular, we have that
$$ (2n-1) (2n-3) \cdots 1 = \frac{(2n) (2n-1)\cdots 1}{(2n)(2n-2) \cdots 2} =2^{-n} \frac{(2n) (2n-1)\cdots 1}{n(n-1) \cdots 1} = 2^{-n} \frac{(2n)!}{n!}$$
and thus
$$ r_n = \frac{(2n-1)(2n-3) \cdots 1}{(2n) (2n-2) \cdots 2}=2^{-n}\frac{(2n-1)(2n-3) \cdots 1}{n (n-1) \cdots 1}=4^{-n} \frac{(2n)!}{(n!)^2} = 4^{-n} \binom{2n}{n}\,.$$
It can be shown that
$$ \binom{2n}{n} \leq \frac{4^n}{\sqrt{\pi n}}$$
for $n\geq 1$.
Thus, we have that
$$r_n \leq \frac{1}{\sqrt{\pi n}}.$$
Specialising for $n=50$, we obtain
$$ \frac{99!!}{100!!} \leq \frac{1}{\sqrt{50 \pi} } < \frac{1}{\sqrt{144}} = \frac{1}{12} \,.$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Why is $|x^2 - a^2| = |x - a||x + a|$? By the definition of absolute value, we have
\begin{equation}
|x^2 - a^2| =
\begin{cases}
(x - a)(x + a) \ if \ x^2 \geq a^2\\
-(x - a)(x + a) \ if \ x^2 \leq a^2.
\end{cases}
\end{equation}
How do we conclude that the right hand side equals $|x - a||x + a|$?
|
Assume that $a\ge0$.
Note that $x^2\ge a^2$ if and only if $x\le -a$ or $x\ge a$.
If $x\le-a$, then $x+a\le0$ and $x-a\le0$. So $|x+a||x-a|=[-(x+a)][-(x-a)]=(x+a)(x-a)$.
If $x\ge a$, then $x+a\ge0$ and $x-a\ge0$. So $|x+a||x-a|=(x+a)(x-a)$.
We also have $x^2\le a^2$ if and only if $-a\le x\le a$. In such case, $x+a\ge0$ and $x-a\le 0$. So $|x+a||x-a|=(x+a)[-(x-a)]=-(x+a)(x-a)$.
Therefore, $|x^2-a^2|=|x+a||x-a|$.
If $a<0$, then $-a>0$. So we have $|x^2-a^2|=|x^2-(-a)^2|=|x+(-a)||x-(-a)|=|x-a||x+a|$.
|
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|
Integral with fourth power in denominator How can I find the antiderivate of $$\int \frac{1}{(x-a)^4+(x-b)^4} dx$$ with $$a<b$$
My try was to rewrite the integral as $$\int \frac{\frac{1}{(x-a)^2(x-b)^2}}{\frac{(x-a)^2}{(x-b)^2}+\frac{(x-b)^2}{(x-a)^2}} dx$$ which is equal to $$\int \frac{\frac{1}{(x-a)^2(x-b)^2}}{(\frac{(x-a)}{(x-b)}+\frac{(x-b)}{(x-a)})^2-2} dx $$now since $$\frac{d}{dx}\left ( \frac{(x-a)}{(x-b)}+\frac{(x-b)}{(x-a)} \right )$$ have $$(x-a)^2(x-b)^2$$ in the denominator i suspect it might have a nice form in the end, but I fail to finish it.
|
Let $d=\frac{a+b}{2}$ and substitute $x=u+d$ to make the integrand
$$\frac{1}{(u+c)^4+(u-c)^4}.$$
Where $c=(b-a)/2.$ Factor out a $c^4$:
$$\frac{1}{c^4((\frac{u}{c}+1)^4 + (\frac{u}{c}-1)^4)}.$$
Substitute $v=\frac{u}{c}$ to get
$$\frac{c}{c^4((v+1^4)+(v-1)^4)}.$$
Ignoring the $c$'s, we need to integrate
$$\frac{1}{(v+1^4)+(v-1)^4} = \frac{1}{2v^4+12v^2+2}.$$
The biquadratic is easily factored and partial fractions give
$$\frac{1}{16}\left( \frac{\sqrt{2}}{v^2+3-2\sqrt{2}} - \frac{\sqrt{2}}{v^2+3+2\sqrt{2}}\right).$$
Complete the square in both denominators and trig sub in the usual way to get:
$$\frac{1}{8} \left(\frac{\tan^{-1}\left(\frac{v}{\sqrt{2}-1}\right)}{2-\sqrt{2}} - \frac{\tan^{-1}\left(\frac{v}{\sqrt{2}-1}\right)}{2+\sqrt{2}} \right).$$
Backsubstitute and recall that there's a $c^3$ in the denominator.
|
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|
Difference in total number of positive integers $x, y, z$ determined from an equation.
Find all positive integers $x, y, z$,such that $$x+2y+3z=12$$
Using the formula (got in almost similar thread) I got ${k−1 \choose n−1}=55$ solutions are available from the equation where $k=12$ and $n=3$
After reading some similar examples from similar threads I tried to solve it using $\text{Trial and error }$ method-
$12=1.2.2.3$ $$\frac{x+2y+3z}{12}=m$$ $$\therefore 1+2.1+3.3=12 \\1+2.4+3.1=12\\ 2+2.2+2.3=12 \\ 3+2.3+3.1=12 \\ 4+2.1+3.2=12 \\ 5+2.2+3.1=12 \\ 7+2.1+3.1=12$$.
So, out of $55$ solutions I can figure out only 7 solutions.
What is the reason for this difference? If I am wrong anywhere please guide me to a right way. Any help is appreciated.
|
Maybe you can get to fifty-five solutions if you ditch the restriction that $x, y, z$ have to be positive:
$$\ldots \\
8 + 2 \times 2 + 3 \times 0 = 12 \\
0 + 2 \times 0 + 3 \times 4 = 12 \\
0 + 2 \times 3 + 3 \times 2 = 12 \\
0 + 2 \times (-3) + 3 \times 6 = 12 \\
-1 + 2 \times 5 + 3 \times 1 = 12 \\
\ldots$$
Oops... it turns out that removing the stricture enables infinitely many solutions. For instance, if you fix $x = -1$, then $y = 5 - 3n$ and $z = 2n + 1$ for all $n \in \mathbb{Z}$ are solutions.
There are only seven solutions satisfying the requirement that $x, y, z$ have to be positive, and you've found them all. If $7 < x < 12$, then you probably need $y$ or $z$ equal to $0$, and for $x = 12$ then certainly $y = z = 0$. For all $x > 12$, you probably need one of $y$ or $z$, most likely both, to be negative.
|
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|
Incorrect in solving $\frac{a}{b} - \frac{a}{c} = 1$ for $c$ I have this:
$$\frac{a}{b} - \frac{a}{c} = 1$$ Solve for $c$. Then,
$$\frac{a}{b} - \frac{a}{c} = 1 \cdot bc$$
$$ = ac - ab = bc$$
$$ = a(c - b) = bc$$
$$ c = \frac{bc}{a} + b$$ This is my final result.
But the correct result is:
$$c=\frac{ab}{a-b}$$
What I development wrong in this equation ?
|
Your algebra is correct but $$c = \frac{bc}{a} + b$$ is solving $c$ in terms of $a$, $b$, and $c$ which is sort of cyclic. We want to see $c$ in terms of only $a$ and $b$.
You have to isolate $c$ by keeping all terms involving $c$ on the left side and other terms on the right side.
Note that $$ ac - ab = bc\implies c(a-b)=ab \implies c= \frac {ab}{a-b}$$
|
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|
Finding $\frac{\partial^6 f}{\partial x^4 \partial y^2}$
Find $$\frac{\partial^6 f}{\partial x^4 \partial y^2}(0,0)$$ Of the $f(x,y)=\frac{1}{1-x^2y}$
$$\frac{1}{1-x^2y}=\sum_{n=1}^{\infty}(x^2y)^n$$ as $|x^2y|<1$
So we have $\frac{1}{1-x^2y}\approx1+x^2y+x^4y^2$
But how should we continue from here?
|
Use the below statement before starts to differentiate.
$$\frac{\partial^6 f(x,y)}{\partial x^4 \partial y^2}=\frac{\partial^4 \left(\frac{\partial^2 f(x,y)}{\partial y^2}\right)}{\partial x^4}$$
Calculate $\displaystyle \frac{\partial^2 f(x,y)}{\partial y^2}$ for first.
$$\displaystyle \frac{\partial f(x,y)}{\partial y}=\frac{x^2}{(1 - x^2 y)^2}$$
$$\displaystyle \frac{\partial^2 f(x,y)}{\partial y^2}=\frac{(2 x^4)}{(1 - x^2 y)^3}$$
Now you should substitute $y=0$ since you are not required to differentiate by $y$ it takes place only as a constant while you differentiate by $x$.
$$\displaystyle \frac{\partial^2 f(x,0)}{\partial y^2}=\frac{(2 x^4)}{(1 - x^2\cdot 0)^3}=2x^4$$
It is now much easier to differentiate by $x$ four times.
$$\displaystyle \frac{\partial^4 (2x^4)}{\partial x^4}=\frac{\partial^3 (4\cdot 2 x^3)}{\partial x^3}=\frac{\partial^2 (4\cdot 3 \cdot 2 x^2)}{\partial x^2}=\frac{\partial^2 (4\cdot 3 \cdot 2 \cdot 2 x)}{\partial x}=4\cdot 3 \cdot 2 \cdot 2=48.$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
If $5x \equiv 15 \pmod{25}$, then definitely $x \equiv 3 \pmod{25}$. Is this true or false? I don't understand how to prove this statement true or false.
If $5x \equiv 15 \pmod{25}$, then definitely $x \equiv 3 \pmod{25}$.
All I know is that $\frac{5}5x \equiv \frac{15}5 \pmod{\frac{25}5}$ = $x \equiv 3 \pmod{5}$, and $x \cdot 5 \equiv 3 \cdot 5 \pmod{25}$ = $5x \equiv 15 \pmod{25}$, how do you prove this equation?
|
The congruence $5x \equiv 15 \pmod{25}$ means that there exists $t \in \mathbb{Z}$ such that
$$5x = 15 + 25t$$
Dividing each side of the equation by $5$ yields
$$x = 3 + 5t$$
Therefore, if
$$x \equiv 3 \pmod{5}$$
it satisfies the congruence $5x \equiv 15 \pmod{25}$. There are five such equivalence classes modulo $25$. They are
\begin{align*}
x & \equiv 3 \pmod{25}\\
x & \equiv 8 \pmod{25}\\
x & \equiv 13 \pmod{25}\\
x & \equiv 18 \pmod{25}\\
x & \equiv 23 \pmod{25}
\end{align*}
Hence, the statement is false.
|
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|
Verfication of deduction made using the Cauchy-Schwarz inequality Is the following proof correct?
Show that $$16\leq(a+b+c+d)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\right)$$ for all positive numbers $a,b,c,d$.
Proof. Let $\mathbf{R}^4$ be the inner product space with the inner product defined as in the euclidean product in $\textbf{6.4}$. Now let $a,b,c,d$ be arbitrary positive numbers and let $u = (|\sqrt{a}|,|\sqrt{b}|,|\sqrt{c}|,|\sqrt{d}|)$ and $v = (\frac{1}{|\sqrt{a}|},\frac{1}{|\sqrt{b}|},\frac{1}{|\sqrt{c}|},\frac{1}{|\sqrt{d}|})$ then by appealing to the Cauchy Schwartz inequality we have\begin{align*}
|\langle u,v\rangle| &= |4| = |\langle (|\sqrt{a}|,|\sqrt{b}|,|\sqrt{c}|,|\sqrt{d}|),(\frac{1}{|\sqrt{a}|},\frac{1}{|\sqrt{b}|},\frac{1}{|\sqrt{c}|},\frac{1}{|\sqrt{d}|})\rangle|\\
&\leq \sqrt{|\sqrt{a}|^2+|\sqrt{b}|^2+|\sqrt{c}|^2+|\sqrt{d}|^2}\cdot\sqrt{\frac{1}{|\sqrt{a}|^2}+\frac{1}{|\sqrt{b}|^2}+\frac{1}{|\sqrt{c}|^2}+\frac{1}{|\sqrt{d}|^2}}\\
&= \sqrt{(a+b+c+d)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d})} = \|u\|\cdot\|v\|.
\end{align*}
Squaring both sides yields the required result.
NOTE The reference to $6.4$ above is simply to what is commonly understood to be the dot product.
$\blacksquare$
|
While OP's proof is fine, it may be worth noting that the result also follows directly from the AM-HM inequality by simply rewriting the given relation as:
$$
\frac{a+b+c+d}{4} \;\ge\; \frac {4}{\cfrac{1}{a}+\cfrac{1}{b}+\cfrac{1}{c}+\cfrac{1}{d}}
$$
|
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|
Calculating $\lim_{x \rightarrow 0} \frac{\tan x - \sin x}{x^3}$. I have a difficulty in calculating this limit:
$$\lim_{x \rightarrow 0} \frac{\tan x - \sin x}{x^3},$$
I have tried $\tan x = \frac{\sin x}{\cos x}$, then I unified the denominator of the numerator of the given limit problem finally I got $$\lim_{x \rightarrow 0} \frac{\sin x}{x^{3} \cos x} - \lim_{x \rightarrow 0} \frac{ \sin x}{x^3},$$
Then I got stucked, could anyone help me in solving it?
|
$$\frac{\tan{x}-\sin{x}}{x^3}=\frac{x+\frac{1}{3}x^3+o(x^3) - (x-\frac{1}{6}x^3+o(x^3))}{x^3}=\frac{\frac{1}{2}x^3+o(x^3)}{x^3}=\frac{1}{2} +\frac{o(x^3)}{x^3}\stackrel{x\rightarrow 0}{\longrightarrow}\frac{1}{2}$$
|
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|
On finding $\sup\left\{3(-1)^n-\frac{1}{n^2+1}\right\}$ How does one find $\sup(A)$ where $A = \left\{3(-1)^n-\frac{1}{n^2+1}: n \in \mathbb{N}\right\}$?
I've tried as follows, but I'm not so sure.
$\displaystyle 3(-1)^n-\frac{1}{n^2+1} \le 3-\frac{1}{n^2+1} \le 3. $ So $3$ is an upper bound for $A$.
Let $\epsilon >0$. We wish to prove that $3-\epsilon$ is not an upper bound for $A$.
For $N = 2n_0$ where $\displaystyle n_0 > \frac{1}{\sqrt{2\epsilon}} $ we have
$$\displaystyle 3(-1)^N-\frac{1}{N^2+1} = 3-\frac{1}{N^2+1} \ge 3-\frac{1}{2N^2} > 3-\epsilon. $$
Hence $3-\epsilon$ is not an upper bound. Hence $3= \sup(A)$.
EDIT:
For $N = 2n_0$ where $\displaystyle n_0 > \frac{1}{\sqrt{\epsilon}} $ we have
$$\displaystyle 3(-1)^N-\frac{1}{N^2+1} = 3-\frac{1}{N^2+1} \ge 3-\frac{1}{N^2} > 3-\epsilon. $$
Hence $3-\epsilon$ is not an upper bound. Hence $3= \sup(A)$.
|
Looks good to me. EDIT - though as Martiny pointed out in the comments, it's not. You wrote $3-\frac{1}{N^2+1} \ge 3-\frac{1}{2N^2}$ when this is not true.
You can also do this by taking every other term directly: noting that $$\lim_{n \to \infty} \left[3(-1)^{2n} - \frac{1}{(2n)^2+1} \right] = \lim_{n \to \infty} 3 - \lim_{n \to \infty} \frac{1}{(2n)^2+1} = 3 - 0 = 3$$
so the sequence must get arbitrarily close to $3$ through its even terms anyway.
|
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|
Solve the equation $\log_2(\frac{8}{2^x}-1)=x-2$ This equation comes from the book of Prilepko. It looks sufficiently simple but I want to know if my solution is correct.
$\log_2(\frac{8}{2^x}-1)=x-2$
I have only been able to transform into this:
$\log_2(\frac{8-2^x}{2^x})=x-2$
$\iff$ $\log_2(8-2^x)=2x-2$
$\iff$ $8-2^x=2^{2x-2}$
$\iff$ $32-4.2^x-2^{2x}= 0$
Let $t=2^x$
$-t^2-4t+32=0$
This equation will produce $t=-8$ and $t=4$
Therefore $x=2$.
Another question that I wish to ask is when solving these logarithmic equations, one very often substitute a dummy variable to overcome the restriction caused by the order of operations. For example, you cannot do much directly with $5^{2x}-130.5^x+625$. But let $t=5^x$, then you can easily factorize the this equation into $(t-125)(t-5)=0$, hence $(5^x-125)(5^x-5)=0$. Isn't this a sort of manipulating symbols according to certain prescribed rules? Is there any logical difficulty with this kind of operation? Since by substituting a new variable, it means that this variable must inherit all properties which the object of the substitution originally possesses. For example, if $a^x$ is the object of substitution for $t$, then if $a<0$ and x can assumed a form of rational power, doesn't this make no sense at all?
Do you know any example in elementary mathematics where this method of substitution can yield contradictory result?
|
If you want "to stretch" the logarithm method or solution to the very end here it is:
$$\log_{2} \left ( \frac{8}{2^{x}} - 1 \right ) = x-2 $$
$$\log_{2} \left ( \frac{8-2^{x}}{2^{x}} \right ) = x-2 $$
$$\log_{2} \left (8-2^{x} \right ) - \log_{2} \left ( 2^{x} \right ) = x-2 $$
$$\log_{2} \left (2^{3}-2^{x} \right ) - x = x-2 $$
$$\log_{2} \left (2^{3}-2^{x} \right ) = 2x-2 $$
$$\log_{2} \left (2^{3} \left ( 1- 2^{x-3}\right ) \right ) = 2x-2 $$
$$\log_{2} \left (2^{3}\right ) + \log_{2} \left ( 1- 2^{x-3} \right ) = 2x-2 $$
$$3\log_{2} \left (2\right ) + \log_{2} \left ( 1- 2^{x-3} \right ) = 2x-2 $$
$$3 + \log_{2} \left ( 1- 2^{x-3} \right ) = 2x-2 $$
$$ \log_{2} \left ( 1- 2^{x-3} \right ) = 2x-5 $$
Using the $\textrm{Antilogarithm}$ (see here or here for any details)
$$ \textrm{antilog}_{2} \left ( \log_{2} \left ( 1- 2^{x-3} \right ) \right ) = \textrm{antilog}_{2} \left ( 2x-5 \right ) $$
$$1- 2^{x-3}= 2^{2x-5}$$
Rearranging the above equation translates into:
$$1- 2^{x}\times 2^{-3}= 2^{2x}\times 2^{-5}$$
$$2^{2x}\times 2^{-5} - 2^{x}\times 2^{-3} - 1 = 0$$
Multiplying by $2^{5}$ becomes into the expression you had found:
$$2^{2x} - 2^{x}\times 2^{2} - 32 = 0$$
From then on it is consistent with a quadratic equation:
$$A=2^{x}$$
$$A_{1,2}=\frac{-4\pm \sqrt{16+128}}{2}=\frac{-4\pm \sqrt{144}}{2}$$
$$A_{1,2}=\frac{-4\pm 12}{2}\,,A_{1}=\frac{-4+12}{2}=4\,,A_{2}=\frac{-4-12}{2}=-8$$
Since we cannot obtain negative results from a power the second solution is discarded. Therefore:
$$2^{x}=4$$
$$x=2$$
Becoming the result you had just found. So it is correct. As mentioned this is more or less the same of what others before me had just commented.
|
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|
Find the pmf of $Y=X^2$ Let $X$ be a random variable with the following pmf:
$$
\begin{array}{c|ccccc}
x& -2 & -1 & 0 & 1 & 2 & \\ \hline
p(x)
& 3/10 & 3/10 & 1/10 & 2/10 & 1/10 &
\end{array}
$$
Find the pmf of $Y = X^2$ and find $P(Y\ge3)$.
I am struggling to get the idea behind that. Even with a solid background in multivariable calculus.
I think $y=g(X)$, where $g(x)=x^2$.
$$
\begin{array}{c|ccccc}
x& -2 & -1 & 0 & 1 & 2 & \\ \hline
g(x)
& 4 & 1 & 0 & 1 & 4 &
\end{array}
$$
$$P_Y(y) =
\begin{cases}\displaystyle
\sum_{x\in R_x:g(x)=y} P_X(x) , & \text{$y \in R_y$} \\[2ex]
0, & \text{otherwise}
\end{cases}$$
*
*I know that creating this table is somewhat necessary. But what is the meaning of all that?
*Why is making a table like that is the pmf?
*I do not understand the summation sign with the range why does it make sense.
Any hint would be greatly appreciated.
|
We have that the pmf of $X$ is$$p_X(x)=\begin{cases}
\frac{3}{10} & x=-2 \\
\frac{3}{10} & x=-1 \\
\frac{1}{10} & x=0 \\
\frac{2}{10} & x=1 \\
\frac{1}{10} & x=4
\end{cases}$$
Transforming this to get the pmf of $Y$ we get
$$p_Y(y)=\begin{cases}
\frac{3}{10} & y=4 \\
\frac{3}{10} & y=1 \\
\frac{1}{10} & y=0 \\
\frac{2}{10} & y=1 \\
\frac{1}{10} & y=4
\end{cases}$$
Notice that the probabilities remain the same.
Finally, we have repeated $y$ values so we combine them to get
$$p_Y(y)=\begin{cases}
\frac{4}{10} & y=4 \\
\frac{5}{10} & y=1 \\
\frac{1}{10} & y=0 \\
\end{cases}$$
We can now easily compute $$P(Y\geq3)=\frac{4}{10}$$
|
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|
Finding the the derivative of $y=\sqrt{1-\sin x}; 0A question I'm attempting is:
Find the derivative of $ y = \sqrt {1 - \sin x} ; 0 < x <\pi/2$.
I did this:
$y = \sqrt {1 - \sin x} = \sqrt {\cos^2\frac{x}{2} + \sin^2\frac{x}{2} - 2\sin \frac{x}{2}\cos \frac{x}{2}} = \sqrt { (\sin \frac{x}{2}-\cos \frac{x}{2})^2} = \sin \frac{x}{2} - \cos \frac{x}{2}$
So, $\frac{dy}{dx} = \frac{1}{2} \cdot (\cos\frac{x}{2} + \sin\frac{x}{2})$.
But apparently this is wrong. The correct solution is:
$\frac{dy}{dx} = -\frac{1}{2}\cdot(\cos\frac{x}{2} + \sin\frac{x}{2})$.
