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prove that $\frac{b+c}{a^2}+\frac{c+a}{b^2}+\frac{a+b}{c^2}\ge\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$ Assume: $a,b,c >0$ prove that :
$$\frac{b+c}{a^2}+\frac{c+a}{b^2}+\frac{a+b}{c^2}\ge\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$$
|
HINT: One may start proving the inequality:
$$\frac{b+c}{a^2}+\frac{c+a}{b^2}+\frac{a+b}{c^2}\ge\ \frac{9}{a+b+c}+ \frac{1}{a}+\frac{1}{b}+\frac{1}{c}$$
that is easy to prove. I met some time ago this inequality and have just remembered now. Of course, it's easy only if you met it before, otherwise it's rather hard to make such a guess.
|
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|
Prove that: $\frac1{20}\le \int_{1}^{\sqrt 2} \frac{\ln x}{\ln^2x+1} dx$ I'm interested in proving the following integral inequality:
$$\frac1{20}\le \int_{1}^{\sqrt 2} \frac{\ln x}{\ln^2x+1} dx$$
According to W|A the result of this integral isn't pretty nice, and involves the exponential integral.
|
$$\begin{array}{c l} \int_1^{\sqrt2} \frac{\log x}{1+(\log x)^2}dx & =\int_0^{\frac{1}{2}\log2}\frac{u}{1+u^2}e^udu \\ & \ge\int_0^{\frac{1}{2}\log2}\frac{u}{1+u^2}du \\ & = \left[\frac{1}{2}\log(1+u^2)\right]_0^{\frac{1}{2}\log 2} \\ & =\frac{1}{2}\log\left(1+\frac{(\log 2)^2}{4}\right) \\ & \approx 0.056714899 \\ & > 0.05=\frac{1}{20}. \end{array}$$
Edit. We can make it tighter with $e^u\ge1+u$:
$$\begin{array}{c l} \int_1^{\sqrt2} \frac{\log x}{1+(\log x)^2}dx & =\int_0^{\frac{1}{2}\log2}\frac{u}{1+u^2}e^udu \\
& \ge\int_0^{\frac{1}{2}\log2}\frac{u}{1+u^2}(1+u)du \\
& = \int_0^{\frac{1}{2}\log2}\left(1+\frac{u}{1+u^2}-\frac{1}{1+u^2}\right)du \\
& = \left[u+\frac{1}{2}\log(1+u^2)-\arctan u\right]_0^{\frac{1}{2}\log 2} \\
& =\frac{1}{2}\log\left(2+\frac{(\log 2)^2}{2}\right)-\arctan\left(\frac{1}{2}\log 2\right) \\
& \approx 0.06\color{Blue}{96694}. \end{array}$$
|
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|
Computing the derivative from the definition Using the limit definition of the derivative which I know is:
$$f'(x)=\lim_{h\to0}\left(\frac{f(x+h)-f(x)}{h}\right)$$
I am trying to solve this problem
$$f(x)= \frac{x}{x+2} $$
How do I go about properly solving this, I seemed to get
$$\frac{x}{x+2}\ $$
as my answer again? What are the steps I should follow?
I am trying to find the derivative of $f(x)= \frac{x}{x+2}$ using the definition of the derivative.
|
First. The derivative function of $f(x)$ is not $\frac{f(x+h)-f(x)}{h}$. (That would be a function of two variables, $x$ and $h$). Rather, the derivative function of $f(x)$ is
$$\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}.$$
The limit is important. Without the limit, it's just plain wrong.
Second. I don't know how you "got back" $f(x)$. The computations aren't that bad, just lots of algebra; perhaps you substituted incorrectly?
$$\begin{align*}
f(x) &= \frac{x}{x+2}\\
f(x+h) &= \frac{(x+h)}{(x+h)+2} = \frac{x+h}{x+h+2}\\
f(x+h) - f(x) &= \frac{x+h}{x+h+2} - \frac{x}{x+2}\\
&= \frac{(x+h)(x+2)}{(x+h+2)(x+2)} - \frac{x(x+h+2)}{(x+2)(x+h+2)}\qquad\text{(common denominator)}\\
&= \frac{(x+h)(x+2) - x(x+h+2)}{(x+2)(x+h+2)}\\
&= \frac{x(x+2) + h(x+2) - x(x+2) - xh}{(x+2)(x+h+2)}\\
&= \frac{h(x+2-x)}{(x+2)(x+h+2)}\\
&= \frac{2h}{(x+2)(x+h+2)}.
\end{align*}$$
Therefore,
$$\begin{align*}
f'(x) &= \lim_{h\to 0}\frac{f(x+h)-f(x)}{h}\\
&= \lim_{h\to 0}\frac{\quad\frac{2h}{(x+2)(x+h+2)}\quad}{h}\\
&= \lim_{h\to 0}\frac{2h}{h(x+2)(x+h+2)}\\
&= \lim_{h\to 0}\frac{2}{(x+2)(x+h+2)} \qquad\text{(since }h\neq 0\text{, so we can cancel)}\\
&= \frac{2}{(x+2)(x+0+2)} \qquad\text{(since the function is defined at }h=0\text{)}\\
&= \frac{2}{(x+2)^2}.
\end{align*}$$
|
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|
Integration Example How can i find the integration of this example
$$\int \frac{\sin x}{\sin x - \cos x } dx$$
I tried first add cos and then substracting cos but then what about $$\int \frac{\cos x}{\sin x - \cos x } dx\ ?$$
|
Let $f(a) = \int \frac{\sin(ax)}{\sin(x) - \cos(x)}dx$.
Differentiating throughout by a, we get
$$f'(a) = a\int \frac{\cos(ax)}{\sin(x) - \cos(x)}dx$$
Therefore, $$af(a) - f'(a) = a\int \frac{\sin(ax) - \cos(ax)}{\sin(x) - \cos(x)}dx$$
Substituting $a = 1$, we get
$$f(1) - f'(1) = x + C_1$$
Also,
$$af(a) + f'(a) = a\int \frac{\sin(ax) + \cos(ax)}{\sin(x) - \cos(x)}dx$$
Substituting $a = 1$ gives,
$$f(1) + f'(1) = \int \frac{\sin(x) + \cos(x)}{\sin(x) - \cos(x)}dx = \ln(\sin(x) - \cos(x)) + C_2$$
The integral was simply solved using the substitution $t = \sin(x) - \cos(x)$.
Solving the above two equations for $f(1)$ gives,
$$2f(1) = x + \ln(\sin(x) - \cos(x)) + C_1 + C_2$$
Therefore,
$$f(1) = \frac{1}{2}(x + \ln(\sin(x) - \cos(x)) + C_1 + C_2)$$
which is what we sought.
|
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|
To find the square root of a polynomial My question is:
Find the value of $k$ such that $$4x^6 - 24x^5 + 20x^4 + 68x^3 -44x^2 - 40x + k$$ is a perfect square.
hey all i have made an edit. Sorry for the inconvenience.
