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Sum of the series $1+\frac{1\cdot 3}{6}+\frac{1\cdot3\cdot5}{6\cdot8}+\cdots$ Decide if the sum of the series
$$1+\frac{1\cdot3}{6}+\frac{1\cdot3\cdot5}{6\cdot8}+ \cdots$$
is: (i) $\infty$, (ii) $1$, (iii) $2$, (iv) $4$.
|
Well , imagination is the magic and you will see the best and most elegent method that follows as such:
$$1+\frac{1\cdot3}{6}+\frac{1\cdot3\cdot5}{6\cdot8}+ \cdots=1+\frac{1\cdot3\cdot(6-5)}{6}+\frac{1\cdot3\cdot5\cdot(8-7)}{6\cdot8}+\cdots$$
$$=1+\frac{1\cdot3\cdot6}{6}-\frac{1\cdot3\cdot5}{6}+\frac{1\cdot3\cdot5\cdot8}{6\cdot8}-\frac{1\cdot3\cdot5\cdot7}{6\cdot8}\cdots=1+1\cdot3=1+3=4$$
Well , this telescoping method really helps in majority of cases.:)
|
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|
relation between inscribed and circumsdribed circle Let T be the triangle with side lengths $b_1,b_2,b_3$, and $r_{insc}$ and $r_{cir}$
be the radii of the inscribed and circumscribed circles,respectively,
I need to show that
$$ \frac {r_{insc}}{r_{cir}}=\frac {(b_2+b_3-b_1)(b_3+b_1-b_2)(b_2+b_1-b_3)}{2b_1b_2b_3}$$
Maybe it is easy, but I forgot relations on triangles,
Please help me.
Thanks
|
Let's side of the triangle be $a,b,c$ and $R$ and $r$ are radii of circumscribed and inscribed circles, respectively.
The equaiton now looks like
$$\frac{r}{R} = \frac{(a+b-c)(a+c-b)(b+c-a)}{2abc}$$
We multiply both sides by $\frac{a+b+c}{8}$
$$\frac{r(a+b+c)}{8R} = \frac{a+b-c}{2} \times \frac{a+c-b}{2} \times \frac{b+c-a}{2} \times \frac{a+b+c}{2}\times \frac{1}{abc}$$
$$\frac{r(a+b+c)}{8R} = \frac{(s-c)(s-b)(s-a)s}{abc}$$
Where $s$ is the semiperimetar. By Heron's formula we know that
$$K = \sqrt{(s-c)(s-b)(s-a)s}$$
Where K is the are of the triangle.
$$\frac{r(a+b+c)}{8R} = \frac{K^2}{abc}$$
We substitute $\frac{a+b+c}{2} = s$
$$\frac{rs}{4R} = \frac{K^2}{abc}$$
We know that $K = \frac{abc}{4R} = rs$
$$\frac{rs \times abc}{4R} = K^2$$
$$K^2 = K^2$$
Q.E.D.
Another solution is making this substitution:
$$R = \frac{abc}{4\sqrt{s(s-a)(s-b)(s-c)}}$$
$$r = \frac{\sqrt{(s-a)(s-b)(s-c)}}{\sqrt{s}}$$
$$\frac{r}{R} = \frac{\frac{\sqrt{(s-a)(s-b)(s-c)}}{\sqrt{s}}}{\frac{abc}{4\sqrt{s(s-a)(s-b)(s-c)}}} = \frac{4(s-a)(s-b)(s-c)}{abc} = \frac{4 \times \frac{c+b-a}{2} \times \frac{a+c-b}{2} \times \frac{a+b-c}{2}}{abc}$$
$$\frac{r}{R} = \frac{(a+b-c)(a+c-b)(c+b-a)}{2abc}$$
Q.E.D.
|
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Matrices determinants problem Could you please give at least a hint to prove this? I really don't know how to start.
Let $A$ be the matrix
$A= \left( \begin{array}{ccc}
A_{11} & A_{12} \\
A_{21} & A_{22} \end{array} \right)$,
where $A_{11}$ and $A_{22}$ are square matrices of order $k$ and $m$ respectively. Prove that for any two matrices $D$ of order $k \times k$ and $B$ of order $m \times k$ we have
$\left| \begin{array}{ccc}
D \cdot A_{11} & D \cdot A_{12} \\
A_{21} & A_{22} \end{array} \right| = |D|\cdot |A|$,
and
$\left| \begin{array}{ccc}
A_{11} & A_{12} \\
A_{21}+B \cdot A_{11} & A_{22}+B \cdot A_{12} \end{array} \right|=|A|.$
|
Note that
\begin{align*}
\left(\begin{array}{cc}D\cdot A_{11}&D\cdot A_{12}\\A_{21}&A_{22}\end{array}\right)=\left(\begin{array}{cc}D&0\\0&I\end{array}\right)\cdot\left(\begin{array}{cc}A_{11}& A_{12}\\A_{21}&A_{22}\end{array}\right),
\end{align*}
where $I$ is the identity matrix of order $m$. The first matrix in the second expression is a “block diagonal matrix,” whose determinant is just $|D|\cdot|I|=|D|$. Now use the product rule for determinants.
As for your second question (in which I'm pretty sure the $D$ should be $B$):
\begin{align*}
\left(\begin{array}{cc}A_{11}& A_{12}\\A_{21}+B\cdot A_{11}&A_{22}+B\cdot A_{12}\end{array}\right)=\left(\begin{array}{cc}\underset{k\times k}{I}&0\\B&\underset{m\times m}{I}\end{array}\right)\cdot\left(\begin{array}{cc}A_{11}& A_{12}\\A_{21}&A_{22}\end{array}\right).
\end{align*}
The first matrix in the second expression is a triangular matrix, whose determinant is equal to the product of the main diagonal elements, which is $1$. Now, use again the product rule.
|
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|
Compute $ \int_{-\infty}^{\infty} \frac{x^2}{(1+x^2)^2} dx$ Compute $$ \int_{-\infty}^{\infty} \frac{x^2}{(1+x^2)^2} dx$$
Of course we have $$ \int_{-\infty}^{\infty} \frac{x^2}{(1+x^2)^2} dx = 2 \int_{0}^{\infty} \frac{x^2}{(1+x^2)^2} dx = 2 \int_{0}^{\infty} \left( \frac{x}{1+x^2} \right) ^2 dx = \lim_{ A \to \infty } \int_{0}^{A} \left( \frac{x}{1+x^2} \right) ^2 dx $$
I think that firstly I should compute $ \int \left( \frac{x}{1+x^2} \right) ^2 dx $ but I don't have idea.
|
I know I'm a bit late for the party, but here is another solution:
$$\begin{align}
2\int_{0}^\infty \frac{x^2}{(1+x^2)^2}\mathrm dx &=
\frac22\int_{-\infty}^\infty \frac{y^{\frac{2-1}{2}}}{(1+y)^2}\mathrm dx \\
&= B\bigl(\tfrac12 + 1, 2-\tfrac12 -1\bigr) = B\bigl(\tfrac32,\tfrac12\bigr) \\
&= \frac{\Gamma\bigl(\tfrac32\bigr)\Gamma\bigl(\tfrac12\bigr)}{\Gamma(2)} \\
&= \frac{\sqrt\pi \cdot \frac12 \sqrt\pi}{1} = \frac\pi 2
\end{align}
$$
The first step is the substitution $x^2 \mapsto y$, the second and third step are the use of well-known (and very easy to proof) representations of the Beta function.
|
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|
How to show $\sqrt{4+2\sqrt{3}}-\sqrt{3} = 1$ I start with $x=\sqrt{4+2\sqrt{3}}-\sqrt{3}$, then
$\begin{align*}
x +\sqrt{3} &= \sqrt{4+2\sqrt{3}}\\
(x +\sqrt{3})^2 &= (\sqrt{4+2\sqrt{3}})^2\\
x^2 + (2\sqrt{3})x + 3 &= 4+ 2\sqrt{3}\\
x^2 + (2\sqrt{3})x - 1 - 2\sqrt{3} &= 0
\end{align*}$
So I have shown that there is some polynomial whose solution is $x=\sqrt{4+2\sqrt{3}}-\sqrt{3}$, but I have not shown it to be 1.
|
We have $4+2 \sqrt 3= 3+ 2 \sqrt 3+1= \left( \sqrt 3+1\right)^2$. Therefore $\sqrt{4+2 \sqrt 3}= \sqrt 3+1$.
Thus, $\sqrt{4+2 \sqrt 3}- \sqrt 3=1$.
|
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Solving a system of equations using modular arithmetic modulo 5 Give the solution to the following system of equations using modular arithmetic modulo 5:
$4x + 3y = 0 \pmod{5}$
$2x + y \equiv 3 \pmod{5}$
I multiplied $2x + y \equiv 3 \pmod 5$ by $-2$, getting $-4x - 2y \equiv -6 \pmod{5}$.
$-6 \pmod{5} \equiv 4 \pmod 5$
Then I added the two equations:
$4x + 3y \equiv 0 \pmod{5}$
$-4x - 2y \equiv 4 \pmod{5}$
This simplifies to $y \equiv 4 \pmod{5}$.
I then plug this into the first equation: $4x + 3(4) = 0 \pmod{5}$
Wrong work:
Thus, $x = 3$.
But when I plug the values into the first equation, I get $2(3) + 4 \not\equiv 3 \pmod{5}$.
What am I doing wrong?
EDIT:
Revised work:
$x = -3 \pmod{5} = 2 \pmod{5}$.
Now when I plug the values into the first equation, I get $2(2) + 4 \equiv 8 \pmod{5} \equiv 3 \pmod{5}$.
|
You may consult Maple as follows to get that $(2,4)$ is the only solution for the system.
[> msolve({2*x+y = 3, 4*x+3*y = 0}, 5);
{x=2,y=4}
|
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Solving a differential equation with composite functions If $c=f(a+e^b)+g(a-e^b)$ where $f$ and $g$ are functions of $a+b^2$ and $a-b^2$ respectively, find $c$ such that when $b=0$, you find that $c=0$ and $\frac{\partial c}{\partial b}=1+a$.
|
You say, that
$$
c(a,b) = f(a+e^b) + g(a-e^b)
$$
where $f, g : \mathbb R \to \mathbb R$, and you need to find some particular form of $c(a,b)$ so that
\begin{align}
c(a,0) &= 0\\
\left . \frac {\partial c}{\partial b} \right |_{b = 0} &= 1 + a
\end{align}
Use first restriction.
$$
c(a,0) = f(a+1) + g(a-1) = 0 \tag 1
$$
Now, let's find the form of of $c_b(a,b)$
$$
c_b(a,b) = f'(a+e^b)e^b - g'(a-e^b)e^b = e^b \left ( f'(a+e^b) - g'(a+e^b)\right) \tag 2
$$
so
$$
c_b(a,0) = f'(a+1)-g'(a-1) = 1 + a \tag 3
$$
Now, differentiate $(1)$
$$
f'(a+1) + g'(a-1) = 0 \tag 4
$$
and therefore
$$
g'(a-1) = -f'(a+1) \tag 5
$$
substitute $(5)$ to $(3)$
$$
2f'(a+1) = a+1
$$
or
$$
f'(x) = \frac x2
$$
which has obvious solution
$$f(x) = \frac {x^2}4 + C$$
Now, find $g(x)$
$$
g(a-1) = -f(a+1) = -\frac {(a+1)^2}4 - C
$$
or, after some trivial manipulations
$$
g(x) = -\frac {(x+2)^2}4 - C
$$
So, final answer is
$$
c(a,b) = f(a+e^b) + g(a-e^b) = \frac {\left ( a+e^b\right )^2}4 - \frac {\left ( a-e^b+2\right )^2}4
$$
|
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|
Help with minimization problem help me, if $x$ and $y$ are real such that $3x-4y = 12$, determine the minimum value of $z = x ^ 2 + y ^ 2$?$$$$I thought of $$3x-4y = 12\Longrightarrow x=4\frac{y+3}{3}\\z = x ^ 2 + y ^ 2\Longrightarrow z = \left(4\frac{y+3}{3}\right) ^ 2 + y ^ 2$$ and then?
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The quantity $x^2+y^2$ is minimized when $\sqrt{x^2+y^2}$ is minimized, and $\sqrt{x^2+y^2}$ is just the distance from the point $\langle x,y\rangle$ to the origin. Thus, you’re looking for the point on the line $3x-4y=12$ that is closest to the origin. Call this line $\ell$; the point closest to the origin is the point of intersection of $\ell$ with a line through the origin and perpendicular to $\ell$. The slope of $\ell$ is $\frac34$, so the slope of the perpendicular is $-\frac43$: you want the intersection of $\ell$ and the line $y=-\frac43x$.
|
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Doubt in derivative problem I'm in trouble with the following problem:
Give $f(x) = x^3 - 3x + 1$ and $g(x) = x^3 + ax^2 + b$, determine the values of $a$ and $b$ in such a way that both $f(x)$ and $g(x)$ have the same relative maximum and minimum.
I already know that the relative maximum and minimum can be found though the roots of the derivative, so $f'(x) = 3x^2 - 3$ and $g'(x) = 3x^2 + 2ax$. As the roots of $f'(x)$ are both ${-1, 1}$, I imagined to build a system of equations to find $a$ and $b$, but I always end in two 'dummy' equations, do someone know a different way to solve this ?
By the way, the dummy equations are when I substitute ${-1, 1}$ in $g(x)$. The equations are $a + b = 4$ and $a + b = -2$
|
As the comments have pointed out, we must match up the function values at the relative min and max (which, in this case, do not have the same $x$-values).
Since $f'=0$ has $x=\pm1$ as roots, we substitute these critical points into $f$ to obtain $f(-1)=-1+3+1=3$ and $f(1)=1-3+1=-1$. You can verify for yourself that these really are local extrema.
Setting $g'=0$ yields $0=3x^2+2ax=x(3x+2a) \implies x=0,-2a/3$. Substitutng these critical points into $g$ yields $g(0) = 0+0+b=b$ and $g(-2a/3) = -8a^3/27 + 4a^3/9 + b = 4a^3/27 + b$.
For these two function values for $g$, it is a bit difficult to see how they should be matched up with our two function values for $f$, since we can't really tell which one is bigger (since we don't know if $a,b$ are positive or negative). So let's guess.
Case 1: Suppose the local minimum is $f(1)=-1=b=g(0)$. Then the local maximum is $f(-1)=3=4a^3/27 + b=g(-2a/3)$. Substituting $b$ and solving for $a$ yields:
\begin{align*}
\dfrac{4a^3}{27} - 1 &= 3 \\
\dfrac{4a^3}{27} &= 4 \\
\dfrac{a^3}{27} &= 1\\
a^3 &= 27\\
a &= 3\\
\end{align*}
So one possible solution is $a=3, b=-1$. Verify for yourself that $g(0)=-1$ and $g(-2)=3$ really are the local extrema of $g$.
Case 2: Suppose the local minimum is $f(1)=-1=4a^3/27 + b=g(-2a/3)$. Then...?
|
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Proving statements by its contrapositive Prove the following statement by proving its contrapositive:
“If $n^3 + 2n + 1$ is odd then n is even”
Therefore: $\lnot q \rightarrow \lnot p =$ "if $n^3 + 2n + 1$ is even then $n$ is odd.
So for this I began assuming that: $n=2k+1$
$(2k+1)^3 +2(2k+1)+1 = 8k^3+12k^2 +10k+4 = 2k(4k^2 +6k+5)+4$
The last statement: $2k$ is even, therefore $2k(4k^2 -6k+5)$ is also even and 4 is $2\cdot 2$ which is also even.
Now, my question is, when proving the contrapositive, what's your final conclusion? If it works for the contrapositive, then your theorem holds? Or is there something else?
|
$\displaystyle n^3+2n+1=n^3-n^2+n^2+n+n+1$
$=n^2(n-1)+n(n+1)+n+1\equiv n+1\pmod2$ as the product of any two consecutive integer is divisible by $2$
$\implies n^3+2n+1$ and $n$ has opposite parity
|
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Proving inequality $(a+\frac{1}{a})^2 + (b+\frac{1}{b})^2 \geq \frac{25}{2}$ for $a+b=1$ If $a, b$ are positive real numbers and $a+b = 1$, prove that :
$$\left(a+\frac{1}{a}\right)^2 + \left(b+\frac{1}{b}\right)^2 \geq \frac{25}{2}$$
I can see that the value $\frac{25}2$ is attained for $a=b=\frac12$. But I do not know how to show that this is the minimal possible value.
Thank you.
|
Generalization:
If $\sum_{1\le r\le n}a_r=S$ where $a_i$s are positive real numbers
$$\sum_{1\le r\le n}\left(a_r+\frac1{a_r}\right)^2=\sum_{1\le r\le n}a_r^2+\sum_{1\le r\le n}a_r^{-2}+2n$$
We know $$\frac{\sum_{1\le r\le n}a_r^m}n> \text{ or } <\left(\frac{\sum_{1\le r\le n}a_r}n\right)^m$$ according as $m$ lies or does not lie in $(0,1)$
Putting $m=2,$ $$\frac{\sum_{1\le r\le n}a_r^2}n>\left(\frac{\sum_{1\le r\le n}a_r}n\right)^2=\left(\frac Sn\right)^2=\frac{S^2}{n^2}$$
Putting $m=-2,$ $$\frac{\sum_{1\le r\le n}a_r^{-2}}n>\left(\frac{\sum_{1\le r\le n}a_r}n\right)^{-2}=\left(\frac Sn\right)^{-2}=\frac{n^2}{S^2}$$
On simplification, $$\sum_{1\le r\le n}\left(a_r+\frac1{a_r}\right)^2\ge \frac{(n^2+S^2)^2}{S^2n}$$
Here $S=1,n=2$
|
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|
Sum $ \sum\limits_{k=1}^{n} (-1)^k \frac{2k+3}{k(k+1)} $ I have the following Sum
$$ \sum\limits_{n=1}^{\infty} (-1)^n \frac{2n+3}{n(n+1)} $$
and I need to calculate the sums value by creating the partial sums.
I started by checking if $$\sum\limits_{n=1}^{\infty} \left| (-1)^n \frac{2n+3}{n(n+1)} \right|$$ converges.
i tried to check for convergence with the 2 criterias a) decreasing and b) zero sequence but yes then i tried to transform the equotation to
$$ \sum\limits_{n=1}^{\infty} \frac{2}{n+1} + \frac{1}{n(n+1)} $$
the first fraction is "ok" - the 2nd one i did partial fraction decomposition and finally got $$ \sum\limits_{n=1}^{\infty} \frac{2}{n+1} + \frac{1}{n} - \frac{1}{n+1} $$
i then tried to see a pattern by find out the first sequences but im not sure if i'm on the right track.
|
$\dfrac{2n+3}{n(n+1)} = \color{red}{\dfrac{3}{n}} - \color{blue}{\dfrac{1}{n+1}}$,
so
$$
\sum_{n=1}^{\infty} (-1)^n \dfrac{2n+3}{n(n+1)}
=\color{red}{\sum_{n=1}^{\infty} (-1)^n \dfrac{3}{n}} - \color{blue}{\sum_{n=1}^{\infty} (-1)^n \dfrac{1}{n+1}}.
$$
(all series are convergent here).
A)
$$
\color{red}{\sum_{n=1}^{\infty} (-1)^n \dfrac{3}{n}} = 3 \sum_{n=1}^{\infty} \dfrac{(-1)^n}{n} = -3\ln(2), \tag{A}
$$
(see Mercator series, Taylor series);
B)
$$
\color{blue}{\sum_{n=1}^{\infty} (-1)^n \dfrac{1}{n+1}} = \sum_{k=2}^{\infty} \dfrac{(-1)^{k-1}}{k} = -1+\sum_{k=1}^{\infty} \dfrac{(-1)^{k-1}}{k} = -1+\ln(2). \tag{B}
$$
Applying $(A)-(B)$, we get:
$$
\sum_{n=1}^{\infty} (-1)^n \dfrac{2n+3}{n(n+1)} = -3 \ln(2) - (-1+\ln(2)) = 1-4\ln(2).
$$
|
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Limit $\lim\limits_{x\to\infty} \frac{5x^2}{\sqrt{7x^2-3}}$ Evaluate the following limit: $$\lim_{x\to\infty} \frac{5x^2}{\sqrt{7x^2-3}}$$
I'm not really sure what to do when there is a square root for an infinity limit.
Please Help!
|
You can divide numerator and denominator by $x$ to find that the limit $\to + \infty$.
$$ \large \frac{5x^2}{\sqrt{7x^2-3}}\cdot \frac{\frac {1}{x}}{\frac{1}{\sqrt{x^2}}} = \frac {\frac {5x^2}{x}}{\sqrt {\frac {7x^2}{x^2} - \frac 3{x^2}}} = \frac {5x}{\sqrt{7 - \frac 3{x^2}}} = \frac {5x}{7}$$
This gives us that $$\lim_{x\to \infty} \frac{5x^2}{\sqrt{7x^2-3}} =\lim_{x \to \infty}\frac {5x}{7}$$
and clearly, $\dfrac {5x}{7} \to +\infty$ as $x \to \infty$.
|
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Making $R(x)=\frac{x^3+8x^2+18x+15}{x+5}$ continuous from $x = -5$ point The function $R(x)=\frac{x^3+8x^2+18x+15}{x+5}$ is not defined at the point $x = -5$. How should it be defined to make it continuous at this point.
I'm pretty sure you have to factor the top but for some reason I'm having a hard time with this problem.
Please Help.
|
This is where polynomial long division comes in handy.
We want to remove the removable singularity at $x = -5$, so we test to determine the quotient when dividing the numerator by the factor $(x + 5)$.
Doing so gives us: $$x^3 + 8x^2 + 18x + 15 = (x+5) (x^2 + 3x + 3)$$
This gives us $$\frac{x^3+8x^2+18x+15}{x+5} \implies \frac{(x+5)(x^2 + 3x + 3)}{(x + 5)} = x^2 + 3x + 3$$
which is continuous at $x = -5$.
|
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find this limit $\lim_{x\to0^{+}}\frac{\tan{(\tan{x})}-\tan{(\sin{x})}}{\tan{x}-\sin{x}}$ find the limit.
$$\lim_{x\to0^{+}}\dfrac{\tan{(\tan{x})}-\tan{(\sin{x})}}{\tan{x}-\sin{x}}$$
my try:
$$\tan{x}=x+\dfrac{1}{3}x^3+o(x^3),\sin{x}=x-\dfrac{1}{6}x^3+o(x^3)$$
so
$$\tan{(\tan{x})}=\tan{x}+\dfrac{1}{3}(\tan{x})^3+o(\tan^3{x})=x+\dfrac{1}{3}x^3+\dfrac{1}{3}(x+\dfrac{1}{3}x^3)^3+o(x^3)$$
$$\tan{(\sin{x})}=\sin{x}+\dfrac{1}{3}(\sin{x})^3+o(\sin^3{x})=x-\dfrac{1}{6}x^3+\dfrac{1}{3}(x-\dfrac{1}{6}x^3)^3+o(x^3)$$
so
$$\tan{(\tan{x})}-\tan{(\sin{x})}=\dfrac{1}{2}x^3+o(x^3)$$
$$\tan{x}-\sin{x}=\dfrac{1}{2}x^3+o(x^3)$$
Have other metods? Thank you
|
Since all the functions involved are continuosly differentiable we can apply the Mean Value Theorem to get
$$
\frac{\tan(\tan x)-\tan(\sin x)}{\tan x-\sin x}=\frac1{\cos^2 z}
$$
for some $z(x)$ with $\sin x\leq z(x)\leq\tan x$. By the Squeeze Rule, $z\to0$ when $x\to0$, and so the limit is $1/\cos^20=1$.
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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|
Compute $\int_0^n \left[\frac x {x+1} + \frac x {2x+4} + \frac x {3x+9} + \cdots\right] \, dx$ for $x>0$ I want to compute $\displaystyle \int_0^n \left[\frac x {x+1} + \frac x {2x+4} + \frac x {3x+9} + \cdots \right] \, dx$ for $x>0$
My attempt: The integrand can be written as a sum of $\displaystyle f_k(x)=\frac x{kx+k^2}$ which is positive for $x>0$, can we interchange sum and integral here? If so, then $\displaystyle \int_0^n \frac x{kx+k^2}=1-\log(2)$, so the given integral diverges?
|
\begin{align}
\sum_{k = 1}^{n}{x \over kx + k^{2}}
&=
\sum_{k = 1}^{n}\left({1 \over k} - {1 \over k + x}\right)
=
\sum_{k = 0}^{n - 1}{1 \over k + 1} - \sum_{k = 0}^{n - 1}{1 \over k + x + 1}
\\[3mm]&=
\left\lbrack\Psi\left(1 + n\right) - \Psi\left(1\right)\right\rbrack
-
\left\lbrack\Psi\left(1 + n + x\right) - \Psi\left(x + 1\right)\right\rbrack
\end{align}
\begin{align}
&\int_{0}^{n}\sum_{k = 1}^{n}{x \over kx + k^{2}}\,{\rm d}x
=
\left\lbrack\Psi\left(1 + n\right) - \Psi\left(1\right)\right\rbrack n
-
\left\lbrack%
\ln\Gamma\left(1 + 2n\right)
-
\ln\Gamma\left(1 + n\right)
\right\rbrack
\\[3mm]&+
\\[3mm]&
\left\lbrack%
\ln\Gamma\left(1 + n\right)
-
\ln\Gamma\left(1\right)
\right\rbrack
=
1 + \left\lbrack\Psi\left(n\right) + \gamma\right\rbrack n
-
\ln\left(2n\right)
-
\ln\Gamma\left(2n\right)
+
2\ln\left(n\right)
+
2\ln\Gamma\left(n\right)
\end{align}
$$
\begin{array}{|c|}\hline\\
\color{#ff0000}{%
\int_{0}^{n}\sum_{k = 1}^{n}{x\,{\rm d}x \over kx + k^{2}}
=
1 - \ln\left(2\right) + \left\lbrack\Psi\left(n\right) + \gamma\right\rbrack n
-
\ln\Gamma\left(2n\right)
+
\ln\left(n\right)
+
2\ln\Gamma\left(n\right)}
\\ \\ \hline
\end{array}
$$
\begin{align}
&\left.