So I want to know what I have done wrongly here. Why is my answer not right?
|
$$y = \sqrt {1 - \sin x} ; 0 < x <\pi/2$$
$$ y' = \frac {-\cos x}{2 \sqrt {1 - \sin x}}$$
$$=\frac {-\cos x\sqrt {1 + \sin x }}{2 \cos x}$$
$$=-\frac {\sqrt {1 + \sin x}}{2}=-\frac12(\sin x/2+\cos x/2)$$
Note that $$(\sin x/2+\cos x/2)^2=1+2\sin x/2\cos x/2=1+\sin x$$
|
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|
Factoring $x^2-y^2-z^2+2yz+x+y-z$ This is a factorization problem on polynomials. I can't find a way to solve it, neither can a math program called MathWay. Help me, please.
$$x^2-y^2-z^2+2yz+x+y-z$$
My book says that the answer is $(x+y-z)(x-y+z+1)$.
|
You can consider $\,x^2+x-y^2-z^2+2yz+y-z\,$ as a quadratic in $\,x\,$, and find its roots using the quadratic formula, the discriminant being:
$$
\begin{align}
\Delta_x = 1 - 4(-y^2-z^2+2yz+y-z) &= 4y^2 - 4(2z+1)y +4z^2+4z+1 \\
&= \left(2y\right)^2- 2 \cdot 2y(2z+1)+\left(2z+1\right)^2 \\
&= \left(2y-2z-1\right)^2
\end{align}
$$
Therefore the roots are $\displaystyle x_{1,2} = \frac{-1 \pm \left(2y-2z-1\right)}{2}=\begin{cases}y-z-1 \\ -y+z\end{cases}\;$ and the original polynomial factors as $\,(x-x_1)(x-x_2)\,$.
|
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|
Change of Basis Given 2 Vectors and Transition Matrix
Given v$_1=\left( \begin{array}{ccc} 3\\
-4 \end{array} \right)$, v$_2=\left( \begin{array}{ccc} 2\\ 5 \end{array} \right)$, $S=\left( \begin{array}{ccc}
-1 & 7\\ 2 &-5 \end{array} \right)$
Find the vectors u$_1$, u$_2$ such that $S$ will be a
transition matrix from [v$_1$, v$_2$] -> [u$_1$,
u$_2$]
So transition matrices work like this: $V = U*S$
Then $V*S^{-1}=U$
Where $V=\left( \begin{array}{ccc}
3 & 2\\
-4 & 5\end{array} \right)$ and $S^{-1} = \frac{1}{-9}\left( \begin{array}{ccc}
5 &-7\\
-2 & 1 \end{array} \right)=\left( \begin{array}{ccc}
-5/9 &7/9\\
2/9 & -1/9 \end{array} \right)$
Then $V*S^{-1} = \left( \begin{array}{ccc}
-11/9 &19/9\\
30/9 & -33/9 \end{array} \right)$ Which is supposed to be $U$ but it's wrong if you check the answer by plugging it into $V=U*S$
What am I doing incorrectly?
|
Note that the inverse should be
$$S^{-1} = \left( \begin{array}{ccc}
5/9 &7/9\\
2/9 & 1/9 \end{array} \right)$$
|
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|
Playing cards: Probability pairs An ordinary deck of $52$ playing cards ($4$ suits, $13$ values) are shuffled. Then player $A$ takes the top two cards. Player $B$ takes the next to top cars. Let $Pr[A]$ denote the probability for player $A$ to have a pair (i.e. two cards with the same value) and $Pr[B]$ the probability for player $B$ to have a pair. Task: Calculate $Pr[A], Pr[B], Pr[A \cap B]$ and $Pr[A|B]$.
My attempt:
$$Pr[A] = 1 \cdot \frac{3}{51} = \frac{3}{51}$$ because $A$ can take any first card but then there are only $3$ out of the remaining $51$ cards left, that match up to a pair.
$$Pr[B] = \frac{3}{49}$$
$$Pr[A \cap B] = Pr[A] \cdot Pr[B]$$ (are they independent?)
$$Pr[A|B] = \frac{Pr[A \cap B]}{Pr[B]}$$
I'm pretty sure that my $Pr[B]$ is wrong (or isn't it??). And because I don't have the exact $Pr[B]$ I can't calculate $Pr[A \cap B]$ and $Pr[A|B]$ just yet.
So, could somebody explain to me how to calculate $Pr[B]$ and if I can then just use the formulas above to calculate the remaining probabilities.
Thank you!
(If anything is unclear, please let me know!)
|
Here is a perhaps inelegant way to show that $P(B)=P(A)$
Suppose player $A$ gets a pair with probability $\frac{3}{51}$
Then player $B$ gets a pair with probability $$\frac{48}{50}\cdot\frac{3}{49}+\frac{2}{50}\cdot\frac{1}{49}$$
since for his first card, he can either get a different rank than $A$ with probability $\frac{48}{50}$ or get the same rank with probability $\frac{2}{50}$
Suppose player $A$ does not get a pair with probability $\frac{48}{51}$
Then player $B$ gets a pair with probability $$\frac{44}{50}\cdot\frac{3}{49}+\frac{6}{50}\cdot\frac{2}{49}$$
since player $A$ gets one of each rank, and so for his first card, he can either get a different rank than both of $A$'s cards with probability $\frac{44}{50}$ or get the same rank as one of them with probability $\frac{6}{50}$
All together
$$\frac{3}{51}\left(\frac{48}{50}\cdot\frac{3}{49}+\frac{2}{50}\cdot\frac{1}{49}\right)+\frac{48}{51}\left(\frac{44}{50}\cdot\frac{3}{49}+\frac{6}{50}\cdot\frac{2}{49}\right)=\frac{3}{51}$$
From here, the rest is trivial.
We now have
$$P(A\cap B)=\frac{3}{51}\cdot\left(\frac{48}{50}\cdot\frac{3}{49}+\frac{2}{50}\cdot\frac{1}{49}\right)\approx 0.0035$$
and
$$\begin{align*}
P(A \mid B)
&=\frac{P(A\cap B)}{P(B)}\\\\
&=\frac{\frac{3}{51}\cdot\left(\frac{48}{50}\cdot\frac{3}{49}+\frac{2}{50}\cdot\frac{1}{49}\right)}{\frac{3}{51}}\\\\
&\approx0.0596
\end{align*}$$
|
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|
Laurent series for $f(z)=\frac1{z^3 +1}$ on $0<|z-1|<\sqrt3$ I have this complex function $$f(z)=\dfrac{1}{z^3 +1}$$ and I have to find the Laurent series for $$0<|z+1|<\sqrt{3}$$
First, I consider that $$f(z) = \dfrac{1}{(z+1)(z^2 - z +1)} = \dfrac{2-z}{3(z^2 - z+1)} + \dfrac{1}{3(z+1)}$$
The second term is ok but I don't know how to expand the second term. Can anyone help me? Thanks a lot.
|
Let $a$ and $b$ be the complex cubic roots of $-1$, so $z^3+1=(z+1)(z-a)(z-b)$. Now you can find the partial fraction decomposition
$$
\frac{1}{z^3+1}=\frac{A}{z+1}+\frac{B}{z-a}+\frac{C}{z-b}
$$
Then
$$
\frac{1}{z-c}=\frac{1}{z+1-(c+1)}=-\frac{1}{c+1}\frac{1}{1-\dfrac{z+1}{c+1}}
$$
and then
$$
\frac{1}{z-c}=-\frac{1}{c+1}\frac{1}{1-\dfrac{z+1}{c+1}}=
-\frac{1}{c+1}\sum_{k\ge0}\frac{(z+1)^k}{(c+1)^k}=
-\sum_{k\ge0}\frac{(z+1)^k}{(c+1)^{k+1}}
$$
Do this for $c=a$ and $c=b$, then multiply by the coefficients $B$ and $C$ you found.
|
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|
Prove $2 \cdot \sum\limits_{k = 0}^{n} \binom{2n}{2k} = 2^{2n}$ without induction I am aware of
$$
2^{2n} = (1 + 1)^{2n} =
\sum_{k = 0}^{2n} \binom{2n}{k}
$$
I then tried grouping the terms and using that $\binom{n}{k} = \binom{n}{n-k}$, to obtain
$$
\begin{align}
\sum_{k = 0}^{2n} \binom{2n}{k} &=
\binom{2n}{0} + \binom{2n}{1} + \ldots + \binom{2n}{n - 1} + \binom{2n}{n} + \binom{2n}{n + 1} + \ldots + \binom{2n}{2n} \\
&=
2 \left[ \binom{2n}{0} + \binom{2n}{1} + \ldots + \binom{2n}{n - 1}\right] + \binom{2n}{n} =
\left( 2 \sum_{k = 0}^{n-1} \binom{2n}{k} \right) + \binom{2n}{n}
\end{align}
$$
But I don't know if I'm on the right track or how to progress from here.
|
Note that
$$
\begin{align}
(1+1)^{2n}&=\sum_{k=0}^{2n}\binom{2n}{k}\tag{1}\\
(1-1)^{2n}&=\sum_{k=0}^{2n}\binom{2n}{k}(-1)^k\tag{2}
\end{align}
$$
by the binomial theorem. Subtracting (2) from (1) we get that
$$
2^{2n}=\sum_{k=0}^{2n}\binom{2n}{k}(1-(-1)^k)=2\sum_{0\leq k\leq2n, \, \text{k even}}\binom{2n}{k}=2\sum_{k=0}^{n}\binom{2n}{2k}.
$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
How can I solve this inequality? $ \frac{x+14|x|-10}{|4x-6|-21}>3$ First I looked the x that doesnt belong to this function.
$$|4x -6| - 21 \neq 0$$
$$ x \neq \frac{-15}{4}$$ and $$ x \neq \frac{27}{4}$$
Then I found the roots of the x
$$x = 0$$
$$x = \frac{3}{2}$$
After I found the roots I wrote the inequality like this:
$$ x + 14|x| - 10 > 3|4x-6| - 63$$
$$ x + 14|x| > 3|4x-6| - 53$$
to finish we may write differents function for each values of x based on roots we found.
for $$ x<0$$ the function is $$ x - 14x > 3(6-4x) - 53$$
for $$x<\frac{3}{2}$$ we have $$ x + 14x > 3(6-4x) - 53$$ for $$x \geq \frac{3}{2}$$ the function is $$ x + 14x > 3(4x-6) - 53$$
But using this inequalities I couldnt find the solutions for x!! If we calculate that on wolfram we can see the solutions for x are $$ x < \frac{-15}{4}$$ and $$ x > \frac{27}{4}$$. Can anyone explain me why?
|
Your way is good, first consider
*
*$|4x-6|-21> 0$
and we have
*
*$x<\frac32\implies -4x+6-21> 0 \implies x<-\frac {15} 4$
*$x\ge\frac32\implies 4x-27> 0 \implies x>\frac {27} 4$
then
*
*$|4x-6|-21> 0$ for $x<-\frac {15} 4$ and $x>\frac {27} 4$
*$|4x-6|-21< 0$ for $-\frac {15} 4<x<\frac {27} 4$
*$|4x-6|-21= 0$ for $x=-\frac {15} 4$ and $x=\frac {27} 4$
Then consider the cases
1) $-\frac {15} 4<x<\frac {27} 4$
*
*$\frac{x+14|x|-10}{ |4x-6|-21 }>3\implies x+14|x|-10<3|4x-6|-63$
and
*
*$x<0 \implies -13x-10<3(-4x+6)-63$
*$0\le x<\frac32 \implies 15x-10<3(-4x+6)-63$
*$x\ge\frac32 \implies 15x-10<3(4x-6)-63$
2) $x<-\frac {15} 4$ and $x>\frac {27} 4$
*
*$\frac{x+14|x|-10}{ |4x-6|-21 }>3\implies x+14|x|-10>3|4x-6|-63$
and
*
*$x<-\frac {15} 4 \implies -13x-10<3(-4x+6)-63$
*$x>\frac {27} 4 \implies 15x-10<3(4x-6)-63$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Evaluate $\int_{0}^{\frac{\pi}{2}}\frac{x^2}{ \sin x}dx$ I want to evaluate $$\int_{0}^{\frac{\pi}{2}}\frac{x^2}{ \sin x}dx$$
First,I tried to evaluate like this:
$$\int_{0}^{\frac{\pi}{2}}\frac{x^2}{ \sin x}dx=\int_{0}^{\frac{\pi}{2}}x^2\left(\frac{1+\cos x}{\sin x}\right)\frac{dx}{1+\cos x}=\int_{0}^{\frac{\pi}{2}}x^2\left(\frac{1+\cos x}{\sin x}\right)d\left(\frac{\sin x}{1+\cos x}\right)$$
$$=\int_{0}^{\frac{\pi}{2}}x^2d\log\left(\frac{\sin x}{1+\cos x}\right)=x^2\log\left(\frac{\sin x}{1+\cos x}\right)|_{0}^{\frac{\pi}{2}}-2\int_{0}^{\frac{\pi}{2}}x\log\left(\frac{\sin x}{1+\cos x}\right)dx$$
$$=0+2\int_{0}^{\frac{\pi}{2}}x\log\left(\frac{1+\cos x}{\sin x}\right)dx=2\int_{0}^{\frac{\pi}{2}}x\log\left(1+\cos x\right)dx-2\int_{0}^{\frac{\pi}{2}}x\log\left(\sin x\right)dx$$
$$=2\int_{0}^{\frac{\pi}{2}}x\log\cot \left(\frac{x}{2}\right)dx=8\int_{0}^{\frac{\pi}{4}}x\log\cot xdx$$
but I can't proceed next step,help me,thanks.
|
We can adapt the formula derived in $(2)$ of this answer:
$$
\log(2\cos(x/2))=\sum_{k=1}^\infty\frac{(-1)^{k-1}}k\cos(kx)\tag{1a}
$$
Substituting $x\mapsto\pi-x$ in $\text{(1a)}$, we get
$$
\log(2\sin(x/2))=\sum_{k=1}^\infty\frac{-1}k\cos(kx)\tag{1b}
$$
Subtracting $\text{(1a)}$ from $\text{(1b)}$, the even terms cancel and we get
$$
\bbox[5px,border:2px solid #C0A000]{\log(\tan(x/2))=\sum_{k=0}^\infty\frac{-2}{2k+1}\cos((2k+1)x)}\tag2
$$
Therefore,
$$
\begin{align}
\int_0^{\pi/2}\frac{x^2}{\sin(x)}\,\mathrm{d}x
&=\int_0^{\pi/2}x^2\,\mathrm{d}\log(\tan(x/2))\tag3\\
&=-2\int_0^{\pi/2}x\log(\tan(x/2))\,\mathrm{d}x\tag4\\
&=\sum_{k=0}^\infty\frac4{2k+1}\int_0^{\pi/2}x\cos((2k+1)x)\,\mathrm{d}x\tag5\\
&=\sum_{k=0}^\infty\frac4{(2k+1)^2}\int_0^{\pi/2}x\,\mathrm{d}\sin((2k+1)x)\tag6\\
&=\sum_{k=0}^\infty\frac4{(2k+1)^2}\left[x\sin((2k+1)x)+\frac{\cos((2k+1)x)}{2k+1}\right]_0^{\pi/2}\tag7\\
&=\sum_{k=0}^\infty\frac4{(2k+1)^2}\left[\frac\pi2(-1)^k-\frac1{2k+1}\right]\tag8\\
&=\bbox[5px,border:2px solid #C0A000]{2\pi\mathrm{G}-\frac72\zeta(3)}\tag9
\end{align}
$$
Explanation:
$(3)$: prepare to integrate by parts
$(4)$: integrate by parts
$(5)$: apply $(2)$
$(6)$: prepare to integrate by parts
$(7)$: integrate by parts
$(8)$: apply the limits of integration
$(9)$: evaluate, where $\mathrm{G}$ is Catalan's Constant
|
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|
Show convergence by using partial fractions Show that $\sum\limits_{n=1}^{\infty}{\frac{1}{n(n+1)(n+2)}}=\frac{1}{4}$
Answer:
$\sum\limits_{n=1}^{\infty}{\frac{1}{n(n+1)(n+2)}}=$ $\sum\limits_{n=1}^{\infty}{(\frac{1}{2n}-\frac{1}{n+1} + \frac{1}{2(n+2)})}$
= $1/2 -1/2 + 1/6 +1/4 -1/3 +1/8+1/6 -1/4 +1/10 +1/8 -1/5 +1/12 +....$
How do I proceed further? Can anyone give hints/guide?
Thank you.
|
We can rewrite the sum as
$$ \sum_{n=1}^{\infty} \frac{1}{2}\left( \left[ \frac{1}{n} - \frac{1}{n+1} \right] + \left[ \frac{1}{n+2} - \frac{1}{n + 1} \right] \right) $$
From this the way the sum telescopes should be clear.
|
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|
Is the field extension $\mathbb{Q}(\sqrt{5 + \sqrt{7}})$ over $\mathbb{Q}$ a Galois extension? Let $K= \mathbb{Q}(\sqrt{5 + \sqrt{7}}) $. Is the field extension $\mathbb{Q}(\sqrt{5 + \sqrt{7}})$ over $\mathbb{Q}$ a Galois extension?
Let $x = \sqrt{5 + \sqrt{7}} \implies x^2 = 5 + \sqrt{7} \implies (x^2-5)^2 = 7 \implies x^4 - 10x^2 +18 = 0$
Let $f(x) = x^4 - 10x^2 +18 $. $f$ is Eisenstein at 2, so $f$ is irreducible and monic. The minimal polynomial for $\sqrt{5 + \sqrt{7}}$ is then $f$.
$(x-\sqrt{5 + \sqrt{7}})(x+\sqrt{5 + \sqrt{7}})(x-\sqrt{5 - \sqrt{7}})(x+\sqrt{5 - \sqrt{7}}) = f(x)$.
So $K$ is Galois if and only if all the roots of $f$ are in $K$.
$\sqrt{5 + \sqrt{7}} \in K \implies (\sqrt{5 + \sqrt{7}})^{-1} \in K \implies \frac{\sqrt{5 - \sqrt{7}}}{\sqrt{18}} \in K \implies \frac{\sqrt{2}}{3} \sqrt{5 - \sqrt{7}} \in K$
So, $\sqrt{5 - \sqrt{7}} \in K \iff \sqrt{2} \in K$
I can't prove either. So maybe it's not Galois???
|
Hint: $K$ is Galois over $F=\Bbb{Q}[\sqrt{7}]$. Suppose $\sqrt{5-\sqrt{7}}$ belonged to $K$. What would the action of $\operatorname{Gal}(K/F)$ on $\sqrt{5+\sqrt{7}}\cdot\sqrt{5-\sqrt{7}}$ be?
|
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|
Find $\frac{dy}{dx}$ if $y=\sin^{-1}[x\sqrt{1-x}-\sqrt{x}\sqrt{1-x^2}]$, $0
Find derivative of $f(x)=\sin^{-1}[x\sqrt{1-x}-\sqrt{x}\sqrt{1-x^2}]$, $0<x<1$
Let $x=\sin a$ and $\sqrt{x}=\cos b$
Then I'll get:
$$
y=\sin^{-1}[\sin a\cos b-\cos a\sin b]=\sin^{-1}[\sin(a-b)]\\
\implies\sin y=\sin(a-b)\\
\implies y=n\pi+(-1)^n(a-b)=n\pi+(-1)^n(\sin^{-1}x-\sin^{-1}\sqrt{x})
$$
Thus,
$$
y'=\frac{d}{dx}\big[n\pi+(-1)^n(a-b)\big]=\begin{cases}\frac{1}{\sqrt{1-x^2}}-\frac{1}{2\sqrt{x}\sqrt{1-x}}\text{ if }n\text{ is even}\\
-\bigg[\frac{1}{\sqrt{1-x^2}}-\frac{1}{2\sqrt{x}\sqrt{1-x}}\bigg]\text{ if }n\text{ is odd}
\end{cases}
$$
Is it the right way to solve this problem and how do I check the solution is correct ?
Note: I think there got to be two cases for the derivative as the graph of the function is
|
First let $g(x) = x\sqrt{1-x} + \sqrt{x}\sqrt{1-x^2}$. Then $f(x) = \sin^{-1}(g(x))$. Now we use the chain rule, so $f'(x) = \frac{1}{\sqrt{1-g(x)^2}}g'(x)$. I let you finish, by finding $g'$.
|
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|
What is the number of elements in $\{1,2,...,2018\}$ such that it can be represented by $ x^{6} + x^{5} + x^{4} + x^{3} + x^{2} + x + 1 \:\:? $ What is the number of elements in $\{1,2,...,2018\}$ such that it can be represented by
$$ x^{6} + x^{5} + x^{4} + x^{3} + x^{2} + x + 1 \:\:? $$
$x \in \mathbb{Z}$
Attempt :
First, $x=1$ then the sum will be $7$, which is an element of the set in concern. Of course $x=0$ then the sum will be $1$, also in the set.
For $x=-1$, the sum will also be $1$.
So far we have 2 elements of the set.
For $|x|>1$, we can view the summation as a finite geometric series :
$$ S = 1 + x + .... + x^{6} $$
$$ Sx = x + x^{2} + .... + x^{7} $$
subtract both of the above then we can have
$$ S(x) = \frac{1 - x^{7}}{1-x}, \:\:\: |x| > 1 $$
The key is to find $x*$ so that $S(x*)$ is the maximum just below $2018$.
I can do this by checking manually from $x=\pm 2, \pm 3, \pm 4 $. Without a calculator, this is quite tedious. Found out that $x = \pm 4$ do not satisfy the condition.
So the answer is $6$ numbers, is this sufficient? (I was thinking that we may also have to check the monotonicity of $S(x)$)
What techniques are better than this one? Thanks.
|
For brevity let $x^6 +x^5+ x^4 +x^3+ x^2+x+1=f(x).$... For $x\in \Bbb Z$ we have:
$$(Ia).\quad |x|\geq 4\implies f(x)=(x^6+x^4+x^2)(1+1/x)+1\geq$$ $$\geq (x^6+x^4+x^2)(1-1/|x|)+1>$$ $$> x^6(1-1/|x|)\geq$$ $$\geq 4^6(1-1/2)=64^2/2=2048>2018.$$
$$(Ib).\quad |x|\leq 3\implies |f(x)|\leq |x|^6+|x|^5+|x|^4|+|x|^3+|x|^2+|x|+1\leq$$ $$\leq f(3)=(3^7-1)/2=(3^8/3-1)/2=$$ $$=(81^2/3-1)/2=1093<2018.$$
$(II). $ Obviously $x\geq 0\implies f(x)\geq 1.$ And $x<0\implies 1+1/x\geq 0\implies f(x)=(x^6+x^4+x^2)(1+1/x)+1\geq 1.$
$(III). $ So the solution set is $\{0,\pm 1,\pm 2,\pm 3\}.$
|
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|
Calculating the largest possible area of a rectangle inscribed in an ellipse So i got the equation $4x^2 + 9y^2 = 3600$
What i've done so far is:
$A= (2x)(2y) = 4xy$
Then I find the expression of $y$
$9y^2= 3600 -4x^2$
$y = \pm \sqrt{3600 -4x^2 / 9} = 2/3(\sqrt {900 - x^2} \quad 2/3(900 -x^2)^{1/2}$
Then i set
$A = 4x(2/3(900 -x^2)^1/2 = (8/3)x(900 -x^2)^1/2$
Taking the derivative
$A'(x) = 8/3(900 -x^2)^{1/2} + (8/3)x(1/2)(900 -x^2)^{-1/2}(-2x)
= (2400 - (16/3)x^2)/(\sqrt{900-x^2})$
Set the $A' = 0$
$2400 - (16/3)x^2 = 0$
$(16/3)x^2 = 2400$
$(16/3)x = \sqrt{2400} = (20\sqrt{6}) / 3$
$x = (5\sqrt{6})/12$
Then i put the value of x in the equation and get
$A = (8/3)((5 \sqrt 2)/12)(900 - ((5\sqrt 2)/12)^2)^{1/2} = 81.6...$
Is this right or?
|
There is no need of any complicated algebra. Following is a geometric way to get the answer. One advantage of this approach is you don't need to assume the largest rectangle is axis aligned with the ellipse.
Given any circle, it is well known the largest quadrilateral inscribed in it is a square. Furthermore, the area of the square is $\frac{2}{\pi}$ of that of the circle.
Given an ellipse of the form $\frac{x^2}{a^2}+\frac{y^2}{b^2} = 1$ and any rectangle inscribed in it. Under the linear transform $(x,y) \mapsto \left(\frac{x}{a},\frac{y}{b}\right)$, the ellipse and rectangle get mapped to a circle and quadrilateral. Since under such a linear transform, the ratio of area of different geometric shapes is invariant, we can deduce the area of the rectangle is at most $\frac{2}{\pi}\times \pi a b = 2ab$.