Any help to solve this question would be greatly appreciated.
|
If the goal is to find $k$ so that the polynomial is a perfect square, start by noting that it must be the square of a cubic polynomial:
$$4x^6 - 24x^5 + 20x^4 + 68x^3 -44x^2 - 40x + k=(ax^3+bx^2+cx+d)^2\;.$$
Clearly this immediately require that $a=2$. Now the square of $(2x^3+bx^2+cx+d)^2$ is
$$4x^6+4bx^5+(4c+b^2)x^4+(4d+2bc)x^3+(2bd+c^2)x^2+2cdx+d^2\;,$$
so you have the following system of equations:
$$\left\{\begin{align*}
&4b=-24\\
&4c+b^2=20\\
&4d+2bc=68\\
&2bd+c^2=-44\\
&2cd=-40\\
&d^2=k
\end{align*}\right.$$
Clearly $b=-6$; the second equation then allows you to find $c$, and you can then use the third, fourth, or fifth to find $d$ and then $k$. (To play safe, you should verify that the third, fourth, and fifth equations all yield the same value of $d$.)
|
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|
Will the following expression be irrational, rational or integer? Will the following expression be irrational, rational or integer?
$$\sqrt[3]{\sqrt a +b} - \sqrt[3]{\sqrt a -b}$$
where $a$ = $52$ and $b$ = $5$ .
By intuition, I think this will be an integer.
|
Let's use the identity $(\alpha + \beta)^3 = \alpha^3+\beta^3+3\alpha\beta(\alpha+\beta)$
Set $\alpha =\sqrt[3]{b+\sqrt{a}} \text{ and } \beta= \sqrt[3]{b - \sqrt{a}} \text { and } \alpha+\beta=x$
We know that $\alpha\beta = \sqrt[3]{b+\sqrt{a}} \times \sqrt[3]{b - \sqrt{a}} = \sqrt[3]{b^2-a} = \sqrt[3]{25-52} = -3$
So $x^3= b+\sqrt a + b-\sqrt a - 9x = 2b-9x = 10-9x$
You know that you are looking for a real answer, because $a$ is positive and the two cube roots are therefore cube roots of real numbers. Arturo has factorised the cubic, but it is easy to see that 1 is a root (integer roots must be divisors of 10).
|
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|
Compute $\sum_0^{n-1}2^i11^{n-i-1}\bmod10^9$ when $n=13^{17}$ Given the following function $f$
$f(1)=1$
$f(n)=11\cdot f(n-1)+2^{n-1}$
I would like to compute $f(13^{17})\mod 10^9$ and ended up using the following :
$f(n)=\sum_{i=0}^{n-1}({11^{n-(i+1)}\cdot 2^i})$
though I am able to quickly compute a single term using modular exponentiation, the loop over $13^{17}-1$ takes ages.
So do you have any suggestion to simplify $f(n)$ to avoid the loop ?
|
Clearly $f(n)$ will be of the form $a \cdot 11^n+b \cdot 2^n$ where $a,b$ are integers.
Given $f(1)=1$, but $f(1)=11a+2b$.
Similarly $f(2)=11f(1)+2=13$, but $f(2)=121a+4b$.
Solving for $a,b$, we get $f(n)= \frac{11^n-2^n}9$.
$f(13^{17})=\frac{11^{13^{17}}-2^{13^{17}}}9$.
$11^{13^{17}} = (10+1)^{13^{17}}=1+{13^{17}} \cdot 10^{13^{17}} + \dots \equiv 1 \pmod {10^{13^{17}}} \equiv 1 \pmod{10^9}$
Let $2^{13^{17}} \equiv c \pmod{10^9} = k10^9+c$
Clearly,$\left(2^{13^{17}},\ 10^9\right)\bigg|c
\Rightarrow c$ is of the form $2^9d \Rightarrow 2^{13^{17}-9} \equiv d \pmod{2^9}$
As $2$ is a primitive root of $5$ and of $5^2$ ,s o it'll be of $5^n$, $n \geq 1$.
$\phi(5^9)=4 \cdot 5^8$.
So, taking discrete logarithm in base $2$, ${13^{17}-9} \equiv \operatorname{ind}_2d \pmod{4 \cdot 5^8}$
=>$\operatorname{ind}_2d\equiv{(13^{17}-9)}\pmod {4 \cdot5^8}$.
and so on.
|
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Given that $\tan^{-1}(x)+\tan^{-1}(y)+\tan^{-1}(xy)=11/12π$, prove that when $x=1, dy/dx=-1-\sqrt{3}/2$ Given that $x$ and $y$ satisfy the equation:
$$\arctan(x)+\arctan(y)+\arctan(xy)=11/12π$$
Prove that, when $x=1, dy/dx=-1-\sqrt{3}/2$.
I tried to differentiate both sides:
$$1/(1+x^2)+y/(1+y^2)+(y+x\,dy/dx)/(1+(xy)^2)=0$$
and I know that when $x=1, y=\sqrt{3}$ by putting $x=1$ into the given equation.
so I got $1/2+√3/4+(√3+dy/dx)/4=0$
$$\implies dy/dx=-2-2√3$$
Thanks for pointing out the mistake. but the answer is still wrong..
|
$x=1\implies \arctan 1+2\arctan y=11\pi/12=\arctan y=\pi/3\implies y=\sqrt 3$. Thus, taking derivative on both sides gives,
$$
\begin{align}
& \frac{1}{1+x^2}+\frac{1}{1+y^2}\frac{dy}{dx}+\frac{1}{1+x^2y^2}\left(y+x\frac{dy}{dx}\right)=0 \\[10pt]
& \implies \frac{1}{2}+\frac{1}{4}\left(\frac{dy}{dx}\right)+\frac{1}{4}\left(\sqrt 3+\frac{dy}{dx}\right)=0 \\[10pt]
& \implies 1/2\frac{dy}{dx}+(1/2+\sqrt 3/4)=0 \\[10pt]
& \implies \frac{dy}{dx}=2(-1/2-\sqrt 3/4)=-1-\sqrt 3/2.