\int_{0}^{n}\sum_{k = 1}^{n}{x \over kx + k^{2}}\,{\rm d}x
\right\vert_{n\ \gg\ 1}
\!\!\!\!\!\!\!\!\!\!\sim
1 - \ln\left(2\right) + \left\lbrack\ln\left(n\right) - {1 \over 2n} + \gamma\right\rbrack n
+
\ln\left(n\right)
+
\left({1 \over 2} - 2n\right)\ln\left(2\right) - {1 \over 2}\,\ln\left(n\right)
\\[3mm]&=
n\,\ln\left(n\right) -
\left\lbrack 2\ln\left(2\right) - \gamma\right\rbrack n
+
{1 \over 2}\,\ln\left(n\right)
+
{1 \over 2}\left\lbrack 1 - \ln\left(2\right)\right\rbrack
\end{align}
$$
\begin{array}{|c|}\hline\\
\color{#ff0000}{\large%
\int_{0}^{n}\sum_{k = 1}^{n}{x \over kx + k^{2}}\,{\rm d}x\
\sim\
n\,\ln\left(n\right)\,,
\qquad
n \gg 1}
\\ \\ \hline
\end{array}
$$
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 2
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|
Relationships between the elements $(a,b,c,d)$ of a solution to $A^2+B^2+4=C^2+D^2$ I have reduced a certain equation (in positive integers) to the equation
$$A^2 + B^2 + 4 = C^2 + D^2. \quad(\star)$$
Assume the positive integers $(a,b,c,d)$ are any solution to $(\star)$. Are there any algebraic restrictions on [i.e., relationships between] the elements which are well-known or easy to prove? I'm thinking of things like Pythagorean means, or other relative size or congruence restrictions.
Thank you,
Kieren.
|
A subset of integer solutions to,
$$A^2+B^2+4 = C^2+D^2\tag{1}$$
can be reduced to solving the Pell equation $x^2-ny^2 = 1$ for any non-square $n$ (where $n=5$ will involve the Fibonacci numbers). Consider the more general,
$$x_1^2+x_2^2+x_3^2 = y_1^2+y_2^2+y_3^2\tag{2}$$
The complete solution is,
$$(a+b)^2+(c+d)^2+(e+f)^2 = (a-b)^2+(c-d)^2+(e-f)^2\tag{3}$$
where,
$$ab+cd+ef = 0\tag{4}$$
It is easy to make the last term of $(3)$ vanish. Let $e=f=t$. Thus,
$$(a+b)^2+(c+d)^2+(2t)^2 = (a-b)^2+(c-d)^2\tag{5}$$
However, if you set $a,b,c = x+y,\, -x+y,\, y^2$, then condition $(4)$ becomes,
$$x^2-(d+1)y^2 = t^2\tag{6}$$
If you set $t = 1$, then any Pell equation will give integer solutions to $(1)$. For a nice example, let $d = 4$ so one has to solve,
$$x^2-5y^2 = 1$$
The solution is $x_n = \tfrac{1}{2}F_{6n}+F_{6n-1}$ (A023039) and $y_n = \tfrac{1}{2}F_{6n}$ (A060645), where $F_n$ are the Fibonacci numbers. After some tweaking so that all terms won't be even, a Fibonacci identity which satisfies $(1)$ is then,
$$(F_{6n+3})^2+( \tfrac{1}{4}F_{6n+3}^2+4)^2+4 = (F_{6n+2}+F_{6n+4})^2+(\tfrac{1}{4}F_{6n+3}^2-4)^2\tag{7}$$
where $n=1$ is,
$$34^2+293^2+4 = 76^2 +285^2$$
and so on.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/494534",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
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|
Calculate $(2013)^4 - 4(2011)^4+6(2009)^4-4(2007)^4+(2005)^4 = $ Calculate $(2013)^4 - 4(2011)^4+6(2009)^4-4(2007)^4+(2005)^4 = $
Try:: Let $x = 2009$, Then expression convert into $(x+4)^4-4(x+2)^4+6x^4-4(x-2)^4+(x-4)^4$
$\left\{(x+4)^4+(x-4)^4\right\}-4\left\{(x+2)^4+(x-2)^4\right\}+6x^4$
But This is very Complicated for calculation.
can anyone explain me better idea for that expression
If yes plz explain me
Thanks
|
Hint: This is a finite differences problem.
If $x(n)$ is any function, then $$(\Delta^4 x)(n)= x(n+4)-4x(n+3)+6x(n+2)-4x(n+1)+x(n)$$
Where $\Delta$ is the finite difference operator.
Now, since $x(n)=(2n+1)^4$, is a polynomial of degree $4$ with lead coefficient $2^4$ we know that $(\Delta^4x)(n)$ is constant, equal to $2^4\cdot 4!$.
Generalizing with $5$ instead of $4$, for example:
$$2013^5-5\cdot 2011^5 + 10\cdot 2009^5-10\cdot 2007^5 + 5\cdot 2005^5 -2003^5= 2^5\cdot 5!$$
or:
$$2013^1-2011^1 = 2^1\cdot 1!$$
$$2013^2-2\cdot 2011^2 + 2009^2 = 2^2\cdot 2!$$
$$2013^3-3\cdot 2011^3+3\cdot 2009^3 - 2007^3 = 2^3\cdot 3!$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/494686",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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|
Different Ways of Integrating $3\sin x\cos x$ I am asking this question for my son who is in (equivalent) twelfth
grade and I failed to answer his query.
When he tries to integrate $3\sin x\cos x$, he finds that
this can be done in at least following three ways.
And these three ways do not produce equivalent results.
ONE
Let us assume, $\sin x = z$.
This gives,
\begin{align*}
\cos x &= \frac{dz}{dx}\\
\cos x dx &= dz
\end{align*}
So, we can write,
\begin{align*}
\int 3\sin x\cos x dx &=3 \int zdz\\
&=3 \frac{z^2}{2}\\
&=\frac{3}{2} \sin^2 x\\
&=\frac{3}{4}\times 2\sin^2 x\\
&=\frac{3}{4} (1 -\cos 2x)\\
\end{align*}
TWO
Let us assume, $\cos x = z$.
This gives,
\begin{align*}
-\sin x &= \frac{dz}{dx}\\
\sin x dx &= -dz
\end{align*}
So, we can write,
\begin{align*}
\int 3\sin x\cos x dx &=-3 \int zdz\\
&=-3 \frac{z^2}{2}\\
&=-\frac{3}{2} \cos^2 x\\
&=-\frac{3}{4}\times 2\cos^2 x\\
&=-\frac{3}{4} (1 +\cos 2x)\\
\end{align*}
THREE
\begin{align*}
\int 3\sin x\cos x dx &=\frac{3}{2}\int 2\sin x\cos x dx\\
&=\frac{3}{2}\int \sin 2x dx\\
&=-\frac{3}{2}\times\frac{1}{2} \cos 2x\\
&=-\frac{3}{4} \cos 2x\\
\end{align*}
The results found in above three methods are not the same.
If we try a simple approach of evaluating the integration results at, $x = \frac{\pi}{6}$, we get as follows.
From the first one,
$\frac{3}{4} (1 -\cos 2x) = \frac{3}{4} (1 -\cos \frac{2\pi}{6})
= \frac{3}{4} (1 -\cos \frac{\pi}{3})
= \frac{3}{4} (1 - \frac{1}{2})
= \frac{3}{4}\times\frac{1}{2}
= \frac{3}{8}$
From the second one,
$-\frac{3}{4} (1 +\cos 2x) = -\frac{3}{4} (1 +\cos \frac{2\pi}{6})
= -\frac{3}{4} (1 +\cos \frac{\pi}{3})
= -\frac{3}{4} (1 + \frac{1}{2})
= -\frac{3}{4}\times\frac{3}{2}
= -\frac{9}{8}$
From the third one,
$-\frac{3}{4} \cos 2x=-\frac{3}{4} \cos \frac{2\pi}{6}
= -\frac{3}{4} \cos \frac{\pi}{3}
= -\frac{3}{4} \times \frac{1}{2}
= -\frac{3}{8} $
Clearly, we are getting some nonequivalent results. We have failed to find
the mistakes or explanations behind this. Your help will be appreciated.
|
Actually, we are to find the antiderivative of the given function, and all the functions you have found out to be potential answers are equally correct because any of them is just a constant shifted version of another.
And these are not the only candidate to be the answers, we can find infinitely many more. When we differentiate all the results you have found out, we only get $3sinxcosx $ . And this is why when we try to find out the antiderivative of $3sinxcosx $ , we can achieve so many answers having a interesting relationship among them that, any two of them are just a constant value shifted from the other.
NB: Desmos is used to plot the graphs
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/495159",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "11",
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"answer_id": 2
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|
Finding the equation of a circle given two points and the radius Can't seem to figure this out - the question is:
There are exactly two circles of radius $r = \sqrt{5}$ through the points $(6,3)$ and $(7,2)$. Find the equations of both circles.
I was thinking that I would find the equation of the line passing through these two points which would give me a chord on the circle. I could then find a line perpendicular to this by taking the negative reciprocal. This perpendicular bisector of the chord would pass through the center of the circle (which I'm assuming I need to find).
Am I making this more complicated than it actually is...? I don't know where to go from here
|
Here is a mostly 'square free' approach:
The two constraints are $\|x-(6,3)\|^2 = 5$ and $\|x-(7,2)\|^2 = 5$. Expanding gives $\|x\|^2-2 \langle (6,3), x \rangle +45 = 5$ and $\|x\|^2-2 \langle (7,2), x \rangle +53 = 5$,
subtracting and simplifying gives $\langle (1,-1), x \rangle = 4$. Note that this is a line at $45 °$.
The bisector of the two points is $(\frac{13}{2}, \frac{5}{2})$. The distance between the two points is $\sqrt{2}$, hence the distance from the points to the bisector is $\frac{1}{\sqrt{2}}$. Pythagoras gives the distance from the bisector to the centre of the circles as $\sqrt{5-\frac{1}{2}} = \sqrt{\frac{9}{2}} = \frac{3}{\sqrt{2}}$.
If we draw a line at $45 °$ through $(\frac{13}{2}, \frac{5}{2})$, and then mark off points that are $\frac{3}{\sqrt{2}}$ away on either side, we will find the centres. The corresponding displacements are $\pm (\frac{3}{2}, \frac{3}{2})$, hence the centres are $(\frac{13}{2}, \frac{5}{2}) \pm (\frac{3}{2}, \frac{3}{2})$, which gives the centres $(5,1), (8,4)$.
Hence the equations are
$$ (x_1-5)^2+(x_2-1)^2 = 5, \ \ (x_1-8)^2+(x_2-4)^2 = 5$$
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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|
LU decomposition by hand Can someone show me a step by step solution to calculate the $LU$ decompisition of the following matrix:
$A = \begin{bmatrix}
5 & 5 & 10 \\
2 & 8 & 6 \\
3 & 6 & -9
\end{bmatrix}$
I am getting the correct U by I am having difficult getting the correct L.
Here is the correct Answer
|
$A = \begin{pmatrix}
5 & 5 & 10 \\
2 & 8 & 6 \\
3 & 6 & -9
\end{pmatrix}$
First step (working on the first column):
a) Multiply the first row with $l_{21}= \frac{2}{5}$ and subtract it from row 2
b) Multiply the first row with $l_{31}= \frac{3}{5}$ and subtract it from row 2
This leads to
$U^{(1)} = \begin{pmatrix}
5 & 5 & 10 \\
0 & 6 & 2 \\
0 & 3 & -15
\end{pmatrix}$
Second step (working on the second column):
c) Multiply row 2 with $l_{32} = \frac{1}{2}$ and subtract it from row 3. This gives
$U = U^{(2)} = \begin{pmatrix}
5 & 5 & 10 \\
0 & 6 & 2 \\
0 & 0 & -16
\end{pmatrix}$
And we have
$L = \begin{pmatrix}
1 & 0 & 0 \\
l_{21} & 1 & 0 \\
l_{31} & l_{32} & 1
\end{pmatrix}
= \begin{pmatrix}
1 & 0 & 0 \\
\frac{2}{5} & 1 & 0 \\
\frac{3}{5} & \frac{1}{2} & 1
\end{pmatrix}
$
So note that $l_{ij}$ was used to create a zero in the $i$-th row and $j$-th column by multiplying (in the $j$-th step) the $j$-th row with $l_{ij}$ and subtract the result from the $i$-th row.
|
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|
How do I solve this Partial Fractions question? $$\frac{x^4}{(x^2-3)(x^2+3)}\;$$
How would I do this?
My attempt started with this:
$$\frac{A+Bx}{(x^2-3)}\; + \frac{C+Dx}{(x^2+3)}$$
But when I start working it all out, A, B, C and D all go to 0. What am I doing wrong?
|
Now that the $x$ in the numerator is really supposed to be $x^4$, we first need the degree in the numerator to be less than that of the denominator before employing partial fraction decomposition. Note that the denominator can be expressed as the difference of squares: $x^4 - 9$, which makes polynomial division very straightforward. We can also see the factor $x^2 - 3$ as a difference of squares and factor accordingly knowing $x^2 - 3 = (x + \sqrt 3)(x - \sqrt 3)$:
$$\begin{align} \frac{x^4}{(x^2-3)(x^2+3)} & = \dfrac{(x^4 - 9) + 9}{(x^2 - 3)(x^2 + 3)} \\ \\\dfrac {x^4 - 9}{x^4 - 9} + \dfrac 9{(x^2 - 3)(x^2 + 3)} &= 1 + \dfrac 9{(x+ \sqrt 3)(x - \sqrt 3)(x^2 + 3)} \\ \\ &= 1 + \dfrac A{x + \sqrt 3} + \frac B{x - \sqrt 3} + \frac{Cx + D}{x^2 + 3}\end{align}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Improving bound on $\sqrt{2 \sqrt{3 \sqrt{4 \ldots}}}$ An old challenge problem I saw asked to prove that $\sqrt{2 \sqrt{3 \sqrt{4 \ldots}}} < 3$. A simple calculation shows the actual value seems to be around $2.8$, which is pretty close to $3$ but leaves a gap. Can someone find $C < 3$ and prove $\sqrt{2 \sqrt{3 \sqrt{4 \ldots}}} < C$?
|
Reusing my answer on Quora and showing how it can be used to improve bounds
Write the expression as
$$\displaystyle 2\sqrt[2]{\frac{3}{2}}\sqrt[4]{\frac{4}{3}}\sqrt[8]{\frac{5}{4}}\sqrt[16]{\frac{6}{5}}...$$
$$\displaystyle \prod\limits_{n=1}^{+\infty} \left ( 1 + \frac{1}{n} \right )^{\frac{1}{2^{n-1}}}$$
$$\displaystyle \prod\limits_{n=1}^{+\infty} \left ( 1 + \frac{1}{n} \right )^{\frac{n}{n2^{n-1}}} < 2\prod\limits_{n=2}^{+\infty} e^{\frac{1}{n2^{n-1}}}$$
Now
$$\displaystyle \prod\limits_{n=1}^{+\infty} e^{\frac{1}{n2^{n-1}}}=4$$
because
$$\displaystyle \sum\limits_{n=1}^{+\infty} \frac{1}{n2^{n-1}}=\log(4)$$
$$\displaystyle 2\prod\limits_{n=2}^{+\infty} e^{\frac{1}{n2^{n-1}}}=2\frac{4}{e}<3$$
The expression can be used to get more convergents, simply take first few terms and set the rest to the exponent.
$$\frac{8\sqrt[4]{3}}{e^{\frac{4}{3}}}\approx 2.77$$
$$4\frac{2^\frac{3}{4}\sqrt[4]{3}\sqrt[8]{5}}{e^{\frac{131}{96}}} \approx 2.766$$ already
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/498774",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "9",
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|
Multiplication of summations The length of the following equation depends on the value of the variable n. How can I write this equation? How can I reformulate it?
$$f(n)=x+y\cdot\sum_{a=1}^{n}\left(x+y\cdot\sum_{b=1}^{n-1}\left(x+y\cdot\sum_{c=1}^{n-2}\left(x+y\cdot\sum_{d=1}^{n-3}....\right)\right)\right)$$
Moreover I'd need that the sum $\sum_{1}^{1}$ equals 0!
for example:
$$f(3)=x+y\cdot\sum_{a=1}^{3}\left(x+y\cdot\sum_{b=1}^{2}\left(x+y\cdot\sum_{c=1}^{1}\left(x+y\cdot0\right)\right)\right)$$
UPDATE
Re-written differently I think it gives. But still, the length of my equation depends on the value of n
$$f(n)=x+y \cdot (n \cdot x + n \cdot y \cdot ((n-1)\cdot x + (n-1) \cdot y \cdot ((n-2)\cdot x + (n-2)\cdot y \cdot (....)) ))$$
for example:
$$f(3)=x+y \cdot (3 \cdot x + 3 \cdot y \cdot ((2)\cdot x + (2) \cdot y \cdot ((1)\cdot x + (1)\cdot y \cdot 0)))$$
Thank you!
|
Given that summing $m$ times the same thing amounts tom multiplying it by $m$, you get
$$
f(n)=x+yn(x+y(n-1)(x+y(n-2)(\cdots(x+y.1(x+0y))\ldots)))
$$
which resembles a polynomial in $y$ expressed by a Horner scheme; you get
$$
\begin{align}
f(n)&=x+xny+xn(n-1)y+n(n-1)(n-2)y^3x+\cdots+n(n-1)\ldots2xy^{n-1}+n!xy^n
\\ &=x\sum_{i=0}^nn^{\underline i}y^i
\end{align}
$$
where $n^{\underline i}$ denotes the falling factorial power.
|
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|
steady state solution to differential equation - checking my work EDIT: fixed a stray negative sign.
The problem as given:
$y'' + 2y' + 5y = 10\cos t$
We want to find the general solution and the steady-state solution. We're using $\mu y'' + c y' + k y = F(t)$ as our general form.
OK, so I first want the general solution to the 2nd order DE. That should be easy: the characteristic equation is $\lambda^2 + 2\lambda + 5$ and it has imaginary roots at $-1 \pm \sqrt{-4}$ which is $-1\pm 2i$.
So the general solution should be $y=C_1e^{-t} \cos 2t + C_2e^{-t}\sin2t$
The steady state solution is given by the Fourier series $y_p(t) = \alpha_0 + \sum_{n=0}^\infty \left( \alpha_n\cos \frac{n\pi}{p} t + \beta_n \sin \frac {n\pi}{p}t\right)$
We know the constants $c, \mu, k$ from the original equation, they are 1, 2, 5, respectively. So $\alpha_0 = a_0/k$ and $a_0 = \frac{1}{2\pi}\int_{-\pi}^\pi 10 \cos tdt$ and since $\cos t$ is an even function it goes to zero on any symmetric interval. so $\alpha_0=0$.
To get $\alpha_n$ and $\beta_n$: $$\alpha_n = \frac{A_n a_n - B_nb_n}{A_n^2+B_n^2},\ A_n = k-\mu\left(\frac{n\pi}{p}\right)^2, \beta_n = \frac{A_n b_n + B_na_n}{A_n^2+B_n^2},\ A_n = k-\mu\left(\frac{n\pi}{p}\right)^2$$ and the latter expression is $(5-1)(\frac{n\pi}{2\pi})^2$ which makes $A_n = n^2$. $B_n = c\frac{n\pi}{2\pi} = n$. That gets us $\alpha_n= \frac{n^2 a_n - nb_n}{n^4+n^2}$. Now to find $a_n$ and $b_n$.
From the other Fourier formulas $$a_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \cos nx dx, \ b_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \sin nx dx$$ or $$a_n = \frac{1}{\pi} \int_{-\pi}^{\pi} 10 \cos t dt \cos nt, \ b_n = \frac{1}{\pi} \int_{-\pi}^{\pi} 10\cos t \sin nt dt$$
We can make the integration simpler using some trig identities because integration by parts gets you a bunch of repeating integrals. $\cos a \sin b = \frac{1}{2}[\sin (a+b) - \sin (a-b)$ and $\frac{1}{2}[\cos(a+b)+\cos(a-b)] = \cos a\cos b$. That gets us $$a_n = \frac{10}{\pi} \int_{-\pi}^{\pi} \frac{1}{2}[\cos ((1+n)t) +\cos ((1-n)t)]dt, \ b_n = \frac{10}{\pi} \int_{-\pi}^{\pi} \frac{1}{2}[\sin ((1+n)t) -\sin ((1-n)t)]dt$$
Integrating all this stuff I got:
$$a_n=\frac{5}{\pi}\left[\frac{\sin (1+n)t}{1+n}+\frac{\sin (1-n)t}{1-n} \right]_ {-\pi}^{\pi}, \ b_n=\frac{5}{\pi}\left[\frac{-\cos (1+n)t}{1+n}+\frac{-\cos (1-n)t}{1-n} \right]_ {-\pi}^{\pi}$$
$a_n$ goes to zero since any integer multiple of $\sin \pi$ is zero. That leaves $b_n$, and leaves us with $b_n =\frac{5}{\pi} \left[\frac{2}{1-n^2}\right]$. We go all the way back to the equations for $\alpha_n$ and $\beta_n$ and when I plug the relevant values into the expression for $\alpha_n$ I got $\frac{-10}{\pi(n - n^5)}$. Meanwhile I got the value for $\beta_n = \frac{-10}{\pi - \pi n^4}$.
After all that I end up with: $$y_p(t) = \sum_{n=0}^\infty \left( \frac{-10}{\pi(n - n^5)}\cos \frac{n}{2} t + \frac{-10}{\pi - \pi n^4} \sin \frac {n}{2}t\right)$$
Is that my steady-state solution? Or did I really screw things up someplace? I am sorry for the long post but I like to make sure I work through all the steps.
And anyone brave enough to read this thanks a bunch for your time and patience.
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Answering my own question and taking Artem's advice, I tried this....
$y'' + 2y' + 5y = 10\cos t$
We want to find the general solution and the steady-state solution. We're using $\mu y'' + c y' + k y = F(t)$ as our general form.
the characteristic equation is $\lambda^2 + 2\lambda + 5$ and it has imaginary roots at $-1 \pm \sqrt{-4}$ which is $-1\pm 2i$.
So the general solution should be $y=C_1e^{-t} \cos 2t + C_2e^{-t}\sin 2t$
The steady state solution is given by the Fourier series $y_p(t) = \alpha_0 + \sum_{n=0}^\infty \left( \alpha_n\cos \frac{n\pi}{p} t + \beta_n \sin \frac {n\pi}{p}t\right)$
But we needn't be that complicated, it seems to me, since could try looking for solutions in the form of $y_p = A \sin t + B \cos t$.
We differentiate and we get $y_p' = A \cos t - B \sin t$ and $y_p'' = -A \sin t - B \cos t$.
Substitute in $y_p$ to the original equation. We get $$-A \sin t - B \cos t + 2(A \cos t - B \sin t)+ 5(A \sin t + B \cos t) = 10 \cos t$$
And simplifying we have
$$(4A - 2B) \sin t - (4B+ 2A) \cos t = 10 \cos t$$
And that leaves us with $(4A - 2B) = 0$ and $(-4B - 2A) = 10$. Solving for both we get $B = -2$ and $A =-1$.
So the steady state solution is
$$y=C_1e^{-t} \cos 2t + C_2e^{-t}\sin 2t - \sin t - 2 \cos t $$
Now here's the part where everyone tells me I messed this up :-) I just feel I have done something way off here. But it seems ok...
Anyhow, thanks for bearing with me.
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|
Skew operator Squared Proof I have a linear algebra question that is a proof and I am unsure how to approach this problem.
For any vector $s$, show that $$(s\times)^2 = ss^T-s^TsI$$ where the skew operator $s\times$ is defined by $$s\times\triangleq\begin{bmatrix}0&-c_s&b_s\\c_s&0&-a_s\\-b_s&a_s&0\end{bmatrix}$$
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So you want to show
$$\begin{bmatrix} 0 & -c & b \\ c & 0 & -a \\ -b & a & 0\end{bmatrix}^2=\begin{bmatrix}a \\ b \\ c\end{bmatrix}\begin{bmatrix} a & b & c\end{bmatrix}-(a^2+b^2+c^2)\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}.$$
What's wrong with simplifying both sides and seeing that they're the same?
You should already understand the following:
$$\begin{bmatrix}a \\ b \\ c\end{bmatrix}\begin{bmatrix} a & b & c\end{bmatrix}=\begin{bmatrix}a^2 & ab & ac \\ ba & b^2 & bc \\ ca & cb & c^2\end{bmatrix}$$
$$(a^2+b^2+c^2)\begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}=\begin{bmatrix}a^2+b^2+c^2 & 0 & 0 \\ 0 & a^2+b^2+c^2 & 0 \\ 0 & 0 & a^2+b^2+c^2\end{bmatrix}$$
This is simply how matrix multiplication and scalar multiplication work.
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|
How prove this $ \sqrt{\frac{a}{a+3b+5bc}}+\sqrt{\frac{b}{b+3c+5ca}}+\sqrt{\frac{c}{c+3a+5ab}}\geq 1.$ Let $a,b,c$ be nonnegative real numbers such that $a+b+c=3$, Prove that
$$ \sqrt{\frac{a}{a+3b+5bc}}+\sqrt{\frac{b}{b+3c+5ca}}+\sqrt{\frac{c}{c+3a+5ab}}\geq 1.$$
This problem is from http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=555716
@Calvin Lin Thank you
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Let $\displaystyle A = \sqrt{\frac{a}{a+3b+5bc}}+\sqrt{\frac{b}{b+3c+5ca}}+\sqrt{\frac{c}{c+3a+5ab}}$ and $\displaystyle B = \sum_{cyc}a^2(a+3b+5bc)$.