For the ellipse at hand, $a = 30$ and $b = 20$ and it is obvious how to find a rectangle which will get mapped to a square. This means the area of largest rectangle inscribed in that ellipse is $2ab = 2(30)(20) = 1200$.
|
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|
How to solve the limit $\lim_{(x,y)\to(2,3)}\frac{\sin(x)+2}{(x^2-y^2+5)^2}$ How can I solve the following limit or prove that the limit does not exist in $\Bbb R$:
$$
\lim_{(x,y)\to(2,3)}\frac{\sin(x)+2}{(x^2-y^2+5)^2}?
$$
I tried to prove the limit does not exist with $y=3x/2$, $y=x+1$, but it looks like a dead end.
|
Note that since $-1 \le \sin x \le 1$, we have:
$$\frac{1}{\left(x^2-y^2+5\right)^2} \le \frac{\sin x +2}{\left(x^2-y^2+5\right)^2} \le \frac{3}{\left(x^2-y^2+5\right)^2}$$
And the denominator clearly tends to $0$ (but stays positive) when $(x,y) \to (2,3)$, so...
|
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|
Solving for $u(x,y)$ for $u_x + y u_y = -u$, $u(x,1) =x^2$ I wish to solve for $u(x,y)$ from the PDE $u_x + y u_y = -u$, $u(x,1) =x^2$. Since this is a first order linear PDE, I think I should try the method of characteristics. Let $u = u(x(s), y(s))$.
Then the char equations are $\frac{dx}{ds} = 1$, $\frac{dy}{ds} = y$ and $\frac{du}{ds} = -u$.
Solving: $x = s+c_1$, $y = c_2 e^{s}$, and $u = c_3e^{-s}$. Now I eliminate the $s$ parameter: $s = x - c_1$ which implies $y = c_2 e^{x- c_1}$ and $u = c_3 e^{-x + c_1}$. If I let $x(0) = x_0$, then $x_0 = c_1$. But i'm a little confused on how to proceed. Could anyone advize on this?
|
Using the fact that $x(0) = x_0$ and $y(0) = 1$, then we see that
\begin{align}
x= x_0 +s, \ \ y= e^s, \ \ \ u(x, y) = u(x_0, 1)e^{-s} = x_0^2e^{-s}.
\end{align}
Finally, since $s= \log y$ and $x_0 = x-s = x-\log y$, then we have that
\begin{align}
u(x, y) = (x-\log y)^2\exp(-\log y) = \frac{(x-\log y)^2}{y}.
\end{align}
Check: Observe
\begin{align}
u_x= \frac{2(x-\log y)}{y}, \ \ u_y = \frac{-2(x-\log y)-(x-\log y)^2}{y^2} = \frac{-(x-\log y)(2+x-\log y)}{y^2}
\end{align}
which means
\begin{align}
u_x+yu_y = \frac{2(x-\log y)-(x-\log y)(2+x-\log y)}{y} = -\frac{(x-\log y)^2}{y}=-u.
\end{align}
|
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|
Using the Euclidean algorithm, deduce that $\gcd(x^3+2x^2+x +4;x^2+1)=1$ So, I've tried it but I keep getting to $$\frac{x^2+1} 2$$ and don't know how to proceed.
Question is Using the Euclidean algorithm, deduce that $$\gcd(x^3+2 x^2+x +4,\;x^2+1)=1.$$
|
\begin{array}{c|cccc}
& x^3+2x^2+x+4 & 1 & 0 \\
-x-2 & x^2+1 & 0 & 1 \\
\hline
-\frac 12x^2 & 2 & 1 & -x-2 \\
& 1 & -\frac 12x^2 &\frac 12x^3+x^2+1
\end{array}
$$-\frac 12x^2\color{red}{(x^3+2x^2+x+4)}+
\left(\frac 12x^3+x^2+1\right)\color{red}{(x^2+1)} = 1$$
|
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|
Show that the product of any $m$ consecutive positive integers is divisible by $m!$
Show that the product of any $m$ consecutive positive integers is divisible by $m!$.
Note that we have that $\frac{n!}{m!(n-m)!} \in \mathbb{Z}$ for $0 \leq m \leq n$.
$\require{cancel} \frac{n!}{m!(n-m)!} = \frac{n\cdot(n-1)\cdot\cdot\cdot(n-m+1)(n-m)!}{m!(n-m)!} = \frac{n\cdot(n-1)\cdot\cdot\cdot(n-m+1)\cancel{(n-m)!}}{m!\cancel{(n-m)!}} = \frac{n\cdot(n-1)\cdot\cdot\cdot(n-m+1)}{m!}$ where $n\cdot(n-1)\cdot\cdot\cdot(n-m+1)$ are $m$ consecutive integers divisible by $m!$
Of course, I'm using the fact that I know the combinations formula is an integer but given that, is the proof this simple? Or did I make an error here. Appreciate help!
|
Both answers posted so far assume that $n \geq m$, which is not given in the problem statement.
However, this assumption can be justified as a WLOG assumption. Indeed, fix a nonnegative integer $m$. We want to prove that
\begin{align}
m! \mid n\left(n-1\right)\left(n-2\right)\cdots\left(n-m+1\right)
\qquad \text{for each } n \in \mathbb{Z} .
\label{darij1.eq.1}
\tag{1}
\end{align}
Assume that we have proven that
\begin{align}
m! \mid n\left(n-1\right)\left(n-2\right)\cdots\left(n-m+1\right)
\qquad \text{for each integer } n \geq m .
\label{darij1.eq.2}
\tag{2}
\end{align}
(The currently existing answers prove this.) Now, let $n \in \mathbb{Z}$ be arbitrary. Since $m!$ is a positive integer, the arithmetic sequence $\left(n, n+m!, n+2m!, n+3m!, n+4m!, \ldots\right)$ of integers grows unboundedly, and thus eventually surpasses the number $m$. In other words, there exists some nonnegative integer $k$ such that $n+km! \geq m$. Consider this $k$. Let $N = n+km!$. Then, $N$ is an integer satisfying $N = n+km! \equiv n \mod m!$ and $N = n+km! \geq m$. Hence, \eqref{darij1.eq.2} (applied to $N$ instead of $n$) yields
\begin{align}
m! \mid N\left(N-1\right)\left(N-2\right)\cdots\left(N-m+1\right) .
\end{align}
In other words,
\begin{align}
N\left(N-1\right)\left(N-2\right)\cdots\left(N-m+1\right) \equiv 0 \mod m! .
\end{align}
But we have $N \equiv n \mod m!$. Thus, $N - i \equiv n - i \mod m!$ for each $i \in \left\{0,1,2,\ldots,m-1\right\}$. Multiplying all these $m$ congruences together, we find
\begin{align}
N\left(N-1\right)\left(N-2\right)\cdots\left(N-m+1\right) \equiv
n\left(n-1\right)\left(n-2\right)\cdots\left(n-m+1\right) \mod m! .
\end{align}
Hence,
\begin{align}
& n\left(n-1\right)\left(n-2\right)\cdots\left(n-m+1\right) \\
& \equiv
N\left(N-1\right)\left(N-2\right)\cdots\left(N-m+1\right) \\
& \equiv 0 \mod m! .
\end{align}
In other words, $m! \mid n\left(n-1\right)\left(n-2\right)\cdots\left(n-m+1\right)$. Thus, \eqref{darij1.eq.1} is proven.
|
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|
Find the inverse of $5+4\sqrt[3]{2}+3\sqrt[3]{4}$? I'm trying to find the inverse of $5+4\sqrt[3]{2}+3\sqrt[3]{4}$ in $\mathbb{Z}[\sqrt[3]{2}]$. I know it is a unit, so there is an inverse, but I feel like I may be doing too much work in the wrong direction. Here's what I have so far:
Let $\alpha = 5+4\sqrt[3]{2}+3\sqrt[3]{4}$ and $\alpha^{-1} = a+b\sqrt[3]{2}+c\sqrt[3]{4}$ for some $a,b,c \in \mathbb{Z}$.
$(a+b\sqrt[3]{2}+c\sqrt[3]{4})(5+4\sqrt[3]{2}+3\sqrt[3]{4})=1$
$=5a+6b+8c+4a\sqrt[3]{2}+5b\sqrt[3]{2}+6c\sqrt[3]{2}+3a\sqrt[3]{4}+4b\sqrt[3]{4}+5c\sqrt[3]{4}=1$
$=a(5+4\sqrt[3]{2}+3\sqrt[3]{4})+b(6+5\sqrt[3]{2}+4\sqrt[3]{4})+c(8+6\sqrt[3]{2}+5\sqrt[3]{4})=1$
and trying to solve for a,b, and c, but I don't know how?
Edit: Regrouping to $(5a+6b+8c)+(4a+5b+6c)\sqrt[3]{2}+(3a+4b+5c)\sqrt[3]{4}=1$
|
Mostly an observation:
$$ 5 + 4 \sqrt[3]{2}+3\sqrt[3]{4}=(1+\sqrt[3]{2}+\sqrt[3]{4})^2$$
so
$$(5 + 4 \sqrt[3]{2}+3\sqrt[3]{4})^{-1}=((1+\sqrt[3]{2}+\sqrt[3]{4})^{-1})^2=(\sqrt[3]{2}-1)^2= 1- 2\sqrt[3]{2}+\sqrt[3]{4}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Positive integers $a,b,$ and $c$ satisfy the equations. Find the value of $a+b+c$
Question: Positive integers $a,b,$ and $c$ satisfy the equations$$29^2-a^2=28^2-b^2=27^2-c^2$$If $a<20$, what is the value of $a+b+c$?
I first tried factoring the three equations so that$$(29-a)(29+a)=(28-b)(28+b)=(27-c)(27+c)$$But I don't really see a way to continue from there. I then looked at the difference of two squares and drew three right triangles of hypotenuse $29,28,27$ and a leg of length $a,b,c$ respectively.
The third side is then the square root of what we want, but I'm not sure what value still goes into $a$ and $b$ and $c$.
Any ideas on where I can begin?
|
Take $a$ and $b$ first, we will have:
$a^2-b^2=29^2-28^2=57 \Rightarrow (a-b)(a+b)=57$
Because $57>0, a>0, b>0$, we will have $0<a-b<a+b<57$.
There are only two positive cases:
$57=1 \times 57= 3 \times 19$.
Because $a-b<a+b$, we will have two set of equations:
*
*Case one: ${\begin{cases}a-b=1\\a+b=57\end{cases}}\Leftrightarrow {\begin{cases}a=29\\b=28\end{cases}}$, this result is eliminated because $a<20$.
*Case two: ${\begin{cases}a-b=...\\a+b=...\end{cases}}\Leftrightarrow {\begin{cases}a=...\\b=...\end{cases}}$
With $a=...;b=...$, you can solve for $c$, but I won't give it away (or you can see the spoiler below).
$a=11;b=8;c=3\Rightarrow a+b+c=11+8+3=22$
|
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|
Inverse Laplace transformation that is slightly different from known transformation. I have the following inverse laplace transformation:
$L^{-1} =\frac{s}{(s-3)(s-4)(s-12)}$
After looking at the laplace transformations the closest I've found is:
$\frac{ae^{at}-be^{bt}}{a-b} = \frac{s}{(s-a)(s-b)}$
I've been working out a solution for having three variables, but I cant seem to get the correct solution. Is there an identity for this type of transformation?
|
Using partial-fraction expansion, we have
$$
\frac{s}{(s-3)(s-4)(s-12)} = \frac{A}{(s-3)} + \frac{B}{(s-4)} + \frac{C}{(s-12)}
$$
where $A, B$ and $C$ are obtained as follows:
$$
\begin{align}
A &= \frac{s}{(s-4)(s-12)} \Big|_{s=3} = \frac{1}{3}, \\
B &= \frac{s}{(s-3)(s-12)} \Big|_{s=4} = -\frac{1}{2}, \\
C &= \frac{s}{(s-3)(s-4)} \Big|_{s=12} = \frac{1}{6}. \\
\end{align}
$$
As a result, we get
$$
\frac{s}{(s-3)(s-4)(s-12)} = \frac{1}{3(s-3)} - \frac{1}{2(s-4)} + \frac{1}{6(s-12)}
$$
Now, it is straightforward to use the Laplace inverse table, specifically, $\mathscr{L}^{-1} \left[ \frac{1}{s+a} \right] = e^{-at}u(t)$, so we get
$$
\begin{align}
\mathscr{L}^{-1}\left[ \frac{s}{(s-3)(s-4)(s-12)} \right] &=
\mathscr{L}^{-1}\left[ \frac{1}{3(s-3)} \right] - \mathscr{L}^{-1}\left[ \frac{1}{2(s-4)} \right] + \mathscr{L}^{1}\left[ \frac{1}{6(s-12)} \right] \\
&=
\frac{1}{3} \mathscr{L}^{1}\left[ \frac{1}{(s-3)} \right] - \frac{1}{2}\mathscr{L}^{1}\left[ \frac{1}{(s-4)} \right] + \frac{1}{6}\mathscr{L}^{1}\left[ \frac{1}{(s-12)} \right] \\
&=
\frac{1}{3} e^{3t} - \frac{1}{2} e^{4t} + \frac{1}{6} e^{12t}
\end{align}
$$
|
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|
Finding the minimum of $λ$ such that $|a-b|^p\le λ(2-|a-b|)\left|\frac{a|a|^{p-1}}{1-|a|}-\frac{b|b|^{p-1}}{1-|b|}\right|$ for $a,b\in(-1,1)$
Let $p\ge 2$ be a give real number. Find the minimum of $λ$ such that for any $a,b\in(-1,1)$,
$$|a-b|^p\le λ(2-|a-b|)\left|\dfrac{a|a|^{p-1}}{1-|a|}-\dfrac{b|b|^{p-1}}{1-|b|}\right|.\tag{1}$$
I have been thinking about this inequality for a long time, and I have not seen the answer. Maybe Dunkl-Williams’ inequality is involved?$$
\|x-y\| \ge \frac 12 (\|x\|+\|y\|)\left\|\dfrac x{\|x\|}- \dfrac y {\|y\|} \right\|. \quad \forall 0\ne x, y \in X$$
Even with this:
The proof by some remarks on the triangle inequality for norms, L.Maligranda, Banach J. Math. Anal. 2 (2008), no. 2, 31–41. https://arxiv.org/pdf/1109.1773.pdf
But (1) has a power $p$.
|
First, take $(a, b) = \left( \dfrac{1}{2}, -\dfrac{1}{2} \right)$, then $λ \geqslant 2^{p - 2}$. Now it will be proved that for $λ = 2^{p - 2}$, the inequality holds. Note that $f(x) = \dfrac{x^p}{1 - x}$ is increasing on $[0, 1)$. If $ab \geqslant 0$, without loss of generality, suppose $a \geqslant b \geqslant 0$, then$$
(1) \Longleftrightarrow \frac{(a - b)^p}{2 - (a - b)} \leqslant 2^{p - 2} \left( \frac{a^p}{1 - a} - \frac{b^p}{1 - b} \right).
$$
If $ab < 0$, without loss of generality, suppose $a > 0 > b$, then$$
(1) \Longleftrightarrow \frac{(a + (-b))^p}{2 - (a + (-b))} \leqslant 2^{p - 2} \left( \frac{a^p}{1 - a} + \frac{(-b)^p}{1 - (-b)} \right).
$$
Thus it suffices to prove that for any $0 \leqslant b \leqslant a < 1$,$$
\frac{(a - b)^p}{2 - (a - b)} \leqslant 2^{p - 2} \left( \frac{a^p}{1 - a} - \frac{b^p}{1 - b} \right),\\
\frac{(a + b)^p}{2 - (a + b)} \leqslant 2^{p - 2} \left( \frac{a^p}{1 - a} + \frac{b^p}{1 - b} \right).
$$
Note that$$
f''(x) = (p - 1)p · \frac{x^{p - 2}}{1 - x} + 2p · \frac{x^{p - 1}}{(1 - x)^2} + 2 · \frac{x^p}{(1 - x)^3}, \quad \forall 0 < x < 1
$$
thus $f$ is convex on $[0, 1)$. By Jensen's inequality,$$
\frac{1}{2} \left( \frac{a^p}{1 - a} + \frac{b^p}{1 - b} \right) \geqslant \frac{\left( \dfrac{a + b}{2} \right)^p}{1 - \dfrac{a + b}{2}},
$$
i.e.$$
\frac{(a + b)^p}{2 - (a + b)} \leqslant 2^{p - 2} \left( \frac{a^p}{1 - a} + \frac{b^p}{1 - b} \right).
$$
Now for any fixed $0 \leqslant c < 1$, define$$
g_c(x) = f(x + c) - f(x) = \frac{(x + c)^p}{1 - (x + c)} - \frac{x^p}{1 - x}. \quad 0 \leqslant x < 1 - c
$$
Because $f''(x) \geqslant 0$ implies $f'$ is increasing on $[0, 1)$, then $g_c'(x) = f'(x + c) - f'(x) \geqslant 0$, and $g_c'$ is increasing on $[0, 1 - c)$. Thus$$
\frac{a^p}{1 - a} - \frac{b^p}{1 - b} = g_{a - b}(b) \geqslant g_{a - b}(0) = \frac{(a - b)^p}{1 - (a - b)}.
$$
Since$$
2^{p - 2} \frac{(a - b)^p}{1 - (a - b)} \geqslant \frac{(a - b)^p}{2 - (a - b)} \Longleftrightarrow \frac{1 - (a - b)}{2 - (a - b)} \leqslant 2^{p - 2},
$$
and $0 \leqslant a - b < 1$ implies $\dfrac{1 - (a - b)}{2 - (a - b)} \leqslant \dfrac{1}{2} < 2^{p - 2}$, then$$
\frac{(a - b)^p}{2 - (a - b)} \leqslant 2^{p - 2} \left( \frac{a^p}{1 - a} - \frac{b^p}{1 - b} \right).
$$
Therefore,$$
|a - b|^p \leqslant 2^{p - 2} (2 - |a - b|) \left| \frac{a |a|^{p - 1}}{1 - |a|} - \frac{b |b|^{p - 1}}{1 - |b|} \right|. \quad \forall a, b \in (-1, 1)
$$
|
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|
Show that $\int_0^{\infty} \frac{dx}{(x+\sqrt{1+x^2})^n}$ converges and is equal to $\frac{n}{n^2-1}$ Testing convergence of $$I=\int_0^{\infty} \frac{dx}{(x+\sqrt{1+x^2})^n},$$ show that $$\int_0^{\infty} \frac{dx}{(x+\sqrt{1+x^2})^n}=\frac{n}{n^2-1}$$
I thought to put $$x=\tan{z}, \text{ then } \quad I=\int_0^{\pi/2}\frac{\sec^2zdz}{(\tan{z}+\sec{z})^n}$$
It is difficult to investigate convergence. Help is earnestly solicited.
|
Using twice the fact that $\begin{aligned} & \frac{d}{d x} \frac{1}{(\tan \theta+\sec \theta)^n} = -\frac{n \sec \theta}{(\tan \theta+\sec \theta)^n}\end{aligned} $ via integration by parts, we have a beautiful recursive relation for $I$ as below:
$$
\begin{aligned}
I& = \int_0^{\infty} \frac{d x}{\left(x+\sqrt{1+x^2}\right)^n}\\& \stackrel{x=\tan \theta }{=} \int_0^{\frac{\pi}{2}} \frac{\sec ^2 \theta d \theta}{(\tan \theta+\sec \theta)^n} d x . \\&=-\frac{1}{n}\int_0^{\frac{\pi}{2}} \sec \theta \, d\left(\frac{1}{(\tan \theta+\sec \theta)^n}\right) \\
& =\left[-\frac{\sec \theta}{n(\tan \theta+\sec \theta)^n}\right]_0^{\frac{\pi}{2}}+\frac{1}{n} \int_0^{\frac{\pi}{2}} \frac{\sec \theta\tan \theta}{(\tan \theta+\sec \theta)^n} d \theta \\
& =\frac{1}{n}-\frac{1}{n^2} \int_0^{\frac{\pi}{2}} \tan \theta d\left(\frac{1}{(\tan \theta+\sec \theta)^n}\right) \\
& =\frac{1}{n}-\left[\frac{1}{n^2} \frac{\tan \theta}{(\tan \theta+\sec \theta)^n}\right]_0^{\frac{\pi}{2}}+\frac{1}{n^2} \int_d^{\frac{\pi}{2}} \frac{\sec ^2 \theta}{(\tan \theta+\sec \theta)^n} d \theta \\
& =\frac{1}{n}+\frac{1}{n^2}I
\end{aligned}
$$
Now we can conclude that
$$\boxed{I_n=\frac{n}{n^2-1}}$$
|
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|
Finding the area of circles in triangle In the figure there are infinitely many circles approaching the
vertices of an equilateral triangle, each circle touching other circles and sides of the triangle. If the triangle has sides of
length $1$, how can i find the total area occupied by the circles?
|
In a equilateral triangle the ratio $\frac{A_{\text{incircle}}}{A_{\text{triangle}}}$ equals $\frac{\pi}{3\sqrt{3}}$. Step 0: we draw the incircle of the original triangle.
Step 1: we draw three incircles for equilateral triangles with side length $\frac{1}{3}$. Step $n$: we draw three incircles for equilateral triangles with side length $\frac{1}{3^n}$. The area occupied by the circles is so
$$ \frac{\pi}{3\sqrt{3}}\cdot \frac{\sqrt{3}}{4}\cdot 1^2 + \sum_{n\geq 1}\frac{\pi}{3\sqrt{3}}\cdot \frac{\sqrt{3}}{4}\cdot 3\left(\frac{1}{3^n}\right)^2 $$
which simplifies to
$$ \frac{\pi}{12}+\frac{\pi}{12}\sum_{n\geq 1}\frac{3}{9^n}=\frac{\pi}{12}+\frac{\pi}{12}\cdot\frac{3}{8}=\frac{11\pi}{96}. $$
|
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|
Solving $\sqrt{a +\sqrt{a-x}}+\sqrt{a-\sqrt{a+x}}=2x, \ a\in\Bbb{R}$
Given
$$\sqrt{a +\sqrt{a-x}}+\sqrt{a-\sqrt{a+x}}=2x$$
and $a\in\Bbb{R}$, express $x$ in terms of $a$.
I rationalised the above expression and then again rationalised which gave me :
$$\sqrt{a+\sqrt{a-x}}-\sqrt{a-\sqrt{a+x}}=\frac{1}{\sqrt{a+x}-\sqrt{a-x}}$$
Now, what should I do?
If didn't rationalised the original Equation, or if the squared both sides twice then too bi-quadratic or higher degree polynomial will be obtained.
What should be done?
|
Here is a simplification, which leads to a parametrized solution.
Write the original equation as two equations
$$
\sqrt{a +\sqrt{a-x}} = x +q\\
\sqrt{a-\sqrt{a+x}}= x -q
$$
where $q$ is a function of $a$.
Double squaring both equations gives
$$
(x+q)^4-2a(x+q)^2+x+a^2-a = 0 \\
(x-q)^4-2a(x-q)^2-x+a^2-a = 0
$$
Subtracting and adding gives the two equations
$$
x^2+q^2-a -\frac{1}{4q}= 0 \\
x^4+ 6 q^2x^2 + q^4-2a(x^2+q^2)+a^2-a = 0
$$
The first equation gives the result
$$
x = \sqrt{-q^2+a +\frac{1}{4q}}= \sqrt{a + f(q)}
$$
with the offset $f(q)$. In here, $q$ has to be inserted, where $q = q(a)$ is obtained by replacing all terms with $x^2$ in the second equation with the first equation, giving
$$
(-q^2+a +\frac{1}{4q})^2+ 6 q^2(-q^2+a +\frac{1}{4q}) + q^4-2a(a +\frac{1}{4q})+a^2-a = 0
$$
Expanding this expression gives an implicit formula for $q(a)$ as follows:
$$
- 64 q^6 + 64 a q^4 + 16 q^3 - 16 a q^2 + 1 = 0
$$
I wasn't able to solve this further for $q$. However, the equations directly lead to a parametrized solution. Let $q$ be a free parameter. The range where $q$ is to be chosen from is discussed below. Then for each $q$, we have $a$ and $x$ according to
$$
a = \frac{64 q^6 - 16 q^3 - 1}{64 q^4 - 16 q^2} \\
x = \sqrt{-q^2+a +\frac{1}{4q}}
$$
The following observations can be made:
For large $a$, the equation for $a$ gives $q^2 \to \frac{1}{4}$. This can also be observed by regarding the very first set of two equations and treating $q$ as a small disturbance, which can in first order be neglected if all other terms grow exceedingly large. We then have
$$
\sqrt{a +\sqrt{a-x_1}} = x_1\\
\sqrt{a-\sqrt{a+x_2}}= x_2
$$
which are solved (see Solve the equation: $x=\sqrt{a-\sqrt{a+x}},(a\geq 1)$) by
$$
x_1 = \frac12 + \sqrt{a -3/4}\\
x_2 = -\frac12 + \sqrt{a -3/4}
$$
So indeed, the $x_{1,2}$ grow exceedingly large, and $q = -\frac12$ gives both the correct offset term $-\frac34$ under the previous root $
x = \sqrt{-q^2+a +\frac{1}{4q}}
$, as well as the two deviations $x - x_{1,2}$. For small $a$, correction terms must be added to $q(a)$ (in the sense of an expansion).