\end{align}
$$
|
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|
Equation of right circular cylinder with radius of the base as 2 units. Obtain the equation of right circular cylinder with radius of the base as 2 units. Its axis passes through $(1, 2, 3)$ and direction cosines are given as $(2, -3, 6)$
I got $45x^2+40y^2+13z^2+12xy-36yz-24zx-42x-280y-126z+294 = 0$
|
Put
$$
\begin{gathered}
A = \left( {1,2,3} \right) \hfill \\
X = \left( {x,y,z} \right) \hfill \\
\mathbf{n} = \frac{1}
{7}\left( {2, - 3,6} \right) \hfill \\
\end{gathered}
$$
Then the required cylinder is the lieu of points such that:
$$
2 = \left| {\mathop {AX}\limits^ \to \times \mathbf{n}} \right|
$$
that is:
$$
14 = \left| {\left( {6\left( {y - 2} \right) + 3\left( {z - 3} \right),\; - 6\left( {x - 1} \right) + 2\left( {z - 3} \right),\; - 3\left( {x - 1} \right) - 2\left( {y - 2} \right)} \right)} \right|
$$
i.e.
$$
\left( {6\left( {y - 2} \right) + 3\left( {z - 3} \right)} \right)^2 + \left( { - 6\left( {x - 1} \right) + 2\left( {z - 3} \right)} \right)^2 + \left( { - 3\left( {x - 1} \right) - 2\left( {y - 2} \right)} \right)^2 - 14^2 = 0
$$
which confirm your equation, except for the sign of
$36yz$ which shall be positive, not negative.
|
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Can $a^2+b^2+2ac$ be a perfect square if $c\neq \pm b$?
Can $a^2+b^2+2ac$ be a perfect square if $c\neq \pm b$?
$a,b,c \in \mathbb{Z}$.
I have tried some manipulations but still came up with nothing. Please help.
Actual context of the question is:
Let say I have an quadratic equation $x^2+2xf(y)+25$ that I have to make a perfect square somehow. So can I conclude that $f(y)=\pm5$
$($i.e $x^2+2xf(y)+25$ is perfect square only if $f(y)=\pm5)$, or are there other possibilities for $f(y)$?
Note:$x$ and $y$ are not related in any other way.
|
A small manipulation changes the problem into a more familiar one. We are interested in the Diophantine equation $a^2+b^2+2ac=y^2$. Complete the square. So our equation is equivalent to $(a+c)^2+b^2-c^2=y^2$. Write $x$ for $a+c$. Our equation becomes
$$x^2+b^2=y^2+c^2.\tag{$1$}$$
In order to get rid of trivial solutions, let us assume that we are looking for solutions of the original equation in positive integers. Then $x=a+c\gt c$. The condition $b\ne c$ means that we are in essence trying to express integers as a sum of two squares in two different ways.
The smallest positive integer that is a sum of two distinct positive squares in two different ways is $65$, which is $8^2+1^2$ and also $7^2+4^2$. So we can take $x=a+c=8$, $b=1$, and $c=7$, giving the solution $a=1$, $b=1$, $c=7$. Or else we can take $c=4$, giving the solution $a=3$, $b=1$, $c=4$. Or else we can take $x=a+c=7$.
The next integer which is the sum of two distinct positive squares in two different ways is $85$. We can use the decompositions $85=9^2+2^2=7^2+6^2$ to produce solutions of our original equation.
General Theory: Suppose that we can express $m$ and $n$ as a sum of two squares, say $m=s^2+t^2$ and $n=u^2+v^2$. Then
$$mn=(su\pm tv)^2+(sv\mp tu)^2.\tag{$2$}$$
Identity $(2)$ is a very important one, sometimes called the Brahmagupta Identity. It is connected, among other things, with the multiplication of complex numbers, and the sum identities for sine and cosine.
Identity $(2)$ can be used to produce infinitely many non-trivial solutions of Equation $(1)$, and therefore infinitely many solutions of our original equation. For example, any prime of the form $4k+1$ can be represented as a sum of two squares. By starting from two distinct primes $m$ and $n$ of this form, we can use Identity $(2)$ to get two essentially different representations of $mn$ as a sum of two squares, and hence solutions of our original equation.
|
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Inequality $\frac{a^2+b^2+c^2}{a^5+b^5+c^5}+\cdots+\frac{d^2+a^2+b^2}{d ^5+a^5+b^5}\le\frac{a+b+c+d}{abcd}$ Let:$a,b,c,d>0$ be real numbers ,how to prove that :
$$\frac{a^2+b^2+c^2}{a^5+b^5+c^5}+\frac{b^2+c^2+d^2}{b^5+c^5+d^5}+\frac{c^2+d^2+a^2}{c^5+d^5+a^5}+\frac{d^2+a^2+b^2}{d ^5+a^5+b^5}\le\frac{a+b+c+d}{abcd}$$.
Edit : I think I proved it. From Cauchy inequality we have
$$
\left(\sum\limits_{cyc}x^5\right)\left(\sum\limits_{cyc}x\right)\geq \left(\sum\limits_{cyc}x^3\right)^2
$$
From Chebyshev inequality it follows
$$
\left(\sum\limits_{cyc}x\right)\left(\sum\limits_{cyc}x^2\right)\leq 3\left(\sum\limits_{cyc}x^3\right)^2
$$
hence
$$
\frac{\left(\sum\limits_{cyc}x^5\right)}{\left(\sum\limits_{cyc}x^2\right)}=
\frac{\left(\sum\limits_{cyc}x^5\right)\left(\sum\limits_{cyc}x\right)}{\left(\sum\limits_{cyc}x\right)\left(\sum\limits_{cyc}x^2\right)}\leq
\frac{3}{\left(\sum\limits_{cyc}x^3\right)}\leq
\frac{1}{xyz}
$$
In the last step I used AM-GM inequality. The rest is clear.
Is there a different way to prove it ?
|
Also, by Muirhead(or Chebyshov) and AM-GM we obtain:
$$\sum_{cyc}\frac{a^2+b^2+c^2}{a^5+b^5+c^5}\leq\sum_{cyc}\frac{3}{a^3+b^3+c^3}\leq\sum_{cyc}\frac{1}{abc}=\frac{a+b+c+d}{abcd}.$$
|
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|
Analytic expression for the primitive of square root of a quadratic Can an analytic expression be given for
$$\int \sqrt{ax^2 + bx +c} \, dx$$
I think substitution doesn't work in this case (I need to compute the integral $\int_0^t \ldots$).