Then by Hölder's inequality we have $A^2B \ge (a+b+c)^3 = 27$.
So it is sufficient to prove that $B \le 27$
$$B = \sum_{cyc}a^3 + 3 \sum_{cyc}a^2b+5\sum_{cyc}a^2bc$$
As $\displaystyle \sum_{cyc}ab^2 \ge 3abc$ by AM-GM, we have
$$B \le \left(\sum_{cyc}a^3 + 3 \sum_{cyc}a^2b + 3 \sum_{cyc}ab^2 + 6abc\right) - 15abc + 5\sum_{cyc}a^2bc \\
= (a+b+c)^3 - 5abc (3-\sum_{cyc}a) = 27$$
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|
Common linear and Quadratic factors (i) If $ax^5+bx^2+c$ has a factor of the form $x^2+px+1$ prove that:
$(a^2-c^2)(a^2-c^2+bc)=(ab)^{2}$
(ii)In this case prove that:
$ax^5+bx^2+c$ and $cx^5+bx^3+a$ have a common quadratic factor
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Part (i) is a continuation of Part (ii).
m is a root of g(x) = 0 and hence is a root of f(x) = 0.
Therefore, $am^5 + bm^2 + c = 0$……(4)
Using n = 1/m as root, from part (ii), we have $a + bm^3 +cm^5 = 0$…….(5)
Eliminating a from (4) & (5) by performing $(5)*m^5 – (4)$, we have
$bm^2(m^6 – 1) + c[m^T – 1] = 0$……(6) [T = 10 because of the display matter.]
Using the same technique to eliminate b and then c, we get
$a(m^6 – 1) – cm(m^4 – 1) = 0$……..(7)
$a(m^T – 1) + bm^3(m^4 – 1) = 0$ …...(8)
(6), (7), (8) combined yield the ratio of a : b : c such that, for some k
$a = –m^3(m^4 – 1) k$;
$b = (m^T – 1)k$; and
$c = –m^2(m^6 – 1)k$.
Using these values can verify that LHS = … = $m^6k^4(m^T – 1)^2(1 – m^4)^2$ = … = RHS.
Note-1 Who designs this question that involves so many calculations?
Note-2 This work is just doing the verification. How to get the required product is still unknown.
Note-3 Would like to know any other shorter method or better approach.
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integrating $\int \frac{dt}{(t+2)^2(t+1)}$ I'm practicing to solve a whole, and I am not able to solve this one, could you help me? $$\int \frac{dt}{(t+2)^2(t+1)}$$I tried $$\frac{1}{(t+2)^2(t+1)}=\frac{A}{(t+2)^2}+\frac{B}{(t+2)}+\frac{C}{(t+1)}\\1=A(t+1)+B(t+2)(t+1)+C(t+2)^2\\t=-2\Longrightarrow 1=-A\Longrightarrow \fbox{$A=-1$}\\t=-1\Longrightarrow1=C(-1)^2\Longrightarrow\fbox{$C=1$}$$Making $t = 0$ and substituting $A$ and $C$ we have $$1=A+2B+4C=-1+2B+4=2B+3\\\fbox{$B=-1$}$$THEN$$\int \frac{dt}{(t+2)^2(t+1)}=\int -\frac{1}{(t+2)^2}-\frac{1}{(t+2)}+\frac{1}{(t+1)}\;dt\\=-\int \frac{1}{(t+2)^2}-\int\frac{1}{(t+2)}+\int\frac{1}{(t+1)}\;dt=-\int u^{-2\;}du-\ln|t+2|+\ln|t+1|\\=-\frac{u^{-1}}{-1}-\ln|t+2|+\ln|t+1|=\\\fbox{$\frac{1}{t+2}-\ln|t+2|+\ln|t+1|+c$}$$Only I could not do the derivative to "take the test", can you help me? Or contains an error in my resolution?
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Differentiating $$\frac{1}{t+2} - \ln |t + 2| + \ln|t+1| + c$$ gives you $$-\frac{1}{(t+2)^2} - \frac{1}{t + 2} + \frac{1}{t+1}$$ which is certainly what you got when you split the fraction initially. To check that your A, B and C were correct, see that this is just $$\frac{-(t + 1) - (t + 2)(t+1) + (t+2)^2}{(t+2)^2(t+1)}.$$ Simplifying, we get $$-(t+1)-(t+2)(t+1)+(t+2)^2 = -(t+1) - (t^2 + 3t + 2) + (t^2 + 4t +4) = 1,$$ as required.
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minimal polynomial of a matrix with some unknown entries Question is to prove that :
characteristic and minimal polynomial of $ \left( \begin{array}{cccc}
0 & 0 & c \\
1 & 0 & b \\
0 & 1 & a \end{array} \right) $ is $x^3-ax^2-bx-c$.
what i have done so far is :
characteristic polynomial of a matrix $A$ is given by $\det(A-xI)$
in case of $A= \left( \begin{array}{cccc}
0 & 0 & c \\
1 & 0 & b \\
0 & 1 & a \end{array} \right)$ we have $\det(A-xI)=\det\left( \begin{array}{cccc}
-x & 0 & c \\
1 & -x & b \\
0 & 1 & a-x \end{array} \right)=-(x^3-ax^2-bx-c)$
So, i have got the characteristic polynomial as $x^3-ax^2-bx-c$.
Now, the problem is how do i find minimal polynomial.
As $a,b,c$ are arbitrary, I can not factorize $x^3-ax^2-bx-c$ so as to see which factor gives me minimal polynomial.
I am confused.
please suggest me some hint.
EDIT : This is just after Mr.Will Jagyy's hint :
I have $A= \left( \begin{array}{cccc}
0 & 0 & c \\
1 & 0 & b \\
0 & 1 & a \end{array} \right)$ then, $A^2= \left( \begin{array}{cccc}
0 & 0 & c \\
1 & 0 & b \\
0 & 1 & a \end{array} \right)\left( \begin{array}{cccc}
0 & 0 & c \\
1 & 0 & b \\
0 & 1 & a \end{array} \right)=\left( \begin{array}{cccc}
0 & c & ac \\
0 & b & c+ab \\
1 & a & b+a^2 \end{array} \right)$
Now,
$A^2+rA+sI=\left( \begin{array}{cccc}
0 & c & ac \\
0 & b & c+ab \\
1 & a & b+a^2 \end{array} \right)+r\left( \begin{array}{cccc}
0 & 0 & c \\
1 & 0 & b \\
0 & 1 & a \end{array} \right)+s\left( \begin{array}{cccc}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1 \end{array} \right)=\left( \begin{array}{cccc}
s & c & * \\
r & b+s & * \\
1& * & * \end{array} \right)$
As element of $3^{rd}$ row $1^{st}$ column is $1$ in above matrix, this can never be $0$
i.e., $A^2+rA+sI$ can never be $0$.
Now, $A+rI=\left( \begin{array}{cccc}
0 & 0 & c \\
1 & 0 & b \\
0 & 1 & a \end{array} \right)+r\left( \begin{array}{cccc}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1 \end{array} \right)=\left( \begin{array}{cccc}
r & *& * \\
\color{magenta}{1} & * & * \\
* & \color{magenta}{1} & * \end{array} \right)\neq 0$ if $r\neq 0$
Thus, $A^2+rA+sI\neq 0$ and $A+rI\neq 0$ for any $r,s$.
Thus, minimal polynomial for $A$ can not be of order less than $3$.
Thus, minimal polynomial for $A$ has to be $x^3-ax^2-bx-c$.
I have written this just to make sure i have tried in correct way as i can not write this in a comment.
I would be thankful if there is any other way to proceed further.. Thank you :)
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Here's how I would argue the point: clearly the minimal polynomial $m_A(x)$ of $A$,
where
$A = \begin{bmatrix} 0 & 0 & c \\ 1 & 0 & b \\ 0 & 1 & a \end{bmatrix}, \tag{1}$
must satisfy $\deg m_A(x) \le 3$, since the characteristic poynomial $p_A(x)$ of $A$ satisfies $\deg p_A(x) = 3$. Furthermore, we also have $m_A(x) \mid p_A(x)$. Thus $p_A(x) = q(x)m_A(x)$ for some polynomial $q(x)$. Now if $\deg m_A(x) = 3$, we must have $\deg q(x) = 0$ since $\deg p_A(x) = \deg q(x) + \deg m_A(x)$. This, combined with the fact that $m_A(x), p_A(x)$ are both monic, leads to the conclusion that $q(x) = 1$ and hence $m_A(x) = p_A(x)$; we are done in this case. Now suppose that $\deg m_A(x) = 1$; then $m_A(x)$ must be of the form $m_A(x) = x - d$ for some constant $d$. Then since $m_A(A) = 0$, we obtain $A = dI$, where $I$ is the $3 \times 3$ identity matrix; then $A$ is diagonal; but this contradicts the fact that $A_{21} = A_{32} = 1$ for any choice of $a, b, c$; thus the possibility that $\deg m_A(x) = 1$ may be excluded. The case $\deg m_A(x) = 2$ remains; here we have $m_A(x) = x^2 - dx - e$ for some $d, e$. Then
$m_A(A) = 0$ implies $A^2 = dA + eI$. We compute $A^2$:
$A^2 = \begin{bmatrix} 0 & 0 & c \\ 1 & 0 & b \\ 0 & 1 & a \end{bmatrix} \begin{bmatrix} 0 & 0 & c\\ 1 & 0 & b \\ 0 & 1 & a \end{bmatrix} =\begin{bmatrix} 0 & c & ac \\ 0 & b & ab + c \\ 1 & a & a^2 + b \end{bmatrix}, \tag{2}$
and observe that $A^2_{31} = 1$ for all $a, b, c$; on the other hand, inspection of (1) shows that, for any $d, e$,
$(dA + e)_{31} = 0; \tag{3}$
we thus conclude that $\deg m_A(x) = 2$ is impossible. This leaves us with $\deg m_A(x) = 3$, and as we have seen, $m_A(x) = p_A(x)$ in this situation. QED.
Additional Observations: Some curiosities of the matrix $A$: Suppose we write $A$ in the form
$A = A_1 + N, \tag{4}$
where
$A_1 = \begin{bmatrix} 0 & 0 & c \\ 0 & 0 & b \\ 0 & 0 & a \end{bmatrix}, \tag{5}$
and
$N = \begin{bmatrix} 0 & 0 & 0 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{bmatrix}; \tag{6}$
then we have
$A_1^2 = \begin{bmatrix} 0 & 0 & ac \\ 0 & 0 & ab \\ 0 & 0 & a^2 \end{bmatrix}, \tag{7}$
and
$N^2 = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 1 & 0 & 0 \end{bmatrix}; \tag{8}$
then
$A^2 = (A_1 + N)^2 = A_1^2 + A_1 N + N A_1 + N^2, \tag{9}$
with
$A_1 N = \begin{bmatrix} 0 & c & 0 \\ 0 & b & 0 \\ 0 & a & 0 \end{bmatrix} \tag{10}$
and
$N A_1 = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & c \\ 0 & 0 & b \end{bmatrix} \tag{11}$
and
$N^3 = 0, \tag{12}$
i.e. $N$ is nilpotent of degree $3$; furthermore, (7) shows that the third column of $A_1$, $(c, b, a)^T$, is in fact an eigenvector of $A$ with eigenvalue $a$; thus
$A_1^m = \begin{bmatrix} 0 & 0 & a^{m - 1}c \\ 0 & 0 & a^{m - 1}b \\ 0 & 0 & a^m \end{bmatrix} \tag{13}$
readily follows; in contrast to (12), $A_1^m \ne 0$ as long as $a$ does not vanish.
Not sure where all this leads, but it seems that the peculiar form of $N$, $N^2$ etc.
has something to do with the above argument that $\deg m_A(x) \ne 2$. My guts tell me there is a generalization waiting to appear. Like William Buther Yeats' "rough beast" (see his poem The Second Coming), something is "slouching towards Bethlehem to be born."
END: Additional Observations: Some curiosities of the matrix $A$.
Hope at least some of this spiel helps. Cheerio,
and as always,
Fiat Lux!!!
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If $ k = \sqrt{n\cdot (n+1)\cdot(n+2)\cdot (n+3)}$. Then $\lfloor k \rfloor =$ If $ k = \sqrt{n\cdot (n+1)\cdot(n+2)\cdot (n+3)}$, where $n\in \mathbb{N}$. Then $\lfloor k \rfloor = $
$\underline{\bf{My\; Try}}::$ We can write the expression $n\cdot (n+1)\cdot(n+2)\cdot (n+3) = (n^2+3n).(n^2+3n+2)$
$ = (n^2+3n)^2+2\cdot (n^2+3n)+1-1 = (n^2+3n+1)^2-1<(n^2+3n)^2$
Now I Did not Understand How can i solve further
Help Required.
Thanks
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You are using the right approach, we need to establish upper and lower bounds on k. The lower bound needs to be an integer and the upper bound must be no greater than 1 plus the lower bound. This will result in the lower bound being the floor of k. jim established the correct upper bound by adding 1 under the radical.
To see the lower bound, note that $n(n+1)(n+2)(n+3) = n^4 + 6n^3 + 11n^2 + 6n$. Looking at the upper bound, we should try to get a lower bound of $n^2+3n$. So we need $(n^2+3n)^2$ to be a lower bound of $n(n+1)(n+2)(n+3)$. We can see that $(n^2+3n)^2 = n^4 + 6n^3 + 9n^2 < n^4 + 6n^3 + 11n^2 + 6n = n(n+1)(n+2)(n+3)$, which establishes our lower bound.
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Finding the product of real and imaginary roots separately If I am given a polynomial of nth degree and asked to fond the product of real and imaginary roots what steps should I take?
I know how to calculate the sum or product of all roots of a polynomial of nth degree but how to separately find the product of real and imaginary roots?
For example if I'm asked to find the product of real roots of $2x^4 + 3x^3 - 11x^2 - 9x + 15 = 0$.
What steps should I take?
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This is too long for a comment. Just found a partial answer for $x^4+a_3x^3+a_2x^2+a_1x+a_0=0$. Let $y$ and $x$ denote the product of the real and complex roots and $b$ and $a$ the sums of the real and complex roots resp. Playing Vieta with the coefficients and those sums and product, you will find the following system of equations.
$$\begin{align*}
yx&=a_0\\
a+b&=a_3\\
y+x+ab&=a_2\\
xb+ya&=a_1
\end{align*}
$$
As this system is symmetric in $a\leftrightarrow b$ and in $x\leftrightarrow y$ we expect six solutions. For the given polynomial Mathematica confronts us with:
$$
\left\{\left\{a\to 0,b\to \frac{3}{2},x\to -3,y\to
-\frac{5}{2}\right\},\left\{a\to \frac{3}{2},b\to 0,x\to -\frac{5}{2},y\to
-3\right\},\left\{a\to \frac{1}{2} \left(5-2 \sqrt{3}\right),b\to
\sqrt{3}-1,x\to -\frac{5 \sqrt{3}}{2},y\to -\sqrt{3}\right\},\left\{a\to
-1-\sqrt{3},b\to \frac{1}{2} \left(5+2 \sqrt{3}\right),x\to \sqrt{3},y\to
\frac{5 \sqrt{3}}{2}\right\},\left\{a\to \sqrt{3}-1,b\to \frac{1}{2}
\left(5-2 \sqrt{3}\right),x\to -\sqrt{3},y\to -\frac{5
\sqrt{3}}{2}\right\},\left\{a\to \frac{1}{2} \left(5+2 \sqrt{3}\right),b\to
-1-\sqrt{3},x\to \frac{5 \sqrt{3}}{2},y\to \sqrt{3}\right\}\right\}
$$
Only the first solution fits to the given equation. At the moment I haven't a clue to de-symmetrize the conditions. Alas, it's a beginning.
Michael
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How find this equation $\prod\left(x+\frac{1}{2x}-1\right)=\prod\left(1-\frac{zx}{y}\right)$ let $x,y,z\in(0,1)$, find the pairs of $(x,y,z)$ such
$$\left(x+\dfrac{1}{2x}-1\right)\left(y+\dfrac{1}{2y}-1\right)\left(z+\dfrac{1}{2z}-1\right)=\left(1-\dfrac{xy}{z}\right)\left(1-\dfrac{yz}{x}\right)\left(1-\dfrac{zx}{y}\right)$$
my try:use
$$x+\dfrac{1}{2x}-1\ge 2\sqrt{\dfrac{1}{2}}-1=\sqrt{2}-1$$
so
$$LHS\ge (\sqrt{2}-1)^3$$
so I guess
$$RHS\le (\sqrt{2}-1)^3$$
But I can't prove it.Thank you
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It is fairly obvious that, since the expressions on both sides are symmetrical with regards to each of the three variables, they can easily be reduced to $$\left(t + \frac1{2t} - 1\right)^3 = \left(1 - \frac{t\ \cdot\ t}t\right)^3 \qquad\iff\qquad \left(t + \frac1{2t} - 1\right) = (1 - t) \quad | \cdot 2t$$ $$2t^2 + 1 - 2t\ =\ 2t - 2t^2 \qquad\iff\qquad 4t^2 - 4t + 1\ =\ 0 \quad\iff\quad (2t - 1)^2\ =\ 0$$ whose only solution is $t = \frac12$ , which is the only point of intersection between the hyperbola on the left, and the linear polynomial on the right.
Another approach, also exploiting symmetry: Let $$f_t(u,v) = t + \frac1{2t} - 1 \qquad ; \qquad g_t(u,v) = 1 - \frac{u\ \cdot\ v}t$$ Then our equation becomes $$f_x(y,z) \cdot f_y(x,z) \cdot f_z(x,y)\ =\ g_x(y,z) \cdot g_y(x,z) \cdot g_z(x,y)$$ which, due to its inherent permutability, is ultimately reduced to $f_t(u,v) = g_t(u,v)$ , for all three possible cases. This, in its turn, yields $$t + \frac1{2t} - 1 = 1 - \frac{uv}t\ \iff\ 2t^2 + 1 - 2t = 2t - 2uv\ \iff\ 2t^2 - 4t + (2uv + 1) = 0$$ from where we get the formula $t = 1 - \sqrt{\frac12 - uv}$ , which, when spelled out for all three variables, becomes $$x = 1 - \sqrt{\tfrac12 - yz} \qquad,\qquad y = 1 - \sqrt{\tfrac12 - xz} \qquad,\qquad z = 1 - \sqrt{\tfrac12 - xy}$$ By extracting $x$ from the latter two in terms of $y$ and $z$, we get $$x = \frac{1 - 2(1-y)^2}z \qquad,\qquad x = \frac{1 - 2(1-z)^2}y$$ $$\iff\quad y\cdot[1 - 2(1-y)^2]\ =\ z\cdot[1 - 2(1-z)^2] \quad\iff\quad y = z$$ Analogously, we show that $x = y$ and $x = z$ , ultimately leading to $x = y = z$ , for which the only possible solution is $\frac12$ .
|
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Diophantus math Find two numbers such that their difference and also the difference of their cubes are given numbers; say, their difference is 6 and the difference of their cubes are 504. Call the numbers $x + 3$ and $x - 3$.
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Well, using the difference of cubes formula $a^3-b^3=(a-b)(a^2+ab+b^2),$ we get $$\begin{align}504 &= (x+3)^3-(x-3)^3\\ &= \bigl((x+3)-(x-3)\bigr)\left((x+3)^2+(x+3)(x-3)+(x-3)^2\right)\\ &= 6\bigl((x^2+6x+9)+(x^2-9)+(x^2-6x+9)\bigr)\\ &= 6(3x^2+9)\\ &= 18x^2+54.\end{align}$$ You can take it from there, yes?
|
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How many even numbers of four digits can be formed with the digits 0,1,2,3,4,5 and 6 no digit being used more? My attempt to solve this problem is:
First digit cannot be zero, so the number of choices only $6 (1,2,3,4,5,6)$
The last digit can be pick from $0,2,4,6$, so the number of choices only 4
Second digit can be only pick from the rest, so the number of choices only 5
Third digit can be only pick from the rest, so the number of choices only 4
The total number of choices is $6\cdot 4\cdot 5\cdot 4= 480$
So, is my solution true? Or I miss something?
Thanks
|
Consider two cases:
Case 1: first digit even
There are three ways to find the first digit, if the first digit is even.
There are 2,4, and 6.
There are three ways to find the last digit, because one is taken for the first digit.
There are 5 ways to find the second digit and 4 ways to find the third digit.
So, the total ways is: 3*5*4*3= 180 ways
Case 2: first digit odd
There are three ways to choose the first digit, there are: 1,3,5
There are 4 ways to choose the last digit, there are 0,2,4,6
There are 5 ways to choose the second digit and 4 ways to choose the third digit.
So, the total ways is: 3*5*4*4= 240 ways
Total ways: case 1+ case 2= 180+240= 420 ways
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Inequality....(RMO $1994$...question $8$) If $a$, $b$, $c$ are positive real numbers such that $a+b+c = 1$, prove that
$$(1+a)(1+b)(1+c) ≥ 8(1-a)(1-b)(1-c)\text{.}$$
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One quite natural start is to replace $1-a$ by $b+c$ as there are more inequalities for nonnegative reals than reals. (In this case all terms are positive but minus signs are ugly in this problem so let's remove them.) So we have to prove that $$(1+a)(1+b)(1+c)\geq 8(b+c)(a+c)(a+b).$$ Also, there are number eight in the inequality so that makes me guess that one might be able to prove the inequality by taking three times arithmetic-geometric inequality. So I have to modify terms somehow. Lets test: $$1+a=a+b+c+a=2a+b+c,1+b=2b+a+c,1+c=2c+a+b.$$ But arithmetic-geometric inequality means that we have to take square root of expressions so squares would be nice. There are something similar terms in $2a+b+c$ and $2b+a+c,$ namely $a+b$. So lets try to write the inequality as $$((a+b)+(a+c))((a+b)+(b+c))((a+c)+(b+c))\geq 8(b+c)(a+c)(a+b).$$ But this is just the arithmetic-geometric inequality of the numbers $a+b,a+c,$ and $b+c.$
|
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Prove that if $\gcd(a,b)=1$, then $\gcd(a^2,b^2)=1$ So, if $\gcd(a,b)=1$, then $\gcd(a^2,b^2)=1$ means $1=ax+by$, and want to show $a^2x+b^2y=1$.
By squaring $1=ax+by$ both sides, I get, $1=(ax)^2+2(ax)(by)+(by)^2$, but this doesn't help my proof.
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First show
$\gcd(a, b^2) = \gcd(a^2, b) = 1$
$\gcd(a,b) = 1$ means that there are $x$, $y$ ($x$, $y$ are integers) such that $ax + by =1$
$ax + by(ax+by) =1$
$a(x+bxy) + b^2y^2 =1$
That means $\gcd(a,b^2)=1$
By a similar derivation, $\gcd(a^2,b)=1$.
$ar_1(ar_2+b^2s_2) +b^2s_1 = 1$
$a^2r_1^2+b^2(ar_1s_1+s_1)=1$
Therefore,
$\gcd(a^2, b^2) = 1$
|
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How to solve this system of equation? I need to solve the following system of $(x,y)$:
\begin{cases}
3y^3+3x\sqrt{1-x}=5\sqrt{1-x}-2y\\
x^2-y^2\sqrt{1-x}=\sqrt{2y+5}-\sqrt{1-x}
\end{cases}
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The first equation can be written as
$x=1-y^{2}$
because it is manipulated in this way.
We bring to the first member the terms in x and to the second member the terms in y:
$(5-3x)\sqrt{1-x}=y(3y^{2}+2)$.
We see this equation as the equality of two products, that is:
$\sqrt{1-x}=y$,
$-3x+5=3y^{2}+2$.
The solution of this system is:
$x=1-y^{2}$.
We take the second equation and replace the value of $x$, obtaining an equation in y of the eighth degree:
$y^{8}-2y^{7}-4y^{6}+6y^{5}+4y^{4}-6y^{3}-3y^{2}-4=0$,
which splits into the product of two polynomies:
$(y-2)(y^{7}-3y^{5}+4y^{3}+2y^{2}+y+2)=0$.
From here we deduce that $y=2$, value that replaced in the equation $x=1-y^2$, as a result $x=-3$.
|
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How do I prove $2^{n+1} + 2n + 1 = 2^{n+2} - 1$ I am attempting to prove using induction:
$\sum_0^n 2i = 2^{n + 1} - 1$
I have gotten to the point where I need to show:
$2^{n+1} + 2n + 1 = 2^{n+2} - 1$
How do I prove this? Or should I be proving the initial question a different way
|
You missed the power of $2$ in the Left hand side
and the range of $i$ should start from $0$ instead of $1$
We find $\displaystyle\sum_0^1 2^i =1+2=3 $ and $ 2^{1 + 1} - 1=4-1=3$
So, $\displaystyle\sum_0^n 2^i = 2^{n + 1} - 1$ is true for $n=1$
Let $\displaystyle\sum_0^n 2^i = 2^{n + 1} - 1$ is true for $n=m$
$$\implies \sum_0^m 2^i = 2^{m + 1} - 1$$
$$\implies \sum_0^{m+1}2^i =\sum_0^m 2^i +2^{m+1}= 2^{m + 1} - 1 +2^{m+1}=2^{m+1}(1+1)-1=2^{m+2}-1$$
|
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Determine $p$ for one to have $\displaystyle\frac{1}{5}+\frac{1}{6}=\frac{1+1}{5+6}=\frac{2}{11}$, in $\mathbb{Z_p}$. Determine $p$ for one to have $\displaystyle\frac{1}{5}+\frac{1}{6}=\frac{1+1}{5+6}=\frac{2}{11}$, in $\mathbb{Z_p}$.