Addendum:
1. As a side observation, one can use the two solutions $x_{1,2}$ to establish that the modified equation
$$
\sqrt{a +\sqrt{a-x -\frac12}}+\sqrt{a-\sqrt{a+x-\frac12}}=2x
$$
is solved by $x = \sqrt{a -3/4}$.
*Note that in the general solution $x =\sqrt{a + f(q)}$, the offset $f(q) = -q^2 +\frac{1}{4q}$ actually has a maximum at $q = -0.5$. This can be seen by
$f'(q) = \frac{-1 - 8 q^3}{4 q^2}$ which is zero at $q = -0.5$. Hence, the offset has its maximum $f(q= -0.5) = -0.75$ for $\alpha \to \infty$. It will be lower for other $q$. The behavior of $q(a)$ is given in the following plot. For low $a=1.5$, for example, we have $q\simeq -0.81$ and $f(q) \simeq -0.965$. The plot $q(a)$, together with the offset behavior $f(q)$, establish that the offset is monotonously rising with $a$ in a small range of $f$-values.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Define $f: \Bbb R^2 \to \Bbb R$ by $f(x,y) =\begin{cases} \frac{xy^2}{x^2 + y^4}, & \text{if $(x,y) \ne 0$} \\ 0, & \text{if $(x,y)= 0$} \end{cases}$ Define $f: \Bbb R^2 \to \Bbb R$ by
$f(x,y) =\begin{cases}
\frac{xy^2}{x^2 + y^4}, & \text{if $(x,y) \ne 0$} \\
0, & \text{if $(x,y)= 0$}
\end{cases}$
I'm to show that $f$ is continuous at any straight line THROUGH the origin, but not continuous AT the origin. I'm confused how this is possible. Is it a hole?
How would I do this?
|
To show that it's continuous along straight lines:
Approach the origin along the half line $(x,y) = (r\cos\theta,r\sin\theta)$, where $r>0$ and $r\to 0^+$.
\begin{eqnarray*}
\frac{xy^2}{x^2+y^2} &\rightsquigarrow&
\frac{r^3\cos\theta\sin^2\theta}{r^2\cos^2\theta+r^2\sin^2\theta} \\ \\
&\equiv& \frac{r^3\cos\theta\sin^2\theta}{r^2} \\ \\
&\equiv& r\cos\theta\sin^2\theta
\end{eqnarray*}
As $r\to 0$, $r\cos\theta\sin^2\theta \to 0$ for all $\theta$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Adding $k^2$ to $1^2 + 2^2 + \cdots + (k - 1)^2$. In his book, Calculus Vol. 1, Tom Apostol mentions that adding $k^2$ to the predicate
$$A(k): 1^2 + 2^2 + \cdots + (k - 1)^2 < \frac{k^3}{3}$$
gives the inequality
$$ 1^2 + 2^2 + \cdots + k^2 < \frac{k^3}{3} + k^2.$$
Why does the RHS $$1^2 + 2^2 + \cdots + (k - 1)^2 $$ become $$ 1^2 + 2^2 + \cdots + k^2 $$ and not $$ 1^2 + 2^2 + \cdots + (k - 1)^2 + k^2$$
when adding $k^2$?
Thank you.
|
I will give you an example.
1+2+3+........+99+100 is same as 1+2+.....+100.
And, I believe this is quite obvious.
Hence, both of them have the same value whatever way you decide to solve it by.
|
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|
Mini-Challenge on a condition Hello I would like to purpose to you an enigma this is the following :
Let $a,b,c$ be positive real number find the condition on $abc$ and $ab+bc+ca$ and $a+b+c$ to have $a^9+b^9+c^9=3$
I have a solution using the following identity :
$$a^9+b^9+c^9= 3a^3b^3c^3−45abc(ab+bc+ca)(a+b+c)^4+ 54abc(ab+bc+ca)^2(a+b+c)^2−27a^2b^2c^2(ab+bc+ca)(a+b+c) + (a+b+c)^9−9(ab+bc+ca)(a+b+c)^7+ 9(ab+bc+ca)^4(a+b+c)−30(ab+bc+ca)^3(a+b+c)^3+ 18a^2b^2c^2(a+b+c)^3+ 27(ab+bc+ca)^2(a+b+c)^5+ 9abc(a+b+c)^6−9abc(ab+bc+ca)^3$$
But I would like a solution without this because it's too ugly .
So could you help me ?
|
To make it less ugly, calculate the lower amounts instead, or setting new variables, for example:
${\begin{cases}a+b+c=x\\ab+bc+ca=y\\abc=z\\S_{n}=a^n+b^n+c^n\end{cases}}$
In fact, I have encountered this problem in a test that gives $x=3;y=-10;z=11$ and it asks me to find $S_{5}$. Here's how I attemped it (I have generalized it for $x;y;z$):
Step one:
$(a+b+c)(ab+bc+ca)=xy$
$\Rightarrow a^2(b+c)+b^2(c+a)+c^2(a+b)+3abc=xy$
$\Rightarrow a^2(b+c)+b^2(c+a)+c^2(a+b)=xy-3z$
Step two:
$a^2+b^2+c^2=(a+b+c)^2-2(ab+bc+ca)=x^2-2y$
Step three:
$(a+b+c)(a^2+b^2+c^2)=x(x^2-2y)$
$\Rightarrow a^3+b^3+c^3+a^2(b+c)+b^2(c+a)+c^2(a+b)=x(x^2-2y)$
$\Rightarrow a^3+b^3+c^3+xy-3z=x^3-2xy$
$\Rightarrow a^3+b^3+c^3=x^3-3xy+3z$
Step four:
$a^4+b^4+c^4$
$=(a^2+b^2+c^2)^2-2(a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2})$
$=(a^2+b^2+c^2)^2-2((ab+bc+ca)^2-2a^{2}bc-2ab^{2}c-2abc^{2})$
$=(a^2+b^2+c^2)^2-2((ab+bc+ca)^2-2abc(a+b+c))$
$=(x^2-2y)^2-2(y^2-2zx)$
$=x^4-4x^{2}y+2y^2+4zx$
Step five:
$(a^3+b^3+c^3)(ab+bc+ca)=y(x^3-3xy+3z)$
$\Rightarrow a^4(b+c)+b^4(c+a)+c^4(a+b)+abc(a^2+b^2+c^2)=x^3y-3xy^2+3yz$
$\Rightarrow a^4(b+c)+b^4(c+a)+c^4(a+b)+z(x^2-2y)=x^3y-3xy^2+3yz$
$\Rightarrow a^4(b+c)+b^4(c+a)+c^4(a+b)+x^2z-2yz=x^3y-3xy^2+3yz$
$\Rightarrow a^4(b+c)+b^4(c+a)+c^4(a+b)=x^3y-x^2z-3xy^2+5yz$
Step six:
$(a+b+c)(a^4+b^4+c^4)=x(x^4-4x^{2}y+2y^2+4zx)$
$\Rightarrow a^5+b^5+c^5+a^4(b+c)+b^4(c+a)+c^4(a+b)=x^5-4x^3y+2xy^2+4x^2z$
$\Rightarrow a^5+b^5+c^5+x^3y-x^2z-3xy^2+5yz=x^5-4x^3y+2xy^2+4x^2z$
$\Rightarrow S_{5}=a^5+b^5+c^5=x^5-5x^3y+5xy^2+5x^2z-5yz$
You can continue similarly to $S_{9}$ to write the expression in terms of $x,y,z$, this will make the result much less ugly and also much easier to check. Thanks for taking time reading this long post.
|
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|
Determining the eigenvalues for the linear operator $T$ on $V$ $V=M_{2x2}$
$T\begin{bmatrix}a&b\\c&d\end{bmatrix} = \begin{bmatrix}c&d\\a&b\end{bmatrix}; \tag 1$
Attempt:
I used the standard basis to find
$A = \begin{bmatrix}0&0&1&0\\0&0&0&1\\1&0&0&0\\0&1&0&0\end{bmatrix}; \tag 2$
then
$\det(A-\lambda I)=\begin{bmatrix}1-λ&0&0&0\\0&1-λ&0&0\\0&0&1-λ&0\\0&1&0&-λ\end{bmatrix}, \tag 3$
and
$4(1-λ)(1-λ)\begin{bmatrix}1-λ&0\\0&-λ\end{bmatrix}; \tag 4$
so
$(1-λ)(1-λ)(1-λ)(-λ); \tag 5$
In my solutions manual, it says the values of $\lambda$ are $-1,-1,1$ and $1$. I dont understand what Im doing wrong
|
First, a few remarks on our OP Essie Stern's work; then I'll provide my own take on this one.
Apparently we are using the mapping
$\begin{bmatrix} a & b \\ c & d \end{bmatrix} \mapsto \begin{pmatrix} a \\b \\ c \\ d \end{pmatrix} \tag 1$
to express members of $M_{2 \times 2}$ as column vectors; the map is clearly a linear isomorphism between vector spaces. It is easy to see that in this representation, the map
$T\begin{bmatrix} a & b \\ c & d \end{bmatrix} = \begin{bmatrix} c & d \\ a & b \end{bmatrix} \tag 2$
corresponds to the matrix
$A = \begin{bmatrix} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \end{bmatrix}; \tag 3$
the eigenvalues of $A$ then satisfy
$0 = \det(A - \lambda I) = \det \left (\begin{bmatrix} -\lambda & 0 & 1 & 0 \\ 0 & -\lambda & 0 & 1 \\ 1 & 0 & -\lambda & 0 \\ 0 & 1 & 0 & -\lambda \end{bmatrix} \right ); \tag 4$
comparing this to equation (3) in the body of the question indicates that our OP Essie Stein erred in constructing the matrix $A - \lambda I$, apparently replacing some of the $1$s in off-diagonal positions with $1$s on the diagonal, which then give rise to the $1 - \lambda$ entries occurring in (3) of the question.
To continue, we may expand (4) in minors along the first column, yielding
$\det(A - \lambda I) = -\lambda \det \left ( \begin{bmatrix} -\lambda & 0 & 1 \\ 0 & -\lambda & 0 \\ 1 & 0 & - \lambda \end{bmatrix} \right ) + 1 \cdot \det \left (\begin{bmatrix} 0 & 1 & 0 \\ -\lambda & 0 & 1 \\ 1 & 0 & -\lambda \end{bmatrix} \right )$
$= -\lambda(-\lambda^3 + \lambda) + 1 \cdot (1 - \lambda^2) = \lambda^4 - 2 \lambda^2 + 1 = (\lambda^2 - 1)^2; \tag 5$
the roots of
$(\lambda^2 - 1)^2 = 0 \tag 6$
are easily seen to be $-1, -1, 1, 1$; these then are the eigenvalues of the $4 \times 4$ matrix $A$; it is from here but a short step to find eigvectors corresponding to these values by solving the equation
$\begin{bmatrix} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \end{bmatrix}\begin{pmatrix} a \\b \\ c \\ d \end{pmatrix} = \begin{pmatrix} c \\ d \\ a \\ b \end{pmatrix} = \pm 1 \begin{pmatrix} a \\b \\ c \\ d \end{pmatrix}; \tag 7$
we in fact must have
$(a, b) = \pm (c, d); \tag 8$
it is from here but a short step to see that the $+1$ and $-1$ eigenspaces of $A$, and hence $T$, are each two-dimensional, and to work out which $2 \times 2$ matrices lie in each eigenspace; the details are straighforward and left to the reader.
Having shown how the approach suggested by our OP Ellie Stein is supposed to work, I will now provide the promised simpler solution:
We observe from (2) that
$T^2\begin{bmatrix} a & b \\ c & d \end{bmatrix} = \begin{bmatrix} a & b \\ c & d \end{bmatrix}; \tag 9$
thus
$T^2 = I \tag{10}$
on $M_{2 \times 2}$; it follows then that the eigenvalues of $T$ must satisfy
$\lambda^2 - 1 = 0, \tag{11}$
that is,
$\lambda \in \{-1, 1 \}; \tag{12}$
the eigenvector matrices of $T$ must then obey
$T\begin{bmatrix} a & b \\ c & d \end{bmatrix} = \begin{bmatrix} c & d \\ a & b \end{bmatrix} = \lambda \begin{bmatrix} a & b \\ c & d \end{bmatrix}; \tag{13}$
it is then easy to see by inspecting (13) that we must have
$\lambda = 1 \Longleftrightarrow (c, d) = (a, b), \tag{14}$
$\lambda = -1 \Longleftrightarrow (c, d) = -(a, b), \tag{15}$
which together show that both $-1$ and $1$ are in fact eigenvalues of $T$; the eigenspaces of each are in fact two-dimainsional, being in fact matrices of the form
$\begin{bmatrix} a & b \\ a & b \end{bmatrix} \; \text{when} \; \lambda = 1, \tag{16}$
and
$\begin{bmatrix} a & b \\ -a & -b \end{bmatrix} \; \text{when} \; \lambda = -1. \tag{17}$
|
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|
Alternate solutions to similar integrals? So, when I started this problem I couldn't see any reasonable trig substitutions that would make it simpler, and I don't know of a better method, so I attempted to solve it using repeated use of integration by parts. Is there a much easier, better, alternate solution to integrals of this nature in general? I'd also like to know where I went wrong with this solution.
$$
\begin{align*}
\int\sin^4x\,\mathrm{d}x&\rightarrow\\
u_1&=\sin^4x\\
u_1'&=4\cos x\sin^3x\\
v_1&=x\\
v_1'&=1\\
\leftarrow\int\sin^4x\,\mathrm{d}x&=x\sin^4x-4\int x\cos x\sin^3x\,\mathrm{d}x\\
\rightarrow\int x\cos x\sin^3x\,\mathrm{d}x&\rightarrow\\
u_2&=\sin^3x\\
u_2'&=3\cos x\sin^2x\\
v_2&=\int x\cos x\,\mathrm{d}x\\
v_2'&=x\cos x\\
\rightarrow\int x\cos x\,\mathrm{d}x\rightarrow\\
u_3&=x\\
u_3'&=1\\
v_3&=\sin x\\
v_3'&=\cos x\\
\leftarrow\int x\cos x\,\mathrm{d}x&=x\sin x-\int\sin x\,\mathrm{d}x\\
&=x\sin x+\cos x\\
\leftarrow\int x\cos x\sin^3x\,\mathrm{d}x&=\left(\sin^3x\right)\left(x\sin x+\cos x\right)-\int\left(3\cos x\sin^2x\right)\left(x\sin x+\cos x\right)\,\mathrm{d}x\\
&=x\sin^4x+\cos x\sin^3x-3\int x\cos x\sin^3 x\,\mathrm{d}x+3\int\cos^2
x\sin^2x\,\mathrm{d}x\\
4\int x\cos x\sin^3x\,\mathrm{d}x&=x\sin^4
x+\cos x\sin^3 x+3\int\cos^2x\sin^2x\,\mathrm{d}x\\
&=x\sin^4
x+\cos x\sin^3 x+3\int\left(1-\sin^2x\right)\sin^2x\,\mathrm{d}x\\
&=x\sin^4
x+\cos x\sin^3 x+3\int\sin^2x\,\mathrm{d}x-\int\sin^4x\,\mathrm{d}x\\
\rightarrow\int\sin^2x\,\mathrm{d}x\rightarrow\\
\cos2x&=1-2\sin^2x\\
\sin^2x&=\frac{1}{2}(1-\cos2x)\\
\leftarrow\frac{1}{2}\int1-\cos2x\,\mathrm{d}x&=\frac{1}{2}x-\frac{1}{4}\sin{2x}\\
\leftarrow4\int x\cos x\sin^3x\,\mathrm{d}x&=x\sin^4
x+\cos x\sin^3 x+3\left[\frac{1}{2}x-\frac{1}{4}\sin{2x}\right]-\int\sin^4x\,\mathrm{d}x\\
&=x\sin^4 x+\cos x\sin^3 x+\frac{3}{2}x-\frac{3}{4}\sin{2x}-\int\sin^4x\,\mathrm{d}x\\
\leftarrow\int\sin^4x\,\mathrm{d}x&=x\sin^4x-\left[x\sin^4 x+\cos x\sin^3 x+\frac{3}{2}x-\frac{3}{4}\sin{2x}-\int\sin^4x\,\mathrm{d}x\right]\\
&=x\sin^4x-x\sin^4x-\cos x\sin^3x-\frac{3}{2}x+\frac{3}{4}\sin2x+\int\sin^4x\,\mathrm{d}x\\
\end{align*}
$$
As you can see, I can't continue because the integral cancels itself.
|
Hint
$$I=\int sin^4(x)dx=\int \sin^2(x)-\sin^2(x)\cos^2(x)dx$$
$$I=\int \sin^2(x)-\frac 14\sin^2(2x)dx$$
Note that
$$\sin^2(x)=1-\cos^2(x)=1-\frac 12(\cos(2x)+1)=\frac 12(1 -\cos(2x))$$
$$I=\frac 12\int (1 -\cos(2x)) -\frac 18\int (1 -\cos(4x))dx$$
$$\boxed{I=\frac 38x-\frac 14 \sin(2x)+\frac 1{32} \sin(4x)+K} $$
|
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|
How to find $y^{(y^2-6)}$? $$\frac{3}{1-3^{x-2}} + \frac{3}{1-3^{2-x}} = y$$ $$x≠2$$ $$y^{(y^2-6)} = ?$$What is $y^{(y^2-6)}$? Could you please explain to me how to solve this question step by step?
|
You have
$$\begin{align}
y & = \frac{3}{1-3^{x-2}} + \frac{3}{1-3^{2-x}} \\
& = \frac{3(1-3^{2-x}) + 3(1-3^{x-2})}{(1-3^{x-2})(1-3^{2-x})} \\
& = \frac{3(1-3^{2-x} + 1-3^{x-2})}{(1 - 3^{2-x}-3^{x-2}+3^{x-2+2-x})} \\
& = \frac{3(2-3^{2-x} - 3^{x-2})}{(1 - 3^{2-x}-3^{x-2}+3^{0})} \\
& = \frac{3(2-3^{2-x} - 3^{x-2})}{(2 - 3^{2-x}-3^{x-2})} \\
& = 3 & \text{since } x \not =2\\
\end{align}$$
So
$$y^{y^2-6} = 3^{3^2-6}=3^{9-6}=3^3=27$$
And you're done.
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Baffled with $\lim\limits_{x\to 0}{e^x-e^{\sin x} \over x-\sin x}$
Calculate $$\lim\limits_{x\to 0}{e^x-e^{\sin x} \over x-\sin x}$$
Personal work:
$$\lim\limits_{x\to 0}{e^x-e^{\sin x} \over x-\sin x}=^{0 \over 0}\lim\limits_{x\to 0}{e^x-e^{\sin x}\cdot\cos x \over 1-\cos x}=^{0 \over 0}\lim\limits_{x\to 0}{{e^x-(e^{\sin x}\cdot\cos x-\sin x}\cdot e^{\sin x}\over \sin x})=\cdots$$
This gets to nowhere. Also, I substituted $t=e^{\sin x}$ but I could not replace the $e^x$.
|
Use the third-order Maclaurin formulae
$$
e^x = 1+x+\frac{x^2}{2}+\frac{x^3}{6}+o(x^3),
$$
$$
\sin x = x-\frac{x^3}{6}+o(x^3),
$$
$$
e^{\sin x} = e^{x-\frac{x^3}{6}+o(x^3)} =
1+\left(x-\frac{x^3}{6}+o(x^3)\right)+\frac{1}{2} \left(x-\frac{x^3}{6}+o(x^3)\right)^2+
\frac16 \left(x-\frac{x^3}{6}+o(x^3)\right)^3 + o\left(\left( x-\frac{x^3}{6}+o(x^3) \right)^3 \right)=
$$
$$
1+x-\frac{x^3}{6} + \frac{x^2}{2}+\frac{x^3}{6} + o(x^3);
$$
when expanding, we took into account only the summands up to the 3rd order, everything else gone into $o(x^3)$. Hence
$$
\frac{e^{\sin x}-e^x}{x-\sin x} = \frac{\frac{x^3}{6}+o(x^3)}{\frac{x^3}{6}+o(x^3)} = \frac{\frac16+o(1)}{\frac16+o(1)} \rightarrow 1.
$$
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
How to factor $a^{3} + b^{3} + c^{3} - 3abc$ into a product of polynomials The question is in the title.
This question is from "Algebra" by Gelfand.
My initial thought is that if $a$, $b$ and $c$ are $1$ or $-1$, then the polynomial evaluates to $0.$ So, maybe two of the factors will be $(a + b + c - 3)$ and $(a + b + c + 3)$. An alternative option that combines these two might be $a^{2} + b^{2} + c^{2} - 3$.
*
*Is the thought process correct here, and would trial and error be a good way to decide between the linear and the quadratic options I described above?
*As you can tell, I am largely doing guess work here. Is there a more systematic way of deciding what terms to add and subtract in orders to factor the polynomial?
Note: The factoring need not be done all the way to linear factors. All that is needed is a product of polynomials.
|
Factor $a^3+b^3+c^3-3abc$ to a product of polynomials.
Think when $a=b=c$.
$$a^3+b^3+c^3-3abc=a^3+a^3+a^3-3a^3=0.$$
This only fits when $a=b=c$, but not when $a=b$ or $b=c$ or $c=a$.
So, you can think about $(a-b)^2+(b-c)^2+(c-a)^2$, which is $0$ if and only if $a=b=c$.
Therefore, you can reason $a^2+b^2+c^2-ab-bc-ca$ as the factor of $a^3+b^3+c^3-3abc$.
Also, think when $a=-b-c$.
$$a^3+b^3+c^3-3abc=(-b-c)^3+b^3+c^3-3(-b-c)bc \\ =-b^3-3b^2c-3bc^2-c^3+b^3+c^3+3(b+c)bc=0.$$
ISW, you can reason $a+b+c$ as the factor of $a^3+b^3+c^3-3abc.$
Since the degree of $a^2+b^2+c^2-ab-bc-ca$ is $2$, while $a+b+c$ has $1$, You can reason $(a+b+c)(a^2+b^2+c^2-ab-bc-ca)=a^3+b^3+c^3-3abc$. checking this, you can find that you got the right one.
|
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|
Evaluation of $\int_{0}^{\pi/6}\tan^{-1}\sqrt{2-\tan^2(x)}\mathrm dx $ $$\int_{0}^{\pi/6}\tan^{-1}\sqrt{2-\tan^2(x)}\mathrm dx $$
$1+\tan^2(x)=\sec^2(x)$
$u=\sqrt{2-\tan^2(x)}, dx={udu\over \sqrt{2-u}+(\sqrt{2-u^2})^3}$
$$\int_{\sqrt{2-\sqrt{3}/3}}^{\sqrt{2}}{udu\over \sqrt{2-u}+(\sqrt{2-u^2})^3}$$
Very messy, how do we evaluate this question in a easy manner?
|
We can approximate the value using composition of Taylor series for the integrand around $x=0$.