|
To deal with the integral, we first use method of completing square and then the well-know result
$$
\int \sqrt{x^2-a^2} d x=\frac{1}{2}\left[x \sqrt{x^2-a^2}-\ln \left|x+\sqrt{x^2-a^2}\right| \right ]+C
$$
Case 1: $a>0$ and $b^2>4ac$
$$
\begin{aligned}
I &=\int \sqrt{a x^2+b x+c} d x\\&=\int \sqrt{\left(\sqrt{a} x+\frac{b}{2 \sqrt{a}}\right)^2-\left(\frac{\sqrt{b^2-4 a c}}{2 \sqrt{a}} \right)^2}d x \\
&= \frac{1}{2 \sqrt{a}}\left[\left(\sqrt{a} x+\frac{b}{2 \sqrt{a}}\right) \sqrt{a x^2+b x+c}-\frac{b^2-4 a c}{4 a}\ln \left(\sqrt{a x^2+b x+c}+\sqrt{a} x+\frac{b}{2 \sqrt{a}}\right)\right]+C\\&=\frac{1}{4 a}(2 a x+b) \sqrt{a x^2+b x+c}-\frac{b^2-4 a c}{8 a^{\frac{3}{2}}} \ln \left(2 \sqrt{a} \sqrt{a x^2+b x+c}+2 a x+b\right)+C
\end{aligned}
$$
Case 2: $a>0$ and $b^2<4ac$
$$I=\int \sqrt{a x^2+b x+c} d x=\int \sqrt{\left(\sqrt{a} x+\frac{b}{2 \sqrt{a}}\right)^2-\left(\frac{i\sqrt{4 a c-b^2}}{2 \sqrt{a}} \right)^2}d x $$
Replacing $\frac{\sqrt{b^2-4 a c}}{2 \sqrt{a}} $ by $\frac{i\sqrt{4 a c -b^2}}{2 \sqrt{a}} $ in $(1)$ yields
$$I=\frac{1}{4 a}(2 a x+b) \sqrt{a x^2+b x+c}-\frac{b^2-4 a c}{8 a^{\frac{3}{2}}} \ln \left(2 \sqrt{a} \sqrt{a x^2+b x+c}+2 a x+b\right)+C$$
Conclusively, when $a>0$,
$$\boxed{I=\frac{1}{4 a}(2 a x+b) \sqrt{a x^2+b x+c}-\frac{b^2-4 a c}{8 a^{\frac{3}{2}}} \ln \left(2 \sqrt{a} \sqrt{a x^2+b x+c}+2 a x+b\right)+C}$$
|
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The minimum value of $(\frac{1}{x}-1)(\frac{1}{y}-1)(\frac{1}{z}-1)$ if $x+y+z=1$ $x, y, z$ are three distinct positive reals such that $x+y+z=1$, then the minimum possible value of $(\frac{1}{x}-1) (\frac{1}{y}-1) (\frac{1}{z}-1)$ is ?
The options are: $1,4,8$ or $16$
Approach: $$\begin{align*}
\left(\frac{1}{x} -1\right)\left(\frac{1}{y}-1\right)\left(\frac{1}{z}-1\right)&=\frac{(1-x)(1-y)(1-z)}{xyz}\\
&=\frac{1-(x+y+z)+(xy+yz+zx)-xyz}{xyz}\\
&=\frac{1-1+(xy+yz+zx)-xyz}{xyz}\\
&=\frac{xy+yz+zx}{xyz} - 1
\end{align*}$$
Now by applying $AM≥HM$, I got the least value of $(xy+yz+zx)/xyz$ as $9$, so I got final answer as $8$. Is it correct?
|
If we put no constraint on $x$, $y$, and $z$ apart from $x$, $y$, $z$ positive and $x+y+z=1$, then indeed your calculation, and the one by Patrick Da Silva, show that the minimum value is $8$, attained at $x=y=z=\frac{1}{3}$.
However, the problem specifies that $x$, $y$ and $z$ are distinct real numbers. If we take that constraint into account, there is no minimum. We can get arbitrarily close to $8$ (but above $8$) by choosing $x$, $y$, and $z$ distinct and close to $\frac{1}{3}$, but we cannot attain $8$ with distinct $x$, $y$ and $z$.
|
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|
Factorise the determinant $\det\Bigl(\begin{smallmatrix} a^3+a^2 & a & 1 \\ b^3+b^2 & b & 1 \\ c^3+c^2 & c &1\end{smallmatrix}\Bigr)$ Factorise the determinant $\det\begin{pmatrix} a^3+a^2 & a & 1 \\ b^3+b^2 & b & 1 \\ c^3+c^2 & c &1\end{pmatrix}$.
My textbook only provides two simple examples.
Really have no idea how to do this type of questions..
|
Regard $b$ and $c$ as constants, and the determinant as a polynomial in $a$. Then find ways of making the determinant equal to 0, and by the Factor Theorem you'll get a factor of the determinant.
Obvious choices: set $a=b$ or $a=c$ and you'll have two identical rows, so $(a-b)$ and $(a-c)$ are factors.
We can clearly see that permuting the variables is the same as permuting the rows, and hence only changes the determinant by a sign. Hence permuting the variables in any factor gives another factor, so $(b-c)$ is a factor.
What's left? Well, we can see without much difficulty that the determinant is cubic in $a$, and the coefficient of $a^3$ is $(b-c)$. We've already taken out a factor of $(b-c)$ and two monic linear factors, so whatever's left is linear and monic in $a$ and, by similar arguments, must also be so in $b$ and $c$. Hence it pretty much has to be $a + b + c + k$ for some constant $k$. So we've got as far as the following:
$$\det\begin{pmatrix} a^3+a^2 & a & 1 \\ b^3+b^2 & b & 1 \\ c^3+c^2 & c &1\end{pmatrix}=(a-b)(a-c)(b-c)(a+b+c+k)$$
for some $k$, and all we need to do is find $k$. I first tried setting $b=c=0$ but that just leaves you with $0k = 0$, so instead let's set $b=1$, $c=0$:
$$\begin{align}
\det\begin{pmatrix} a^3+a^2 & a & 1 \\ 2 & 1 & 1 \\ 0 & 0 &1\end{pmatrix}&=a(a-1)(a+1+k) \\
\det\begin{pmatrix} a^3+a^2 & a \\ 2 & 1\end{pmatrix}&=a(a-1)(a+1+k) \\
a^3 + a^2 - 2a &=a(a-1)(a+1+k) \\
a^2 + a - 2 &= (a-1)(a+1+k)\end{align}$$
Comparing constant terms, $-2 = -1-k$ so $k=1$, and we're done!
|
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|
Inequality for cosines Is the following inequality in a triangle known?
$$4(\cos A + \cos B + \cos C) \le 3 + \cos \left(\frac{B-C}{2}\right) + \cos \left(\frac{C-A}{2}\right) + \cos \left(\frac{A-B}{2}\right)$$
It looks correct to me but I would appreciate if someone confirm it.
|
Let $a=y+z$, $b=x+z$ and $c=x+y$.