I know that $p$ must, $13, 43, 61, 101,103$.
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In addition to $p=121-60=61$ we can use the extended Euclidean algorithm to obtain $5^{-1}=-12$, $6^{-1}=-10$ and $11^{-1}=-11$ in $\mathbb{Z}/61 \mathbb{Z}$. Hence
$$
\frac{1}{5}+\frac{1}{6}=-12-10=-22=2\cdot (-11)=\frac{2}{11}.
$$
|
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How to solve this symmetric system of equations? How many solutions are there to this equation?
$$\begin{align*}
x^2-y^2&=z\\
y^2-z^2&=x\\
z^2-x^2&=y
\end{align*}$$
|
We have
\begin{align}
x^2 - y^2 & = z\\
y^2 - z^2 & = x\\
z^2 - x^2 & = y
\end{align}
Adding the first two equations, we get that
$$x^2-z^2 = z+x \implies z=-x \text{ or }x-z=1$$
Similarly, we get that
$$x=-y \text{ or }y-x=1$$
$$y=-z \text{ or }z-y=1$$
Hence, there are $8$ possible choices of getting $3$ equations. But the choice
$$x-z=1 \text{ and }y-x=1 \text{ and }z-y=1$$ gives no solution. The rest of the seen choices give the following solutions
$$(0,0,0); (-1,0,1); (0,1,-1); (1,-1,0)$$
|
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The sum of three square roots bounded below by $\sqrt{82}$
Let $a,b,c >0$ and $a+b+c \le 1$. Prove: $$\sqrt{a^2+1/a^2}+\sqrt{b^2+1/b^2}+\sqrt{c^2+1/c^2} \ge \sqrt{82}$$
Progress: I tried to have 3 vectors $(a,1/a)$, $(b,1/b)$ and $(c,1/c)$, played around with the vector inequalities but no success.
|
since $f（x)=x^2+\dfrac{1}{x^2}$ is mono decreasing function when $x \in (0,1] $, we can prove $a+b+c=1$ case first, for any $a'+b'+c' <1$, let $p(a'+b'+c')=1 \implies p>1$, then $a=pa'>a',b=pb'>b',c=pc'>c' \implies \sum \sqrt{a'^2+\dfrac{1}{a'^2}}>\sum\sqrt{a^2+\dfrac{1}{a^2}}$
to prove the case $a+b+c=1$, it is same as Martin's post:
$\sum \sqrt{x_i^2+y_i^2} \ge \sqrt{(\sum x_i)^2+ (\sum y_i)^2}$,that is:
LHS$\ge \sqrt{(a+b+c)^2+\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2}$
$\dfrac{3}{\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}} \le \dfrac{a+b+c}{3} \implies \dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c} \ge 9$
QED.
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$x,y,z$ positive real numbers , $x+y+z=3$ $\implies x^4y^4z^4(x^3+y^3+z^3)≤3$ If $x,y,z$ are positive real numbers with $x+y+z=3$ then how to prove (without using calculus) that $\space$ $x^4y^4z^4(x^3+y^3+z^3)≤3$ ?
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We have,
$$(x+y+z)^3=x^3+y^3+z^3+3(x+y)(y+z)(x+z) \ge x^3+y^3+z^3+\frac{8}{3}(x+y+z)(xy+yz+xz) = x^3+y^3+z^3+(xy+yz+xz)+(xy+yz+xz)..[8 \text{times}] \ge 9\sqrt[9]{(x^3+y^3+z^3)(xy+yz+xz)^8}$$
Since $$(xy+yz+xz)^2 = x^2y^2+y^2z^2+x^2z^2+2xyz(x+y+z) \ge 3xyz(x+y+z)$$ we have,
$$ 9\sqrt[9]{(x^3+y^3+z^3)(xy+yz+xz)^8} \ge 9\sqrt[9]{3^8(x^3+y^3+z^3)x^4y^4z^4}$$
And so we have $$ 3 \ge x^4y^4z^4(x^3+y^3+z^3) \Box $$
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How would one discover this finite product identity? I recently found the following finite product identity in a table of products:
\begin{align}
\prod_{k=0}^{n-1}\left[\sinh^2y+\sin^2\left(x+\frac{k\pi}{n}\right)\right]=2^{1-2n}(\cosh(2ny) -\cos(2nx))
\end{align}
I'd like to know how one would have gone about discovering this product identity (not just proving the identity) if one were handed the product.
This question is related to my other question Limit of a sequence of determinants. and would effectively answer that question as well.
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Discovering such identities often is just recognising things one has already seen elsewhere.
The factors of the product are easily recognised as
$$\left\lvert\sin \left(x+\frac{k\pi}{n} +iy\right) \right\rvert^2$$
when one has some experience with complex analysis. Euler's formulae make the connection between the trigonometric and hyperbolic functions obvious,
$$\begin{align}
\sin z = \frac{e^{iz} - e^{-iz}}{2i};\; \sinh z = \frac{e^z - e^{-z}}{2}&\Rightarrow \sinh (iz) = i\sin z;\, \sin (iz) = i\sinh z,\\
\cos z = \frac{e^{iz}+e^{-iz}}{2};\; \cosh z = \frac{e^z + e^{-z}}{2} &\Rightarrow \cosh (iz) = \cos z;\; \cos (iz) = \cosh z.
\end{align}$$
Thus by the addition formula
$$\sin (x+iy) = \sin x \cos (iy) + \cos x\sin (iy) = \sin x \cosh y + \cos x \sinh y$$
and
$$\lvert\sin (x+iy)\rvert^2 = \sin^2 x \cosh^2 y + \cos^2 x \sinh^2 y,$$
which can be simplified to either $\sin^2 x + \sinh^2 y$ or $\cosh^2 y - \cos^2 x$ using the identities $\sin^2 x + \cos^2 x \equiv 1$ and $\cosh^2 y - \sinh^2 y \equiv 1$.
We can also write the right hand side in similar form, we have by the addition formulae $\cos (2\xi) = \cos^2\xi - \sin^2 \xi = 2\cos^2\xi - 1$ and $\cosh (2\eta) = \cosh^2 \eta + \sinh^2 \eta = 2\cosh^2\eta - 1$, whence
$$\cosh (2\eta) - \cos (2\xi) = 2(\cosh^2 \eta - \cos^2\xi) = 2\lvert \sin (\xi + i\eta)\rvert^2.$$
So if one knows the formula linked by Hans Lundmark,
$$\prod_{k=0}^{n-1} \sin \left(x + \frac{k\pi}{n}\right) = \frac{\sin (nx)}{2^{n-1}},\tag{1}$$
it directly follows from that (note that by the identity theorem, it holds for all $x \in\mathbb{C}$, not only for real $x$),
$$\prod_{k=0}^{n-1} \left\lvert \sin\left( z + \frac{k\pi}{n}\right)\right\rvert^2 = \left\lvert \frac{\sin (nz)}{2^{n-1}}\right\rvert^2 = \frac{\lvert \sin (nz)\rvert^2}{2^{2n-2}}.$$
If one doesn't know formula $(1)$ (and it's not one that is very well-known), one can work it out from the product without too much pain (knowing what should come out definitely helps, but even without that, one can find the right hand side from the product).
It is not a far-fetched idea to start with Euler's formula for the sine, and then to separate the $e^{ix}$ from the $e^{ik\pi/n}$ to arrive at
$$\begin{align}
\prod_{k=0}^{n-1} \sin \left(x + \frac{k\pi}{n}\right)
&= \prod_{k=0}^{n-1} \frac{e^{ix}e^{ik\pi/n} - e^{-ix}e^{-ik\pi/n}}{2i}\\
&= \frac{e^{-inx}e^{i\pi(n-1)/2}}{(2i)^n}\prod_{k=0}^{n-1}\left(e^{2ix} - e^{-2\pi ik/n}\right).\tag{2}
\end{align}$$
Writing $X = e^{2ix}$ and $\zeta_n = e^{2\pi i/n}$ for the primitive $n$-th root of unity, the last product reads
$$\prod_{k=0}^{n-1}\left(X - \zeta_n^k\right),$$
and that is well-known to be the factorisation of $X^n - 1$. Plugging that into $(2)$ and noting that $e^{i\pi(n-1)/2} = i^{n-1}$, we obtain
$$\prod_{k=0}^{n-1} \sin \left( x + \frac{k\pi}{n}\right) = \frac{e^{-inx}\left(e^{2inx}-1\right)}{2^ni} = \frac{e^{inx} - e^{-inx}}{2^ni} = \frac{\sin (nx)}{2^{n-1}}.$$
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Prove that x is rational Let $x$ be a real number with the properties that $x^3+x$ and $x^5+x$ are rational.
Prove that $x$ is rational. Denote $a=x^3+x$; $b=x^5+x$. We can multiply and add them together until we get the desired result. I also know some non-elementary proofs of this, but have you some nice elementary proofs?
Thank you.
|
We may take $x>0$. $\frac{x^5+x}{x^3+x}$ is rational, so $bx^4+b = ax^2 +a$ for some integers $a,b$ and $x = \sqrt{k}$, where $k=c+\sqrt{d}$ is the root of a quadratic equation, with $c,d\in\mathbb{Q}$, $d=0$ or $d>0$ not a square.
If $k$ is rational ($d=0$) but not a square, $x^2-1$ is rational and $x^5+x = (x^3+x)(x^2-1)+2x$ is irrational, a contradiction.
If $k$ is irrational, ($d\neq 0$), then expanding $x^3+x$ we get
$$x^3+x = c^3+3cd+c + (3c^2+d+1)\sqrt{d}$$
is irrational, a contradiction.
|
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How prove this $\overrightarrow{r_{i}}\cdot\overrightarrow{r_{j}}\ge\frac{1}{2}$ let $a,b,c$ are real numbers,and such $a+b+c=0,a^2+b^2+c^2=1$, we define:
$\overrightarrow{r}=(x_{i},y_{i},z_{i})(i=1,2,3,4,5,6)$,where $\{x_{i},y_{i},z_{i}\}=\{a,b,c\}$,
show that: there are exst $\overrightarrow{r_{i}}\neq\overrightarrow{r_{j}}$,such $$\overrightarrow{r_{i}}\cdot\overrightarrow{r_{j}}\ge\dfrac{1}{2}$$
My try: $$a+b+c=0,\Longrightarrow c=-a-b$$
then $$a^2+b^2+c^2=1\Longrightarrow a^2+b^2+(a+b)^2=1\Longrightarrow a^2+b^2+ab=\dfrac{1}{2}$$
let $$a=x+y,b=x-y\Longrightarrow (x+y)^2+(x-y)^2+x^2-y^2=\dfrac{1}{2}$$
$$\Longrightarrow 3x^2+y^2=\dfrac{1}{2}$$
then I can't ,Thank you
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The points $r_i$ lie on a unit circle (the intersection of the plane $x+y+z=0$ and the sphere $x^2 + y^2 + z^2 = 1$). There isn't enough "room" on a circle for 6 points that are all more than 60 degrees apart. So, there must be two of them (say $r_i$ and $r_j$) whose angles differ by 60 degrees or less. Then $\vec{r_i} \cdot \vec{r_j} \ge \cos 60^\circ = \tfrac12$.
|
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On $n! + k^2 = m^2$ Brocard's problem is the case $k=1$ of,
$$n!+k^2 = m^2\tag{1}$$
However, for general $k$,
$$4!+1 = 5^2\tag{2}$$
$$6!+3^2 = 27^2\tag{3}$$
$$8!+9^2 = 201^2\tag{4}$$
and using the first two, we can observe two polynomial identities,
$$(x+1)(x+2)(x+3)(x+4) + 1 = (x^2+5x+5)^2\tag{5}$$
$$(2x+1)(2x+2)(3x+3)(2x+4)(2x+5)(3x+6)+3^2 = (12x^3+54x^2+72x+27)^2\tag{6}$$
$$(a_1x+1)(a_2x+2)\dots(a_8x+8)+k^2 \overset{\bf{?}}{=} P(x)^2\tag{7}$$
(The first is by Alain Verghote.)
Question: Anybody knows how to find non-zero $a_i$ to $(7)$ using $(4)$, or using any of the 36 solutions to $8!+k^2=m^2$?
|
Let's pretend I haven't seen any of the identities above. I just know the fact that $4!+1=5^2$ and I want to find an identity of this form:
$$
(ax+1)(bx+2)(cx+3)(dx+4) + 1 = P(x)^2
$$
Where $a, b, c, d \in \mathbb{N}$ and $P\in \mathbb{N}[x]$. Well, since $4!+1=5^2$, we know that the constant term of $P$ is $5$. So now let's rewrite the equation using polynomial $Q$, where $P(x) = Q(x) + 5$ (so the $Q$ doesn't have the constant term).
$$
(ax+1)(bx+2)(cx+3)(dx+4) + 1 = (Q(x)+5)^2
$$
$$
(ax+1)(bx+2)(cx+3)(dx+4) = (Q(x)+4)(Q(x)+6)
$$
Now because of the prime factorization of $6$, the factor $Q(x)+6$ must be a product of $(bx+2)$ and $(cx+3)$ and so $Q(x)+4$ must be a product of $(ax+1)$ and $(dx+4)$.
$$
\begin{array}{rcl}
(ax+1)(dx+4) &=& adx^2 + (4a+d)x+4 \\
(bx+2)(cx+3) &=& bcx^2 + (3b+2c)x+6 \\
\end{array}
$$
Notice that $Q(x)+4$ and $Q(x)+6$ differ only in the constant term so the following equations must hold:
$$
\begin{array}{rcl}
ad &=& bc \\
4a+d &=& 3b+2c \\
\end{array}
$$
Now, every such solution $(a, b, c, d)$ should form a suitable polynomial. For example $(a, b, c, d) = (7, 7, 7, 7)$ is a solution. For this quadruple we get the equality:
$$
(7x+1)(7x+2)(7x+3)(7x+4) + 1 = (49x^2 + 35x + 5)^2
$$
Quite amusing but I think it can be aplicable to the higher degrees. Again, for the fact $8!+9^2=201^2$ we use the same technique and the equation will now look like this:
$$
(ax+1)(bx+2)(cx+3)(dx+4)(ex+5)(fx+6)(gx+7)(hx+8) + 9^2 = (Q(x)+201)^2
$$
$$
(ax+1)(bx+2)(cx+3)(dx+4)(ex+5)(fx+6)(gx+7)(hx+8) = (Q(x)+192)(Q(x)+210)
$$
From the prime factorisation of $192$ and $210$ we see that $Q(x)+210$ can be only equal either to $(ax+1)(ex+5)(fx+6)(gx+7)$ or $(bx+2)(cx+3)(ex+5)(gx+7)$. And now we would just need to create that set of (four) equations and find any solution $(a, b, c, d, e, f, g, h)$.
But that was too much pain for me... :-/
|
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|
Evaluating $\int_0^1 \frac{\log x \log \left(1-x^4 \right)}{1+x^2}dx$ I am trying to prove that
\begin{equation}
\int_{0}^{1}\frac{\log\left(x\right)
\log\left(\,{1 - x^{4}}\,\right)}{1 + x^{2}}
\,\mathrm{d}x = \frac{\pi^{3}}{16} - 3\mathrm{G}\log\left(2\right)
\tag{1}
\end{equation}
where $\mathrm{G}$ is Catalan's Constant.
I was able to express it in terms of Euler Sums but it does not seem to be of any use.
\begin{align}
&\int_{0}^{1}\frac{\log\left(x\right)
\log\left(\,{1 - x^{4}}\,\right)}{1 + x^{2}}
\,\mathrm{d}x
\\[3mm] = &\
\frac{1}{16}\sum_{n = 1}^{\infty}
\frac{\psi_{1}\left(1/4 + n\right) -
\psi_{1}\left(3/4 + n\right)}{n} \tag{2}
\end{align}
Here $\psi_{n}\left(z\right)$ denotes the polygamma function.
Can you help me solve this problem $?$.
|
$$I=\int_0^1 \frac{\ln x \log \left(1-x^4 \right)}{1+x^2}dx$$
Let,
\begin{align*}
\displaystyle A&=\int_0^1 \dfrac{x\arctan x\ln x}{1+x^2}dx\\
\displaystyle B&=\int_0^1 \dfrac{\ln x \ln(1+x^2)}{1+x^2}dx\\
\displaystyle C&=\int_0^1 \dfrac{\arctan x\ln x}{1+x}dx\\
\end{align*}
From Evaluating $\int_0^{\pi/4} \ln(\tan x)\ln(\cos x-\sin x)dx=\frac{G\ln 2}{2}$ ,
it follows that,
$$\displaystyle \int_0^1\dfrac{\ln x\ln(1+x)}{1+x^2}dx=A-\dfrac{1}{2}B-C-2G\ln 2+\beta(3)$$
and,
$$(1)\boxed{A=\dfrac{1}{64}\pi^3-B-G\ln 2}$$
with,
$$\displaystyle \beta(3)=\sum_{n=1}^{\infty} \dfrac{(-1)^n}{(2n+1)^3}$$
Thus,
$$(2)\boxed{\displaystyle \int_0^1\dfrac{\ln x\ln(1+x^2)}{1+x^2}dx=\dfrac{\pi^3}{64}-\dfrac{3}{2}B-C-3G\ln 2+\beta(3)}$$
$G$, being the Catalan constant,
From Evaluating $\int_0^{\pi/4} \ln(\tan x)\ln(\cos x-\sin x)dx=\frac{G\ln 2}{2}$ ,
it follows that,
$$\displaystyle\int_0^{1}\dfrac{\ln x\ln(1-x)}{1+x^2}dx=A+\dfrac{1}{2}B+\beta(3)+\int_0^1\dfrac{\ln x\arctan x }{1-x}dx$$
and,
$$\displaystyle \int_0^1\dfrac{\ln x\arctan x}{1-x}dx=B-C+2G\ln 2-\dfrac{1}{16}\pi^3$$
Thus, using (1),
$$(3)\boxed{\displaystyle\int_0^{1}\dfrac{\ln x\ln(1-x)}{1+x^2}dx=\dfrac{1}{2}B-C+G\ln 2-\dfrac{3\pi^3}{64}+\beta(3)}$$
Therefore,
\begin{align}
\int_0^1 \frac{\ln x \log \left(1-x^4 \right)}{1+x^2}dx&=\int_0^1\dfrac{\ln x\ln(1+x^2)}{1+x^2}dx+\int_0^1\dfrac{\ln x\ln(1+x)}{1+x^2}dx+\int_0^1\dfrac{\ln x\ln(1-x)}{1+x^2}dx\\
&=B+\dfrac{\pi^3}{64}-\dfrac{3}{2}B-C-3G\ln 2+\beta(3)+\dfrac{1}{2}B-C+G\ln 2-\dfrac{3\pi^3}{64}+\beta(3)\\
&=2\beta(3)-\dfrac{\pi^3}{32}-2G\ln 2-2C
\end{align}
$C$ have been already evaluated (see Evaluating $\int_0^1 \frac{\arctan x \log x}{1+x}dx$ )
\begin{equation}
\boxed{\displaystyle C=\dfrac{G\ln 2}{2}-\dfrac{\pi^3}{64}}
\end{equation}
and, knowing that,
\begin{equation}
\beta(3)=\dfrac{\pi^3}{32}
\end{equation}
it follows that,
$$\boxed{I=\dfrac{\pi^3}{16}-3G\ln 2}$$
|
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|
Prove that $\log^25 + \log^27 > \log12$. Prove that $\log^25 + \log^27 > \log12$.
What I tried so far:
$\log^25 + \log^27 > \log3 + \log4$
$(\log5 + \log7)^2 - 2 \cdot \log5 \cdot\log7 > \log3 + \log4$
But it seems that I'm not even near the result.
Every suggestion / hint would be appreciated :)
Thanks in advance.
EDIT: $\log$ means $\log_{10}$
|
Without actually computing exact logs,...
$$\log_{10}5 \cdot \log_{10}5 + \log_{10}7\cdot \log_{10}7 > \log_{10}12$$
$$\iff \frac{\log_{10}5}{\log_{10}7} + \frac{\log_{10}7}{\log_{10}5} > \frac{\log_{10}12}{\log_{10}5 \cdot \log_{10}7}$$
Now LHS $>2$ as it is the sum of a positive number ($\neq 1$) and its reciprocal. So it is sufficient to show that RHS $< 2$, which is equivalent to:
$$\log_{10}12 < 2\log_{10}5 \cdot \log_{10}7 \iff \log_{5}12 < \log_{10}49 \iff 3\cdot\log_{5}12 < 3\cdot \log_{10}49$$
But $12^3 = 1728 < 5^5$, while $49^3 > 10^5$ shows $3\cdot\log_{5}12< 5$, while $3\cdot \log_{10}49> 5$.
|
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|
How to find the determinant of this matrix I have the following matrix:
$
\begin{bmatrix}
a & 1 & 1 & 1 \\
1 & a & 1 & 1 \\
1 & 1 & a & 1 \\
1 & 1 & 1 & a \\
\end{bmatrix}
$
My approach is to rref the matrix so that i can find the determinant by multiplying along the diagonals.
I attempted to do an rref and ended up with
$
\begin{bmatrix}
1 & a & 1 & 1 \\
1-a & a-1 & 0 & 0 \\
0 & 1-a & a-1 & 1 \\
0 & 0 & 1-a & a-1 \\
\end{bmatrix}
$
I then factored out (1-a) to get this
$
\begin{bmatrix}
1 & a & 1 & 1 \\
1 & -1 & 0 & 0 \\
0 & 1 & -1 & 1 \\
0 & 0 & 1 & -1 \\
\end{bmatrix}
$
which will then make my determinant a multipile of $(1-a)^3$:
$\det(A) = (1-a)^3x$
But now here I don't know what to do. I have a feeling that my approach is wrong. Any help or guide please?
|
How about the LU decomposition (modulo a rearrangements of rows)?
\begin{align*}
&\left[\begin{array}{cccc}0&1&0&0\\0&0&0&1\\0&0&1&0\\1&0&0&0\end{array}\right]\times\left[\begin{array}{cccc}
a&1&1&1\\
1&a&1&1\\
1&1&a&1\\
1&1&1&a\end{array}\right]\\=&\left[\begin{array}{cccc}1&0&0&0\\1&1&0&0\\1&1&1&0\\a&1+a&-1&1\end{array}\right]\times\left[\begin{array}{cccc}1&a&1&1\\0&1-a&0&a-1\\0&0&a-1&1-a\\0&0&0&(3+a)(1-a)\end{array}\right].
\end{align*}
It follows that the determinant is $(a-1)^3(3+a)$.
|
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|
Likelihood of sum of dice roll is exactly 1 million A board game is set up such that there is a number line with squares numbered 0..1 million. You roll a standard 6 sided die and move forward the number of spaces that you roll. Eventually you will either land on, or pass the 1 millionth spot. What is the probability you land exactly on the 1 millionth spot?
In other words, what is the probability that after a number of rolls the sum is exactly 1 million
|
Let $S(\omega) \subset \mathbb{N}$ be the spaces that are landed on during trial $\omega$. That is, if $d_k(\omega) \in \{1,...,6\}$ are the die rolls, then $S(\omega) = \{ \sum_{k=1}^n d_k(\omega) \}_{n=1}^\infty $.
For $k \ge 1$, let $r_k(\omega) = \min \{ j-k | j \ge k, \ j \in S(\omega) \}$. Note that $r_k(\omega) = 0$ iff space $k$ was landed on.
Also, note that $r_k(\omega) \in \{0,...,5\}$, and that if $r_k(\omega) >0$, then $r_{k+1}(\omega) = r_k(\omega) +1$.
To simplify notation, for $n \in \{0,...,5\}$ let $\pi_k(n) = p\{ \omega | r_k(\omega) = n \}$.
We have $\pi_1(n) = \frac{1}{6}$ (initial roll must land on one of $\{1,...6\}$ with probability $\frac{1}{6}$).
Then for $k >1$, we can compute $\pi_k(n) = \sum_{m=0}^{5} p \left(\{ \omega | r_k(\omega) = n \} | \{ \omega | r_{k-1}(\omega) = m \} \right) \pi_{k-1}(m)$.
If $m \in \{1,...,5\}$, then $p \left(\{ \omega | r_k(\omega) = n \} | \{ \omega | r_{k-1}(\omega) = m \} \right) = \begin{cases} 1, & n=m-1 \\ 0, & \text{otherwise}\end{cases}$.
If $m = 0$, then $p \left(\{ \omega | r_k(\omega) = n \} | \{ \omega | r_{k-1}(\omega) = 0 \} \right) = \frac{1}{6}$.
Letting $\pi_k = (\pi_k(0),...,\pi_k(5))^T$, we can write the above as $\pi_k = M \pi_{k-1}$, where
$M= \begin{bmatrix}
\frac{1}{6} & 1 & 0 & 0 & 0 & 0 \\
\frac{1}{6} & 0 & 1 & 0 & 0 & 0 \\
\frac{1}{6} & 0 & 0 & 1 & 0 & 0 \\
\frac{1}{6} & 0 & 0 & 0 & 1 & 0 \\
\frac{1}{6} & 0 & 0 & 0 & 0 & 1 \\
\frac{1}{6} & 0 & 0 & 0 & 0 & 0 \\
\end{bmatrix}$.
The solution is then given by $\pi_{1000000}(0) = e_1^T M^{999999} \pi_1$.
Solving $M \hat{\pi} = \hat{\pi}$, gives the solution $\hat{\pi} = c(1, \frac{5}{6}, \frac{4}{6}, \frac{3}{6},
\frac{2}{6}, \frac{1}{6})^T$. Choosing $c= \frac{7}{2}$ gives the stationary distribution $\hat{\pi} = \frac{2}{7}(1, \frac{5}{6}, \frac{4}{6}, \frac{3}{6},
\frac{2}{6}, \frac{1}{6})^T$, hence we get the limiting $\hat{\pi}(0) = \frac{2}{7}$, as Ross pointed out with much less work.