This would give
$$\tan ^{-1}\left(\sqrt{2-\tan ^2(x)}\right)=\tan ^{-1}\left(\sqrt{2}\right)-\frac{x^2}{6 \sqrt{2}}-\frac{23 x^4}{144
\sqrt{2}}-\frac{3727 x^6}{25920 \sqrt{2}}-\frac{124627 x^8}{967680
\sqrt{2}}+O\left(x^{10}\right)$$ Using the expansion to $O\left(x^{2n}\right)$ and computing the decimal representation of the result, we should get
$$\left(
\begin{array}{cc}
n & \text{result} \\
1 & 0.494564 \\
2 & 0.493675 \\
3 & 0.493518 \\
4 & 0.493488 \\
5 & 0.493482 \\
6 & 0.493481
\end{array}
\right)$$ quite close to the result from Wolfram Alpha.
|
{
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Matrix Notation Form of Roots of a Quadratic Equation We know that the quadratic equation
$$f(x)=ax^2+bx+c=0$$
has roots
$$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=-\frac b{2a}\pm \frac 1a\sqrt{-\left(ac-\frac {b^2}4\right)}$$
Also, $f(x)$ can be written in matrix notation as follows:
$$f(x)=
\left(\begin{matrix}x&1\\\end{matrix}\right)
\left(\begin{matrix}a&\frac b2\\\frac b2&c\end{matrix}\right)
\left(\begin{matrix}x\\1\end{matrix}\right)=\mathbf{x^T Q x}$$
where the determinant of $\mathbf Q$ is $\left(ac-\frac {b^2}4\right)=-\frac 14\left(b^2-4ac\right)$, where coincidentally the familiar $(b^2-4ac)$ is the discriminant of the quadratic $f(x)$.
Hence the roots of the quadratic $f(x)=0$ may be written as
$$x=-\frac b{2a}\pm \frac 1a\sqrt{-\det(\mathbf Q)}$$
This is equivalent to
$$\left(x+\frac b{2a}\right)^2=\frac {-\det(\mathbf Q)}{a^2}$$
Or in neater form,
$$\left(ax+\frac b{2}\right)^2={-\det(\mathbf Q)}$$
Question
Can the roots of $f(x)=0$ be derived and written completely in matrix notation, given the link between the determinant and discriminant as shown above?
|
I have obtained some formula with a little changed notation comparing the notation used in the question.
I've interchanged the components of $\mathbf x$ (the new vector is denoted as $\mathbf{ \hat{x}}$ ) and I've interchanged the entries on the main diagonal of $ \mathbf Q$ (new matrix is denoted as $ \mathbf M$ - interchanging entries on the diagonal for $2 \times 2$ matrices doesn't change the determinant) .
Then $f(x)$ can be written down as
$$f(x)=
\begin{bmatrix}1&x\\\end{bmatrix}
\begin{bmatrix}c&\frac b2\\\frac b2&a\end{bmatrix}
\begin{bmatrix}1\\x\end{bmatrix} =\mathbf{\hat{x}^T M \hat{x}}=0$$
It can be calculated that
$ \begin{bmatrix}1 & x\\0 & 1\end{bmatrix}
\begin{bmatrix}c & \frac{b}{2}\\\frac{b}{2} & a\end{bmatrix}
\begin{bmatrix}1 & 0\\x & 1\end{bmatrix} =
\begin{bmatrix}\frac{b x}{2} + c + x \left(a x + \frac{b}{2}\right) & a x + \frac{b}{2}\\a x + \frac{b}{2} & a\end{bmatrix} = \begin{bmatrix} 0 & a x + \frac{b}{2}\\a x + \frac{b}{2} & a \end{bmatrix} $
We know also that determinant of a matrix product is equal to the product of determinants of multiplied matrices.
Additionally we have $\det( \mathbf Q)=\det(\mathbf M)$.
From this follows that
$ \det( \mathbf Q)=-\left( ax+ \frac{b}{2}\right)^2$.
|
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|
Find the remainder when $(x+1)^n$ is divided by $(x-1)^3$. I know this question has been answered before, but I have a slightly different different question.
I saw the solution of this question in my book and the author has solved it by substituting $x-1=y$ and then equating the coefficients of $y^2$, $y^1$ and $y^0$ to $A(y+1)^2$, $B(y+1)^1$ and $C$.
My question is why do we have to substitute $x-1=y$ and why can't equate coefficients of $x^2$, $x^1$ and $x^0$ to $A$, $B$ and $C$ without substituting? Thanks in advance.
|
$$(x+1)^n=(x-1)^3Q (x)+ax^2+bx+c $$
for $x=1$, we get
$$\boxed {2^n=a+b+c} $$
differentiating,
$$n (x+1)^{n-1}=3 (x-1)^2R (x )+2ax+b $$
with $x=1$,
$$\boxed {2^{n-1}=2a+b} $$
differentiating again
$$n (n-1)(x+1)^{n-2}=6 (x-1)S (x)+2a $$
with $x=1$, we find
$$\boxed {a=n (n-1)2^{n-3}}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Why A.M.$\geq $ G.M. not works here. Find the minimum value of $f(x)= x^2+4x+(4/x)+(1/x^2)$ for $x>0$
The minimum value of given function $f(x)= x^2+4x+(4/x)+(1/x^2)$ where $x>0$
(A) 9.5
(B) 10
(C) 15
(D) 20
My try
By A.M G.M inequality
$\frac {x^2+4x+(4/x)+(1/x^2)}{4} \geq \left(x^2.4x. \frac {4}{x}. \frac{1}{x^2}\right)^\frac{1}{4} $
thus minimum value of $f(x)=8$
But by calculus approach the answer is 10.
I am unable to find my mistake where am i doing wrong in A.M. G.M. inequality
|
AM-GM works but a bit differently:
$$f(x)= x^2+4x+(4/x)+(1/x^2) = x^2+\frac{1}{x^2} + 4(x+\frac{1}{x}) \geq 2 + 4\cdot 2 = 10$$
|
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|
$\lim_{n\to\infty}\frac{n -\big\lfloor\frac{n}{2}\big\rfloor+\big\lfloor\frac{n}{3}\big\rfloor-\dots}{n}$, a Brilliant problem I encounter a question when visiting Brilliant:
Find
$\space\space\space\space\lim_{n\to\infty}s_n$
$=\lim_{n\to\infty}\frac{n - \big \lfloor \frac{n}{2} \big \rfloor+ \big \lfloor \frac{n}{3} \big \rfloor - \big \lfloor \frac{n}{4} \big \rfloor + \dots}{n}$
$=\lim_{n\to\infty}\frac{\sum_{k=1}^n(-1)^{k+1}\lfloor\frac{n}{k}\rfloor}{n}$
The answer in the website above doesn't really satisfies me, as the answer does not tell how the sequence converge and I doesn't understand how we can take subsequence $n_k=k!$ to solve the problem.
I had some idea but doesn't seems to work:
1) It is easy to show $$s_n=\frac{\sum_{k=1}^{\big\lceil\frac{n}{2}\big\rceil}(\big\lfloor\frac{n}{2k-1}\big\rfloor-\big\lfloor\frac{n}{2k}\big\rfloor)}{n}$$
2) On the other hand,
$$s_n\approx\sum_{k=1}^n(-1)^{k+1}\frac{1}{k}\to\ln2$$
So I am wondering how $s_n\approx$ the alternate hamonic series
$$\forall(n,k\in\mathbb N:n\ge k),\space\space\frac{n}{k}\in\Bigg[\bigg\lfloor\frac{n}{k}\bigg\rfloor,\bigg\lfloor\frac{n}{k}\bigg\rfloor+\bigg(\frac{k-1}{k}\bigg)\Bigg]$$
I tried to look at the graph, the sequence $s_n$ is very likely to converge to $\ln 2$, and the alternating harmonic series seems to be bounded by the graph of $s_n$ at most of the time.
3) Also I observed that $s_8=\frac{8-4+2-2+1-1+1-1}{8}=\frac{1}{2}$, the terms cancelled nicely, but I am afraid that the anlalogue is not generaly true for all $s_{2^k}$.
4) I have tried to use Stolz–Cesàro Theorem, but doesn't seems useful neither.
5) I know that $\forall x,y\in\mathbb R:x+y\in\mathbb Z, \lfloor x\rfloor+\lceil y\rceil=x+y$, which maybe is useful since we may thus write $s_n$ in a more beautiful manner?
6) If there is no $(-1)^{k+1}$, I think we can treat $s_n$ as a Riemann sum, but well, ... , seems useless.
7) I have tried to think about how many terms of summand of $ns_n$ is integer.
8) I have tried to think $\big\lfloor\frac{n}{k}\big\rfloor$ as the number of positive integer multiple of $k$ that $\lt n$, and I then considered sets of number that is counted and uncounted respectively, but well, the question doesn't seems that easy.
Does this help?
(1)
(2)
(3)
Any help will be appreciate. Thank you!
Remarks: I was wondering is there a deep subject studying this (if so references please). Can this (or variants) be represented as a simpler function?
|
Elementary high school approach
One can show immediately with the squeeze theorem that
$$L_1=\lim_{n\to\infty}\left(\sum_{k=1}^{2\left\lfloor \sqrt{n} \right\rfloor}\frac{1}{2n}\left\lfloor\frac{2n}{k}\right\rfloor-\sum_{k=1}^{\left\lfloor \sqrt{n} \right\rfloor}\frac{1}{n}\left\lfloor\frac{n}{k}\right\rfloor\right)=\log(2),$$
using the simple fact that $\lim_{n\to\infty}(H_{2n}-H_n)=\log(2)$.
But guess what! Your limit is
$$\lim_{n\to\infty}\left(\sum_{k=1}^{2n}\frac{1}{2n}\left\lfloor\frac{2n}{k}\right\rfloor-\sum_{k=1}^{n}\frac{1}{n}\left\lfloor\frac{n}{k}\right\rfloor\right)=L_1,$$
and therefore you can use the first limit to calculate your initial limit and then show that the sum of the remaining terms tends to $0$, which is straightforward.
Adding some steps requested by OP
WLOG, for the comfort of calculations, we can replace in the initial limit $n$ by $2n$, and using the double inequality with floor function $x\ge\left\lfloor x\right\rfloor\ge x-1$, we have
$$H_{2\left\lfloor\sqrt{n}\right\rfloor}\ge\sum_{k=1}^{2\left\lfloor\sqrt{n}\right\rfloor}\frac{1}{2n}\left\lfloor\frac{2n}{k}\right\rfloor\ge H_{2\left\lfloor\sqrt{n}\right\rfloor}-\frac{\left\lfloor\sqrt{n}\right\rfloor}{n}$$
$$-H_{\left\lfloor\sqrt{n}\right\rfloor}+\frac{\left\lfloor\sqrt{n}\right\rfloor}{n}\ge-\sum_{k=1}^{\left\lfloor\sqrt{n}\right\rfloor}\frac{1}{n}\left\lfloor\frac{n}{k}\right\rfloor\ge-H_{\left\lfloor\sqrt{n}\right\rfloor},$$
that give
$$H_{2\left\lfloor\sqrt{n}\right\rfloor}-H_{\left\lfloor\sqrt{n}\right\rfloor}+\frac{\left\lfloor\sqrt{n}\right\rfloor}{n}\ge\sum_{k=1}^{2\left\lfloor\sqrt{n}\right\rfloor}\frac{1}{2n}\left\lfloor\frac{2n}{k}\right\rfloor-\sum_{k=1}^{\left\lfloor\sqrt{n}\right\rfloor}\frac{1}{n}\left\lfloor\frac{n}{k}\right\rfloor\ge H_{2\left\lfloor\sqrt{n}\right\rfloor}-H_{\left\lfloor\sqrt{n}\right\rfloor}-\frac{\left\lfloor\sqrt{n}\right\rfloor}{n}.$$
Letting $n\to\infty$ you get the value of the limit $L_1$. If denoting $\left\lfloor\sqrt{n}\right\rfloor=m$, we have the type of limit mentioned above, $\lim_{m\to\infty}(H_{2m}-H_m)=\log(2)$.
Your limit, after replacing $n$ by $2n$, is
$$\lim_{n\to\infty}\sum_{k=1}^{2n}(-1)^{k-1}\frac{1}{2n}\left\lfloor\frac{2n}{k}\right\rfloor=\lim_{n\to\infty}\left(\sum_{k=1}^{2n}\frac{1}{2n}\left\lfloor\frac{2n}{k}\right\rfloor-2\sum_{k=1}^{n}\frac{1}{2n}\left\lfloor\frac{2n}{2k}\right\rfloor\right)$$
$$=\lim_{n\to\infty}\left(\sum_{k=1}^{2n}\frac{1}{2n}\left\lfloor\frac{2n}{k}\right\rfloor-\sum_{k=1}^{n}\frac{1}{n}\left\lfloor\frac{n}{k}\right\rfloor\right).$$
What's left to do? Using in the last limit that limit $L_1$ calculated above.
$$\lim_{n\to\infty}\left(\sum_{k=1}^{2n}\frac{1}{2n}\left\lfloor\frac{2n}{k}\right\rfloor-\sum_{k=1}^{n}\frac{1}{n}\left\lfloor\frac{n}{k}\right\rfloor\right)$$
$$=\lim_{n\to\infty}\left(\sum_{k=1}^{2\left\lfloor \sqrt{n} \right\rfloor}\frac{1}{2n}\left\lfloor\frac{2n}{k}\right\rfloor+\sum_{k=2\left\lfloor \sqrt{n} \right\rfloor+1}^{2n}\frac{1}{2n}\left\lfloor\frac{2n}{k}\right\rfloor-\sum_{k=1}^{\left\lfloor \sqrt{n} \right\rfloor}\frac{1}{n}\left\lfloor\frac{n}{k}\right\rfloor-\sum_{k=\left\lfloor \sqrt{n} \right\rfloor+1}^n\frac{1}{n}\left\lfloor\frac{n}{k}\right\rfloor\right)$$
$$=\underbrace{\lim_{n\to\infty}\left(\sum_{k=1}^{2\left\lfloor \sqrt{n} \right\rfloor}\frac{1}{2n}\left\lfloor\frac{2n}{k}\right\rfloor-\sum_{k=1}^{\left\lfloor \sqrt{n} \right\rfloor}\frac{1}{n}\left\lfloor\frac{n}{k}\right\rfloor\right)}_{\displaystyle \log(2)}+\underbrace{\lim_{n\to\infty}\left(\sum_{k=2\left\lfloor \sqrt{n} \right\rfloor+1}^{2n}\frac{1}{2n}\left\lfloor\frac{2n}{k}\right\rfloor-\sum_{k=\left\lfloor \sqrt{n} \right\rfloor+1}^n\frac{1}{n}\left\lfloor\frac{n}{k}\right\rfloor\right)}_{\displaystyle 0}$$
$$=\log(2),$$
and the limit tending to $0$ can be done by arranging the sums under the limit under the form of an alternating sum and then squeezing the sum.
Further explanations on the limit tending to $0$
It's clear that
$$0\le\sum_{k=2\left\lfloor \sqrt{n} \right\rfloor+1}^{2n}\frac{1}{2n}\left\lfloor\frac{2n}{k}\right\rfloor-\sum_{k=\left\lfloor \sqrt{n} \right\rfloor+1}^n\frac{1}{n}\left\lfloor\frac{n}{k}\right\rfloor=\frac{1}{2n}\sum_{k=\underbrace{2\left\lfloor \sqrt{n} \right\rfloor+1}_{m}}^{2n}(-1)^{k-1}\left\lfloor\frac{2n}{k}\right\rfloor$$
$$=\frac{1}{2n}\left(\left\lfloor\frac{2n}{m}\right\rfloor-\left\lfloor\frac{2n}{m+1}\right\rfloor+\left\lfloor\frac{2n}{m+2}\right\rfloor-\left\lfloor\frac{2n}{m+3}\right\rfloor+\left\lfloor\frac{2n}{m+4}\right\rfloor-\left\lfloor\frac{2n}{m+5}\right\rfloor+\left\lfloor\frac{2n}{m+6}\right\rfloor-\cdots\right)$$
$$\le\frac{1}{2n}\left(\left\lfloor\frac{2n}{m}\right\rfloor-\left\lfloor\frac{2n}{m+1}\right\rfloor+\left\lfloor\frac{2n}{m+1}\right\rfloor-\left\lfloor\frac{2n}{m+3}\right\rfloor+\left\lfloor\frac{2n}{m+3}\right\rfloor-\left\lfloor\frac{2n}{m+5}\right\rfloor+\left\lfloor\frac{2n}{m+5}\right\rfloor-\cdots\right)$$
$$=\frac{1}{2n}\left(\left\lfloor\frac{2n}{m}\right\rfloor-1\right)=\frac{1}{2n}\left(\left\lfloor\frac{2n}{2\left\lfloor \sqrt{n} \right\rfloor+1}\right\rfloor-1\right),$$
where letting $n\to\infty$, we obviously get $0$ which accounts for the limit tending to $0$ in my solution.
A final remark
One should have dealt from the very beginning with this part of the limit tending to $0$ since we had it in the form with the alternating sum as expected.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Finding line of intersection between two planes by solving a system of equations Ok, so the question is to find the line of intersection between two planes, given their equations.
$x+3y+2z=4$
$x-y-z=4$
I know there's the way of using the vector perpendicular to both normals of the planes as the direction vector of the line of intersection and then finding a specific point on the line.
But I wanted to find the line of intersection by solving the system- so taking one variable as a parameter and putting the two other variables in terms of that variable. So I just randomly chose x to act as the parameter:
$3y+2z=4-x$
$-y-z=4-x$
$y=x+4-z$
and then I substituted this y to $3y+2z=4-x$
$3x+12-3z+2z=4-x$
$z=8+4x$
then
$y=x+4-8-4x$
$y=-4-3x$
and I put x=t and so
$x=t, y=-4-3t, z=8+4t$
But this doesn't correspond with the answer which is $x=4+t, y=-3t, z=4t$. And I think it's because I took the wrong variable as the parameter..? Is there like a set rule for choosing which variable to be the parameter?
|
Having
$$
a_1x+b_1y+c_1z= d_1\\
a_2x+b_2y+c_2z=d_2
$$
or
$$
\left[
\begin{array}{ccc}
a_1 & b_1 & c_1\\
a_2&b_2 & c_2
\end{array}
\right]
\left[
\begin{array}{c}
x\\
y\\
z
\end{array}
\right]
=
\left[
\begin{array}{c}
d_1\\
d_2
\end{array}
\right]
$$
Now choosing an invertible $2\times 2$ submatrix in $\left[
\begin{array}{ccc}
a_1 & b_1 & c_1\\
a_2&b_2 & c_2
\end{array}
\right]$
for instance $\left[
\begin{array}{cc}
a_1 & c_1\\
a_2 & c_2
\end{array}
\right]$
we rearrange the system as
$$
\left[
\begin{array}{cc}
a_1 & c_1\\
a_2 & c_2
\end{array}
\right]
\left[
\begin{array}{c}
x\\
z
\end{array}
\right]
=
\left[
\begin{array}{c}
d_1\\
d_2
\end{array}
\right]
-y
\left[
\begin{array}{c}
b_1\\
b_2
\end{array}
\right]
$$
and then
$$
\left[
\begin{array}{c}
x\\
z
\end{array}
\right]
=
\left[
\begin{array}{cc}
a_1 & c_1\\
a_2 & c_2
\end{array}
\right]^{-1}
\left(
\left[
\begin{array}{c}
d_1\\
d_2
\end{array}
\right]
-t
\left[
\begin{array}{c}
b_1\\
b_2
\end{array}
\right]
\right)\\
y = t
$$
|
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|
Given linear mapping and bases, determine the transformation matrix and the change of basis
Given is linear mapping $f:\mathbb{R}^3 \rightarrow \mathbb{R}^3$. Matrix of $f$ in terms of the ordered standard basis $B_0^3 = \left\{\vec{e_1}, \vec{e_2}, \vec{e_3}\right\}$ of $\mathbb{R}^3$ is $\,\,\,\,
A_{B_0^3 B_0^3}^{f} = \begin{pmatrix}
4 & 0 & -2\\
1 & 3 & -2\\
1 & 2 & -1
\end{pmatrix} \in \mathbb{R}^{3 \times 3}$, moreover
$B=\left\{\vec{b_1}, \vec{b_2}, \vec{b_3}\right\} = \left\{\begin{pmatrix}
2\\
2\\
3
\end{pmatrix};
\begin{pmatrix}
1\\
1\\
1
\end{pmatrix};
\begin{pmatrix}
2\\
1\\
1
\end{pmatrix}\right\}$ is an ordered basis of $\mathbb{R}^3$.
Determine the transformation matrix $A_{BB}^f$ and the change of basis $T_{B_0^3}^{B}$.
I try to calculate the transformation matrix $A_{BB}^f$ first. So I think because we have $A_{B_0^3 B_0^3}^{f} = \begin{pmatrix}
4 & 0 & -2\\
1 & 3 & -2\\
1 & 2 & -1
\end{pmatrix}$, where $B_0^3$ is standard basis, we have that
$A^f = \begin{pmatrix}
4 & 0 & -2\\
1 & 3 & -2\\
1 & 2 & -1
\end{pmatrix}$ because standard basis shouldn't have any changes on it.
But for calculating $A^f_{BB}$, I need to know the mapping, but I just know $\mathbb{R}^3 \rightarrow \mathbb{R}^3$.. I don't know the precise mapping from matrix to another matrix. I need that so I can insert the given basis $B$ and determine the transformation matrix like that. But how is this supposed to work here? :S
And about the change of basis I really have no idea because it requires the previous thing I mentioned : /
|
The change of basis matrix $T_B^{B_0^3}$ is simply$$\begin{pmatrix}2&1&2\\2&1&1\\3&1&1\end{pmatrix}$$and $T_{B_0^3}^B$ is$$\begin{pmatrix}2&1&2\\2&1&1\\3&1&1\end{pmatrix}^{-1}=\begin{pmatrix}0 & -1 & 1 \\ -1 & 4 & -2 \\ 1 & -1 & 0\end{pmatrix}.$$So$$A_{BB}^f=T^B_{B_0^3}A_{B_0^3B_0^3}^fT^{B_0^3}_B=\begin{pmatrix}1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3\end{pmatrix}.$$
|
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|
Suppose $x, y, z$ are positive real number such that $x + 2y + 3z = 1$. Find the maximum value of $xyz^2$ I'm trying to solve this problem using Lagrange's multiplier method but I'm unable to get the value of lambda and also $z.$ kindly help me out with this problem.
I'm getting $x = 2y$
I'm stuck somewhere between the steps
|
Alt. hint (no calculus): the following inequality holds by AM-GM, with equality iff $\displaystyle x = 2y = \frac{3}{2}z\,$:
$$
\frac{\sqrt{3}}{\sqrt[4]{2}} \cdot \sqrt[4]{xyz^2} = \sqrt[4]{x \cdot 2y \cdot \frac{3}{2}z \cdot \frac{3}{2}z} \;\le\; \frac{x+2y+3z}{4} = \frac{1}{4} \;\;\iff\;\;xyz^2 \le \frac{2}{3^2 \cdot 4^4} = \frac{1}{1152}
$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
How do I solve the inequality $x^{12}−x^9+x^4−x+1>0$ using intervals? My book has the question $x^{12}−x^{9}+x^{4}−x+1>0$. The solution given gives three cases, when $x \le 0$, when $0 < x \le 1$ and when $x > 1$. How did they get these intervals? What is the method used?
|
Clearly for $x=0$, We have $$x^{12}-x^{9}+x^4-x+1>0$$
Now for $x\neq 0$, Using Arithmetic Geometric Inequality
$$\frac{x^{12}+x^{12}+x^{12}+1}{4}\geq \sqrt[4]{x^{12}\cdot x^{12}\cdot x^{12}\cdot 1}=|x^9|>x^9$$
$$\frac{x^4+1+1+1}{4}\geq \sqrt[4]{x^4\cdot 1 \cdot 1\cdot 1}=|x|>x$$
So $\displaystyle \frac{3x^{12}}{4}-x^9+\frac{x^4}{4}-x+\frac{1}{4}+\frac{3}{4}>0$(Equality hold when $x=1$)
So $$ x^{12}-x^9+x^4-x+1>\frac{3x^{12}}{4}-x^9+\frac{x^4}{4}-x+1>0\forall x\in\mathbb{R}.$$
|
{
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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|
Simplifying the derivative of $x^{\frac{2}{3}} \cdot (6-x)^{\frac{1}{3}}$ $$x^{\frac{2}{3}} \cdot (6-x)^{\frac{1}{3}}$$
So I get:
$$-x^{\frac{2}{3}} \cdot \frac{1}{3} (6-x) ^{\frac{-2}{3}} + (6-x) ^{\frac{1}{3}} \cdot \frac{2}{3} x ^ {\frac{-1}{3}}$$
How does one go about simplifying this?