Hence, we need to prove that
$$2\sum_{cyc}\frac{a^2+b^2-c^2}{ab}\leq3+\sum_{cyc}\left(\sqrt{\frac{(a+b+c)^2(b+c-a)(a+c-b)}{16c^2ab}}+\sqrt{\frac{(a+b-c)^2(b+c-a)(a+c-b)}{16c^2ab}}\right)$$ or
$$2\sum_{cyc}c(a^2+b^2-c^2)\leq3abc+\sum_{cyc}\frac{a+b}{2}\sqrt{ab(a+c-b)(b+c-a)}$$ or
$$4\sum_{cyc}(x+y)(z^2+xz+yz-xy)\leq$$
$$\leq3(x+y)(x+z)(y+z)+\sum_{cyc}(x+y+2z)\sqrt{xy(x+z)(y+z)}$$ or
$$\sum_{cyc}(x+y+2z)\sqrt{xy(x+z)(y+z)}\geq\sum_{cyc}(x^2y+x^2z+6xyz),$$
which is C-S and AM-GM:
$$\sum_{cyc}(x+y+2z)\sqrt{xy(x+z)(y+z)}\geq\sum_{cyc}(x+y+2z)\sqrt{xy}(\sqrt{xy}+z)=$$
$$=\sum_{cyc}(x^2y+x^2z+2xyz)+\sum_{cyc}(x+y+2z)z\sqrt{xy}\geq$$
$$=\sum_{cyc}(x^2y+x^2z+2xyz)+\sum_{cyc}4xyz=$$
$$=\sum_{cyc}(x^2y+x^2z+6xyz).$$
Done!
|
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|
How to prove that $\frac{10^{\frac{2}{3}}-1}{\sqrt{-3}}$ is an algebraic integer As the title says, I'm trying to show that $\frac{10^{\frac{2}{3}}-1}{\sqrt{-3}}$ is an algebraic integer.
I suppose there's probably some heavy duty classification theorems that give one line proofs to this but I don't have any of that at my disposable so basically I'm trying to construct a polynomial over $\mathbb{Z}$ which has this complex number as a root.
My general strategy is to raise both sides of the equation
$$x = \frac{10^{\frac{2}{3}}-1}{\sqrt{-3}}$$
to the $n^{th}$ power and then break up the resulting sum in such a way as to resubstitute back in smaller powers of $x$. Also since this root is complex I know it must come in a conjugate pair for the coefficients of my polynomial to be real, thus I know that
$$x = -\frac{10^{\frac{2}{3}}-1}{\sqrt{-3}}$$
must also be a root of my polynomial. Hence from this I obtain:
$$3x^2 = (10^{\frac{2}{3}} -1 )^2$$
However since my root is pure imaginary I don't really get any more information from this, so I'm a bit stumped, I tried raising both sides of $3x^2 -1 = 10^{\frac{4}{3}} - (2)10^{\frac{2}{3}}$ to the third power but it doesn't look like it's going to break up correctly, can anyone help me with this? Thanks.
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If we start with $\sqrt{-3}x + 1 = 10^{2/3}$, we get
$$(\sqrt{-3}x+1)^3 = 100$$
hence
$$100 = -3\sqrt{-3}x^3 - 9x^2 + 3\sqrt{-3}x + 1.$$
Therefore,
$$3\sqrt{-3}x^3 + 9x^2 - 3\sqrt{-3}x + 99 = 0.$$
Dividing through by $3\sqrt{-3}$ we obtain
$$x^3 + \frac{3}{\sqrt{-3}}x^2 - x + \frac{33}{\sqrt{-3}}=0$$
and rationalizing we get
$$x^3 + \frac{3\sqrt{-3}}{-3}x^2 - x + \frac{33\sqrt{-3}}{-3} = 0$$
or
$$x^3 - \sqrt{-3}x^2 - x -11\sqrt{-3}=0.$$
This is a monic polynomial with coefficients in $\mathbb{Z}[\sqrt{-3}]$; hence $x$ is integral over $\mathbb{Z}[\sqrt{-3}]$, which in turn is integral over $\mathbb{Z}$, so $x$ is integral over $\mathbb{Z}$, as desired.
To go from this to the polynomial in Robert Israel's answer, we multiply by the conjugate:
$$\begin{align*}
&\Bigl((x^3-x) - \sqrt{-3}(x^2+11)\Bigr)\Bigl((x^3-x)+\sqrt{-3}(x^2+11)\Bigr)\\
&\qquad = (x^3-x)^2 + 3(x^2+11)^2\\
&\qquad = x^6 - 2x^4 + x^2 + 3x^4 + 66x^2 + 363\\
&\qquad = x^6 + x^4 + 67x^2 + 363,
\end{align*}$$
and we are done.
|
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|
Evaluating $\int ^\frac{\pi}{2}_{0} \sin\left(2x+\frac{\pi}{4}\right)\ dx$
Find the exact value of the following definite integral:
$$\int ^\frac{\pi}{2}_{0} \sin\left(2x+\frac{\pi}{4}\right)\:dx=\left[-\frac{1}{2}(2x+\frac{\pi}{4})\right]^\frac{\pi}{2}_{0}$$
$$=-\frac{1}{2}\left(2\frac{\pi}{2}+\frac{\pi}{4}\right)+\frac{1}{2}\left(2\cdot 0+\frac{\pi}{4}\right)$$
$$=-\frac{1}{2}\left(\pi+\frac{\pi}{4}\right)+\frac{1}{2}\left(\frac{\pi}{4}\right)=-\frac{\pi}{2}$$
but the right answer is:
$$\int_{0}^{\frac{\pi}{2}}{\sin{\left(2x+\frac{\pi}{4}\right)\:dx}}=\frac{\sqrt{2}}{2}$$
Help me out! thanks!
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You have forget $\cos$ after you have done antiderivative.
Solve-
$$\int^\frac{\pi}{2}_{0}\sin\left(2x+\frac{\pi}{4}\right)\:dx=\left[-\frac{1}{2}\cos\left(2x+\frac{\pi}{4}\right)\right]^\frac{\pi}{2}_{0}$$
$$=-\frac{1}{2}\cos\left(2\frac{\pi}{2}+\frac{\pi}{4}\right)+\frac{1}{2}\cos\left(2\cdot 0+\frac{\pi}{4}\right)$$
$$=-\frac{1}{2}\cos\left(\pi+\frac{\pi}{4}\right)+\frac{1}{2}\cos\left(\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2} \quad \blacksquare$$
|
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|
Poisson summation formula (in general) Define Poisson kernel as $$ P_r ( \theta) := \frac{1}{2\pi} \frac{1-r^2}{1- 2r \cos \theta + r^2} $$ Then I want to prove the Poisson summation formula which is $$ P_r (2\pi x) = \sum_{n=-\infty}^\infty P_y (x+n)\;\;\;\;\text{(here $r = e^{-2 \pi y} $}) $$
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We will use the result that
$$\sum_{n=-\infty}^{\infty} \frac{y}{(x+n)^2+y^2}= \frac{1}{2} \frac{1 - e^{-4 \pi y }}{1 - 2 e^{-2 \pi y} \cos ( 2 \pi x ) + e^{-4 \pi y}} = P_y(2\pi\,x)\,, $$
Recalling Poisson formula in the upper half plane for $y>0$,
$$ P_{y}(x) = \frac{y}{x^2+y^2}\,. $$
We construct the sum
$$ \sum_{n=-\infty}^{\infty} P_y(x+n) = \sum_{n=-\infty}^{\infty} \frac{y}{(x+n)^2+y^2} =\frac{1}{2} \frac{1 - e^{-4 \pi y }}{1 - 2 e^{-2 \pi y} \cos( 2 \pi x ) + e^{-4 \pi y}} = P_y(2\pi\,x) $$
Substituting $r= {\rm e}^{-2\pi y}$ in the above result gives the desired result.