(Thanks to my friend Takis Konstantopolos for his 'random set' approach ($S(\omega)$ in this instance) and introducing the $r_k$ to simplify my life.)
|
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|
Did I answer this probability question correctly? Hello I have the following question to answer:
Imagine three boxes, each of which has three slips of paper in it each with a number
marked on it. The numbers for box A are 2, 4 and 9, for box B 1, 6 and 8, and for box C
3, 5 and 7. One slip is drawn, independently and with equal probability, from each box.
Compute P{A Slip > B slip} (a slip taken from A is greater than a slip from B)
My attempted solution:
First note the possible outcomes: (highlighted are outcomes that satisfy A>B)
{2,1} || {2,6} || {2,8}
{4,1} || {4,6} || {4,8}
{9,1} || {9,6} || {9,8}
Since there are 5 possible outcomes where A>B and since the events are independent from one another then the probability is simply:
(5/9) * P{B|A}, where B is a slip chosen from B that is smaller than A
The probability P{B|A} is P{B|2} + P{B|4} + P{B|9} = 0+1/3+1 = 4/3
So finally P{Aslip>Bslip} = (5/9)(4/3) = 20/27 or 74%
|
There are $3$ equally likely possibilities for A's number, $2$, $4$, and $9$.
If A got $2$, then the probability that A beat B is $\frac{1}{3}$, for B must get $1$.
If A got $4$, then again the probability that A beat B is $\frac{1}{3}$.
If A got $9$, then the probability that A beat B is $1$, or more prettily $\frac{3}{3}$.
Thus the probability that A beats B is $\frac{1}{3}\cdot\frac{1}{3}+\frac{1}{3}\cdot\frac{1}{3}+\frac{1}{3}\cdot\frac{3}{3}$.
Remark: Your counting argument gave the same result, but then got spoiled.
The problem will become interesting when you compute the probability that B beats C, and the probability that C beats A.
|
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|
How to find the $f^{-1}(x)$ of $f(x)=x^{3}-12x+\frac{48}{x}-\frac{64}{x^{3}}$ It is a question from a quiz.
The following is the whole question.
Let
\begin{eqnarray}
\\f(x)=x^{3}-12x+\frac{48}{x}-\frac{64}{x^{3}} , \space x\in (-\infty ,0),
\end{eqnarray}
find $f^{-1}(x)$. Hint : $f(x)$ can be written in the form, like$(A+B)^{3}$.
The first thing I think is $(A+B)^{3}=A^3+3A^2B+3AB^2+B^3$, then try to make it become the the form of $A^3+3A^3B+3AB^3+B^3$. However, it it so difficult to obtain this form.
I need help.
Update :
Now I have $\left(x - \frac 4x\right)^3$ but how to find the $f^{-1}(x)$ of $f(x)=\left(x - \frac 4x\right)^3$?
Thank you for your attention
|
First of all, $f$ does not have a inverse function because you always have two root of $f - y$. One positive root, one negative root.
If you add the constraints that $x$ is positive and $y$ is real, then there will be an inverse function.
$$(x - \frac4x)^3=y$$
Since $y$ is real:
$$x - \frac4x=\sqrt[3]y$$
Since $x$ is not zero.
$$x^2 - 4=\sqrt[3]yx$$
Solve the quadratic equation:
$$x^2 - \sqrt[3]yx = 4 $$
$$(x-\frac{\sqrt[3]y}2)^2=4+(\frac{\sqrt[3]y}2)^2$$
$$ x - \frac{\sqrt[3]y}2 = \pm \sqrt{4+(\frac{\sqrt[3]y}2)^2}$$
$$ x = \frac{\sqrt[3]y}2 \pm \sqrt{4+(\frac{\sqrt[3]y}2)^2}$$
Since $x$ is positive, and the magnitude of the second term is greater than the first term, so you have to take the plus sign. This is your inverse function:
$$f^{-1}(y) = \frac{\sqrt[3]y}2 + \sqrt{4+(\frac{\sqrt[3]y}2)^2}$$
If, however, you constrain $x$ to be negative, then take the minus sign above.
|
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|
Prove $u_n$ has the limit given Let $u_n = \sqrt{u_n+1}~ and ~u_1=1$. Prove that $\displaystyle\lim_{n\to\infty} u_n = \frac{1}{2}(1+\sqrt5)$
Here is what I did.
First, to show the sequence converges:
$u_1 = 1, \space
u_2 = \sqrt 2 = 2^{(1/2)}, \space
~u_3 = \sqrt {1 + 2^{(1/2)}} < \sqrt{2*2^{(1/2)}} = 2^{(3/4)} $
therefore
$u_4 < \sqrt {1 + 2^{(3/4)}}< \sqrt{2*2^{(3/4)}} = 2^{(7/8)}, \space
~u_5 < \sqrt {1 + 2^{(7/8)}} < \sqrt{2*2^{(7/8)}} = 2^{(15/16)} $
... by weak induction:
$u_n < 2^((2^{(n-1)} - 1)/2^{(n-1)}) < 2, \space ∀n $
now to find the limit:
$u_{n+1} = \sqrt {1 + u_n} $
squaring both sides
$u²_{n+1} = 1 + u_n$
so
$u²_{n+1} - u_n - 1 = 0$
let $φ = \lim_{n → ∞} u_n = \lim_{n → ∞} u_{n+1}$
and substitute into the equation above:
$φ² - φ - 1 = 0$
now solve for φ with the quadratic formula
$\displaystyle\lim_{n\to\infty} u_n = \frac{1}{2}(1+\sqrt5)$
|
$$\left. \begin{align} u_{_{\infty\ +\ 1}} &= \sqrt{u_\infty + 1} \\ \\ \infty + 1 &= \infty \end{align}\ \right\} \iff u_\infty = \sqrt{u_\infty + 1} \iff u_\infty^2 = u_\infty + 1 \iff u_\infty^2 - u_\infty - 1 = 0$$ $$u_\infty = \frac{1\pm\sqrt5}2$$ Obviously, since a radical is always positive, the only possible solution is $u_\infty = \frac{1\ +\ \sqrt5}2 = \varphi > 0$ .
|
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|
Show that $\frac{(1-y^{-1})^2}{(1-y^{-1 -c})(1-y^{-1+c} )}$ is decreasing in $y > 1 $. I am interested in the function
$f(y) = \frac{(1-y^{-1})^2}{(1-y^{-1 -c})(1-y^{-1+c} )},$
for values of $c \in (0,1)$, and $y > 1$, and have been trying to show that the function is decreasing.
I have tried differentiating the function, but this does not yield a particularly amenable formula to decide whether $f'(y) < 0$. Similarly I have considered the ratios $f(y)/f(\tilde y)$, but again have been unable to ascertain an inequality for this.
I'd appreciate any other approaches to the problem.
|
Let $f_c(x)=\frac{(1-x^{-1})^2}{(1-x^{-1-c})(1-x^{-1+c})}$. Then $\frac{\partial}{\partial c}(\ln f_c(x))=-\ln x(\frac{1}{x^{c-1}-1}+\frac{1}{x^{c+1}-1}+1)$, so since $f_0(x)=1$, it suffices to show that $\frac{1}{x^{c-1}-1}+\frac{1}{x^{c+1}-1}$ is increasing with respect to $x > 1$. $\frac{1}{x^{c-1}-1}$ is negative and increasing, $\frac{1}{x^{c+1}-1}$ is positive and decreasing, and $\frac{\partial (x^{c-1}-1)^{-1}}{\partial x} = \frac{c-1}{c+1}\left(\frac{x^{c+1}-1}{x-x^c}\right)^2 \frac{\partial (x^{c+1}-1)^{-1}}{\partial x}$, so we must show that $\frac{x^{c+1}-1}{x-x^c} > \sqrt{\frac{1+c}{1-c}}$. By L'Hopital's rule $\frac{x^{c+1}-1}{x-x^c}$ approaches $\frac{1+c}{1-c} > \sqrt{\frac{1+c}{1-c}}$ at $x=1$, so it suffices to show that $\frac{x^{c+1}-1}{x-x^c}$ is increasing on $x>1$.
$\frac{\partial}{\partial x}(\ln\frac{x^{c+1}-1}{x-x^c}) = \frac{cx^c(x^2-1)-x(x^{2c}-1)}{x(x-x^c)(x^{c+1}-1)}$, so since the denominator is positive, we must show that $cx^c(x^2-1)>x(x^{2c}-1)$. Since both sides are zero when $c=0$, taking $\frac{\partial}{\partial c}$ of both sides and simplifying shows that it is sufficient to show that $\phi_c(x) > \psi_c(x)$, where $\phi_c(x)=x^{1-c}-x^{-1-c}$ and $\psi_c(x)=\frac{2}{c + (\ln x)^{-1}}$. Now $\frac{\partial \phi_k}{\partial c}(x)=-\phi_k(x)\ln x$ and $\frac{\partial \psi_k}{\partial c}(x) = -\frac{1}{2}(\psi_k(x))^2$, so since $\ln x > \frac{1}{2}\psi_k(x)$, if $\phi_c(x) > \psi_c(x)$ for a given $c$ then $\frac{\partial \phi_c}{\partial c}(x) < \frac{\partial \psi_c}{\partial c}(x)$. Therefore it suffices to show that $\phi_1(x) > \psi_1(x)$, i.e. that $x^2-1+\frac{x^2-1}{\ln x} > 2$.
By L'Hopital's rule $x^2-1+\frac{x^2-1}{\ln x}$ approaches $2$ at $x=1$, so it suffices to show that $\frac{x^2-1}{\ln x}$ is increasing. Now $(\frac{x^2-1}{\ln x})'=\frac{x^2(2\ln x - 1)+1}{x\ln^2 x}$. Since the derivative of $x^2(2\ln x - 1)+1$ is $4x\ln x$, whose only zero (for $x > 0$) is at $x=1$, it follows that the minimum value of $x^2(2\ln x - 1)+1$ is at $x=1$ where it equals $0$, and thus for $x>1$ we have $\frac{x^2(2\ln x - 1)+1}{x\ln^2 x} > 0$. Therefore $\frac{x^2-1}{\ln x}$ is increasing for $x>1$. QED
|
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|
Prove that there are no positive integers $a,b,c,d$ such that $a^2 + b^2 = c^2$ and $a^2 - b^2 = d^2$ Prove that there are no positive integers $a,b,c,d$ such that $a^2 + b^2 = c^2$ and $a^2 - b^2 = d^2$?
Could you show me the proof?
|
Summing and subtracting, you get $$\begin{cases} 2a^2=c^2+d^2\\2b^2=c^2-d^2\end{cases}$$ Thus $c,d$ have the same parity. If they are both even, we obtain $2a^2=4c'^2+4d'^2$ and $2b^2=4c'^2-4d'^2$, so $$\begin{cases} a^2=2(c'^2+d'^2)\\b^2=2(c'^2-d'^2)\end{cases}$$
Since $2\mid a^2,b^2$, we must have $a^2=4a'^2$, $b^2=4b'^2$, so we get that $$\begin{cases} 2a'^2=c'^2+d'^2\\2b'^2=c'^2-d'^2\end{cases}$$
And now you're back to the same, but with smaller numbers. And thus we descend.
It remains the case that both are odd, which I couldn't solve.
|
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"url": "https://math.stackexchange.com/questions/530003",
"timestamp": "2023-03-29T00:00:00",
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|
None exact first order ODE i have to solve the following $1^{st}$ order differential equation
$(xy+1)dx+(2y-x)dy=0$
i am in the elementary differential class,and have not learned multivariate functions,
the equation below is none exact,since
$M_y=x\ne N_x=-1$
so i am looking for a substitution that can make it exact because the intergration factor has both $x$ and $y$ ...
$\large \frac{dy}{dx}=\frac{xy+1}{x-2y}$
|
Approach $1$:
$(xy+1)~dx+(2y-x)~dy=0$
$(xy+1)~dx=(x-2y)~dy$
$(x-2y)\dfrac{dy}{dx}=xy+1$
Let $u=\dfrac{x}{2}-y$ ,
Then $y=\dfrac{x}{2}-u$
$\dfrac{dy}{dx}=\dfrac{1}{2}-\dfrac{du}{dx}$
$\therefore2u\left(\dfrac{1}{2}-\dfrac{du}{dx}\right)=x\left(\dfrac{x}{2}-u\right)+1$
$u-2u\dfrac{du}{dx}=\dfrac{x^2}{2}-xu+1$
$2u\dfrac{du}{dx}=(x+1)u-\dfrac{x^2}{2}-1$
$u\dfrac{du}{dx}=\dfrac{(x+1)u}{2}-\dfrac{x^2+2}{4}$
This belongs to an Abel equation of the second kind.
Let $s=x+1$ ,
Then $\dfrac{du}{dx}=\dfrac{du}{ds}\dfrac{ds}{dx}=\dfrac{du}{ds}$
$\therefore u\dfrac{du}{ds}=\dfrac{su}{2}-\dfrac{(s-1)^2+2}{4}$
Let $t=\dfrac{s^2}{4}$ ,
Then $s=\pm2\sqrt t$
$\dfrac{du}{ds}=\dfrac{du}{dt}\dfrac{dt}{ds}=\dfrac{s}{2}\dfrac{du}{dt}$
$\therefore\dfrac{su}{2}\dfrac{du}{dt}=\dfrac{su}{2}-\dfrac{(s-1)^2+2}{4}$
$u\dfrac{du}{dt}=u-\dfrac{(s-1)^2+2}{2s}$
$u\dfrac{du}{dt}-u=\dfrac{(\pm2\sqrt t-1)^2+2}{\pm4\sqrt t}$
$u\dfrac{du}{dt}-u=\pm\sqrt t+1\pm\dfrac{3}{4\sqrt t}$
This belongs to an Abel equation of the second kind in the canonical form.
Please follow the method in https://arxiv.org/ftp/arxiv/papers/1503/1503.05929.pdf
Approach $2$:
$(xy+1)~dx+(2y-x)~dy=0$
$(xy+1)~dx=(x-2y)~dy$
$(yx+1)\dfrac{dx}{dy}=x-2y$
Let $u=x+\dfrac{1}{y}$ ,
Then $x=u-\dfrac{1}{y}$
$\dfrac{dx}{dy}=\dfrac{du}{dy}+\dfrac{1}{y^2}$
$\therefore yu\left(\dfrac{du}{dy}+\dfrac{1}{y^2}\right)=u-\dfrac{1}{y}-2y$
$yu\dfrac{du}{dy}+\dfrac{u}{y}=u-\dfrac{2y^2+1}{y}$
$yu\dfrac{du}{dy}=\dfrac{(y-1)u}{y}-\dfrac{2y^2+1}{y}$
$u\dfrac{du}{dy}=\dfrac{(y-1)u}{y^2}-\dfrac{2y^2+1}{y^2}$
This belongs to an Abel equation of the second kind.
In fact all Abel equation of the second kind can be transformed into Abel equation of the first kind.
Let $u=\dfrac{1}{v}$ ,
Then $\dfrac{du}{dy}=-\dfrac{1}{v^2}\dfrac{dv}{dy}$
$\therefore-\dfrac{1}{v^3}\dfrac{dv}{dy}=\dfrac{y-1}{y^2v}-\dfrac{2y^2+1}{y^2}$
$\dfrac{dv}{dy}=\dfrac{(2y^2+1)v^3}{y^2}-\dfrac{(y-1)v^2}{y^2}$
Please follow the method in http://www.hindawi.com/journals/ijmms/2011/387429/#sec2
|
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|
Prove that $\sum_{k=0}^{m}\frac{\binom{m}{k}}{\binom{n}{k}}=\frac{n+1}{n+1-m}$ Prove that $\sum_{k=0}^{m}\dfrac{\binom{m}{k}}{\binom{n}{k}}=\dfrac{n+1}{n+1-m}$.
We know, n > m. From the right side. we have $\dfrac{n+1}{n+1-m}=\dfrac{1}{1-\dfrac{m}{n+1}}$. since n > m. $0<\dfrac{m}{n+1}<1$. Then
$\dfrac{1}{1-\dfrac{m}{n+1}}=1+\dfrac{m}{n+1}+(\dfrac{m}{n+1})^2+...$
I don't know what to do next? Did I choose the right path?
|
Partial Fractions
Using the Heaviside Method for Partial Fractions, we get
$$
\begin{align}
\sum_{k=0}^m\frac{\binom{m}{k}}{\binom{n}{k}}
&=1+\sum_{k=1}^m\frac{m(m-1)(m-2)\dots(m-k+1)}{n(n-1)(n-2)\dots(n-k+1)}\tag{1}\\
&=1+m\sum_{k=1}^m\sum_{j=0}^{k-1}\frac{(-1)^{k-j-1}\binom{m-1}{k-1}\binom{k-1}{j}}{n-j}\tag{2}\\
&=1+m\sum_{k=1}^m\sum_{j=0}^{k-1}\frac{(-1)^{k-j-1}\binom{m-1}{j}\binom{m-j-1}{k-j-1}}{n-j}\tag{3}\\
&=1+m\sum_{j=0}^{m-1}\sum_{k=j+1}^m\frac{(-1)^{k-j-1}\binom{m-1}{j}\binom{m-j-1}{k-j-1}}{n-j}\tag{4}\\[6pt]
&=1+\frac{m}{n-m+1}\tag{5}\\[6pt]
&=\frac{n+1}{n-m+1}\tag{6}
\end{align}
$$
Explanation:
$(1)$: break out the $k=0$ term and expand numerator and denominator
$(2)$: apply the Heaviside Method to the fraction in the sum
$(3)$: $\binom{m-1}{k-1}\binom{k-1}{j}=\binom{m-1}{j}\binom{m-j-1}{k-j-1}$
$(4)$: switch the order of summation
$(5)$: $\sum\limits_{k=j+1}^m(-1)^{k-j-1}\binom{m-j-1}{k-j-1}=0^{m-j-1}=\big[j=m-1\big]$
$(6)$: addition
Hockey-Stick Identity
$$
\begin{align}
\sum_{k=0}^m\frac{\binom{m}{k}}{\binom{n}{k}}
&=\sum_{k=0}^m\frac{\frac{m!}{(m-k)!}}{\frac{n!}{(n-k)!}}\tag7\\
&=\frac{m!(n-m)!}{n!}\sum_{k=0}^m\frac{(n-k)!}{(m-k)!(n-m)!}\tag8\\
&=\frac1{\binom{n}{m}}\sum_{k=0}^m\binom{n-k}{n-m}\tag9\\
&=\frac1{\binom{n}{m}}\binom{n+1}{n-m+1}\tag{10}\\[3pt]
&=\frac{n+1}{n-m+1}\tag{11}
\end{align}
$$
Explanation:
$\phantom{1}(7)$: expand the binomial coefficients as factorials and cancel
$\phantom{1}(8)$: shuffle factors and throw in $\frac{(n-m)!}{(n-m)!}$
$\phantom{1}(9)$: recognize binomial coefficients
$(10)$: Hockey-stick identity
$(11)$: simplify
|
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|
How can I find the value of $\ln( |x|)$ without using the calculator? I want to know if there is a way to find for example $\ln(2)$, without using the calculator ?
Thanks
|
We can represent the logarithm of positive rational numbers as follows.
First, consider the following null conditionally convergent series (cancelled harmonic series):
$$0=(1-1)+\left(\frac{1}{2}-\frac{1}{2}\right)+\left(\frac{1}{3}-\frac{1}{3}\right)+\left(\frac{1}{4}-\frac{1}{4}\right)+\left(\frac{1}{5}-\frac{1}{5}\right)+...$$
Note that we are computing $0=\log(1)=\log\left(\frac{1}{1}\right)$ by adding consecutive terms with 1 positive fraction and 1 negative fraction each, taken from the inverses of non-zero integers. This observation may sound trivial now, but it is interesting for what comes next.
We can rearrange the terms of this series to compute $\log(2)$ by taking two positive fractions and one negative for each term.
$$\log\left(2\right)=\left(1+\frac{1}{2}-1\right)+\left(\frac{1}{3}+\frac{1}{4}-\frac{1}{2}\right)+\left(\frac{1}{5}+\frac{1}{6}-\frac{1}{3}\right)+\left(\frac{1}{7}+\frac{1}{8}-\frac{1}{4}\right)+...$$
This can be easily seen to be the Mercator series in disguise, so we have discovered nothing new yet.
But there is more. Similarly, we have
$$\log\left(3\right)=\left(1+\frac{1}{2}+\frac{1}{3}-1\right)+\left(\frac{1}{4}+\frac{1}{5}+\frac{1}{6}-\frac{1}{2}\right)+\left(\frac{1}{7}+\frac{1}{8}+\frac{1}{9}-\frac{1}{3}\right)+\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{4}\right)+...$$
This pattern holds for all positive integers, so the next step is applying the property that $\log(p/q)=\log(p)-\log(q)$ on these representations.
This leads to $\log(p/q)$ by adding $p$ positive fractions and $q$ negative fractions at each step. For example, we have
$$\log\left(\frac{3}{2}\right)=\left(1+\frac{1}{2}+\frac{1}{3}-1-\frac{1}{2}\right)+\left(\frac{1}{4}+\frac{1}{5}+\frac{1}{6}-\frac{1}{3}-\frac{1}{4}\right)+\left(\frac{1}{7}+\frac{1}{8}+\frac{1}{9}-\frac{1}{5}-\frac{1}{6}\right)+...$$
as illustrated in http://oeis.org/A166871.
See also Do these series converge to logarithms? and Series for logarithms for further discussion of generalized Mercator series.
|
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|
$123^{561}$ find last $2$ digits Modular Exponentiation mod $100$? $123^{561}$
Find the last $2$ digits
Can't I work this out using modular exponentiation working mod $100$?
$123^2 = 29\pmod{100}$
$123^4 = (123^2)^2 = 41\pmod{100}$
$123^8 = (123^4)^2 = 81\pmod{100}$
$123^{16} = (123^8)^2 = 61\pmod{100}$
$123^{32} = (123^16)^2 = 21\pmod{100}$
$123^{64} = (123^32)^2 = 41\pmod{100}$
$123^{128} = (123^64)^2 = 81\pmod{100}$
$123^{256} = (123^128)^2 = 61\pmod{100}$
$123^{512} = (123^256)^2 = 21\pmod{100}$
Then $561 = 512 \cdot 32 \cdot 16 \cdot 1$
so $21 \cdot 21 \cdot 61 \cdot 29 = 29\pmod{100}$
The answer is meant to be $21$!
What am I doing wrong?
Do I have to use Eulers?
|
While dealing with modulo of composite number, Carmichael function is general more, if not equally efficient than Euler's Totient Theorem .
Here $\lambda(100)=$lcm$(\lambda(5^2),\lambda(2^2))=$lcm$(20,4)=20$
$\implies a^{20}\equiv1\pmod{100}\ \ \ \ (1)$ if $(a,100)=1\iff (a,10)=1$
whereas $\phi(10^2)=10\cdot\phi(10)=40$
$\implies a^{40}\equiv1\pmod{100} \ \ \ \ (2)$ if $(a,100)=1\iff (a,10)=1$
Now using $(1),$ as $561\equiv1\pmod{20}, a^{561}\equiv a\pmod{100}$
Using $(2),$ as $561\equiv1\pmod{40}, a^{561}\equiv a\pmod{100}$
|
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|
How to prove $\sum_{k=1}^n \frac{2^k}{k}< 3\frac{2^n}{n}$? How to prove $$\sum_{k=1}^n \frac{2^k}{k}< 3\frac{2^n}{n}$$ and further $$\lim_{n\rightarrow \infty}\frac{n}{2^n}\sum_{k=1}^n
\frac{2^{k}}{k} = 2$$?
These results are verified by computer, yet I can't figure out a neat proof.
|
I'll answer the second part of the question. It is helpful to notice that the sum
$$
S_n = \sum_{k=1}^n \frac{2^k}{k}
$$
is dominated by only a few terms near $k=n$. In a sense, the terms near the end constitute the principal contribution to the whole sum, while the terms near the beginning are negligible. We can therefore hope for a good estimate for the whole sum by splitting it apart into these two pieces and estimating them individually.
In this case it turns out that we can split the sum evenly.
For the first half we can get away with the crude estimate
$$
\begin{align}
\sum_{k=1}^{\lfloor n/2 \rfloor} \frac{2^k}{k} &< \sum_{k=1}^{\lfloor n/2 \rfloor} 2^k \\
&= 2\left(2^{\lfloor n/2 \rfloor}-1\right) \\
&< 2^{n/2+1} \\
&= O\left(2^{n/2}\right).
\end{align} \tag{1}
$$
To help us estimate the second half of the sum we will introduce the new variable $h=n-k$. Note that when $\lfloor n/2 \rfloor + 1 \leq k \leq n$ we have $0 \leq h < n/2$.