I guess I can pull out common terms like this:
$$\frac{1}{3} x ^{-\frac{1}{3}} (6-x)^{\frac{-2}{3}} ( -x + (6-x) \cdot 2)$$
Is that right?
|
Pulling out this common factor is fine. We can simplify the last expression slightly more and obtain
\begin{align*}
\color{blue}{-x^{\frac{2}{3}}}&\color{blue}{ \cdot \frac{1}{3} (6-x) ^{-\frac{2}{3}} + (6-x) ^{\frac{1}{3}} \cdot \frac{2}{3} x ^ {-\frac{1}{3}}}\\
&=\frac{1}{3}x^{-\frac{1}{3}}(6-x)^{-\frac{2}{3}}(-x+(6-x)2)\\
&=\frac{1}{3}x^{-\frac{1}{3}}(6-x)^{-\frac{2}{3}}(12-3x)\\
&\,\,\color{blue}{=x^{-\frac{1}{3}}(6-x)^{-\frac{2}{3}}(4-x)}
\end{align*}
|
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"timestamp": "2023-03-29T00:00:00",
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|
Calculate $\sum\limits_{k=0}^{20}(-1)^k\binom{k+2}2$ without calculating each term separately Is it possible to calculate $\sum_{k=0}^{20}(-1)^k\binom{k+2}2$ without calculating each term separately?
The original question was find the number of solutions to $2x+y+z=20$ which I calculated to be the coefficient of $x^{20}$ in $(1+x^2+x^4\dots)(1+x+x^2\dots)^2$ which simplified to the term above.
I know $\sum_{k=0}^{20}\binom{k+2}{2}=\binom{23}3$ but the $(-1)^k$ is ruining things.
|
Alternatively:
$$\begin{align}\sum_{k=0}^{20}(-1)^k\binom{k+2}2=&\sum_{k=0}^{20}\binom{k+2}2-2\cdot \sum_{k=0}^{9}\binom{2k+3}2=\\
&{23\choose 3}-\sum_{k=0}^9 (2k+3)(k+1)=\\
&1771-2\cdot 825=121.\end{align}$$
|
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"language": "en",
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|
Determinant of a matrix of ones, whose anti diagonal elements are zero I'm trying to prove a formula I have constructed for the determinant of a general $n\times n $ real matrix $A$, given here in the case $n=5$:
$$ A = \begin{bmatrix}
1 & 1 & 1 & 1 & 0 \\
1 & 1 & 1 & 0 & 1 \\
1 & 1 & 0 & 1 & 1 \\
1 & 0 & 1 & 1 & 1 \\
0 & 1 & 1 & 1 & 1
\end{bmatrix}.
$$
That is, the matrix containing all 1's apart from the anti-diagonal which consists of zeros.
Using a simple matlab code I've computed the determinants for the first few values of $n$, and have come to the formula
$$ \det A =
\begin{cases}
-(n-1) \hspace{1em}\mbox{ if }\,\,\, n \equiv -1,0\mod 4, \\
n-1 \hspace{2.2em}\mbox{ otherwise,}
\end{cases}
$$
but I'm unsure where to start to prove this.
|
This is easy to calculate by row reduction:
Add all rows to row 1.
$$\det(A) = \begin{vmatrix}
1 & 1 & ... & 1 & 0 \\
1 & 1 & ... & 0 & 1 \\
... & ... & ... & ... & ... \\
1 & 0 & ... & 1 & 1 \\
0 & 1 & ... & 1 & 1
\end{vmatrix}=\begin{vmatrix}
n-1 & n-1 & ... & n-1 & n-1 \\
1 & 1 & ... & 0 & 1 \\
... & ... & ... & ... & ... \\
1 & 0 & ... & 1 & 1 \\
0 & 1 & ... & 1 & 1
\end{vmatrix}\\=(n-1)\begin{vmatrix}
1 & 1 & ... & 1 & 1 \\
1 & 1 & ... & 0 & 1 \\
... & ... & ... & ... & ... \\
1 & 0 & ... & 1 & 1 \\
0 & 1 & ... & 1 & 1
\end{vmatrix}=(n-1)\begin{vmatrix}
1 & 1 & ... & 1 & 1 \\
0 & 0 & ... & -1 & 0 \\
... & ... & ... & ... & ... \\
0 & -1 & ... & 0 & 0 \\
-1 & 0 & ... & 0 & 0
\end{vmatrix}$$
where in the last row we subtracted row 1 from all rows. Now add again all rows to row 1:
$$\det(A)=(n-1)\begin{vmatrix}
0 & 0 & ... & 0 & 1 \\
0 & 0 & ... & -1 & 0 \\
... & ... & ... & ... & ... \\
0 & -1 & ... & 0 & 0 \\
-1 & 0 & ... & 0 & 0
\end{vmatrix}$$
The last determinant is easy now to calculate.
|
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|
Condition for rank to be 2. Suppose that
$$\mbox{rank}\begin{equation}
\begin{pmatrix}
1 & 1 & 2 & 2\\
1 & 1 & 1 & 3\\
a & b & b & 1
\end{pmatrix}
\end{equation} = 2$$
for some real numbers $a$ and $b$. What is the value of $b$?
*
*$0.3$
*$3$
*$1$
*$0.5$
I'm unable to get the condition on $b$ only.
|
$$\begin{equation}
\begin{pmatrix}
1 & 1 & 2 & 2\\
1 & 1 & 1 & 3\\
1/3 & 1/3 & 1/3 & 1
\end{pmatrix}
\end{equation}$$
has rank $2$ thus the closest answer is $ b=0.3$
|
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|
Evaluate $\int \frac {dx}{\sin \frac x2\sqrt {\cos^3 \frac x2}}$
Evaluate $$\int \frac {dx}{\sin \frac x2\sqrt {\cos^3 \frac x2}}$$
My try
Write $t=\frac x2$ and hence $dx=2dt$
To change the integral to $$\int \frac {\csc t dt}{\cos^{\frac 32} t}$$
Multiplying both bottom and top by $\csc t$ and then using $\csc^2 t=1+\cot^2 t$ in the numerator the problem simplifies to $$2\int (\sin t)(\cos ^{\frac {-3}{2}} t) dt+2\int \frac {\cot^3 t dt}{\sqrt {\cos t}}$$
Now the first integral is easy to go but I am not getting any idea for the second one. Any help would be very beneficial. New methods are also welcome.
|
Elaborating on answer of @JoseCarlosSantos:
By performing the substitution $t=u^2$,
$$\int\frac{dt}{(1-t^2)t\sqrt t}=\int\frac{2udu}{(1-u^4)(u^3)}=2\int\frac{du}{(1-u^4)u^2}$$
Performing partial fraction decomposition,
$$\frac1{(1-u^4)u^2}$$
$$=\frac1{u^2}+\frac{u^2}{1-u^4}$$
$$=\frac1{u^2}+\frac{u^2}{2(1+u^2)}+\frac{u^2}{2(1-u^2)}$$
$$= \frac1{u^2}+\frac{u^2+1-1}{2(1+u^2)}+\frac{u^2-1+1}{2(1-u^2)}$$
$$=\frac1{u^2}+\frac12-\frac{1}{2(1+u^2)}-\frac12+\frac{1}{2(1-u^2)}$$
$$= \frac1{u^2}-\frac{1}{2(1+u^2)}+\frac{1}{2(1-u^2)} $$
$$= \frac1{u^2}-\frac{1}{2(1+u^2)}+\frac{1}{4(1-u)}+\frac{1}{4(1+u)} $$
which can be integrated easily.
|
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|
Show that $\binom{2n}{n}\geq2^n$ for all $n\in\mathbb{N}_0$. Show that $\binom{2n}{n}\geq2^n$ for all $n\in\mathbb{N}_0$.
First, observe that $\binom{2n}{n} = \frac{(2n)!}{n!(2n-n)!}=\frac{(2n)!}{(n!)^2}=\frac{1\cdot 2\cdot 3\cdot...\cdot 2n-1\cdot 2n}{(n!)^2}=\frac{(1\cdot 3\cdot 5\cdot ... \cdot 2n-1)\cdot(2\cdot 4\cdot 6\cdot ... \cdot 2n)}{(n!)^2}=\frac{(1\cdot 3\cdot 5\cdot ... \cdot 2n-1)\cdot(2\cdot 1\cdot 2\cdot 2\cdot 2\cdot 3\cdot ... \cdot 2n)}{(n!)^2}=\frac{(1\cdot 3\cdot 5\cdot ... \cdot 2n-1)\cdot 2^n\cdot n!}{(n!)^2}=2^n\cdot\frac{1\cdot 3\cdot 5\cdot ... \cdot 2n-1}{(n!)}.$
For $n=0$ we have:
$\frac{(2*0)!}{(0!)^2}=1=2^0.$
Now to our inductive step $n+1$:
\begin{align*}
& \frac{(2(n+1))!}{((n+1)!)^2}\geq2^{n+1}\\
\Leftrightarrow\,\, & \frac{(2n+2)!}{((n+1)!)^2}\geq2^{n+1}\\
\Leftrightarrow\,\, & \frac{1\cdot 2\cdot 3\cdot...\cdot 2n+1 \cdot 2n+2}{((n+1)!)^2}\geq2^{n+1}\\
\Leftrightarrow\,\, & \frac{(1\cdot 3\cdot 5\cdot ... \cdot 2n+1)\cdot(2\cdot 4\cdot 6\cdot ... \cdot 2n \cdot 2n+2)}{((n+1)!)^2}\geq2^{n+1}\\
\Leftrightarrow\,\, & \frac{(1\cdot 3\cdot 5\cdot ... \cdot 2n+1)\cdot 2^{n+1} \cdot (n+1)!}{((n+1)!)^2}\geq2^{n+1}\\
\Leftrightarrow\,\, & 2^{n+1}\cdot \frac{(1\cdot 3\cdot 5\cdot ... \cdot 2n+1)}{((n+1)!)}\geq2^{n+1}\\
\end{align*}
Is this proof correct/sufficient? Was induction even needed here or should I have argued instead that $\frac{1\cdot 3\cdot 5\cdot ... \cdot 2n-1}{(n!)}>0$? Is this step necessary even for the induction?
|
Another approach: Consider a line of $2n$ people, and we want to put them in two teams of $n$ people say $\color{red}{red}$ and $\color{blue}{blue}$ then what if you restrict to the first $n$ people that you put in the line, they can be whatever because regardless of the choice at most you picked $n$ of them in one team and so $2^n\leq \binom{2n}{n}.$
Example: Say $n = 2,$ then the possibilities are:
$$\underbrace{\color{red}{**}}_{\text{first 2}}\color{blue}{**},\hspace{1mm} \underbrace{\color{red}{*}\color{blue}{*}}_{\text{first 2}}\color{blue}{*}\color{red}{*},\hspace{1mm}\underbrace{\color{red}{*}\color{blue}{*}}_{\text{first 2}}\color{red}{*}\color{blue}{*},\hspace{1mm}\underbrace{\color{blue}{*}\color{red}{*}}_{\text{first 2}}\color{red}{*}\color{blue}{*},\hspace{1mm}\underbrace{\color{blue}{*}\color{red}{*}}_{\text{first 2}}\color{blue}{*}\color{red}{*},\hspace{1mm}\underbrace{\color{blue}{*}\color{blue}{*}}_{\text{first 2}}\color{red}{*}\color{red}{*},\hspace{1mm}.$$
Notice that restricting to the first $2$ we get all possibilities of red and blue.
|
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|
Determinant of $\left(\begin{smallmatrix} -2a &a+b &a+c \\ b+a& -2b &b+c \\ c+a&c+b & -2c \end{smallmatrix}\right)$ Evaluate $$D=\begin{vmatrix}
-2a &a+b &a+c \\
b+a& -2b &b+c \\
c+a&c+b & -2c
\end{vmatrix}$$
My try:
Applying $R_1 \to R_1+R_2$ we get
$$D=\begin{vmatrix}
b-a&a-b &a+b+2c \\
b+a& -2b &b+c \\
c+a&c+b & -2c
\end{vmatrix}$$
Now apply $$C_1 \to C_1+C_2$$
$$D=\begin{vmatrix}
0&a-b &a+b+2c \\
a-b& -2b &b+c \\
2c+a+b&c+b & -2c
\end{vmatrix}$$
Now apply $C_2 \to C_2 +C_3$
$$D= \begin{vmatrix}
0&2a+2c &a+b+2c \\
a-b& c-b &b+c \\
2c+a+b&b-c & -2c
\end{vmatrix}$$
Now use $R_3 \to R_3+R_2$
$$D= \begin{vmatrix}
0&2a+2c &a+b+2c \\
a-b& c-b &b+c \\
2c+2a&0 & b-c
\end{vmatrix}$$
any way to proceed here using elementary operations?
|
Let $p(a,b,c) $ be the determinant. Note that each term of $p$ has degree 3 (sum of degrees of $a,b,c$).
Note that $p(a,-a,c) = 0$, hence $a+b$ divides $p$.
Similarly we see that $a+c, b+c$ divide $p$.
Hence $p$ has the form $p(a,b,c) = k (a+b)(b+c)(a+c)$ for some constant $k$.
Compute the determinant for $a=b=c={1\over 2}$ to get $k = 4$.
|
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|
Line-integral: 2-form integrated over cube
Let $$\omega=x\cos(xy)\cos(2\pi x)\ \ \ \text{d}x\wedge\text{d}y$$ Calculate $$\int_{[0,\frac{1}{4}]\times[0,2\pi]}\omega$$
Now we have:
\begin{align} \int_{[0,\frac{1}{4}]\times[0,2\pi]}x\cos(xy)\cos(2\pi x)\ \text{d}x\wedge\text{d}y & = \int_0^{\frac{1}{4}}\Big(\int_0^{2\pi}x\cos(xy)\cos(2\pi x) \text{d}y\Big)\ \text{d}x\\
&= \int_0^{\frac{1}{4}}\cos(2\pi x)\sin(2\pi x)\\
&=\frac{1}{4}\pi
\end{align}
I am not sure if this is correct, could anyone have a look at it and spot any mistakes? I would appreciate it a lot!
|
It should be:
$$
\begin{align}
\int\limits_0^\frac{1}{4} \cos(2\pi x)\sin(2\pi x) \mathrm{d}x &= \int\limits_0^\frac{1}{4}\frac{1}{2}\sin(4\pi x)\mathrm{d}x \\
&= \frac{1}{2}\left[\frac{-1}{4\pi}\cos(4\pi x)\right]_0^\frac{1}{4} \\
&= \frac{1}{2}\cdot\frac{1}{4\pi}(1 - (-1)) \\
&= \frac{1}{4\pi}
\end{align}
$$.
|
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|
This function has two different integrals? $f(x)=∫\frac{1}{x^2}dx$
Integrating by u-substitution:
$u=x^2$
$du=2dx$
$\frac{1}{2}du = dx$
$∫\frac{1}{x^2}dx=$ $∫\frac{1}{u}\times\frac{1}{2}du$
$\frac{1}{2}$∫ $\frac{1}{u}du$
$=\frac{1}{2}ln u+c$
$=\frac{1}{2}ln x^2+c$
$=lnx+c$
Another way:
$∫\frac{1}{x^2}dx=∫x^{-2}dx $
$∫x^{-2}dx$
$=\frac{x^{-1}}{-1} + c$
$=-\frac{1}{x} + c$
Where have I gone wrong?
|
If $u=x^2$, then $\mathrm du=2x\,\mathrm dx$.
|
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|
How I find $\gcd(\frac{a^{2m+1}+1}{a+1}, a+1)$? $$\gcd\left(\frac{a^{2m+1}+1}{a+1}, a+1\right)$$
My answer until this moment is:
$$
For: a^{2m+1}+1^{2m+1} = (a + 1)^{m+1} - 2a = (a+1)^m(a+1)-2a
$$ where
\begin{align}
(a + 1)^{m} = a^{2m}+a^{2m-1}+\cdots+1
\end{align}
So,
\begin{align}
a^{2m+1}+1 = [a^{2m}+a^{2m-1}+\cdots+1](a+1)-2a
\end{align}
But, I don't know how to divide :
\begin{align}
\frac{[a^{2m}+a^{2m-1}+\cdots+1](a+1)-2a}{a+1}
\end{align}
|
Here is one approach. Let $b = a + 1$ so $a = b - 1$. Then, we can apply the binomial theorem to obtain the modular equivalence
$$\begin{align*} \frac{a^{2m + 1} + 1}{a + 1} & = \frac{(b - 1)^{2m + 1} + 1}{b}\\
& = \frac{1}{b} \left(1 + \sum_{k = 0}^{2m + 1} \binom{2m + 1}{k} b^{k}(-1)^{2m + 1 - k} \right)\\
& = \frac{1}{b} \left(1 + (-1) + \sum_{k = 1}^{2m + 1} \binom{2m + 1}{k} b^{k}(-1)^{2m + 1 - k}\right) \\
& = \sum_{k = 1}^{2m + 1} \binom{2m + 1}{k} b^{k - 1}(-1)^{2m + 1 - k} \\
& = \binom{2m + 1}{1}(-1)^{2m} + \sum_{k = 2}^{2m + 1} \binom{2m + 1}{k} b^{k - 1}(-1)^{2m + 1 - k} \\
& \equiv 2m + 1 \pmod{b}.\end{align*}$$
Thus, we have $$\frac{a^{2m + 1} + 1}{a + 1} = p(a + 1) + 2m + 1$$ for an integer $p$. Finally, we can use the fact that $\gcd(x,y) = \gcd(x - ty,y)$ for any integer $t$, to obtain $$\gcd\left( \frac{a^{2m + 1} + 1}{a + 1}, a + 1 \right) = \gcd \left(\frac{a^{2m + 1} + 1}{a + 1} - p(a + 1), a + 1 \right) = \gcd(2m + 1, a + 1).$$
|
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|
Need Help Finding Solution to These 5 Simple Exponent Exercises In this document, I have 5 exercises involving exponents and their five corresponding solutions. I am trying to find out why these are the solutions for these equations, as I cannot seem to get the same results as the provided solutions. The exercises ask to simplify the following equations. Can anyone assist in explaining how to solve for these five equations?
Thank you!
Exponent Equations
http://i349.photobucket.com/albums/q368/Chaplive/Screenshot%2035_zpsx2sjpp5f.png
|
$1)$ Use the law of exponents ie $a^{-n}=\frac1{a^n}$
so $(-3)^4\div3^5=\frac{(-1)^43^4}{3^5}=3^{4-5}=3^{-1}$
$2)$ $\dfrac{2^{-1}+2^{-2}}{2^{-1}\cdot2^{-2}}=\frac{2^{-1}}{2^{-3}}+\frac{2^{-2}}{2^{-3}}= 2^2+2=6$
$3)$ note that $(a\cdot b)^c = a^c\cdot b^c$
$\dfrac{81^4\cdot4^9}{6^{16}}= \dfrac{3^{4\cdot4}\cdot2^{2\cdot9}}{3^{16}\cdot2^{16}}=\dfrac{3^{16}\cdot2^{18}}{3^{16}\cdot2^{16}}= 2^2 =4$
$4)$ remember that $(-1)^{even}= 1$ and $(-1)^{odd}=-1$
$-(-1)^2-(-1^5)-(-1)^7-(-1)^{100} = -(1)-(-1)-(-1)-(1) =-1+1+1-1 = 0$
$5)$ $a^0 =1$ so,
$\dfrac{(5^{-1}+5^{-2})^0}{5^{1}+5^{-2}}= \dfrac{1}{\frac15+\frac1{25}}=\frac{1}{\frac{5+1}{25}}=\dfrac{25}6$
|
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|
Can all natural numbers be expressed as $\lceil \frac{3^a}{2^b} \rceil$ or $\lfloor \frac{3^a}{2^b} \rfloor$? Can any natural number $n$ be expressed as
$n=\lceil \frac{3^a}{2^b} \rceil$ or $n=\lfloor \frac{3^a}{2^b} \rfloor$ where $ a,b \in \mathbb{N}$ ?
I have found no approach to this problem. All suggestions are appeciated.
Examples:
$0=\lfloor \frac{3^0}{2^1} \rfloor$
$1=\lceil \frac{3^0}{2^1} \rceil$
$2=\lceil \frac{3^1}{2^1} \rceil$
$3=\lceil \frac{3^1}{2^0} \rceil$
$4=\lfloor \frac{3^2}{2^1} \rfloor$
$5=\lceil \frac{3^2}{2^1} \rceil$
|
Yes, all natural numbers $n$ can be represented as $n=\left\lceil \frac{3^a}{2^b} \right\rceil$ or as $n=\left\lfloor \frac{3^a}{2^b} \right\rfloor$ (whichever one you want). Let's do floors, for concreteness.
For $n=\left\lfloor \frac{3^a}{2^b} \right\rfloor$ you want positive integers $a$ and $b$ such that $n \le \frac{3^a}{2^b} < n + 1$, or equivalently $\log n \le a \log 3 - b \log 2 < \log(n + 1)$, or in other words
$$\frac{\log n}{\log 2} \le \left(a \frac{\log 3}{\log 2} - b\right) < \frac{\log(n+1)}{\log 2}.$$
That is, if you call the ratio $\frac{\log 3}{\log 2}$ as $\theta$, then you want to find an integer $a$ such that the fractional part of $a\theta$ lies between the fractional part of $\frac{\log n}{\log 2}$ and that of $\frac{\log(n+1)}{\log 2}$.
This problem is known as “inhomogeneous diophantine approximation”. As $\theta$ is irrational, its fractional parts are dense and equidistributed in $[0, 1)$ (this is a theorem), so you can make the fractional part of $a\theta$ arbitrary close to any number you choose. There are many questions on this site about it; two I've answered are at Fractional part of $b \log a$ and at Prefix of Fibonacci number but you can also click on the “Linked” and “Related” on those questions for more.
|
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|
Indefinite integral of $\sqrt{x^2-x}$ i was trying to compute the indefinite integral:
$$
\int\sqrt{x^2-x}dx
$$
but i got stuck: after a few (unsuccessful) attempts for some $u$-substitution, i tried integration by parts:
$$
\int\sqrt{x^2-x} \ dx=\int(x)'\sqrt{x^2-x} \ dx= \\ x\sqrt{x^2-x}-\int x(\sqrt{x^2-x})'dx= \\
=x\sqrt{x^2-x}-\frac{1}{2}\int x\frac{2x-1}{\sqrt{x^2-x}}dx= \\
=x\sqrt{x^2-x}-\int \frac{x^2}{\sqrt{x^2-x}}dx+\frac{1}{2}\int \frac{x}{\sqrt{x^2-x}}dx=...
$$
and now what? Can anybody help?
|
complete the square
$$\begin{align}
\int\sqrt{x^2-x} \ dx &=I \\
x^2-x&=\left(x-\frac{1}{2}\right)^2-\frac{1}{4} \\
\text{set} \quad u&=x-\frac{1}{2} \quad \text{then} \quad dx=du\\
I&=\int\sqrt{u^2-a} \ du \quad \text{such that} \quad a=\frac{1}{4}\\
\end{align}$$
This type of integral is well known. You should now use a trig substitution.
|
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|
Limit of the ratio of two modified Bessel functions Could you help me evaluate the following limit:
$$
\lim_{x\to \infty} \frac{K_1(x)}{K_0(x)}
$$
where $K_\nu$ is the modified Bessel function of the second kind of order $\nu$. Both $K_1$ and $K_0$ approaches $0$ as $x\to\infty$, so we have an indeterminate form. L'Hospital doesn't seem to help because the derivative of $K_\nu$ is $K_{\nu +1}$, and all the orders of the modified Bessel function of the second kind approach $0$ as $x\to\infty$.