$$ P_r(2\pi x) = \frac{1}{2} \frac{1 - r^2}{1 - 2 r \cos( 2 \pi x ) + r^2} $$
The whole idea was to exploit the Poisson integral formula in the upper half plane.
|
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|
Indefinite Integral of $\sqrt{\sin x}$ $$\int \sqrt{\sin x} ~dx.$$
Does there exist a simple antiderivative of $\sqrt{\sin x}$? How do I integrate it?
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Since $\sqrt{\sin(x)} = \sqrt{1 - 2 \sin^2\left(\frac{\pi}{4} -\frac{x}{2}\right)}$, this matches with the elliptic integral of the second kind:
$$\begin{align*}
\int \sqrt{\sin(x)} \, \mathrm{d} x &\stackrel{u = \frac{\pi}{4}-\frac{x}{2}}{=} -2 \int \sqrt{1-2 \sin^2(u)} \,\mathrm{d} u\\ &= -2 E\left(u\mid 2\right) + c = -2 E\left(\frac{\pi}{4}-\frac{x}{2}\middle|\, 2\right) + c
\end{align*}$$
where $c$ is an integration constant.
|
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hints on solving DE How to solve this DE?
$$ {dx \over x} = {dy \over y} = {dz \over z - a \sqrt{x^2+y^2+z^2}}$$
From the first part, I get $y = c_1x$. How to find the other solution?
The answer according to answer sheet is $ z + \sqrt{x^2 + y^2 + z^2} = c_2$. Thank you for help.
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Let
\begin{equation}
{\frac{dx}{x}} = {\frac{dy}{y}} = {\frac{dz}{z - a \sqrt{x^2+y^2+z^2}}} = K
\end{equation}
\begin{equation}
{\frac{2xdx}{2x^{2}}} = {\frac{2ydy}{y^{2}}} = {\frac{2zdz}{2z^{2} - 2az \sqrt{x^2+y^2+z^2}}} = K
\end{equation}
Then
\begin{equation}
\frac{dx^{2}}{2x^{2}} = \frac{dy^{2}}{2y^{2}} = \frac{dz^{2}}{2z^{2} - 2az \sqrt{x^2+y^2+z^2}} = K
\end{equation}
Adding all the three terms, we get
\begin{equation}
\frac{dx^{2} + dy^{2} + dz^{2}}{2x^{2} + 2y^{2} + 2z^{2} - 2az \sqrt{x^2+y^2+z^2}} = K
\end{equation}
\begin{equation}
\frac{dw^{2}}{2w^{2} - 2az \sqrt{w^{2}}} = {\frac{2zdz}{2z^{2} - 2az \sqrt{x^2+y^2+z^2}}}
\end{equation}
Hence,
\begin{equation}
\frac{dw^{2}}{2w^{2} - 2az \sqrt{w^{2}}} = {\frac{dz}{z - a \sqrt{w^2}}}
\end{equation}
And then,
\begin{equation}
\frac{2w dw}{2w^{2} - 2az w} = {\frac{dz}{z - a w}}
\end{equation}
And so,
\begin{equation}
\frac{dw}{dz}= {\frac{w - az}{z - a w}}
\end{equation}
|
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|
The probability of one Gaussian larger than another. For two Gaussian-distributed variables, $ Pr(X=x) = \frac{1}{\sqrt{2\pi}\sigma_0}e^{-\frac{(x-x_0)^2}{2\sigma_0^2}}$ and $ Pr(Y=y) = \frac{1}{\sqrt{2\pi}\sigma_1}e^{-\frac{(x-x_1)^2}{2\sigma_1^2}}$. What is probability of the case X > Y?
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I assume that $X$ and $Y$ are independent. Let $Z=X-Y$ then $Z\sim\cal{N}(x_0-y_0,\sigma_0^2+\sigma_1^2)$. Accordingly
$$P(Z>0)=\int_0^\infty\frac{1}{\sqrt{2\pi(\sigma_0^2+\sigma_1^2)}}\exp\left(\frac{-(z-x_0+y_0)^2}{2(\sigma_0^2+\sigma_1^2)}\right)\mathrm{d}z$$
if we use the complementary error function
$$\operatorname{erf}c(x)=\frac{2}{\sqrt\pi}\int_x^\infty e^{-t^2}dt$$
with $t=\frac{z-x_0+y_0}{\sqrt{2(\sigma_0^2+\sigma_1^2)}}$, we have $\sqrt{2(\sigma_0^2+\sigma_1^2)}dt=dz$
$$P(Z>0)=\frac{2}{2\sqrt{\pi}\sqrt{2(\sigma_0^2+\sigma_1^2)}}\int_{t=\frac{y_0-x_0}{\sqrt{2(\sigma_0^2+\sigma_1^2)}}}^\infty e^{-t^2}\sqrt{2(\sigma_0^2+\sigma_1^2)}dt$$
and we get finally
$$P(Z>0)=\frac{1}{2}\operatorname{erfc}\left(\frac{y_0-x_0}{\sqrt{2(\sigma_0^2+\sigma_1^2)}}\right)$$
|
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|
Evaluating $\int_1^3\frac{\ln(x+2)}{x^2+2x+15} \ dx$ Could you please give me a hint on how to compute:
$$
\int_1^3\frac{\ln(x+2)}{x^2+2x+15}dx
$$
Thank you for your help
|
It is not really a simple integral (even if nothing special happens in the range $(1,3)$). Sasha gave a fine approximation (+1) let's provide the dilogarithm answer...