The power series for $1/(1-x)$ converges uniformly and absolutely on the interval $[0,1/2]$, so we can write $1/(1-x) = 1 + O(x)$ there. We then have
$$
\begin{align}
\frac{1}{k} &= \frac{1}{n-h} \\
&= \frac{1}{n} \cdot \frac{1}{1-h/n} \\
&= \frac{1}{n} \left[1 + O\left(\frac{h}{n}\right)\right] \\
&= \frac{1}{n} + O\left(\frac{h}{n^2}\right)
\end{align} \tag{$\star$}
$$
for $n \geq 1$ and $\lfloor n/2 \rfloor + 1 \leq k \leq n$. This allows us to rewrite the second half of $S_n$ as
$$
\sum_{k=\lfloor n/2 \rfloor + 1}^{n} \frac{2^k}{k} = \frac{1}{n} \sum_{k=\lfloor n/2 \rfloor + 1}^{n} 2^k + \sum_{k=\lfloor n/2 \rfloor + 1}^{n} 2^k O\left(\frac{h}{n^2}\right). \tag{2}
$$
The first sum on the right-hand side is
$$
\begin{align}
\sum_{k=\lfloor n/2 \rfloor + 1}^{n} 2^k &= 2^{\lfloor n/2 \rfloor + 1} \sum_{k=0}^{\lceil n/2 \rceil - 1} 2^k \\
&= 2^{\lfloor n/2 \rfloor + 1} \left(2^{\lceil n/2 \rceil}-1\right) \\
&= 2^{n+1} + O\left(2^{n/2}\right),
\end{align}
$$
where we have used the fact that $\lfloor n/2 \rfloor + \lceil n/2 \rceil = n$. To estimate the second sum we first rewrite it in terms of $h$ and $n$ only:
$$
\begin{align}
\frac{1}{n^2} \sum_{k=\lfloor n/2 \rfloor + 1}^{n} 2^k h &= \frac{2^n}{n^2} \sum_{k=\lfloor n/2 \rfloor + 1}^{n} 2^{k-n} h \\
&= \frac{2^n}{n^2} \sum_{h=0}^{\lceil n/2 \rceil - 1} 2^{-h}h.
\end{align}
$$
The quantity $\sum_{h=0}^{\lceil n/2 \rceil - 1} 2^{-h}h$ is bounded, so for the second sum we get the estimate
$$
\sum_{k=\lfloor n/2 \rfloor + 1}^{n} 2^k O\left(\frac{h}{n^2}\right) = O\left(\frac{2^n}{n^2}\right).
$$
Combining these and absorbing the first error term into the second we find that $(2)$ becomes
$$
\begin{align}
\sum_{k=\lfloor n/2 \rfloor + 1}^{n} \frac{2^k}{k} &= \frac{2^{n+1}}{n} + O\left(\frac{2^{n/2}}{n}\right) + O\left(\frac{2^n}{n^2}\right) \\
&= \frac{2^{n+1}}{n} + O\left(\frac{2^n}{n^2}\right).
\end{align}
$$
By using the estimate in $(1)$ we obtain an estimate for the whole sum,
$$
\begin{align}
S_n &= \sum_{k=1}^{\lfloor n/2 \rfloor} \frac{2^k}{k} + \sum_{k=\lfloor n/2 \rfloor + 1}^{n} \frac{2^k}{k} \\
&= O\left(2^{n/2}\right) + \frac{2^{n+1}}{n} + O\left(\frac{2^n}{n^2}\right) \\
&= \frac{2^{n+1}}{n} + O\left(\frac{2^n}{n^2}\right).
\end{align}
$$
Thus
$$
\frac{n}{2^n} \sum_{k=1}^{n} \frac{2^k}{k} = 2 + O\left(\frac{1}{n}\right).
$$
If you desire more terms in the approximation you can take more terms in the series expansion $(\star)$. For instance, taking
$$
\frac{1}{k} = \frac{1}{n} + \frac{h}{n^2} + O\left(\frac{h^2}{n^3}\right)
$$
yields
$$
\frac{n}{2^n} \sum_{k=1}^{n} \frac{2^k}{k} = 2 + \frac{2}{n} + O\left(\frac{1}{n^2}\right).
$$
|
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|
Improper Integral $\int\limits_0^1\frac{\ln(x)}{x^2-1}\,dx$ How can I prove that?
$$\int_0^1\frac{\ln(x)}{x^2-1}\,dx=\frac{\pi^2}{8}$$
I know that
$$\int_0^1\frac{\ln(x)}{x^2-1}\,dx=\sum_{n=0}^{\infty}\int_0^1-x^{2n}\ln(x)\,dx=\sum_{n=0}^{\infty}\frac{1}{(2n+1)^2}=\frac{\pi^2}{8}$$
but I want another method.
|
We have$$\int_0^1\frac{\ln(x)}{x^2-1}\,\mathrm dx=\int_0^1\frac{\ln(1-x)}{(1-x)^2-1}\,\mathrm dx=\int_0^1\frac{\ln(1-x)}{x(x-2)}\,\mathrm dx$$
We will generalize by introducing parameter $\alpha$ such that
$$I(\alpha )=\int_0^1\frac{\ln(1-\alpha x )}{x(x-2)}\,\mathrm dx$$
And we have $I(0)=0$
Then
$$I'(\alpha )=-\int_0^1\frac{1}{(1-\alpha x)(x-2)}\,\mathrm dx=\frac{1}{2\alpha -1}\left[\ln\left(\frac{x-2}{1-\alpha x}\right)\right]_0^1=\frac{\ln(2-2\alpha )}{1-2\alpha }$$
And we have
$$I'(\alpha )=\frac{\ln(2-2\alpha )}{1-2\alpha }$$
$$I(\alpha )=\frac{1}{2}\text{Li}_2(2\alpha -1)+c$$
$$I(0)=\frac{1}{2}\text{Li}_2(-1)+c=0\implies c=-\frac{1}{2}\text{Li}_2(-1)=\frac{\pi^2}{24}$$
$$I(\alpha )=\frac{1}{2}\text{Li}_2(2\alpha -1)+\frac{\pi^2}{24}$$
$$\begin{align}I(1)&=\frac{1}{2}\text{Li}_2(1)+\frac{\pi^2}{24}\\
&=\frac{\pi^2}{12}+\frac{\pi^2}{24}\\
&=\frac{\pi^2}{8}\\
\end{align}$$
$$I(1)=\int_0^1\frac{\ln(x)}{x^2-1}\,\mathrm dx=\frac{\pi^2}{8}$$
|
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|
Calculating Value Of Trigonometric Formula when $~\tan\theta+\sin\theta=\dfrac{1}{2}$, evaluate $~(\sin^{2}\theta-\sin 2\theta)$
is it possible to get the exact value?
I got $~\sin^{2}\theta-\sin 2\theta=\dfrac{\tan^{2}\theta-2\tan\theta}{1+\tan^{2}\theta}$, but can't calculate it more.
|
$\large \tan \theta + \sin \theta=\frac{1}{2} $; Squaring we get
$\large\tan^2 \theta+2.\sin \theta.\tan \theta+\sin^2 \theta=\frac{1}{4}$
$\large\frac{\sin^2 \theta}{\cos^2 \theta}+2.\frac{\sin^2 \theta}{\cos\theta}+\sin^2 \theta=\frac{1}{4}$
$\large\sin^2\theta[\frac{1}{\cos^2\theta}+2.\frac{1}{\cos\theta}+1]=\frac{1}{4}$
$\large\sin^2\theta[\frac{1+2.\cos\theta+\cos^2\theta}{\cos^2\theta}]=\frac{1}{4}$
$\large\sin^2\theta[\frac{(1+\cos\theta)^2}{\cos^2\theta}]=\frac{1}{4}$
$\large\tan\theta.(1+\cos\theta)=\frac{1}{2}$
$\large(\frac{1}{2}-\sin\theta)(1+\cos\theta)=\frac{1}{2}$
$\cos\theta-2.\sin\theta-\sin2\theta=0$
$\sin2\theta=\cos\theta-2.\sin\theta$
put this expression for $\sin2\theta$ into $\sin^2\theta-\sin2\theta$ to get
$\sin^2\theta-\cos\theta+2.\sin\theta+ \color{red}{1-1}$ (making perfect square)
$(\sin\theta+1)^2-(\cos\theta+1)$
Is this what you need?
|
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|
If $a^3 + b^3 +3ab = 1$, find $a+b$
Given that the real numbers $a,b$ satisfy $a^3 + b^3 +3ab = 1$, find $a+b$.
I tried to factorize it but unable to do it.
|
$$a^3+b^3-1+3ab=(a+b)^3-1-3a^2b-3ab^2+3ab=$$
$$=(a+b-1)((a+b)^2+a+b+1)-3ab(a+b-1)=$$
$$=(a+b-1)(a^2+b^2-ab+a+b+1).$$
Thus, $$a+b=1$$ or
$$a^2+b^2-ab+a+b+1=0$$ or
$$(a-b)^2+(a+1)^2+(b+1)^2=0,$$ which gives $a=b=-1$ and $$a+b=-2.$$
|
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|
How to derive this second derivative using the quotient rule? If a given first derivative is: $\ {dy \over dx} = {-48x \over (x^2+12)^2} $
What are the steps using the quotient rule to derive the second derivative: $\ {d^2y \over dx^2} = {-144(4-x^2) \over (x^2+12)^3} $
My Steps:
\begin{align*}
{d^2y \over dx^2}
&= {-48(x^2 +12)^2 - 2(x^2+12)(2x)(-48x)\over (x^2+12)^4} \\
&=-48{(x^2 +12)^2 - 4x^2(x^2+12)\over (x^2+12)^4} \\
&= -48{(x^2 +12)( - 4x^2 +(x^2+12))\over (x^2+12)^4} \\
&= -48{(x^2 +12)( - 4x^2)\over (x^2+12)^3} \\
&= {???}
\end{align*}
|
Recall the Quotient Rule:
$$(\frac{f}{g})' = \frac{f'\cdot g - g' \cdot f}{g^2}$$
So you have the first derivative, $\frac{dy}{dx}= \frac{-48x}{(x^2+12)^2}$. Just apply the quotient rule:
Let $f(x) = -48x$. So $f'(x) = -48$.
Let $g(x) = (x^2 + 12)^2$. So, by the chain rule, $g'(x) = 2(x^2 + 12) \cdot 2x = 4x(x^2+12)$.
$$\frac{d^2y}{dx^2} = \frac{(-48 \cdot (x^2+12)^2) - (4x(x^2+12) \cdot -48x)}{((x^2+12)^2)^2}$$
Notice that both elements of the numerator contain $-48$. We can factor that out.
$$\frac{d^2y}{dx^2} = \frac{-48 \cdot ((x^2+12) - (4x^2(x^2+12)))}{(x^2+12)^4}$$
We also can factor out the $(x^2+12)^2$ term:
$$\frac{d^2y}{dx^2} = \frac{-48 \cdot (x^2+12) \cdot ((x^2+12) - 4x^2)}{(x^2+12)^4}$$
And now we can cancel:
$$\frac{d^2y}{dx^2} = \frac{-48 \cdot (12 - 3x^2)}{(x^2+12)^3}$$
Observe that we can factor the numerator some more:
$$\frac{d^2y}{dx^2} = \frac{-48 \cdot 3 \cdot (4 - x^2)}{(x^2+12)^3}$$
And there we are:
$$\frac{d^2y}{dx^2} = \frac{-144 \cdot (4 - x^2)}{(x^2+12)^3}$$
|
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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|
How to find the minimum $n$
Find the smallest $n$ such that the sequence
$1000n,1000n+1,1000n+2,\cdots,1000n+2013$ does not contain a perfect square.
My try:
$n=1$ does not work because
$$1000n=1000,1000n+1=1001,\cdots,1000n+4=1024,\cdots,1000n+2013=3013$$
contains the square numbers $32^2=1024,33^2=1089,\cdots\cdots,54^2=2916$
so $n\neq 1$
Similarly, when $n=2$,then
$$1000n=2000,1000n=2001,\cdots,1000n+2013=4013$$
it is easy find this sequence contains square numbers $$45^2=2015,46^2=2116,\cdots,63^2=3969$$
so $n\neq 2$
$\cdots\cdots\cdots$
Thank you for your help
|
Let $x^2$ be the largest perfect square $\leq 1000n-1$. Then $(x+1)^2 \geq 1000n+2014$ so $(2x+1)=(x+1)^2-x^2 \geq 2015$ so $x \geq 1007$.
Put $x=1000+y, y \geq 0$, so that $x^2 \equiv y^2 \pmod{1000}$. If $y \leq 30$, then $x^2 \equiv 0, 1, 4, \ldots ,900 \pmod{1000}$ so $x^2 \leq 1000n-100$. Thus $(2x+1)=(x+1)^2-x^2 \geq 2114$ so $x \geq 1057$ so $y \geq 57$, a contradiction.
Thus $y \geq 31$, so that $x \geq 1031$, and $n \geq \frac{1031^2+1}{1000}=1062.962$, so $n \geq 1063$. When $n=1063$, we have $1031^2=1062961<1063000=1000n<1000n+2013=1065013<1065024=1032^2$ so $n=1063$ is minimal.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Convergence of $\frac{1}{\sqrt{2\pi i\varepsilon}}\exp\left( -\frac{x^2}{2i\varepsilon}\right)$ as $\varepsilon \to 0$ Is it true that $\frac{1}{\sqrt{2\pi i\varepsilon}}\exp\left( -\frac{x^2}{2i\varepsilon}\right)$ converges (in some sense) to $\delta_0$, Dirac delta distribution at point $0$, as $\varepsilon \to 0$ ? Thanks.
|
So the $i$ is intentional. Then let's look how that function acts as a distribution, hence let $\varphi$ an arbitrary test function:
$$\begin{align}
\frac{1}{\sqrt{2\pi i\varepsilon}}\int_{-\infty}^\infty \exp \left(i\frac{x^2}{2\varepsilon}\right)\varphi(x)\,dx &= \frac{1}{\sqrt{2\pi i\varepsilon}}\int_{-\infty}^\infty \left(\cos \frac{x^2}{2\varepsilon} + i\sin \frac{x^2}{2\varepsilon}\right)\varphi(x)\,dx\\
&= \frac{1}{\sqrt{\pi i}}\int_{-\infty}^\infty \left(\cos (y^2)+ i \sin (y^2)\right)\varphi(\sqrt{2\varepsilon}\cdot y)\,dy\\
&= \frac{1}{\sqrt{\pi i}}\int_{-\infty}^0 \left(\cos (y^2)+ i \sin (y^2)\right)\varphi(\sqrt{2\varepsilon}\cdot y)\,dy\\
&\qquad + \frac{1}{\sqrt{\pi i}}\int_{0}^\infty \left(\cos (y^2)+ i \sin (y^2)\right)\varphi(\sqrt{2\varepsilon}\cdot y)\,dy\\
&= \frac{1}{\sqrt{\pi i}} \int_0^\infty \left(\cos (y^2) + i\sin (y^2)\right)\left(\varphi(\sqrt{2\varepsilon}\cdot y) + \varphi(-\sqrt{2\varepsilon}\cdot y) \right)\,dy.
\end{align}$$
Now for $\varepsilon = 0$, we have the Fresnel integral
$$\frac{2\varphi(0)}{\sqrt{\pi i}}\int_0^\infty \cos (y^2) + i \sin (y^2)\,dy = \frac{2\varphi(0)(1+i)}{\sqrt{\pi i}}\sqrt{\frac{\pi}{8}} = \varphi(0),$$
and it remains to argue that taking the limit is legitimate. I'll leave that to handwaving, it can be done.
|
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|
Calculate $a^3+b^3+c^3+d^3$ for the real roots of $x^4+2x^3-3x^2-3x+2$ $a,b,c,d \in \mathbb{R}$ are the real roots of $x^4+2x^3-3x^2-3x+2$.
Calculate $a^3+b^3+c^3+d^3$.
With approximation i found out, that $a^3+b^3+c^3+d^3 = -17$, but how can I proof that without calculating the roots exactly?
Cheers
|
thanks for your help.
$P(x) = x^4 + 2x^3 -3x^2 -3x + 2:$
$\sigma_1 = -2$
$\sigma_2 = -3$
$\sigma_3 = 3$
$\sigma_4 = 2$
$a^3 + b^3 + c^3 + d^3 = \sigma_1^3 - 3*\sigma_1 * \sigma_2 + 3 * \sigma_3$
We get: $a^3 + b^3 + c^3 + d^3 = (-2)^3 - 3 * (-2) * (-3) + 3 * 3 = -17$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Prove: Sum and Difference of two distinct positive integers are both perfect squares. I'm trying to prove that there exists two distinct positive integers whose sum and difference are both perfect squares. I cannot find any pattern or characteristic between the pairs of numbers that work i.e.
*
*4, 5
*6, 10
*8, 17
*10, 6
*10, 26
Any help will be greatly appreciated!
Thanks!
|
Let those positive integers be $a$ and $b$.Then,
$a+b=x^2$
$a-b=y^2$
Adding the two we get $a=\frac{x^2+y^2}{2}$ and $b=\frac{x^2-y^2}{2}$
Since, a and b are integers we must have $x^2+y^2$ and $x^2-y^2$ must be even, for that we must have x and y both even or both odd.
Now for finding such pairs take any even $x,y$ for example let x=8 and y=4
which gives $a=40$ and $b=24$, we have $a+b=64=8^2$ and $a-b=16=4^2$
|
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|
Find asymptotics in a given form $n=(e+o(1))^{f(s)}$
Let $p\to\infty$, $s={\binom {p^4} p}$ and $n={\binom {p^4}{p^2}}$. Find a function $f(s)$ in the following form $$\large n=(e+o(1))^{f(s)}$$
I've tried to use the followinf asymptotics for binomial coeffs:
$$\binom n k = \frac{n^k \exp\left(k^2/2n\right)}{k!}(1+O(1))$$
And I've got the following results:
$$s={\binom {p^4} p} = \exp\left({2p^2 \ln p - \frac{1}{2} (1 + \ln 2\pi) + p^2 + \ln p + O\left(\frac{1}{p^2}\right)}\right)$$
$$n={\binom {p^4}{p^2}} = \exp\left({4p\ln p - \frac{1}{2p^2} - \left(p + \frac{1}{2}\right)\ln p + p - \frac{1}{2}\ln 2\pi + O\left(\frac{1}{p}\right)}\right)$$
What should I do next? A hint would be useful!
|
For the beginning some preparatory computations:
$$
\begin{align}
{n\choose k}
&=\frac{n\ldots (n-k+1)}{k!}\\
&=\frac{n^k}{k!}\left(1-\frac{1}{n}\right)\ldots\left(1-\frac{k-1}{n}\right)\\
&=\frac{n^k}{k!}\exp\left(\sum_{i=1}^{k-1}\ln\left(1-\frac{i}{n}\right)\right)\\
&=\frac{n^k}{k!}\exp\left(\sum_{i=1}^{k-1}\left(-\frac{i}{n}+O\left(\frac{i^2}{n^2}\right)\right)\right)\\
&=\frac{n^k}{k!}\exp\left(-\frac{k(k-1)}{2n}+O\left(\frac{k^3}{n^2}\right)\right)\\
\end{align}
$$
Thus
$$
{n\choose k}=\frac{n^k}{k!}e^{-\frac{k(k-1)}{2n}+O\left(\frac{k^3}{n^2}\right)}
$$
Given this formula and Stirling's approximation we get
$$
\begin{align}
\ln s
&=\ln {p^4\choose p}\\
&=p\ln p^4-\ln p!-\frac{p(p-1)}{2p^4}+O\left(\frac{p^3}{p^8}\right)\\
&=4p\ln p-\left(p\ln p-p+\frac{1}{2}\ln p+\ln\sqrt{2\pi}+O\left(\frac{1}{p}\right)\right)-\frac{1}{2p^2}+\frac{1}{2p^3}+O\left(\frac{1}{p^5}\right)\\
&=3p\ln p+p-\frac{1}{2}\ln p-\frac{1}{2}\ln 2\pi+O\left(p^{-1}\right)\\
&=3p\ln p+O(p)
\end{align}
$$
$$
\begin{align}
\ln n
&=\ln {p^4\choose p^2}\\
&=p^2\ln p^4-\ln (p^2)!-\frac{p^2(p^2-1)}{2p^4}+O\left(\frac{p^6}{p^8}\right)\\
&=4p^2\ln p-\left(p^2\ln p^2-p^2+\frac{1}{2}\ln p^2+\ln\sqrt{2\pi}+O\left(\frac{1}{p^2}\right)\right)-\frac{1}{2}+\frac{1}{2p^2}+O\left(\frac{1}{p^2}\right)\\
&=2p^2\ln p+p^2-\ln p-\frac{1}{2}\ln 2\pi e+\frac{1}{2p^2}+O\left(p^{-2}\right)\\
&=2p^2\ln p+O(p^2)
\end{align}
$$
We claim, that
$$
f(x)=\frac{2\ln^2 x}{9\ln\ln x}
$$
is the desired function. Indeed,
$$
\begin{align}
\ln^2 s
&=(3p\ln p+O(p))^2\\
&=9p^2\ln^2 p+6p\ln p O(p)+O(p)^2\\
&=9p^2\ln^2 p+ O(p^2\ln p)+O(p^2)\\
&=9p^2\ln^2 p+O(p^2\ln p)\\
\end{align}
$$
$$
\begin{align}
\ln\ln s
&=\ln(3p\ln p+O(p))\\
&=\ln(3p\ln p(1+O(\ln^{-1}p)))\\
&=\ln 3+\ln p + \ln\ln p + \ln(1+O(\ln^{-1}p))\\
&=\ln p + \ln\ln p+\ln 3+O(\ln^{-1} p)\\
&=\ln p+O(\ln\ln p)\\
\end{align}
$$
$$
\begin{align}
f(s)
&=\frac{2\ln^2 s}{9\ln\ln s}\\
&=\frac{2(9p^2\ln^2p+O(p^2\ln p))}{9\ln p+O(\ln\ln p)}\\
&=\frac{9\ln p(2p^2\ln p+O(p^2))}{9\ln p\left(1+O\left(\frac{\ln\ln p}{\ln p}\right)\right)}\\
&=(2p^2\ln p+O(p^2))\left(1+O\left(\frac{\ln\ln p}{\ln p}\right)\right)\\
&=2p^2\ln p+O(2p^2\ln\ln p)+O(p^2)+O\left(p^2\frac{\ln\ln p}{\ln p}\right)\\
&=2p^2\ln p+O(p^2)\\
\end{align}
$$
Since
$$
\ln n-f(s)=2p^2\ln p+O(p^2)-(2p^2\ln p+O(p^2))=O(p^2)=o(2p^2\ln p+O(p^2))=o(f(s))
$$
then $n=(e+o(1))^{f(s)}$
|
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|
Stochastics with induction prove that for all $n \in \mathbb{N}$:
$\sum_{r=0}^n \binom{n}{r}(-1)^{r} = 0$.
The base step is easy, i only get lots of problems when i try to mess with the sum boundries....
so far i've tried:
$\sum_{r=0}^{n+1} \binom{n+1}{r}(-1)^{r} = \sum_{r=0}^{n+1}(\binom{n}{r}+\binom{n}{r-1})(-1)^{r} = \sum_{r=0}^{n+1}\binom{n}{r}(-1)^{r}+\sum_{r=0}^{n+1}\binom{n}{r-1}(-1)^{r}$
Don't know how to proceed... Please help?
|
Perhaps this way:
$$
\begin{split}
\sum_{r=0}^{n+1} \binom{n+1}{r} (-1)^r
&= \binom{n+1}{0} + \binom{n+1}{n+1} (-1)^{n+1}
+ \sum_{r=1}^n \binom{n+1}{r} (-1)^r \\
&= 1 + (-1)^{n+1}
+ \sum_{r=0}^{n-1} \binom{n+1}{r+1} (-1)^{r+1} \\
&= 1 + (-1)^{n+1}
- \sum_{r=0}^{n-1} \binom{n+1}{r+1} (-1)^r \\
&= 1 + (-1)^{n+1}
- \sum_{r=0}^{n-1} \binom{n}{r+1} (-1)^r
- \sum_{r=0}^{n-1} \binom{n}{r} (-1)^r \\
&= 1 + (-1)^{n+1}
+ \sum_{r=0}^{n-1} \binom{n}{r+1} (-1)^{r+1}
- \sum_{r=0}^{n-1} \binom{n}{r} (-1)^r \\
&= 1 + (-1)^{n+1}
+ \sum_{r=1}^n \binom{n}{r} (-1)^r
- \sum_{r=0}^{n-1} \binom{n}{r} (-1)^r \\
& \text{cancel terms in the sum indexed $1...n-1$} \\
&= 1 + (-1)^{n+1} + \binom{n}{n}(-1)^n- \binom{n}{0} \\
&= (-1)^{n+1} + (-1)^n \\
&= 0
\end{split}
$$
as desired.
|
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|
Find the Area of the ellipse Given $$\frac{x^2}{a^2} + \frac{y^2}{b^2}=1$$
where $a>0$, $b>0$
I tried to make $y$ the subject from the equation of the ellipse and integrate from $0$ to $a$. Then multiply by $4$ since there are $4$ quadrants.
$$Area=4\int^a_0\left(b^2-x^2\left(\frac{b^2}{a^2}\right)\right)^\frac{1}{2}dx$$
I can't get the answer $\pi ab$
|
Here is my proof if it is any use to anyone.
The equation of an ellipse is given by: $$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$
Rearrange and write this ( the equation of an ellipse ) in terms of y: $$y=b\sqrt{1-\frac{x^2}{a^2}}$$
Lets find the area of one quarter of the ellipse and multiple that by 4 to get the area of the entire ellipse. We will integrate from $x=0$ to $x=a$. So we are looking for: $$4\int_0^a y.dx$$
Which is the same as ... $eq1$: $$4\int_0^a b\sqrt{1-\frac{x^2}{a^2}}.dx$$
Let: $$\sin u=\frac{x}{a}$$
Rearrange to get: $$x=a\sin u$$
Differentiate to get: $$\frac{dx}{du}=a\cos u$$
Therefore: $$dx=a\cos u.du$$
Sub this back into equation $eq1$.
But we also need to find the range of $u$ values.
Using $\sin u = \frac{x}{a}$ we can sub in our original limits of $x=0$ and $x=a$ to get the new limits in terms of $u$ of $u=0$ and $u=\frac{\pi}{2}$.