Another related limit that I am also struggling is:
$$
\lim_{x\to 0} \frac{K_1(x)}{K_0(x)}
$$
|
For large values of $x$
$$K_1(x)=\sqrt{\frac{\pi }{2}}\,e^{-x}\left(\frac 1 {x^{ 1/2}}+\frac 3 8\frac 1 {x^{ 3/2}}-\frac{15}{128}\frac 1 {x^{5/2}}+O\left(\frac 1 {x^{7/2}}\right)\right)$$
$$K_0(x)=\sqrt{\frac{\pi }{2}}\,e^{-x}\left(\frac 1 {x^{ 1/2}}-\frac 1 8\frac 1 {x^{ 3/2}}+\frac{9}{128}\frac 1 {x^{5/2}}+O\left(\frac 1 {x^{7/2}}\right)\right)$$ making
$$\frac{K_1(x)}{K_0(x)}=1+\frac{1}{2 x}-\frac{1}{8 x^2}+O\left(\frac{1}{x^3}\right)$$
For small values of $x$
$$K_1(x)=\frac{1}{x}+\frac{1}{4} x (2 \log (x)+2 \gamma -1-2 \log (2))+O\left(x^3\right)$$
$$K_0(x)=-\log (x)-\gamma +\log (2)+\frac{1}{4} x^2 (-\log (x)-\gamma +1+\log
(2))+O\left(x^3\right)$$ making
$$\frac{K_1(x)}{K_0(x)}=\frac{-1}{x (\log (x)+\gamma -\log (2))}+O\left(x\right)$$
Edit
For large values of $x$
$$K_n(x)=\sqrt{\frac{\pi }{2}}\,e^{-x}\left(\frac 1 {x^{ 1/2}}+\frac {4n^2-1} 8\frac 1 {x^{ 3/2}}+\frac{16 n^4-40 n^2+9}{128}\frac 1 {x^{5/2}}+O\left(\frac 1 {x^{7/2}}\right)\right)$$ making
$$\frac{K_m(x)}{K_n(x)}=1+\frac{(m-n) (m+n)}{2 x}+\frac{(m-n) (m+n) \left(m^2-n^2-2\right)}{8
x^2}+O\left(\frac{1}{x^3}\right)$$
|
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|
Non negative integer triplets $(x,y,z)$ in $x^2+2y^2+4z^2+5=2(x+y+xy)+4z(y-1)$
Non negative integer triplets $(x,y,z)$ in $x^2+2y^2+4z^2+5=2(x+y+xy)+4z(y-1)$
Try: Writting equation as $$4z^2-4(y-1)z+x^2+2y^2-2(x+y+xy)+5=0$$
Now if equation has real roots. Then $$z=\frac{(y-1)\pm \sqrt{2xy+2x-x^2-y^2-4}}{2}$$
So for integer roots $2xy+2x-x^2-y^2-4=k^2$ , Where $k\in\mathbb{Z}$.
So $$2x-(x+y)^2-4=k^2\Rightarrow (x+y)^2=2x-4-k^2$$
Now i did not understand how can i find non negative integer triplets $(x,y,z)$. Could some help me , Thanks
|
There is no way
$(x+y)^2+k^2=2x-4$ is giving you non negative values for $x,y.$
Now looking on RHS we get that $2x-4$ will give least value at $x=2$.
Also we know that $x^2>mx$ iff $x>m$. ---(1)
Now as $y$ cant be negative and $k^2$ too can't be negative.
From $x^2>2x$ for all $x>2$ which is a contradiction as least value of $(x+y)^2$ is $x^2$ (as $x$ can't be less than $2$ but $y$ can be $0$)
Conclusion:It should be $(x-y)^2$ in last line of your problem
|
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|
Can we say that $\frac{a+b}{c+d} \leq \frac{a}{c} + \frac{b}{d} $ If $a,b,c,d$ are positive real numbers, can we say that $\frac{a+b}{c+d} \leq \frac{a}{c} + \frac{b}{d} $ is always true? If no, can you please give insignts on under which conditions this might be true.
Any references to a similar type of inequalities are also welcome
Thank you,
|
After the latest edit:
Wlog. $\frac ac\le\frac bd$. Then $$\frac ac\le\frac{a+b}{c+d}\le\frac bd<\frac ac+\frac bd$$
Indeed, $\frac ac\le \frac bd$ implies $\frac{bc-ad}{cd}\ge 0$, i.e., $bc-ad\ge 0$.
Then
$$ \frac bd-\frac{a+b}{c+d}=\frac{b(c+d)-d(a+b)}{d(c+d)}=\frac{bc-ad}{d(c+d)}\ge0$$
and
$$ \frac{a+b}{c+d}-\frac ac=\frac{c(a+b)-a(c+d)}{(c+d)a}=\frac{bc-ad}{(c+d)a}
\ge 0$$
|
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|
Factorisation of $4x^4-4x^3+4x^2+2x+1$ with real coefficients I have used many mathematical trick to factorise this polynomial : $4x^4-4x^3+4x^2+2x+1$ with real coeffecients but i didn't succeed because as I see all it's root are complex , and I want if there is any suitable method to factorise it ? and thanks in advanced .
Note: The motivation of this question is to compute some integral
|
Let $$ x = \frac{i t }{ \sqrt 2} $$
Your polynomial is now
$$ t^4 + i \sqrt 2 t^3 - 2 t^2 + i \sqrt 2 t + 1 $$
Divide by $t^2$ and then introduce $w = t + \frac{1}{t}$
dividing by $t^2$
$$ t^2 + i \sqrt 2 t - 2 + \frac{i \sqrt 2}{t} + \frac{1}{t^2} $$
which is
$$ w^2 + i \sqrt 2 w - 4 $$
so the roots are
$$ w = \frac{- i \sqrt 2 \pm \sqrt{14}}{2} $$
We get back to four roots in $t$ with two quadratics,
$$ t^2 + \frac{ i \sqrt 2 \pm \sqrt{14}}{2} t + 1 = 0 $$
Finally we get four $x$ roots in conjugate pairs giving real quadratics as $(x - r)(x - \bar{r})$
|
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|
Rearrangement inequality and minimal value of $\frac{\sin^3x}{\cos x} +\frac{\cos^3x}{\sin x}$ For $x \in \left(0, \dfrac{\pi}{2}\right)$, is the minimum value of $\dfrac{\sin^3x}{\cos x} +\dfrac{\cos^3x}{\sin x} = 1$? So considering ($\dfrac{1}{\cos x}$, $\dfrac{1}{\sin x}$) and ($\sin^3x$, $\cos^3x$), is it right to use the rearrangement inequality and conclude that $\dfrac{\sin^3x}{\cos x} +\dfrac{\cos^3x}{\sin x}$ is more than or equal to $\dfrac{\sin^3x}{\sin x} +\dfrac{\cos^3x}{\cos x}$ which is equal to $1$? Thanks.
|
Yes it is fine, indeed for $x\in(0,\pi/2)$ we have in both cases
*
*$\sin x\ge \cos x \implies \sin^3 x\ge \cos^3 x \implies \frac1{\sin x}\le \frac1{\cos x}$
then
$$\frac{\sin^3 x}{\cos x}+\frac{\cos^3 x}{\sin x}\ge \frac{\sin^3 x}{\sin x}+\frac{\cos^3 x}{\cos x}=1$$
*$\cos x\ge \sin x \implies \cos^3 x\ge \sin^3 x \implies \frac1{\cos x}\le \frac1{\sin x}$
then
$$\frac{\sin^3 x}{\cos x}+\frac{\cos^3 x}{\sin x}\ge \frac{\sin^3 x}{\sin x}+\frac{\cos^3 x}{\cos x}=1$$
and equality holds for $\sin x=\cos x$ that is $x=\pi/4$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Evaluate $\lim\limits_{x \to \infty}\frac{(x+\sqrt{2})^{\sqrt{2}}-(x-\sqrt{2})^{\sqrt{2}}}{x^{\sqrt{2}-1}}$. Problem
Evaluate $$\lim\limits_{x \to \infty}\frac{(x+\sqrt{2})^{\sqrt{2}}-(x-\sqrt{2})^{\sqrt{2}}}{x^{\sqrt{2}-1}}.$$
Solution
Notice that $$\lim\limits_{x \to \infty}\frac{(x+\sqrt{2})^{\sqrt{2}}-(x-\sqrt{2})^{\sqrt{2}}}{x^{\sqrt{2}-1}}=\lim\limits_{x \to \infty}x \cdot \left[\left(1+\frac{\sqrt{2}}{x}\right)^{\sqrt{2}}-\left(1-\frac{\sqrt{2}}{x}\right)^{\sqrt{2}}\right].$$
According to Taylor's Formula $(1+x)^{\alpha}=1+\dfrac{\alpha}{1!}x+\mathcal{O}(x)$, we have $$\left(1+\frac{\sqrt{2}}{x}\right)^{\sqrt{2}}=1+\frac{2}{x}+\mathcal{O}\left(\frac{\sqrt{2}}{x}\right)$$and $$\left(1-\frac{\sqrt{2}}{x}\right)^{\sqrt{2}}=1-\frac{2}{x}+\mathcal{O}\left(\frac{\sqrt{2}}{x}\right).$$Therefore,$$\left(1+\frac{\sqrt{2}}{x}\right)^{\sqrt{2}}-\left(1-\frac{\sqrt{2}}{x}\right)^{\sqrt{2}}=\frac{4}{x}+\mathcal{O}\left(\frac{\sqrt{2}}{x}\right).$$As a result,
\begin{align*}
\lim\limits_{x \to \infty}\frac{(x+\sqrt{2})^{\sqrt{2}}-(x-\sqrt{2})^{\sqrt{2}}}{x^{\sqrt{2}-1}}&=\lim\limits_{x \to \infty}x \cdot \left[\left(1+\frac{\sqrt{2}}{x}\right)^{\sqrt{2}}-\left(1-\frac{\sqrt{2}}{x}\right)^{\sqrt{2}}\right]\\
&=\lim\limits_{x \to \infty}x \cdot \left[
\frac{4}{x}+\mathcal{O}\left(\frac{\sqrt{2}}{x}\right)\right]\\
&=4.
\end{align*}
Hope to see another solution. Thanks!
|
Let $a=\sqrt2$. (Other values also work.)
Then we have
$$
\begin{aligned}
&\lim_{x\to\infty}
\frac{(x+a)^a-(x-a)^a}{x^{a-1}}
\\
&\qquad=
\lim_{x\to\infty}
a\cdot \frac xa\left(\ \left(1+\frac ax\right)^a-\left(1-\frac ax\right)^a\ \right)
\\
&\qquad\qquad\text{ after forced division with $x^a$ in both numerator and denominator,}
\\
&\qquad=
\qquad\lim_{y\to0}
a\cdot \frac 1y\left(\ \left(1+y\right)^a-\left(1-y\right)^a\ \right)
\\
&\qquad\qquad\text{ using the substitution $y=a/x$,}
\\
&\qquad=
\lim_{y\to0}
a\cdot \frac 1y\left(\ \left(1+ay+\dots\right)-\left(1-ay+\dots\right)\ \right)
\\
&\qquad\qquad\text{ using the Taylor expansion around zero and neglecting $O(y^2)$,}
\\
&\qquad=2a^2\ .
\end{aligned}
$$
Computer aid and confirmation:
sage: var('x');
sage: a = sqrt(2)
sage: limit( ( (x+a)^a-(x-a)^a ) / x^(a-1), x=oo )
4
|
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|
Arc length of ellipse in polar coordinates I have the equation of an ellipse given in Cartesian coordinates as
$\left(\frac{x}{0.6}\right)^2+\left(\frac{y}{3}\right)^2=1$ .
I need the equation for its arc length in terms of $\theta$, where $\theta=0$ corresponds to the point on the ellipse intersecting the positive x-axis, and so on.
So converting to polar coordinates with the substitutions
\begin{Bmatrix}
x=r\cos\theta\\y=r\sin\theta
\end{Bmatrix}
gives $r(\theta)=\frac{1.8}{\sqrt{9\cos^2\theta+0.36\sin^2\theta}}$ , as is illustrated here on Desmos: https://www.desmos.com/calculator/h27qsdnotm.
Then substituting into the arc length formula for polar equations:
$L=\int\sqrt{r^2+\left(\frac{dr}{d\theta}\right)^2} \ d\theta$
gives the rather ugly integral $L=\int\sqrt{\frac{1.8^2}{9\cos^2\theta+0.36\sin^2\theta}+\frac{1944^2\sin^2\theta\cos^2\theta}{(216\cos^2\theta+9)^3}}$ . The online https://www.integral-calculator.com/ was unable to determine an antiderivative, so my question is whether or not the above expression can be expressed in terms of elementary functions, or if there are other methods of finding the equation of the arc length of an ellipse in polar coordinates with respect to $\theta$.
|
\begin{align}
r &= \frac{a b}{\sqrt{a^2\sin^2 \theta+b^2\cos^2 \theta}} \\
r' &=
\frac{ab(b^2-a^2)\sin \theta \cos \theta}
{(a^2\sin^2 \theta+b^2\cos^2 \theta)^{3/2}} \\
ds &=
\frac{ab\sqrt{a^4\sin^2 \theta+b^4\cos^2 \theta}}
{(a^2\sin^2 \theta+b^2 \cos^2 \theta)^{3/2}} d\theta \\
s &= b E
\left(
\phi,
\sqrt{1-\frac{a^2}{b^2}} \,
\right) \\
\tan \phi &= \frac{a\tan \theta}{b} \\
\end{align}
where $E(\phi,k)$ is the incomplete elliptic integral of the second kind.
It is consistent with the result of $(x,y)=(a\sin \phi, b\cos \phi)$ parametrization.
|
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|
Compute $\int_0^1\frac {\sqrt{x}}{(x+3)\sqrt{x+3}}dx.$
Evaluate $\int_0^1\frac {\sqrt{x}}{(x+3)\sqrt{x+3}}dx.$
I tried so many substitutions but none of them led me to the right answer:
$u=\frac 1{\sqrt{x+3}}$, $u=\frac 1{x+3}$, $u=\sqrt{x}$... I even got to something like $\int_0^1 \frac {u^2}{(u^2+3)^{\frac 32}}du$ or $\int_0^1 \frac {\sqrt{1-3u^2}}{u}du$... and I don't know how to solve these...
|
Hint:
Integrate by parts
$$\int\sqrt x\dfrac1{(x+3)^{3/2}}dx$$
$$=\sqrt x\int\dfrac1{(x+3)^{3/2}}dx-\int\left(\dfrac{d(\sqrt x)}{dx}\int\dfrac1{(x+3)^{3/2}}dx\right)dx=?$$
Now for $\displaystyle\int\dfrac{dx}{\sqrt{x(x+3)}}$
$x(x+3)=\dfrac{(2x+3)^2-3^2}4$
set $2x+3=3\sec y$
|
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|
Orthonormal diagonalizable Let:
$$ A = \begin{pmatrix} 5 &2 & -1 \\ 2 & 2 & 2 \\ -1 & 2 & 5 \end{pmatrix} \in Mat_3(\mathbb{R})$$
1) Show that $0$ and $6$ are eigenvalues for $A$ and find the basis for the corresponding eigenspace.
2) Explain why $A$ is orthonormal diagonalizable and find an orthonormal basis for $\mathbb{R^3}$ consisting of eigenvectors for $A$
1) By solving the characteristic polynomial of $A$ it is possible to show that 0 and 6 are eigenvalues for A.
$det(A-\lambda \cdot I) = det( \begin{pmatrix} 5-t &2 & -1 \\ 2 & 2-t & 2 \\ -1 & 2 & 5-t \end{pmatrix} = (5-t)(t^2-7t+6)-2(-2t+12)-1(6-t)=-t^3+12t^2-36t = -t(t^2-12t+36) = -t((t-6)(t-6))=-t(t-6)^2$
To solve $-t(t-6)^2 = 0$ we either have $0$ or $6$, which means that 0 and 6 are eigenvalues for $A$.
The basis for the eigenspace can be found by calculating the null space and we get that: $E_A(0) = N(A) = span(\begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix})$ and $E_A(6) = N(A-6I) = span(\begin{pmatrix} 2 \\ 1 \\ 0 \end{pmatrix},\begin{pmatrix} -1 \\ 0 \\ 1 \end{pmatrix})$
2) We see that the geometric multiplicity and the algebraic multiplicity are equal to each other, which means that A is diagonalizable.
How do I go on from here? I know that I have an invertible matrix $P$ consisting of the eigenvectors $P=\begin{pmatrix} 1 & 2 & -1 \\ -2 & 1 & 0 \\ 1 & 0 & -1 \end{pmatrix}$ such that $D = P^{-1}AP$. But this only means that it is diagonalizable and not orthogonal diagonalizable.
|
It should be obvious from inspection that the two eigenspaces are orthogonal, so you just need an orthogonal basis for each one. $E_A(0)$ trivially has an orthogonal basis, and one iteration of the Gram-Schmidt process will get you one for $E_A(6)$.
|
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|
IVP Differential Equation I have stumbled across a very old exam question from my linear algebra course and the solutions are not available. I was wondering if my working/logic is correct and if any improvements can be made.
Let $\ B=\begin{pmatrix}
1 & -1 & 4 \\
6 & -7 & 2 \\
-3 & 1 & -6
\end{pmatrix}\ $ and $\ \vec{v_1}=\begin{pmatrix} 1 \\ 1 \\ -1 \end{pmatrix}$.
Part (i): Calculate $(B+5I)\vec{v_1}$ and $(B+5I)^2\vec{v_1}$.
Using simply matrix multiplication, I calculated $(B+5I)\vec{v_1}=\begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix}$ and $(B+5I)^2\vec{v_1}=\vec{0}$.
Part (ii): Find all eigenvalues and eigenvectors of B.
From our calculations in part (i), we can see that $\ \lambda=-5,-5$ are two of the eigenvalues, with corresponding eigenvector $\ \begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix}$ and generalised eigenvector $\ \begin{pmatrix} 1 \\ 1 \\ -1 \end{pmatrix}$. From the trace of B, the remaining eigenvalue is $\lambda=-2$ with corresponding eigenvector $\begin{pmatrix} -2 \\ -2 \\ 1 \end{pmatrix}$.
Part (iii): Giving reasons, write down a basis for the generalised eigenspace $GE_{-5}$ of B.
I am unsure of this part of the question. If I had to guess, I would say a basis for $GE_{-5}=\left\{\begin{pmatrix} 1 \\ 1 \\ -1 \end{pmatrix},\begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix}\right\}$. Is this correct?
Part (iv): Solve the initial value problem $\vec{y'}=B\vec{y}$ where $\ B=\begin{pmatrix}
1 & -1 & 4 \\
6 & -7 & 2 \\
-3 & 1 & -6
\end{pmatrix}\ $ and $\vec{y}(0)=\begin{pmatrix} 1 \\ 1 \\ -1 \end{pmatrix}$.
Well, the solution to this IVP will take the form $\vec{y}=e^{tB}\begin{pmatrix} 1 \\ 1 \\ -1 \end{pmatrix}$. I noticed that $\begin{pmatrix} 1 \\ 1 \\ -1 \end{pmatrix}$ was a generalised eigenvector and hence we can write $\vec{y}=e^{-5t}\begin{pmatrix} 1 \\ 1 \\ -1 \end{pmatrix}+e^{-5t}\begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix}t$. Thus $\vec{y}=e^{-5t}\begin{pmatrix} 1+t \\ 1+2t \\ -1-t \end{pmatrix}$
|
You can rewrite the matrix in its Jordan normal form
$$ \textbf{B} =
\textbf{V}
\textbf{J}
\textbf{V}^{-1}
$$
where
$$ \textbf{J} = \begin{bmatrix} -5 & 1 & 0 \\ 0 & -5 & 0 \\ 0 & 0 & -2 \end{bmatrix}, \quad \textbf{V} = \begin{bmatrix} 1 & 1 & -2 \\ 2 & 1 & -2 \\ -1 & -1 & 1 \end{bmatrix} $$
Applying the matrix exponential gives
$$\vec{y} = e^{t\textbf{B}}\ \vec{y}_0 =
\textbf{V}
\begin{bmatrix} e^{-5t} & te^{-5t} & 0 \\ 0 & e^{-5t} & 0 \\ 0 & 0 & e^{-2t} \end{bmatrix}
\textbf{V}^{-1}\vec{y}_0
$$
Since $\vec{y}_0$ happens to be the second column of $\textbf{V}$, it follows that
$$ \textbf{V} \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} = \vec{y}_0 \implies \textbf{V}^{-1}\vec{y}_0 = \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} $$
and so
$$ \vec{y} = \textbf{V} \begin{bmatrix} te^{-5t} \\ e^{-5t} \\ 0 \end{bmatrix} = \begin{bmatrix} 1 \\ 2 \\ -1 \end{bmatrix}te^{-5t} + \begin{bmatrix} 1 \\ 1 \\ -1 \end{bmatrix}e^{-5t} $$
Your answer is correct.
|
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|
Is there a formula that generalizes $\sin A+\sin B+\sin C = 4\cos\frac{A}{2}\cos\frac{B}{2}\cos\frac{C}{2}$ (where $A+B+C=\pi$) to four angles? If $A+B+C=\pi$ then we have $$\sin A+\sin B+\sin C = 4\cos\left(\frac{A}{2}\right)\cos\left(\frac{B}{2}\right)\cos\left(\frac{C}{2}\right)$$
If $A+B+C+D=\pi$ is there a similar formula?
|
There is this similar result: if $\,A + B + C + D = 2\pi\,$ (not $\pi$), then
$$\sin(A) + \sin(B) + \sin(C) + \sin(D) = 4 \sin\Big(\frac{A+B}{2}\Big) \sin\Big(\frac{B+C}{2}\Big) \sin\Big(\frac{C+A}{2}\Big),$$ or, alternatively, both equal
$$-4 \sin\Big(\frac{A+B}{2}\Big) \sin\Big(\frac{A+C}{2}\Big) \sin\Big(\frac{A+D}{2}\Big).$$
The method I used to discover this was factoring and exponential substitution. That is, let
$\,A := \log(a)/i,\, B := \log(b)/i,\, C := \log(c)/i,\, D := \log(d)/i.\,$
where $\,d := 1/(a b c).\,$ The left side factors as $\,i(ab-1)(ac-1)(bc-1)/(2abc).\,$ The $\,(ab-1)\,$ factor is $\,\sin((A+B)/2)\,$ up to some simple factors, and similarly for the other two factors.