Let's start by factoring the denominator $\ x^2+2x+15$ :
The reduced discriminant is $\Delta=1-1\cdot 15=-14\ $ so that that it will have two complex conjugate solutions : $\ a=-1-i\sqrt{14}\ $ and $\ \overline{a}=-1+i\sqrt{14}$
Let's rewrite a little the integral :
$$I:=\int_1^3\frac{\ln(x+2)}{x^2+2x+15}dx=\int_1^3\frac{\ln(x+2)}{(x-a)(x-\overline{a})}dx$$
$$I=\frac 1{a-\overline{a}}\int_1^3 \left(\frac {\ln(x+2)}{x-a}-\frac {\ln(x+2)}{x-\overline{a}}\right)dx=\frac {I1-I2}{a-\overline{a}}$$
The (promised!) dilogarithm function looks like :
$$\operatorname{Li}_2(z)=-\int_0^z \frac {\ln(1-t)}t dt$$
Let's rewrite the first part of our integral the same way :
$$I1=\int_1^3 \frac {\ln(x+2)}{x-a} dx=\int_1^3 \frac {\ln(x-a+a+2)}{x-a} dx$$
$$I1=\int_1^3 \frac {\ln((a+2)(\frac{x-a}{a+2}+1)}{x-a} dx=\int_1^3 \frac {\ln(a+2)+\ln(1-\frac{a-x}{a+2})}{x-a} dx$$
$$I1=\left[\ln(x-a)\ln(a+2)\right]_1^3 -\int_1^3 \frac {\ln(1-\frac{a-x}{a+2})}{a-x} dx$$
set $\ \displaystyle t:=\frac{a-x}{a+2}$ (so that $\displaystyle \frac {dt}t=-\frac{dx}{a-x}$) to get :
$$I1=\left[\ln(x-a)\ln(a+2)\right]_1^3 -\int_{\frac{a-1}{a+2}}^{\frac{a-3}{a+2}} \frac {\ln(1-t)}{t} (-dt)=\left[\ln(x-a)\ln(a+2)-\operatorname{Li}_2\left(\frac{a-x}{a+2}\right)\right]_1^3$$
Of course the second part of the integral will be :
$$I2=\int_1^3 \frac {\ln(x+2)}{x-\overline{a}} dx=\left[\ln(x-\overline{a})\ln(\overline{a}+2)-\operatorname{Li}_2\left(\frac{\overline{a}-x}{\overline{a}+2}\right)\right]_1^3$$
So that your integral should be (with $\ a=-1-i\sqrt{14}\ $ and $\ \overline{a}=-1+i\sqrt{14}$) :
$$I=\frac {\left[\ln(x-a)\ln(a+2)-\ln(x-\overline{a})\ln(\overline{a}+2)+\operatorname{Li}_2\left(\frac{\overline{a}-x}{\overline{a}+2}\right)-\operatorname{Li}_2\left(\frac{a-x}{a+2}\right)\right]_1^3}{a-\overline{a}}\approx 0.11865036886767$$
EDIT (the last part was corrected $a$ had been replaced by $x$ in $\ln(a+2)$)
|
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|
Evaluate $\lim\limits_{n\to \infty}\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{6n}$ Show that $$\lim_{n\to \infty}\left(\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{6n}\right)=\log 6$$ Here I need to use the definition of integral but I faced problem in range . Please help.
|
Maybe it is intended that you mention Riemann sums explicitly.
Rewrite our sum as
$$\frac{1}{n}\left(\frac{1}{1+\frac{1}{n}} + \frac{1}{1+\frac{2}{n}} + \frac{1}{1+\frac{3}{n}}+\cdots +\frac{1}{1+\frac{5n}{n}} \right).$$
We recognize this as a (right) Riemann sum for the integral
$$\int_0^5 \frac{dx}{1+x},$$
which has value $\log 6$.
For imagine dividing the interval from $x=0$ to $x=5$ into equal subintervals of width $\frac{1}{n}$. The right ends of these subintervals are at $\frac{1}{n}$, $\frac{2}{n}$, $\frac{3}{n}$, and so on up to $\frac{5n}{n}$. So if $f(x)=\frac{1}{1+x}$, then our summands are precisely the function values at these right endpoints. The limit of the Riemann sum as $n \to\infty$ is therefore $\log 6$.
|
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|
Finding positive integer solutions to $n = ax^2 +by^2 - cxy$ How can I find the positive integer solutions to $x$ and $y$, given that $n$, $a$, $b$ and $c$ are all positive integers, in an equation of the form:
$$n = ax^2 + by^2 - cxy.$$
Specifically, I want to find the positive integer solutions to the following equation, given $n$:
$$n=3 x^2+20 y^2-16 x y.$$
|
$ax^2+by^2-cxy=n$
Expressing as a quadratic equation of x, $ax^2-x(cy)+by^2-n=0$
As x is positive integer, $=>(cy)^2-4.a(by^2-n)=(c^2-4ab)y^2 +4an$ must be perfect square.
$c=16, a=3, b=20 =>16y^2+12n=d^2(say)$=>d is even=2e(say)
$=>4y^2+3n=e^2$=>e is odd $>e^2≡1(mod\ 8)$
$=>e^2-4y^2 \equiv 1 \pmod4$ => $3n \equiv 1 \pmod 4$=>$n\equiv -1 \pmod 4$ to admit solution.
So, n=4m-1 for some integer m.
Applying the the approach on y, $(c^2-4ab)x^2 +4bn$ must be perfect square,
or, $16x^2+80n=f^2$=>f must even=2g(say),
$=>4x^2+20n=g^2$=>g is even=2h(say),
$=>x^2+5n=h^2$
If h=5s for integer s, $x^2≡0(mod\ 5)$=>x=5t for some integer t.
If h=5s±1 for integer s, $x^2≡1(mod\ 5)$=>x=5t±1 for some integer t.
If h=5s±2 for integer s, $x^2≡4(mod\ 5)$=>x=5t±2 for some integer t.
n will be $\frac{h^2-x^2}{5}$
$=>x^2+5(4m-1)=h^2$
If x,h are both odd or both even, $5(4m-1)=h^2-x^2≡0(mod\ 4)$ which is impossible.
If x is even, $h^2=x^2+5(4m-1)≡-1(mod\ 4)$ which is impossible.
=> x must be odd and h must be even to admit solution.
y can also calculated from the given once x is known.
|
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|
Inequality. $a^2+b^2+c^2 \geq a+b+c$ Let $a,b,c$ be positive real numbers such that $abc=1$. Prove that
$a^2+b^2+c^2 \geq a+b+c$.