Subbing this back into $eq1$ as well we get $eq2$: $$4\int_0^\frac{\pi}{2} b\sqrt{1-(\sin u)^2}.a\cos u.du$$
From trigonometry we know that: $$\sin^2\theta + \cos^2\theta = 1$$
Therefore $$\sqrt{1-\sin^2u}.\cos u = \sqrt{\cos^2u}.\cos u = \cos^2u$$
Subbing this back into $eq2$ we get: $$4\int_0^\frac{\pi}{2} ab\cos^2u.du$$
Which is the same as: $$4ab\int_0^\frac{\pi}{2} \cos^2u.du$$
But again from trigonometry we know that: $$\cos^2\theta=\frac{1}{2}(1+\cos2\theta)$$
Subbing gives us: $$4ab\int_0^\frac{\pi}{2} \frac{1}{2}(1+\cos2u).du$$
Which is the same as: $$2ab\int_0^\frac{\pi}{2} (1+\cos2u).du$$
Now we can finally integrate to get: $$2ab[u+\frac{\sin2u}{2}]_0^\frac{\pi}{2}$$
Which goes to: $$2ab[(\frac{\pi}{2}+0)-(0+0)]$$
Which finally goes to: $$\pi ab$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Using Newton's method calculate $ \frac{1}{\sqrt{a}} $ without division Suggest algorithm for the numerical calculation $ \frac{1}{\sqrt{a}} \ a > 0 $ without division, use Newton's method.
My idea is:
$$ \frac{1}{\sqrt{a}} = (\sqrt{a})^{-1} = a^{-\frac{1}{2}}$$
$$ x = a^{-\frac{1}{2}} \Rightarrow f(x)=x^2 - a^{-1}$$
And next step is calulate $ f'(x)=2x $, but after substitution I got: $$ x_{n+1} = \frac{x_n^2+a^{-1}}{2x_n}$$ And know I still have divion, I guess. Maybe you've better idea on this task?
|
After using dark magic to make an initial guess $y_0$, the fast inverse square root algoritm essentially approximates $y=1/\sqrt{x}$, $x > 0$, by using the iteration
$$
y_{n+1} = \frac{3}{2}y_n - \frac{1}{2}xy_n^3.
$$
There is no division here, unless you count the multiplication by the constants $3/2$ and $1/2$.
If the limit as $n \to \infty$ exists then it is a solution to the cubic equation
$$
y = \frac{3}{2}y - \frac{1}{2}xy^3.
$$
Of course the only solutions are $y=0$ and $y=\pm 1/\sqrt{x}$. We can do some analysis using the standard techniques for discrete dynamical systems to determine when the iteration will converge and to which point.
Let
$$
f(y) = \frac{3}{2}y - \frac{1}{2}xy^3.
$$
Then $f'(0) = 3/2 > 1$, so $y=0$ is an unstable equilibrium of the system. We also have $f'(1/\sqrt{x}) = 0$, which is $<1$ in absolute value, so that the points $y=\pm 1/\sqrt{x}$ are stable equilibria.
In other words, if $y_0$ is close enough to $\pm 1/\sqrt{x}$ then $y_n$ will converge to $\pm 1/\sqrt{x}$. We can determine how close we need to be by calculating
$$
f'(y) = \frac{3}{2}\left(1-xy^2\right).
$$
We require $|f'(y)| < 1$ in order for $f$ to be a contraction, and this is equivalent to
$$
-\frac{2}{3} < 1 - xy^2 < \frac{2}{3}.
$$
Rearranging, we end up with
$$
\frac{1}{3x} < y^2 < \frac{5}{3x}.
$$
So, if
$$
\sqrt\frac{1}{3x} < y_0 < \sqrt\frac{5}{3x}
$$
then $y_n \to 1/\sqrt{x}$ as $n \to \infty$, and if
$$
-\sqrt\frac{5}{3x} < y_0 < -\sqrt\frac{1}{3x}
$$
then $y_n \to -1/\sqrt{x}$ as $n \to \infty$.
Like the traditional Newton's method, this iterative scheme also converges quadratically. If we let $y_n = 1/\sqrt{x} + \epsilon_n$, then the system becomes
$$
\epsilon_{n+1} = -\frac{3\sqrt{x}}{2} \epsilon_n^2 - \frac{x}{2} \epsilon_n^3.
$$
|
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|
Proving the form of a sequence's terms How do I go about attacking this problem and what is it asking?
Suppose that $\alpha^2 = \alpha + 1$ and suppose $F_n$ denotes the Fibonacci sequence.
Show that $\alpha^3 = 2\alpha +1, \alpha^4 = 3\alpha +2$
|
More explicitly,
since
$x^2 = x+1$,
$x^3 = x(x^2)
=x(x+1)
=x^2+x
=(x+1)+x
=2x+1
$.
Do the same for
$x^4$.
Essentially,
any power of $x$,
say $x^n$ with $n \ge 3$,
can be replaced
by lower powers of $x$
via
$x^n
=x^{n-2}x^2
=x^{n-2}(x+1)
=x^{n-1}+x^{n-2}
$.
Keep on applying this
until only linear and constant terms are left.
You will find an interesting formula
for $x^n$
in terms of $x$ and $1$.
|
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|
Simplifying square root with fraction I'm not sure about this equality $$4(-3+\sqrt {15})/4)^2 = (9-6 \sqrt{15} +15)/4$$
Hope some one can enlighten me. I will be facing more of such fractions, please guide me on how to solve/simplify in easy method.
Thanks :)
|
I’ll even finish the simplification:
$$\begin{align*}
4\left(\frac{-3+\sqrt{15}}4\right)^2&=4\cdot\frac{(-3+\sqrt{15})(-3+\sqrt{15}}{4\cdot4}\\
&=\frac{(-3)^2+2(-3)\sqrt{15}+(\sqrt{15})^2}4\\
&=\frac{9-6\sqrt{15}+15}4\\
&=\frac{24-6\sqrt{15}}4\\
&=6-\frac32\sqrt{15}\;.
\end{align*}$$
In short: cancel a factor of $4$, and multiply out the numerator.
|
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|
Show that $\frac{(2n)!}{(n)!}=2^n(2n-1)!!$ Show that $\frac{(2n)!}{(n)!}=2^n(2n-1)!!$ is the question I am struggling with.
I started by saying: $(2n)!=2n(2n-1)(2n-2)(2n-3)...3*2*1$
But then I'm stuck.
|
$$\frac{2n!}{n!} = \frac{1\cdot2\cdot\ldots\cdot2n}{1\cdot2\cdot\ldots \cdot n}=\frac{1\cdot3\cdot\ldots\cdot(2n-1)\cdot2\cdot4\cdot\ldots\cdot2n}{1\cdot2\cdot\ldots \cdot n} = \frac{1\cdot3\cdot\ldots(2n-1)\cdot2^n\cdot 1\cdot2\cdot\ldots \cdot n}{1\cdot2\cdot\ldots \cdot n} = 2^n\cdot1\cdot3\cdot\ldots \cdot (2n-1)=2^n(2n-1)!!$$
|
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|
Elementary proof that $\pi < \sqrt{5} + 1$ I wanted to show that
$$ \frac{\pi}{4\phi} < \frac{1}{2} $$
Where $\phi$ is the golden ratio. I have confirmed the results numerically, and by
simple algebra the inequality simplifies down to
$$
\pi < \sqrt{5} + 1
$$
This is a weaker relation than what was shown here. Prove that $\dfrac{\pi}{\phi^2}<\dfrac{6}5 $.
By squaring my inequality (valid since both sides are positive), and dividing by $6$ I obtain.
$$ \frac{\pi^2}{6} < 1 + \frac{\sqrt{5}}{3} $$
Where the left handside has a very neat series representation, alas the same does not hold for the right handside. However this is far from an elementary solution. Does someone have a relative simple proof for the equality? To be precise something that is not using advanced knowledge of series. =)
|
Below are two geometric ways to get better approximations than the one requested. Neither of the two involve the expression $\sqrt{5}+1$ directly so it would still take some work to show that these results are in fact stronger. However, a trick combining both results will show that $\pi < \frac{16}{5}$ which is quickly seen to be sufficient.
1. Let $0 < x \leq \frac{\pi}{2}$. Draw two lines through $(1, \frac{x}{2})$ that are tangent to the unit circle. One line touches the unit circle at $(1,0)$ the other one at $$p = \left(\frac{4-x^2}{4+x^2}, \frac{4x}{4+x^2}\right).$$
Since the distance over the unit circle from $(1,0)$ to $p$ is less than $x$ we get the inequality
$$\sin(x) > \frac{4x}{4+x^2}.$$
In particular $\sin(4-2\sqrt{3}) > \frac{1}{2}$ and therefore $\frac{\pi}{6} < 4-2\sqrt{3}$.
2. Let $$\begin{eqnarray}x(t) &=& 1 - \frac{t^2}{2}\\ y(t) &=& t\end{eqnarray}$$ Then $x^2+y^2 \geq 1$. Let $\alpha = \sqrt{3} - 1$ then $x(\alpha) = y(\alpha) = \alpha$. So for $t \in [0, \alpha]$ the curve $(x,y)$ travels from $(1,0)$ to $(\alpha, \alpha)$ in counter clockwise direction. The area of the sector bounded by this piece of curve is
$$
\frac{1}{2}\int_0^{\alpha} \begin{vmatrix} x & \dot{x} \\ y & \dot{y} \end{vmatrix} dt = \frac{1}{2}\int_0^{\alpha} \left(1+\frac{t^2}{2}\right) dt = \sqrt{3} - \frac{4}{3}.
$$
Since the curve is at least at a distance $1$ from the origin we get the inequality $ \frac{\pi}{8} < \sqrt{3} - \frac{4}{3}$.
Trick. Take a convex combination of the inequalities
$$
\begin{eqnarray}
\pi &<& 24 - 12 \sqrt{3} \\
\pi &<& -\frac{32}{3} + 8 \sqrt{3}
\end{eqnarray}
$$
with weights $\frac{2}{5}$ and $\frac{3}{5}$ to get $\pi < \frac{16}{5}$.
|
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|
Find the inverse of a matrix with a variable $$X=
\begin{pmatrix}
2-n & 1 & 1 & 1 & \ldots & 1 & 1 \\
1 & 2-n & 1 & 1 & \ldots & 1 & 1 \\
\vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\
1 & 1 & 1 & 1 & \ldots & 2-n & 1 \\
1 & 1 & 1 & 1 & \ldots & 1 & 2-n\end{pmatrix}_{n\times n}
$$
Which means that the matrix with the size of $n\times n$ have $n-2$ along the diagonal and $1$ everywhere else.
Help please, I am stuck with this problem.
|
Hint: Use the Sherman–Morrison formula on $$X=\text{diag}(1-n)_{n\times n}+\begin{pmatrix} 1\\ \vdots \\1\end{pmatrix}_{n\times 1}\begin{pmatrix}1 & \ldots &1 \end{pmatrix}_{1\times n}.$$
|
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|
Prove that $\sin(x + \alpha)$, $\sin(x + \beta)$ and $\sin(x + \gamma)$ are linearly dependent.
Functions $f$ and $g$ are independent on an interval $D$ if $af(x) + bg(x) = 0$ implies that $a = 0$ and $b = 0$ $\forall x \in D$
let $\alpha$, $\beta$, $\gamma$ be real constants. Prove that
$\sin(x + \alpha)$, $\sin(x + \beta)$ and $\sin(x + \gamma)$
are linearly dependent vectors in $C^0[0, 1]$. Be convincing in your reasoning (argument)
I was researching and found Wronskian. Using the Wronskian for three functions. The determinant of $f$, $g$ and $h$ is $W(f, g, h) = $
$$
\begin{vmatrix}
f & g & h \\
f' & g' & h' \\
f'' & g'' & h'' \\
\end{vmatrix}
$$
If $W(f, g, h) \neq 0$ then $f(x)$, $g(x)$ and $h(x)$ are linearly independent.
If $f(x)$, $g(x)$, and $h(x)$ are linearly dependent then $W(f, g, h) = 0$
My attempt
Let
$f(x) = \sin(x + \alpha)$, $g(x) = \sin(x + \beta)$ and $h(x) = \sin(x + \gamma)$
$W(f, g, h) =$
$$
\begin{vmatrix}
\sin(x + \alpha) & \sin(x + \beta) & \sin(x + \gamma) \\
\cos(x + \alpha) & \cos(x + \beta) & \cos(x + \gamma) \\
-\sin(x + \alpha) & -\sin(x + \beta) & -\sin(x + \gamma) \\
\end{vmatrix}
$$
$= \sin(x + \alpha)[-\sin(x + \gamma)\cos(x + \beta) + \cos(x + \gamma)\sin(x + \beta)] - sin(x + \beta)[-\sin(x + \gamma)\cos(x + \alpha) + \cos(x + \gamma)\sin(x + \alpha)] + \sin(x + \gamma)[-\sin(x+ \beta)\cos(x + \alpha) + \cos(x + \beta)\sin(x + \alpha)]$
$= -\sin(x + \alpha)[\sin((x + \gamma) +(x + \beta))] + \sin(x + \beta)[\sin((x + \gamma) + (x + \alpha))] - \sin(x + \gamma)[\sin((x + \beta) + (x + \alpha))] = 0$
By Wronskian, $f(x)$, $g(x)$ and $h(x)$ are linearly dependent since $W(f, g, h) = 0$
Not sure if this argument is sound?
|
Explain how you got that expression for $W(f,g,h)$ equal to $0$. Otherwise the argument is good.
Another idea: consider rewriting the functions with the sum-angle formulas, then interpret the three functions as three vectors in a two-dimensional vector space spanned by $\sin$ and $\cos$.
|
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|
Proving that a polynomial about the volume of a tetrahedron is irreducible We know that the volume of a tetrahedron $ABCD$ can be represented as
$$144V^2=(a^2b^2d^2+b^2c^2e^2+c^2a^2f^2+b^2a^2e^2+c^2b^2f^2+a^2c^2d^2+c^2e^2f^2+a^2f^2d^2+b^2d^2e^2+c^2d^2f^2+a^2e^2d^2+b^2f^2e^2)-(a^2b^2c^2+a^2e^2f^2+b^2f^2d^2+c^2d^2e^2)$$
where $a=BC, b=CA, c=AB, d=AD, e=BD, f=CD$.
Let $F(a,b,c,d,e,f)$ be the right hand side of the above equation. Then, here is my question.
Question : Is $F(a,b,c,d,e,f)$ irreducible as a polynomial?
The answer must be YES, but I don't know how to prove that. Can anyone help?
|
Notice there are no linear terms in $a$. So it is a polynomial of the form $Aa^2+B$. So, $B/A$ must be a square.
$A=b^2(d^2+e^2+f^2e^2+c^2)+(e^2f^2+c^2f^2+c^2d^2+f^2d^2+c^2d^2f^2+e^2d^2)$
and
$B=b^2(f^2d^2+c^2e^2+c^2f^2+d^2e^2)+(c^2d^2e^2+c^2e^2f^2)$
if I didn't miss some term.
Check that $A$ and $B$ don't have a common factor, which I haven't done. And then it is clear they are not squares themselves, since they have no terms linear in $b$.
|
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|
Easy exponents question I have the GRE Friday... I got hung up on this easy exponents problem (I think it was these exponents, don't recall exactly)
$$\frac{6^{14}}{2^7 \times 3^5} = ? $$
The answer is $2^73^9$, but could anyone double check for me?
|
Essentially yes. You need to use the rules $(xy)^a = x^a y^a$ and also $x^ax^b = x^{a+b}$ (the latter of which sometimes looks like $\frac{x^a}{x^b} = x^a x^{-b} = x^{a-b}$ for division). Here you first factor $6$ and then simplify. Explicitly:
$$
\frac{6^{14}}{2^7\cdot 3^5} = \frac{(2\cdot3)^{14}}{2^7\cdot 3^5} = \frac{2^{14}\cdot3^{14}}{2^7\cdot 3^5} = 2^{14-7}\cdot3^{14-5} = 2^7\cdot 3^{9}.
$$
Good luck on the GRE!
|
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|
Inequality: $ab^2+bc^2+ca^2 \le 4$, when $a+b+c=3$. Let $a,b,c $ are non-negative real numbers, and $a+b+c=3$.
How to prove inequality
$$
ab^2+bc^2+ca^2\le 4.\tag{*}
$$
In other words, if $a,b,c$ are non-negative real numbers, then how to prove inequality
$$
27(ab^2+bc^2+ca^2)\le 4(a+b+c)^3.\tag {**}
$$
$\color{gray}{\mbox{(Without using "universal" Lagrange multipliers method).}}$
Thanks!
|
Let $\{a,b,c\}=\{x,y,z\}$, where $x\geq y\geq z$.
Hence, by Rearrangement and AM-GM we obtain:
$$ab^2+bc^2+ca^2=ab\cdot b+bc\cdot c+ca\cdot a\leq xy\cdot x+xz\cdot y+yz\cdot z=y(x^2+xz+z^2)\leq$$
$$\leq y(x+z)^2=4y\left(\frac{x+z}{2}\right)^2\leq4\left(\frac{y+\frac{x+z}{2}+\frac{x+z}{2}}{3}\right)^3=4.$$
Done!
|
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|
Probability of group I have seen a problem in Probability that I am stuck :
Imagine that we have $12$ students in class and among this students there are $3$ honor students. Say that a teacher wants to assign a group project and wants to balance the groups out by forming 3 groups of students with exactly one honor student.
*
*What is the probability of forming groups of 4 students with exactly one honor student in each group, if the students are selected randomly?
P.S. I have thought that the number of different ways to select groups of $4$ students are $$\binom{12}{4}\binom{8}{4}\binom{4}{4}$$ And I reasoned that the number of ways of selecting groups with exactly one honor student could be $$\binom{9}{3}\binom{6}{3}\binom{3}{3}$$ The answer indicates approximately $.60$ as the answer. What am I doing wrong? Thank you.
|
Imagine that we line up the students at random, and assign the first $4$ to one group, the next $4$ to another, and the last $4$ to another.
We can analyze as follows. We have $3$ copies of the letter $H$, and $9$ copies of the letter $M$. We make a $12$-letter word. What is the probability there will be an $H$ among the first $4$ letters, and an $H$ among the next $4$, and an $H$ among the last $4$?
There are $\binom{12}{3}$ equally likely ways to place the $3$ $H$'s.
The number of choices with one $H$ in the first $4$, another $H$ in the next $4$, another in the next $4$ is $4^3$.
Our required probability is therefore $\dfrac{4^3}{\binom{12}{3}}$.
Remark: Your number $\binom{12}{8}\binom{8}{4}\binom{4}{4}$ counts the number of ways to divide our $12$ people into three uniformed teams, the Blues, the Whites, and the Reds. That's fine.
Now we count the number of ways to divide into uniformed teams with an Honours student on each team. The Honours student of the Blues can be chosen in $\binom{3}{1}$ ways. The rest of the team can be chosen in $\binom{9}{3}$ ways. For every such choice, the Honours student in the Whites can be chosen in $\binom{2}{1}$ ways, and the rest of her team in $\binom{6}{3}$ ways. Now we are finished, although we can tack on a decorative $\binom{1}{1}\binom{3}{3}$ for a total of $\binom{3}{1}\binom{9}{3}\binom{2}{1}\binom{6}{3}\binom{1}{1}\binom{3}{3}$.
Now for the probability, divide.
|
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|
On the equation $(1-x)^2/x + (1-y)^2/y + (1-z)^2/z + 4 = 0$ The problem is to solve the equation,
$$\frac{(1-x)^2}{x} + \frac{(1-y)^2}{y} + \frac{(1-z)^2}{z} + 4 = 0\tag{1}$$
in the rationals. Treating this as an equation in $z$, easy solutions would involve $z = \pm 1$, $z = \pm x, \pm y$. More complicated ones would be to make the discriminant $D$ of $(1)$ a square,
$$D = -4x^2y^2 + (x+y-2xy+x^2y+xy^2)^2 = t^2\tag{2}$$
with one solution being,
$$ x = a/b$$
$$ y = -b/a\, (p_1/p_2)$$
$$p_1 = 7a^6 + 4a^5b - 14a^4b^2 + 12a^3b^3 - 25a^2b^4 + 8a b^5 - 8b^6$$
$$p_2 = 8a^6 - 8a^5b + 25a^4b^2 - 12a^3b^3 + 14a^2b^4 - 4a b^5 - 7b^6$$
and $a,b$ being the legs of the Pythagorean triple $a^2+b^2 = c^2$.
However, can someone find a polynomial parameterization of small degree to $(1)$?
Edit (a few days later):
Courtesy of Allan MacLeod, a simple parameterization to $(1)$ can be given by using $x=1/y$ and the discriminant $(2)$ greatly simplifies to just solving,
$$y^2+1 = w^2$$
Hence, his first answer below can also be expressed in terms of Pythagorean triples $a^2+b^2 = c^2$ as,
$$x=a/b,\;\; y = b/a,\;\; z = \frac{(a+c)(b-c)}{ab}$$
|
After some further thought and computation, if $f=(k^2-1)/2k$ then $u=2(k-1)^2$ gives
\begin{equation*}
v= \pm \frac{(k-1)^2(k^2-2k-1)(k^2-2k+3)}{2k^2}
\end{equation*}
This point gives the parametric form
\begin{equation*}
f=\frac{k^2-1}{2k} \hspace{2cm}
g=\frac{2k}{k^2-1} \hspace{2cm} h=\frac{k(1-k)}{k+1}
\end{equation*}
Doubling this point (with $f=(k^2-1)/2k$) gives
\begin{equation*}
g=\frac{(1-k)(k^2+2k-1)(3k^2+2k+1)}{2k(k+1)(k^2+1)(k^2-2k+3)}
\end{equation*}
\begin{equation*}
h=\frac{k(k+1)(k^2+2k-1)(k^2-2k+3)}{2(k-1)(k^2+1)(3k^2+2k+1)}
\end{equation*}
Further forms can probably be derived using the torsion points.
|
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|
Prove $\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)^{2}\geq (a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)$
Prove that $$\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)^{2}\geq (a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)$$ for $a,b,c>0$
Any hints/solutions?
|
Expanding the terms, we want to show that
$$ \frac{a^2}{b^2} + \frac{b^2}{c^2} + \frac{c^2}{a^2} + \frac{a}{c} + \frac{b}{a} + \frac{c}{b} \geq 1 + 1 + 1 + \frac{a}{b} + \frac{b}{c} + \frac{c}{a} $$
This is true because
$ \frac{a^2}{b^2} + \frac{b}{a} + \frac{b}{a} \geq 3$ and $ \frac{a^2}{b^2} + 1 \geq 2 \frac{a}{b}$ by direct AM-GM.
Add up the cyclic versions of the above inequalities.
|
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|
Evaluate the limit $\lim_{x\rightarrow 0} \frac{\sqrt{1-\sin(5x)}-\sqrt{1+\sin(5x)}}{x^2+x}$ Trying to find
$$\lim_{x\rightarrow 0} \dfrac{\sqrt{1-\sin(5x)}-\sqrt{1+\sin(5x)}}{x^2+x}=\lim_{x\rightarrow 0} \dfrac{(1-\sin(5x))-(1+\sin(5x))}{(x^2+x)(\sqrt{1-\sin(5x)}+\sqrt{1+\sin(5x)})}=\lim_{x\rightarrow 0} \dfrac{-2\sin(5x)}{(x^2+x)(\sqrt{1-\sin(5x)}+\sqrt{1+\sin(5x)})}$$
How to solve it?
|
Recall
$$\sin^2(a) + \cos^2(a) = 1$$
Hence,
$$1 \pm \sin(2a) = ( \cos(a) \pm \sin(a))^2$$
For small $x$, we have
$$\sqrt{1 \pm \sin(5x)} = ( \cos(5x/2) \pm \sin(5x/2))$$
Use this along with the fact that $\lim_{x \to 0} \dfrac{\sin(\alpha x)}{x} = \alpha$ to conclude the limit.
|
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|
How prove this $\sqrt[5]{1782+\sqrt[3]{35+15\sqrt{6}}+\cdots}$ is positive integer numbers. Prove that
$$\sqrt[5]{1782+405\sqrt[3]{35+15\sqrt{6}}+405\sqrt[3]{35-15\sqrt{6}}}-\sqrt[3]{35+15\sqrt{6}}-\sqrt[3]{35-15\sqrt{6}}\in N$$
This problem from this
My try: let
$$x=\sqrt[3]{35+15\sqrt{6}}+\sqrt[3]{35-15\sqrt{6}}$$
then
$$x^3=70+3\sqrt[3]{35^2-(15\sqrt{6})^2}\cdot x=70-3\sqrt[3]{125}x$$
then we have
$$(x^3-70)^3=-27\cdot 125 x$$
$$\Longrightarrow x^6-15x^4-140x^3+225x^2+1050x+4900=0$$
Then I can't ,and I think this is interesting problem, Thank you
|
This isn't a complete answer, but I think it might be enough for you to go on:
Note that $x=-3$ is a root of the polynomial. Therefore $x+3$ divides $x^5-405x-972$. Dividing yields $x^4-3x^3+9x^2-27x-324$. It turns out that $x=-3$ is a root of this quotient as well. Dividing again by $x+3$ yields $x^3-6x^2+27x-108$. The exact roots of the cubic can be found, and $\sqrt[3]{35\pm\sqrt{6}}$ appears in those roots.
|
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|
Evaluate $ \sum\limits_{n=1}^{\infty}\frac{n}{n^{4}+n^{2}+1}$ The question was: Evaluate, ${\textstyle {\displaystyle \sum_{n=1}^{\infty}\frac{n}{n^{4}+n^{2}+1}}}.$
And I go, since $\frac{n}{n^{4}+n^{2}+1}\sim\frac{1}{n^{3}}$ and we know that ${\displaystyle \sum_{n=}^{\infty}\frac{1}{n^{3}}}$ converges. so ${\displaystyle \sum_{n=1}^{\infty}\frac{n}{n^{4}+n^{2}+1}}$ is convergent as well.