The original identity has a geometric application. If $\,A,B,C\,$ are the three angles of a triangle, then $\,A+B+C=\pi\,$ and the identity is equating the sum of the sines of the three angles to four times the product of three cosines of half the angles. Similarly, if $\,A,B,C,D\,$ are the four angles of a quadrilateral, then $\,A + B + C + D = 2\pi\,$ and the new identity is equating the sum of the sines of the four angles to four times the product of three sines of half of sum of two angles.
|
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|
If $2 (\cos \alpha-\cos\beta)+\cos\alpha\cos\beta=1$, then $p\tan\frac{\alpha}{2}+q\tan\frac{\beta}{2}=0$ for which choices of $p$ and $q$? Let $\alpha$ and $\beta$ be nonzero real numbers such that $2 (\cos \alpha-\cos\beta)+\cos\alpha\cos\beta=1$. Then which of the following is/are true ?
a) $\sqrt3\tan(\frac\alpha2)+\tan(\frac\beta2)=0$
b) $\sqrt3\tan(\frac\alpha2)-\tan(\frac\beta2)=0$
c) $\tan(\frac\alpha2)+\sqrt3\tan(\frac\beta2)=0$
d) $\tan(\frac\alpha2)-\sqrt3\tan(\frac\beta2)=0$
I solved the equation by converting all cosines to half angles and the answers options c) and d), seems pretty straightforward from there. But, when the question was asked in an engineering entrance exam (JEE ADVANCED 2017), the question was given as BONUS that would mean either the question was not complete, was wrong or might be some other mistake in understanding the question. I do not understand what was the problem. Can someone help me point out the ambiguity or mistake in the question or give a solution that differs from mine.
|
Given $2(\cos \alpha-\cos \beta)+\cos \alpha\cos \beta = 1$
So $\displaystyle \cos \alpha(\cos \beta+2)=1+2\cos \beta \Rightarrow \cos \alpha =\frac{2\cos \beta+1}{2+\cos \beta}$
So $\displaystyle \frac{1}{\cos \alpha} = \frac{2+\cos \beta}{2\cos \beta +1}$
Using Componendo Dividedno, We have
$\displaystyle \frac{1+\cos \alpha}{1-\cos \alpha} = \frac{2+\cos \beta+2\cos \beta+1}{(2+\cos \beta)-(2\cos \beta +1)} = \frac{3(\cos \beta+1)}{1-\cos \beta}$
So $\displaystyle \frac{2\sin^2\frac{\alpha}{2}}{2\cos^2 \frac{\alpha}{2}} = 3\cdot \frac{2\sin^2\frac{\beta}{2}}{2\cos^2 \frac{\beta}{2}}$
So we have $$\tan \frac{\alpha}{2} = \pm\sqrt{3}\tan \frac{\beta}{2}.$$
|
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|
How to prove $1+\cos2\theta+\cos4\theta+\cos6\theta+\cos8\theta=\frac{(\cos4\theta)(\sin5\theta)}{\sin\theta} $? I need help to prove that the following is true:
$$1+\cos2\theta+\cos4\theta+\cos6\theta+\cos8\theta=\frac{(\cos4\theta)(\sin5 \theta)}{\sin\theta}$$
I realize that I must evaluate the real part of this, but whatever I get I am not quite sure how to get to the required expression. I have multiplied the numerator and denominator of the result of the geometric sum by the conjugate of $ e^{2i\theta} $ and still have no luck.
$$\sum_{i=0}^4 e^{2ni\theta}$$
(Apologies for poor formatting)
|
According to the link posted as a possible duplicate and containing my answer in it, we have:
$$1+\cos{\theta}+\cos{2\theta}+...+\cos{n\theta}=\frac{\cos{\frac{n\theta}{2}}\cdot\sin{\frac{(n+1)\theta}{2}}}{\sin{\frac{\theta}{2}}} \tag{1}$$
Now
$$1 + \cos{2\theta} + \cos{4 \theta} + \cos{6 \theta} + \cos8{\theta}=\\
1 + \cos{(1\cdot \color{red}{2\theta})} + \cos{(2\cdot \color{red}{2\theta})} + \cos{(3 \cdot \color{red}{2\theta})} + \cos{(4\cdot \color{red}{2\theta})}=\\
\frac{\cos{\frac{4\cdot\color{red}{2\theta}}{2}}\cdot\sin{\frac{(4+1)\cdot\color{red}{2\theta}}{2}}}{\sin{\frac{\color{red}{2\theta}}{2}}}=
\frac{\cos{(4\color{red}{\theta})}\cdot\sin{(5\color{red}{\theta}})}{\sin{\color{red}{\theta}}}$$
|
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|
Difference equation solution My question : $y_{n+1}= ay_n+b$ ; $y_0 = \alpha$
We solved this difference equation in a class and got that $y_n = a^nc + \beta$
Can someone please explain the way how to solve it? And what is the was to solve differential equations in general? Is there some good literature with difference equations theory and examples?
|
Note that it must be: $y_n=a^n\color{red}{\alpha}+\beta$.
If $a=1$, it is just an $n$-th term of an Arithmetic Progression.
Assume $a\ne 1$.
Method $1$:
$$\begin{align}y_{n+1}&= ay_n+b=\\
&=a(ay_{n-1}+b)+b= \qquad \qquad \qquad \ \ \ \ \ \ \ \ \ \ \ \ \ \ a^2y_{n-1}+(a+1)b=\\
&=a^2(ay_{n-2}+b)+(a+1)b=\qquad \qquad \qquad a^3y_{n-2}+(a^2+a+1)b=\\
&\ \ \ \vdots \\
&=a^n(ay_{0}+b)+(a^{n-1}+a^{n-2}+\cdots+1)b=a^{n+1}y_{0}+(a^n+a^{n-1}+\cdots+1)b=\\
&=a^{n+1}y_0+\frac{(a^{n+1}-1)b}{a-1}=\\
&=a^{n+1}y_0+a^{n+1}\frac{b}{a-1}-\frac{b}{a-1}=\\
&=a^{n+1}\left(y_0+\frac{b}{a-1}\right)-\frac{b}{a-1} \Rightarrow \\
y_n&=a^n\left(y_0+\frac b{a-1}\right)-\frac{b}{a-1}=a^n\underbrace{y_0}_{\alpha}+\underbrace{(a^n-1)\cdot \frac b{a-1}}_{\beta}
\end{align}$$
Method 2:
$$y_{n+1}=ay_n+b \Rightarrow y_{n+1}+\frac{b}{a-1}=a(y_n+\frac b{a-1})\\
z_n=y_n+\frac b{a-1}, z_0=y_0+\frac b{a-1}\\
z_{n+1}=az_n \Rightarrow z_n=a^nz_0=a^n\left(y_0+\frac b{a-1}\right)=y_n+\frac b{a-1} \Rightarrow \\
y_n=a^n\left(y_0+\frac b{a-1}\right)-\frac b{a-1}=a^n\underbrace{y_0}_{\alpha}+\underbrace{(a^n-1)\cdot \frac b{a-1}}_{\beta}.$$
|
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|
Find a and p in the given question
I would be grateful if someone helps me with this question.
I get $p$ but to get $a$, the term with $a$ vanishes.
I checked continuity at $x=1$ for finding $p$.
$p$ comes out to be $-2$.
|
As $f(x)$ is differentiable for $x\in(0,2)$ we know that the limit $$\lim_{n\to\infty}\dfrac{ax(x-1)\left(\cot^n\left(\dfrac{\pi x}4\right)\right)+px^2+2}{\left(\cot^n\left(\dfrac{\pi x}4\right)\right)+1}$$ is finite, and $f(x)$ is continuous for $x\in(0,2)$.
$$$$
Let $$L=\lim_{n\to\infty}\dfrac{ax(x-1)\left(\cot^n\left(\dfrac{\pi x}4\right)\right)+px^2+2}{\left(\cot^n\left(\dfrac{\pi x}4\right)\right)+1}$$
$$\Rightarrow L=\lim_{n\to\infty}\dfrac{ax(x-1)+(px^2+2)\left(\tan^n\left(\dfrac{\pi x}4\right)\right)}{1+\left(\tan^n\left(\dfrac{\pi x}4\right)\right)}$$
$$$$
For $x\in(1,2)$, $\dfrac\pi4<\dfrac{\pi x}{4}<\dfrac{\pi}2\Rightarrow 0<\cot\left(\dfrac{\pi x}4\right)<1\Rightarrow\cot^n\left(\dfrac{\pi x}4\right)\to 0 $ as $n\to\infty$
$$$$
Hence $$L=\lim_{n\to\infty}\dfrac{ax(x-1)\left(\cot^n\left(\dfrac{\pi x}4\right)\right)+px^2+2}{\left(\cot^n\left(\dfrac{\pi x}4\right)\right)+1}=\dfrac{0(ax(x-1))+px^2+2}{0+1}=px^2+2$$
$$$$
We need that the limit is equal to $0$ as $x\to 1^+$ for the function to be continuous. Hence $\lim_{x\to1^+}px^2+2=0\Rightarrow p=-2$.
$$$$
Let $L_1$ and $L_2$ be the limits (of $f(x)$) from the left and from the right of $x=1$ respectively.
$$$$
Thus $$L_2=\lim_{n\to\infty}\dfrac{ax(x-1)\left(\cot^n\left(\dfrac{\pi x}4\right)\right)+2-2x^2}{\left(\cot^n\left(\dfrac{\pi x}4\right)\right)+1},x\to1^+$$
$$L_1=\lim_{n\to\infty}\dfrac{ax(x-1)+(2-2x^2)\left(\tan^n\left(\dfrac{\pi x}4\right)\right)}{1+\left(\tan^n\left(\dfrac{\pi x}4\right)\right)},x\to1^-$$
$$$$
Now, for $x\in(0,1)$, $\tan^n\left(\dfrac{\pi x}4\right)\to0$ as $n\to\infty$ $$$$For $x\in(1,2)$, $\cot^n\left(\dfrac{\pi x}4\right)\to 0$ as $n\to\infty$.
$$$$
Thus, $$L_2=\dfrac{2-2x^2}{2},x\to 1^+$$
$$\Rightarrow L_2=\lim_{h\to 0}\dfrac{2-2(1+h)^2}{2}=\lim_{h\to0}\dfrac{-2h^2-4h}{2}=\lim_{h\to0}\dfrac{-2h(h-2)}{2}=\lim_{h\to0}\dfrac{2h(2-h)}{2}$$
$$$$
$$L_1=\dfrac{ax(x-1)}{2},x\to1^-$$
$$\Rightarrow L_1=\lim_{h\to 0}\dfrac{a(1-h)(1-h-1)}{2}=\lim_{h\to 0}\dfrac{a(1-h)(-h)}{2}$$
$$$$
Now it is given that $f(x)$ is differentiable for $x\in(0,2)$. Thus, $f(x)$ is continuous for $x\in(0,2)$. Thus $f(x)$ must be continuous at $x=1$. Therefore, $L_2$ must be equal to $L_1$ (and both must be respectively equal to $f(1)=0$). Equating both $L_2$ and $L_1$,
$$$$
$$\lim_{h\to0}\dfrac{2h(2-h)}{2}=\lim_{h\to 0}\dfrac{a(1-h)(-h)}{2}$$
$$\Rightarrow \lim_{h\to0}\dfrac{2h(2-h)}{2}-\lim_{h\to 0}\dfrac{a(1-h)(-h)}{2}=0$$
$$$$We now use the property $\lim_{x\to a}f(x)\pm\lim_{x\to a}g(x)=\lim_{x\to a}(f\pm g)(x)$
$$$$
$$\Rightarrow \lim_{h\to0}\left(\dfrac{2h(2-h)}{2}-\dfrac{a(1-h)(-h)}{2}\right)=0$$
$$\Rightarrow \lim_{h\to0}\dfrac{2h(2-h)+a(1-h)(h)}{2}=0$$
$$\Rightarrow 2(2-h)+a(1-h)=0, h\to0$$
$$\Rightarrow a+4=0$$
Finally, we get $a=-4$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2813637",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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|
Showing that $f(x,y)=(\lvert x\rvert -\lvert y\rvert)\log\left(2x^2+\lvert y\rvert\right)$ has no local extrema I have an exercise asking to find the local extrema of $$ f(x,y)=(\lvert x\rvert -\lvert y\rvert)\log\left(2x^2+\lvert y\rvert\right)$$if they exist... Wolfram Alpha tells me they don't, but I think I've spent too long trying to show it, so I'm wondering what I missed, resulting in a more time-consuming task. Below is my work:
$f$ is continuous on the whole plane except the origin.
For $x,y\ne0$$$f(x,y)=\begin{cases}(x-y)\log\left(2x^2+y\right) &x,y>0 \\ (x+y)\log\left(2x^2-y\right) &x>0>y\\-(x+y)\log\left(2x^2+y\right) &x<0<y \\ -(x-y)\log\left(2x^2-y\right) &x,y<0\end{cases}$$hence $$f_x(x,y)=\begin{cases}\log\left(2x^2+y\right)+\dfrac{4x(x-y)}{2x^2+y} &x,y>0 \\ \log\left(2x^2-y\right)+\dfrac{4x(x+y)}{2x^2-y} &x>0>y \\ -\log\left(2x^2+y\right)-\dfrac{4x(x+y)}{2x^2+y} &x<0<y \\ -\log\left(2x^2-y\right)-\dfrac{4x(x-y)}{2x^2-y} &x,y<0 \end{cases} $$ and
$$f_y(x,y)=\begin{cases}-\log\left(2x^2+y\right)+\dfrac{x-y}{2x^2+y} &x,y>0 \\ \log\left(2x^2-y\right)-\dfrac{x+y}{2x^2-y} &x>0>y \\ -\log\left(2x^2+y\right)-\dfrac{x+y}{2x^2+y} &x<0<y \\ \log\left(2x^2-y\right)+\dfrac{x-y}{2x^2-y} &x,y<0 \end{cases}.$$ Setting both partial derivatives equal to $0$ I got $$\begin{cases}(x-y)(4x+1) &x,y>0 \\ (x-y)(4x+1) &x>0>y \\ (x+y)(4x-1) &x<0<y \\ (x-y)(4x-1) &x,y<0 \end{cases} \iff y=\pm x\ne0.$$ So for $x_0\ne0$ and small $h_1,h_2$ I examined \begin{align}\operatorname{sgn}\left\{f(x_o+h_1,\pm x_0+h_2)-f(x_0,\pm x_o)\right\}&:=\operatorname{sgn}(h_3\log(2x_0^2+|x_0|+h_4)) \\&= \operatorname{sgn}(h_3\log(2x_0^2+|x_0|)) \end{align}which obviously depends on how $h_3$ approaches $0$, so the $(x_0,\pm x_0)$ are saddle points.
EDIT: ok on second thought I forgot to show that $h_3$ doesn't depend on $h_1, h_2$ in a way that would make it approach $0$ only from the left/right, but let's pretend I did show it
Now I should study the behaviour around points $(x_0,0)$ and $(0,y_0)$ , and this is where I've been stuck. After a while I tried looking at things like $f(x_o,h)-f(x_0,0)$ but it didn't help... what am I missing?
|
The function
$$
f(x,y) = \left(|x|-|y|\right)\log(2x^2-|y|)
$$
has symmetries along the $x$ and $y$ axis because
$$
f(x,y) = f(\pm x, \pm y)
$$
Now considering the first quadrant or
$$
f(x,y) =(x-y)\log(2x^2-y)
$$
making the change of variable $u = 2x^2+y$ we have
$$
g(x,u) = (x+2x^2-u)\log u
$$
in this coordinate system the stationary points are given as the solutions for
$$
\nabla g(x,u) = \left\{
\begin{array}{rcl}
(1+4x)\log u & = & 0\\
x+2x^2-u - u\log u & = & 0
\end{array}\right.
$$
and solving we have
$$
\left(
\begin{array}{ccc}
x & u & y \\
-1 & 1 & -1 \\
-\frac{1}{4} & -\frac{1}{8 W\left(-\frac{e}{8}\right)} & -\frac{1}{8}-\frac{1}{8 W\left(-\frac{e}{8}\right)} \\
-\frac{1}{4} & -\frac{1}{8 W_{-1}\left(-\frac{e}{8}\right)} & -\frac{1}{8}-\frac{1}{8 W_{-1}\left(-\frac{e}{8}\right)} \\
\frac{1}{2} & 1 & \frac{1}{2} \\
\end{array}
\right)
$$
Here $W(\cdot)$ is the so called Lambert function. In numbers
$$
\left(
\begin{array}{ccc}
x & u & y \\
-1. & 1. & -1. \\
-0.25 & 0.191569 & 0.0665694 \\
-0.25 & 0.0860153 & -0.0389847 \\
0.5 & 1. & 0.5 \\
\end{array}
\right)
$$
The feasible first quadrant solution gives
$$
x = \frac{1}{2}, u = 1, y = \frac{1}{2}
$$
The qualification for this stationary point is made with the contribution of the $g(x,u)$ hessian
$$
H_g = \left(
\begin{array}{cc}
4 \log (u) & \frac{4 x+1}{u} \\
\frac{4 x+1}{u} & -\frac{2 x^2+x-u}{z^2}-\frac{2}{u} \\
\end{array}
\right)
$$
At $(x,y) = (1/2,1/2)$ the hessian gives
$$
H = \left(
\begin{array}{cc}
0 & 3 \\
3 & -2 \\
\end{array}
\right)
$$
characterizing this point as a saddle point. Considering the four quadrants the surface takes the form shown below.
and we can observe the existence of local minima undetected by the traditional method.
The minima are located at
$$
\{x,y\} = \{\pm\frac{1}{e\sqrt2},0\}
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2813755",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
How to integrate the product of two or more polynomials raised to some powers, not necessarily integral This question is inspired by my own answer to a question which I tried to answer and got stuck at one point.
The question was:
HI DARLING.
USE MY ATM CARD, TAKE ANY AMOUNT OUT, GO SHOPPING AND TAKE YOUR FRIENDS FOR LUNCH.
PIN CODE: $\displaystyle \int_{0}^{1} \frac{3x^3 - x^2 + 2x - 4}{\sqrt{x^2 - 3x + 2}} \, dx $
I LOVE YOU HONEY.
Anyone knows? Are we gonna get an integer number?
My attempt:
Does this help?
$$\frac{3x^3-x^2+2x-4}{x-1}=3x^2+2x+4$$
(long division)
\begin{align*}
I&=\int\frac{3x^3-x^2+2x-4}{[(x-1)(x-2)]^{1/2}} dx = \\
&=\int\frac{(3x^2+2x+4)(x-1)^{1/2}}{(x-2)^{1/2}} dx = \\
&=\int 3(u^4-4u^2-4)(u^2+1)^{1/2}du \times 2
\end{align*}
after the substitution
\begin{gather*}
(x-2)^{1/2}=u\\
du=\frac1{2(x-2)^{1/2}}dx\\
u^2=x-2\\
(x-1)^{1/2}=(u^2+1)^{1/2}
\end{gather*}
Update: This may help us proceed.
I tried to proceed:
$$6\int (u^4-4u^2-4)(u^2+1)^{1/2} du = 6\int ((t-3)^2-8)t \frac{dt}{2u}$$
after $u^2+1=t$ and $dt=2udu$
\begin{align*}
u^4-4u^2-4
&= (u^2+1)^2-(6u^2+5) \\
&= (u^2+1)^2-6(u^2+1)+1 \\
&= ((u^2+1)-3)^2-8
\end{align*}
I wonder whether this question can be solved from here?
Update:
This has been getting a lot of views, and I think most people came for the sort of problem mentioned in the title (where I got stuck) rather than the original problem itself.
Keepin this in mind, I'm reopening the question and here's the kind of answers I expect — Solutions to the original problem are good, but I'd prefer solutions that continue from the part where I got stuck — the polynomial in $u$ — that's the sort of problem mentioned in the title.
|
Noticing that
$3 x^3-x^2+2 x-4=-(1-x)\left(3 x^2+2 x+4\right)$ and $x^2-3 x+2=(1-x)(2-x)$,
we transform the integral into
$$
I=-\int_0^1 \sqrt{\frac{1-x}{2-x}}\left(3 x^2+2 x+4\right) d x
$$
Let $t^2=\frac{1-x}{2-x}$, then $x=\frac{2 t^2-1}{t^2-1}$ and $d x=-\frac{2 t}{\left(t^2-1\right)^2} d t$, which changes the integral into
$$
I=-2 \int_0^{\frac{1}{\sqrt{2}}} \frac{t^2\left(20 t^4-26 t^2+9\right)}{\left(t^2-1\right)^4} d t
$$
Resolving the integrand into partial fractions as
$$ \frac{t^2\left(20 t^4-26 t^2+9\right)}{\left(t^2-1\right)^4} = {-\frac{135}{32} \cdot \frac{1}{t+1}+\frac{185}{32} \cdot \frac{1}{(t+1)^2}-\frac{7}{4} \cdot \frac{1}{(t+1)^3}+\frac{3}{16} \cdot \frac{1}{(t+1)^4}}+\frac{135}{32} \cdot \frac{1}{t-1}+\frac{185}{32} \cdot \frac{1}{(t-1)^2}+\frac{7}{4} \cdot \frac{1}{(t-1)^3}+\frac{3}{16}\cdot \frac{1}{(t-1)^4}$$
Integrating it from $0$ to $\frac{1}{\sqrt 2} $ yields
$$
\begin{aligned}
I=&-2 \left[-\frac{135}{32} \ln |t+1|-\frac{185}{32(t+1)}+\frac{7}{8(t+1)^2}-\frac{1}{16(t+1)^3} +\frac{135}{32} \ln |t-1|\right.\\ &\left.-\frac{185}{32(t-1)}-\frac{7}{8(t-1)^2}-\frac{1}{16(t-1)^3}\right]_0^{\frac{1}{\sqrt{2}}}\\=& \frac{135}{16} \ln (3+2 \sqrt{2})-\frac{101 \sqrt{2}}{8}
\end{aligned}
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2814179",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 3
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|
Prove $\sin^2 a + \sin^2 b + \sin^2 c = 2(1-\cos a \cos b \cos c)$ when $a=b+c$ I want to prove this identity:
$$\sin^2 a + \sin^2 b + \sin^2 c = 2(1-\cos a \cos b \cos c) \qquad\text{when}\;a=b+c$$
Can somebody give a hint in the easiest way possible? I am debugging this for hours and can't get the left side to be the right.
|
Ignoring the condition $a=b+c$, we can see that the equation can be manipulated into an interesting form:
$$\begin{align}
\sin^2 a + \sin^2 b + \sin^2 c &= 2 ( 1 - \cos a\cos b\cos c ) \\
( 1 - \cos^2 a ) + ( 1 - \cos^2 b ) + ( 1 - \cos^2 c ) &=
2 - 2 \cos a\cos b\cos c \\
1 - \cos^2 b - \cos^2 c &=
\cos^2 a - 2 \cos a\cos b\cos c \\
\qquad 1- \cos^2 b - \cos^2 c + \cos^2 b\cos^2 c &= \cos^2 a - 2 \cos a \cos b\cos c + \cos^2 b \cos^2 c \\
(1-\cos^2b)(1-\cos^2 c) &= ( \cos a - \cos b \cos c )^2 \\
\sin^2 b\sin^2 c &= ( \cos a - \cos b \cos c )^2 \\
0 &= ( \cos a - \cos b \cos c )^2 -\sin^2 b\sin^2 c \\
0 &= (\cos a - \cos b \cos c + \sin b \sin c) \\
&\phantom{=}\cdot( \cos a - \cos b \cos c - \sin b\sin c) \\
0 &= (\cos a - \cos(b+c))( \cos a - \cos(b-c))
\end{align}$$
Thus, $\cos a = \cos(b\pm c)$, so that
$$a = \pm b \pm c + 2\pi k \quad\text{(with independent $\pm$s)}$$
In particular, $a=b+c$ is one of those cases.
It's perhaps worth noting that the original equation reduces even further using the "difference-to-product" identity:
$$\cos\theta - \cos\phi = - 2 \sin\frac{\theta+\phi}{2}\cdot\sin\frac{\theta-\phi}{2}$$
to
$$\sin\frac{a+b+c}{2}\cdot\sin\frac{-a+b+c}{2}\cdot\sin\frac{a-b+c}{2}\cdot\sin\frac{a+b-c}{2} = 0$$
which bears an interesting resemblance to Heron's formula for the area of a triangle. Moreover, had the sign of the $2\cos a\cos b\cos c$ term been reversed, then we'd have instead
$$(\cos a + \cos(b+c))( \cos a + \cos(b-c)) = 0$$
which gives rise to another Heron-like form:
$$\cos\frac{a+b+c}{2}\cdot\cos\frac{-a+b+c}{2}\cdot\cos\frac{a-b+c}{2}\cdot\cos\frac{a+b-c}{2} = 0$$
This relation is satisfied, for instance, when $a+b+c=180^\circ$; that is, when $a$, $b$, $c$ are the angles of a triangle.
I guess my point is that the equation in question isn't entirely random.
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/2815136",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
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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.