But I find it hard to calculate the sum. can you give me some hints?
|
HINT:
As $n^4+n^2+1=(n^2+1)^2-n^2=(n^2+1-n)(n^2+1+n)$
and $(n^2+1+n)-(n^2+1-n)=2n$
$$\frac n{n^4+n^2+1}=\frac12\left(\frac{2n}{(n^2+1-n)(n^2+1+n)}\right)$$
$$=\frac12\left(\frac{(n^2+1+n)-(n^2+1-n)}{(n^2+1-n)(n^2+1+n)}\right)$$
$$=\frac12\left(\frac1{n^2-n+1}-\frac1{n^2+n+1}\right)$$
Also observe that $: (n+1)^2-(n+1)+1=n^2+n+1$ inviting cancellations
|
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|
Solution of differential equation $x^3dy -x\ln x dx = (1+x^2) \tan^{-1}x dx -x dy $.... Solution of differential equation
$x^3dy -x \ln x dx = (1+x^2) \tan^{-1}x dx -x dy $ ( options are )
(a) $y = \frac{\ln x}{\tan^{-1}x}+c$
(b) $y = \tan^{-1}x \ln x +c$
(c) $y =\frac{\tan^{-1}x}{\ln x }+c$
(d) none of these
Solution : The given equation can be written as :
$$\frac{dy}{dx} =\frac{(1+x^2)\tan^{-1}x +x \ln x }{x^3+x} \Rightarrow \int dy = \int ( \frac{\tan^{-1}x}{x} +\frac{\ln x}{1+x^2})dx $$
Let $$ \frac{\tan^{-1}x}{x} = I \Rightarrow I = \tan^{-1} \int ( \frac{1}{x})dx + \int ( d (\tan^{-1}x) \int \frac{1}{x}dx$$ .... Please suggest how to integrate this function as it is again coming the same integral. Thanks...
|
It follows from integration by parts that $$\int\frac{\tan^{-1}x}{x}dx =\tan^{-1}x\ln x-\int \frac{\ln x}{1+x^2}dx$$ and hence we have $$\int\frac{\tan^{-1}x}{x}dx+\int \frac{\ln x}{1+x^2}dx=\tan^{-1}x\ln x+c.$$ Therefore, $y=\tan^{-1}x\ln x+c$ is the solution.
|
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|
Find a power series solution centered at 0 (Differential equations Here's the problem: $$(x-1)y''+y'=0$$ This is the work that I've already done: $$y=\sum_{n=0}^{\infty}a_{n}x^n$$
$$y'=\sum_{n=0}^{\infty}(a_{n+1})(n+1)x^n$$
$$y''=\sum_{n=0}^{\infty}(a_{n+2})(n+2)(n+1)x^n$$
I then plug those into the original equation:
$$(x-1)\sum_{n=0}^{\infty}(a_{n+2})(n+2)(n+1)x^n+\sum_{n=0}^{\infty}(a_{n+1})(n+1)x^n=0$$
$$\sum_{n=0}^{\infty}(a_{n+2})(n+2)(n+1)x^{n+1}-\sum_{n=0}^{\infty}(a_{n+2})(n+2)(n+1)x^n+\sum_{n=0}^{\infty}(a_{n+1})(n+1)x^n=0$$
And this is where I'm stuck and don't know what to do from here. If someone could help me out that would be great.
|
Setting
$$
y(x)=\sum_{n=0}^\infty a_nx^n,
$$
we have
\begin{eqnarray}
(x-1)y''(x)+y'(x)&=&(x-1)\sum_{n=2}^\infty n(n-1)a_nx^{n-2}+\sum_{n=1}^\infty na_nx^{n-1}\\
&=&\sum_{n=2}^\infty n(n-1)a_nx^{n-1}-\sum_{n=2}^\infty n(n-1)a_nx^{n-2}+\sum_{n=1}^\infty na_nx^{n-1}\\
&=&\sum_{n=1}^\infty n^2a_nx^{n-1}-\sum_{n=2}^\infty n(n-1)a_nx^{n-2}\\
&=&\sum_{n=0}^\infty(n+1)^2a_{n+1}x^n-\sum_{n=0}^\infty(n+1)(n+2)a_{n+2}x^n\\
&=&\sum_{n=0}^\infty(n+1)[(n+2)a_{n+2}-(n+1)a_{n+1}]x^n.
\end{eqnarray}
Therefore
$$
(x-1)y''(x)+y'(x)=0 \iff a_{n+2}=\frac{n+1}{n+2}a_{n+1} \quad \forall n\ge 0,
$$
i.e.
$$
a_n=\frac{n-1}{n}a_{n-1} \quad \forall n\ge 2.
$$
Thus
$$
a_n=\frac{(n-1)\cdot(n-2)\ldots(2-1)}{n\cdot(n-1)\ldots(2-1)}a_1=\frac{1}{n}a_1 \quad \forall n \ge 2.
$$
The solution of the DE is then given by
$$
y(x)=a_0+a_1x+a_1\sum_{n=2}\frac{1}{n}x^n=a_0+a_1\sum_{n=1}^\infty\frac{1}{n}x^n\equiv a_0-a_1\ln(1-x),
$$
where $a_0,a_1$ are real constants.
|
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|
How prove this Polynomial $g(x)=\sum_{i=1}^{n}a^m_{i}x^i$have only real roots? Question 1:
let Polynomial $f(x)=\displaystyle\sum_{i=0}^{3}a_{i}x^i,$ have three real numbers roots,where $a_{i}>0,i=1,2,3$.
show that:
$$g(x)=\sum_{i=0}^{3}a^m_{i}x^i$$ have only real roots,where $m\in R,m\ge 1$
My try:
case (1):
suppose that $f(x)$ has a zero of multiplicity 3, then we assume
$$f(x)=(x+p)^3=x^3+3x^2p+3xp^2+p^3$$
then
$$g(x)=x^3+(3p)^mx^2+(3p^2)^mx+(p^3)^{m}=(x+p^m)[x^2+(3^mp^m-p^m)x+p^{2m}]$$
then
$$h(x)=x^2+p^m(3^m-1)x+p^{2m}\Longrightarrow \Delta =(p^m(3^m-1))^2-4p^{2m}>0$$
so this case $g(x)$ have only three real roots.
for case (2):
let $f(x)=(x+p)^2(x+q)$,
I can't prove it,
and the case (3):
$$f(x)=(x+p)(x+q)(x+r)$$
and this case I can't prove it too.
I hope someone can help solve this nice problem ;
Thank you very much!
|
As I explained in my comment, it is enough to do case (2).
Let $x,y \ge 0$. Then $(X+x)(X+x)(X+y) = X^3+(2x+y)X^2 + (x^2+2xy)X + x^2y$.
The discriminant of $X^3+bX^2+cX+d$ is, according to wikipedia, $\Delta(b,c,d) = b^2c^2-4c^3-4b^3d-27d^2+18bcd$.
The derivative of $m \mapsto \Delta(b^m,c^m,d^m)$ at $m=1$ is
$$\delta(b,c,d) \\
= (2b^2c^2-12b^3d+18bcd)\log b + (2b^2c^2-12b^3d+18bcd)\log c + (18bcd-4b^3d-54d^2)\log d
$$
Plugging $b = 2x+y,c = x^2+2xy,d = x^2y$ we compute and obtain
$$ \delta(x,y) =
4x^6(\frac yx -1)^3\left(-(2+ \frac yx)\log \frac {(2x+y)^2}{x^2+2xy} + \frac yx \log \frac {(2x+y)^3} {x^2y}\right)$$
$\delta(x,y)/4x^6$ is actually a function of $t = \frac yx$ and has the same sign as $\delta(x,y)$ so it is enough to study the sign of the function $$g(t) = (t-1)\left(-(2+t)\log \frac{(2+t)^2}{1+2t} + t \log \frac {(2+t)^3}t\right) \\
= (t-1)\left((t-4)\log(2+t) + (2+t)\log(1+2t) - t\log t\right)
$$
Letting $g(t) = (t-1)h(t)$, we compute $h'(t) = \log(9+(t- t^{-1})^2) + 2\frac{(t-1)^2}{2t^2+5t+2} \ge \log 9 > 0 $. Since $h(1) = -3\log3+3\log3-\log1 = 0$, we have $h(t)>0$ for $t>1$, $h(t)<0$ for $t<1$, and $h(1)=0$.
Going back to $g$ we learn that $g(t)>0$ except when $t=1$, which means that $\delta(x,y) > 0$ except when $x=y$, which is what we wanted.
|
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|
Finding a recursive definition and computing $B(10)$ For $n \geq 1$, let $B(n)$ be the number of ways to express $n$ as the sum of $1$s and $2$s, taking order into account. Thus $B(4) = 5$ because $4 = 1 + 1 + 1 + 1 = 1 + 1 + 2 = 1 + 2 + 1 = 2 + 1 + 1 = 2 + 2$.
(a) Compute $B(i)$ for $1 \leq i \leq 5$ by showing all the different ways to write these numbers as above.
(b) Find a recursive definition for $B(n)$ and identify this sequence.
(c) Compute $B(10)$.
I've done part (a)
$B(1) = 1$ because $1 = (1)$
$B(2) = 2$ because $2 = (1+1) = (2)$
$B(3) = 3$ because $3 = (1+1+1) = (1+2) = (2+1)$
$B(4) = 5$ because $4 = (1 + 1 + 1 + 1) = (1 + 1 + 2) = (1 + 2 + 1) = (2 + 1 + 1) = (2 + 2)$
$B(5) = 8$ because $$5 = (1+1+1+1+1) = (1+1+1+2) = (1+1+2+1) = (1+2+1+1)\\ = (2+1+1+1) = (1+2+2) = (2+1+2) = (2+2+1)$$
*Note I used parenthesis for clarity
How would I do part B?
|
Hint: There are two questions for you to answer. Suppose that $n\ge 3.$ How many ways are there to obtain $n$ as such a sum where the first number is $1$? What if the first number is $2$? What can you then conclude?
|
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|
How to prove this limit exists: $a_{n+2}=\sqrt{a_{n+1}}+\sqrt{a_{n}}$ Question:
Consider a sequence $\{a_{n}\}$ such that $a_{1},a_{2}>0$, and for all $n \in \mathbb{N}$ we have:
$$a_{n+2}=\sqrt{a_{n+1}}+\sqrt{a_{n}}$$
Prove that :
$\displaystyle\lim_{n\to\infty}a_{n}$ exists and find this limit.
My work: If this limit exists, let $\displaystyle\lim_{n\to\infty}a_{n}=x>0$;
then we have
$$x=\sqrt{x}+\sqrt{x}\Longrightarrow x=4$$
But I can't see how to prove the limit of $\{a_{n}\}$ exists.
Thank you for you help!
|
and I have consider other solution
if $a_{0}>0,a_{1}>0,$,and $a_{n+2}=\sqrt{a_{n+1}}+\sqrt{a_{n}}$,then $a_{n}\to 4$
pf: if $0<a_{0}\le a_{1}\le 1$,then $$a_{2}=\sqrt{a_{1}}+\sqrt{a_{1}}\ge a_{1}$$
we note
$$a_{n+2}-a_{n+1}=(\sqrt{a_{n+1}}+\sqrt{a_{n}})-(\sqrt{a_{n}}+\sqrt{a_{n-1}})=(\sqrt{a_{n+1}}-\sqrt{a_{n}})+(\sqrt{a_{n}}-\sqrt{a_{n-1}})\ge 0$$
so $\{a_{n}\}$is Monotone increasing,
$$a_{n}\le 4,\mbox{since } a_{n+2}=\sqrt{a_{n+1}}+\sqrt{a_{n}}\le 2+2=4$$
(2):if $0<a_{1}\le a_{0}\le 1$,then
$$a_{2}=\sqrt{a_{1}}+\sqrt{a_{0}}\ge a_{0}$$
and
$$a_{n+2}-a_{n+1}=\sqrt{a_{n+1}}-\sqrt{a_{n-1}}\ge 0$$.the simalar we have $\{a_{n}\}$ is increasing and $a_{n}\le 4$.
(3):if$a_{0}>1$(or $a_{1}>1$),then we have
$$a_{2}=\sqrt{a_{1}}+\sqrt{a_{0}}>1,\Longrightarrow a_{n+2}=\sqrt{a_{n+1}}+\sqrt{a_{n}}>1$$
let $x_{n}=|a_{n}-4|$,then we have
$$x_{n+1}=|a_{n+1}-4|<|\sqrt{a_{n}}-2|+|\sqrt{a_{n-1}}-2|=\dfrac{|a_{n}-4|}{\sqrt{a_{n}}+2}+\dfrac{|a_{n-1}-4|}{\sqrt{a_{n-1}}+2}<\dfrac{1}{3}x_{n}+\dfrac{1}{3}x_{n-1}$$
so
$$x_{n+1}-\dfrac{1-\sqrt{13}}{6}x_{n}<\dfrac{1+\sqrt{13}}{6}(x_{n}-\dfrac{1-\sqrt{13}}{6}x_{n-1})<\cdots<\left(\dfrac{1+\sqrt{13}}{6}\right)^n\left(x_{1}-\dfrac{1-\sqrt{13}}{6}x_{0}\right)\to 0,n\to\infty$$
so
$$0<x_{n}<x_{n}+\dfrac{\sqrt{13}-1}{6}x_{n-1}=x_{n}-\dfrac{1-\sqrt{13}}{6}x_{n-1}\to 0$$
|
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|
Find a primitive of $x^2\sqrt{a^2 - x^2}$ I'm stuck on this problem for the last while now and any help would be appreciated. I need to find the indefinite integral of $$\int x^2\sqrt{a^2-x^2} dx$$
and show that it equals $$\frac x8(2x^2-a^2)\sqrt{a^2-x^2}+{a^4\over 8}\arcsin{x\over a}+c$$
I've tried using the substitution $x=a\sin t$ and $x=a\sec t$ Both have looked promising but I just can't seem to finish it out.
Thanks in advance for any help.
|
$$ \int x^2 \sqrt{a^2 - x^2} ~dx ~ = ~ a^4 \int \sin^2 u \cos^2 u ~du ~ = ~ a^4 \int \sin^2 u - \sin^4 u ~ du \quad (x = a \sin u)$$
Using the double angle formulae $\cos 2A=2\cos^2A-1=1-2\sin^2A$
$$\begin{align}
\sin^2u &=\dfrac{1-\cos{2u}}{2}\\
\\
\sin^4u &=\left(\sin^2u\right)^2\\
\\
&=\left(\dfrac{1-\cos{2u}}{2} \right)^2\\
\\
&=\dfrac{1}{4} \left( 1-2\cos 2u+\cos^2 2u \right)\\
\\
&=\dfrac{1}{4} \left( 1-2\cos 2u+\dfrac{1+\cos{4u}}{2} \right)\\
\\
&=\dfrac{1}{8} \left( 3-4\cos 2u+\cos{4u} \right)\\
\end{align}$$
Hence$$a^4 \int \left(\sin^2 u - \sin^4 u \right) ~ du =\frac{a^4}{2} \int \left(1-\cos{2u} \right) \ du \ -\frac{a^4}{8} \int \left( 3-4\cos 2u+\cos{4u} \right) \ du$$
|
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|
Solving $(z+1)^5 = z^5$ The question says to solve this equation: $(z+1)^5 = z^5$
I did. Just want to find out if I did it properly and if my run-around logic makes sense.
First I begin my writing the equations as:
$$ (z+1)^5 = z^5$$
$$ \mathbf{e}^{5 \mathbf{Log}(z+1)} = \mathbf{e}^{5 \mathbf{Log}(z)} $$
So $$ \mathbf{Log}(z+1) = \ln|z+1| + \mathbf{Arg}(z+1)i $$
$$ \mathbf{Log}(z) = \ln|z| + \mathbf{Arg}(z)i $$
Now, because the natural logarithm is one-to-one, I write:
$$ \ln |z| = \ln |z+1| \Rightarrow |z| = |z+1|$$
So assign $ z = a +bi$
So that $|z| = \sqrt{a^2 +b^2} = |z+1| = \sqrt{(a+1)^2 +b^2} \Rightarrow a^2 +b^2 = (a+1)^2 +b^2 \Rightarrow a^2 = (a+1)^2 \Rightarrow a = -\frac12 $
So, $z = -\frac12 + bi$ and $z+1 = \frac12 + bi$ for some $b \in \mathbb R$
Now to find $b$
$$ \mathbf{Arg}(z+1) = \mathbf{Arg}(z)$$
$$ \tan^{-1} \frac{b}{\frac12} = \pi - \tan^{-1}\frac{b}{-\frac12}$$
I have a feeling this last part isn't quite right, so I just want to find out if I'm approaching this question properly?
Ultimately, I get $ z = -\frac12$ which upon inspection...is wrong...
|
How about, let $ u = z-1/2 $. Then in terms of $ u $ you have:
$$
\left(u+\frac{1}{2}\right)^5 = \left(u - \frac{1}{2}\right)^5
$$
Upon expansion:
$$
5 u^4 + \frac{5}{2} u^2 + \frac{1}{16} = 0
$$
A quadratic equation in $ u^2 $. Can you get it from here?
|
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|
Proving for every odd number $x$, $x^2$ is always congruent to $1$ or $9$ modulo $24$ A problem I have been presented with asks the following:
Prove for every odd number $x$, $ x^2$ is always congruent to $1$ or $9$ modulo $24$.
This seems odd and non-intuitive to me. Of course, it must be true other wise they wouldn't be asking for me to prove it.
I know that:
$9$ modulo $24$ $=$ $9$
$1 = 1$
How could every odd number in existence squared be equal to either 1 or 9?
|
$$x^2-1=(x-1)(x+1)$$
is the product of two consecutive even numbers. Thus, one is multiple of $4$ and the other is even. This shows that $x^2-1$is a multiple of $8$.
Moreover, one of $x-1,x,x+1$ is a multiple of three.
Case 1: $x$ is not a multiple of $3$. Then $x^2-1$ must be a multiple of three. Thus
$x^2-1$ is a multiple of $24$.
Case 2: $x$ is a multiple of $3$. Then $x^2$ is a multiple of $9$. In this case we have
$$9|x^2 \, \mbox{and}\, 8 |x^2-1 \,.$$
Then
$$9| x^2-9 \, \mbox{and}\, 8 |x^2-9 \,.$$
thus
$$72|x^2-9 \,.$$
In this case $x^2-1$ is a multiple of $72$ hence of $24$.
|
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|
question about the limit $\lim_{h\to0}\frac{\arcsin(x+h)-\arcsin(x)}{h}$ Because $\sin'(x)=\cos(x)$ we can prove that $\arcsin'(x)=\frac{1}{\sqrt{1-x^2}}$. but, by definition we have
$$\arcsin'(x)=\lim_{h\to0}\frac{\arcsin(x+h)-\arcsin(x)}{h}\tag{1}$$
therefore,
$$\lim_{h\to0}\frac{\arcsin(x+h)-\arcsin(x)}{h}=\frac{1}{\sqrt{1-x^2}}\tag{2}$$
My question: Can we prove $(2)$ without using $(1)$? How?
|
$$\text{Using }\arcsin p-\arcsin q=\arcsin(p\sqrt{1-q^2}-q\sqrt{1-p^2}),$$
$$\arcsin(x+h)-\arcsin x=\arcsin((x+h)\sqrt{1-x^2}-x\sqrt{1-(x+h)^2})$$
$$\displaystyle\lim_{h\to0}\frac{\arcsin(x+h)-\arcsin x}h $$
$$=\lim_{h\to0}\frac{\arcsin((x+h)\sqrt{1-x^2}-x\sqrt{1-(x+h)^2})}h$$
$$=\lim_{h\to0}\frac{\arcsin((x+h)\sqrt{1-x^2}-x\sqrt{1-(x+h)^2})}{(x+h)\sqrt{1-x^2}-x\sqrt{1-(x+h)^2}}\cdot\lim_{h\to0}\frac{(x+h)\sqrt{1-x^2}-x\sqrt{1-(x+h)^2}}h$$
Putting $\displaystyle\arcsin((x+h)\sqrt{1-x^2}-x\sqrt{1-(x+h)^2})=v$
$\displaystyle\implies(x+h)\sqrt{1-x^2}-x\sqrt{1-(x+h)^2}=\sin v$ in the first limit,
$$\lim_{h\to0}\frac{\arcsin((x+h)\sqrt{1-x^2}-x\sqrt{1-(x+h)^2})}{(x+h)\sqrt{1-x^2}-x\sqrt{1-(x+h)^2}}=\lim_{v\to0}\frac v{\sin v}=1$$
$$\text{Again for the second limit, }\lim_{h\to0}\frac{(x+h)\sqrt{1-x^2}-x\sqrt{1-(x+h)^2}}h$$
$$=\lim_{h\to0}\frac{(x+h)^2(1-x^2)-x^2(1-(x+h)^2)}{h\{(x+h)\sqrt{1-x^2}+x\sqrt{1-(x+h)^2}\}}$$
$$=\lim_{h\to0}\frac{(x+h)^2-x^2}h\cdot\lim_{h\to0}\frac1{(x+h)\sqrt{1-x^2}+x\sqrt{1-(x+h)^2}}$$
$$=\lim_{h\to0}\frac{h(2x+h)}h\cdot\frac1{2x\sqrt{1-x^2}}$$
$$=\lim_{h\to0}(2x+h)\cdot\frac1{2x\sqrt{1-x^2}}\text{ as } h\ne0\text{ as } h\to0$$
$$=2x\cdot\frac1{2x\sqrt{1-x^2}}=\frac{1}{\sqrt{1-x^2}}$$
|
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|
Is this matrix positive-semidefinite in general? for the matrix written below I was wondering if one can show that it is positive-semidefinite for $n>3$ and $0< \alpha<1$. (Or not. For $n=2, 3$ it works by showing that all principal minors are non-negative.)
$$
C_{n,n} =
\begin{pmatrix}
1 & \alpha^1& \alpha^2 & \cdots & \alpha^{n-1} \\
\alpha^1 & 1 & \alpha^1&\cdots & \alpha^{n-2} \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
\alpha^{n-1} & \alpha^{n-2} & \alpha^{n-3}& \cdots & 1 \end{pmatrix}
=\begin{pmatrix}
\alpha^0 & \alpha^1& \alpha^2 & \cdots & \alpha^{n-1} \\
\alpha^1 & \alpha^0 & \alpha^1&\cdots & \alpha^{n-2} \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
\alpha^{n-1} & \alpha^{n-2} & \alpha^{n-3}& \cdots & \alpha^0
\end{pmatrix}
$$
|
Taking the first column, and substracting to it $\alpha$ times the column 2, we get $\det C_{n,n}=(1-\alpha^2)\det C_{n-1,n-1}$, hence we can conclude by Sylvester's criterion.
|
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|
Evaluate $\tan\left(2\sin^{-1}\frac{\sqrt{5}}{5}\right)$ without using a calculator Evaluate without using a calculator:
$\displaystyle{\tan\left(2\sin^{-1}\left(\sqrt{5} \over 5\right)\right).}$
So I built my triangle hyp=$5$, adj=$2\sqrt{5}$, opp=$\sqrt{5}$.
$$
\tan\left(2\theta\right) = 2\sin\left(\theta\right)\cos\left(\theta\right)\,,
\qquad\qquad
\tan\left(2\theta\right) = 2\,{\sqrt{5} \over 5}\,{2\sqrt{5} \over 5} = 4.
$$
I used my calculator and it says the answer is $4/3$. I'm not sure what I'm doing wrong
|
$\displaystyle \frac{\sqrt{5}}{5} = \frac{1}{\sqrt{5}}$. So, draw a triangle with
sides $1$, $2$, and $\sqrt{5}$ which should be a right triangle since $1^2 + 2^2 = \left(\sqrt{5}\right)^2$. The sine of one of the angles of this triangle, call
it $\theta$, is $\sin(\theta) = \frac{1}{\sqrt{5}}$, so its tangent $\tan(\theta)$ equals
$\frac{1}{2}$. You want to find $\displaystyle \tan(2\theta) = \frac{2\tan(\theta)}{1-\tan^2(\theta)}$.
|
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|
Find all solutions of the equality $y^2=x^3+23$ for integers $x,y$
Find all solutions of the equality $y^2=x^3+23$ for integers $x,y$
I guess, that x cannot be even. because we can apply mod 4 test
say $x=2k$ then $y^2=8k^3+23\equiv3\mod(4)$
but, this is not possible, since the square of an integer is congruent to $0$ or $1$ $\mod (4)$
so $x$ is of the form $4k+3$ or $4k+1$
$x=4k+3$ is also not possible for the same reason but for the last case the test fails.
any hints ?
|
There are no integer solutions.
You noted already that $x$ cannot be even, because the equation would then
fail mod $4$; so we need only consider $x$ odd. Then $x^3+23$ is even,
so must be a multiple of $4$, whence $x \equiv 1 \bmod 4$.
We cannot obtain a contradiction from further congruence considerations.
But since $23 = 3^3 - 2^2$ we try adding $4$ on both sides, getting
$$
y^2 + 4 = x^3 + 27 = (x+3) (x^2-3x+9),
$$
and observe that $x \equiv 1 \bmod 4$ implies
$x^2 - 3x + 9 \equiv 1 - 3 + 9 \equiv 3 \bmod 4$.
Since also $x^2-3x+9 > 0$ we deduce that $x^2-3x+9$ has a prime factor
$p \equiv 3 \bmod 4$. But this is impossible by quadratic reciprocity:
$y$ would be a square root of $-4 \bmod p$. QED
(In fact mwrank reports that there aren't even any rational solutions,
but that's not an elementary proof.)
[Added later: See Keith Conrad's
Examples
of Mordell's Equation
for further examples of elementary but nontrivial proofs that certain
equations of this form $y^2 = x^3 + k$ have no integer solutions,
and also some examples where a nonempty list of solutions of $y^2 = x^3 + k$
can be proved complete.]
|
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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.