Thanks
|
Since $(2,0,0)\succ\left(\frac{4}{3},\frac{1}{3},\frac{1}{3}\right)$, by Murhead we obtain:
$$a^2+b^2+c^2\geq\sum_{cyc}a^{\frac{a}{3}}b^{\frac{1}{3}}c^{\frac{1}{3}}=a+b+c$$
|
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|
Evaluating the elliptic integral $\int_{-\pi}^\pi\frac{dx}{\sqrt{(t-2\cos x)^2-4}}$ I have the following integral,
$$I(t)=\int_{-\pi}^\pi\frac{dx}{\sqrt{(t-2\cos x)^2-4}},$$
where $t>4$ is a real parameter. I know from messing around numerically and playing with Mathematica that
$$I(t)=\frac{4}{t}K\left(\frac{16}{t^2}\right),$$
where $K$ is the complete elliptic integral of the first kind with parameter $m=k^2=16/t^2$. However, I seek proof of that fact. I have tried a handful of changes of variables which didn't get the job done, and I've searched tables of integrals without finding this integrand or similar. Any suggestions or hints would be appreciated.
|
Since $\cos(-x) = \cos(x)$ we can reduce the integration range to $(0,\pi)$, and then do the change of variable $u = \cos(x)$:
$$
\int\limits_{-\pi}^\pi \frac{\mathrm{d} x}{\sqrt{(t-2 \cos(x))^2-4}} = \int_0^\pi \frac{\mathrm{d} x}{\sqrt{\left(\frac{t}{2} - \cos(x)\right)^2 -1}} = \int_{-1}^1 \frac{2 \mathrm{d} u}{\sqrt{1-u^2} \sqrt{4 u^2 + 4 u t + t^2-4}}
$$
Now perform a change of variables:
$$
u = \frac{v-z}{1- v z}, \qquad \text{where} \quad v = \frac{t - \sqrt{t^2-16}}{4}
$$
that maps $-1<u<1$ into $-1<z<1$, that leads to
$$
\int_{-1}^1 \frac{ v \, \mathrm{d} z}{\sqrt{1-z^2} \sqrt{1-v^4 z^2}} = 2 v \cdot \mathrm{K}\left( v^4\right)
$$
This is not exactly what the OP asked for, but quite elegant nonetheless.
Here is a confirmation of the equivalence in Mathematica:
Added:
The equivalence of the above answer to the one conjectured by OP and established by @J.M. is through the quadratic transformation:
$$
\mathrm{K}(z) = \frac{2}{1+\sqrt{1-z}} \mathrm{K} \left( \left(\frac{1-\sqrt{1-z}}{1+\sqrt{1-z}} \right)^2 \right)
$$
where $z = \frac{16}{t^2}$. Indeed:
$$
\frac{2}{1+\sqrt{1-\frac{16}{t^2}}} = \frac{2t}{t + \sqrt{t^2-16}} = \frac{2t}{t + \sqrt{t^2-16}} \cdot \frac{t - \sqrt{t^2-16}}{t - \sqrt{t^2-16}} = \frac{t}{4} \cdot \frac{t-\sqrt{t^2-16}}{2}
$$
and similarly:
$$
\left(\frac{1-\sqrt{1-\frac{16}{t^2}}}{1+\sqrt{1-\frac{16}{t^2}}} \right)^2 = \left(\frac{t-\sqrt{t^2-16}}{t+\sqrt{t^2-16}} \cdot \color\green{ \frac{t-\sqrt{t^2-16}}{t-\sqrt{t^2-16}} } \right)^2 = \left(\frac{\left(t-\sqrt{t^2-16}\right)^2}{16} \right)^2
$$
Combining, we arrive at the equality:
$$
\frac{4}{t} \mathrm{K}\left(\frac{16}{t^2}\right) = \frac{t-\sqrt{t^2-16}}{2} \mathrm{K}\left(\left( \frac{t-\sqrt{t^2-16}}{4} \right)^4\right)
$$
|
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|
Finite summation
Possible Duplicate:
why is $\sum\limits_{k=1}^{n} k^m$ a polynomial with degree $m+1$ in $n$
What is the proof without induction for :
$(1)$ $\sum_{i=1}^n\ i^2= \frac{n(n+1)(2n+1)}{6}$
$(2)$ $\sum_{i=1}^n\ i^3=\frac{n^2(n+1)^2}{4}$
|
There are probably many ways. For example, if you already know that $\sum_{i=1}^n i = \frac{n(n+1)}{2}$, it follows $$3\sum_{i=1}^n i^2 + 3\sum_{i=1}^n i + n = \sum_{i=1}^n (3i^2 + 3i + 1) = \sum_{i=1}^n ((i+1)^3 - i^3) = (n+1)^3 - 1$$ since the last sum telescopes, so $$\sum_{i=1}^n i^2 = \frac{1}{3}((n+1)^3 - 1 - 3\sum_{i=1}^n i - n) = \frac{1}{3}(n^3 + 3n^2 + 2n - \frac{3}{2}n(n+1))$$ $$=\frac{1}{6}(2n^3 + 3n^2 + n) = \frac{n(n+1)(2n+1)}{6}.$$
The second problem can be done in the same way.
|
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|
How do I solve a Continued Fraction of solution to quadratic equation? I know that it is possible to make a CF (continued fraction) for every number that is a solution of a quadratic equation but I don't know how.
The number I'd like to write as a CF is:
$$\frac{1 - \sqrt 5}{2}$$
How do I tackle this kind of problem?
|
Suppose $x$ is a root of $p(z) = z^2 - b z - c$. Then, diving $p(z)$ over $z$ and solving that for $z$ gets us
$$
z = b+ \frac{c}{z}
$$
Iterating:
$$
z = b + \cfrac{c}{b + \cfrac{c}{z}} = \cfrac{b}{c + \cfrac{c}{b+ \ddots}}
$$
Since $\frac{1-\sqrt{5}}{2}$ is a root of $z^2 - z -1$ we have:
$$
\frac{1-\sqrt{5}}{2} = -\frac{1}{\frac{1+\sqrt{5}}{2}} = - \cfrac{1}{1 + \cfrac{1}{1+ \frac{1}{1+\ddots}}}
$$
Added: Consider a sequence, defined by $x_{n+1} = b + \frac{c}{x_n}$, with $x_0 = \frac{c}{y}$. Few initial terms of the sequence are $\frac{c}{y}$, $b + y$, $b + \cfrac{c}{b+y}$, $b + \cfrac{c}{b + \cfrac{c}{b+y}}$, etc. It is well known that terms of this sequence can also be obtained as a ratio of two solutions, $a_n$ and $b_n$ to the following recurrence equation:
$$
v_{n} = b v_{n-1} + c v_{n-2} \tag{1}
$$
with initial conditions $a_0=y$, $a_1 = c$ and $b_0 = 1-\frac{b}{c} y$, $b_1 = y$. Then $x_n = \cfrac{a_n}{b_n}$. The solution to $(1)$ has the form:
$$
v_{n} = v_0 \frac{z_1 z_2^n - z_2 z_1^n}{z_1-z_2} + v_1 \frac{z_1^n - z_2^n}{z_1-z_2}
$$
where $z_1$ and $z_2$ are the two roots of $z^2 - b z -c = 0$. Assume $z_2>z_1$. In the large $n$ limit,
$$
\lim_{n \to \infty} x_n = \lim_{n \to \infty} \frac{a_n}{b_n}= \lim_{n\to \infty} \frac{a_1 (z_1^n - z_2^n) + a_0 (z_1 z_2^n - z_1^n z_2)}{b_1 (z_1^n - z_2^n) + b_0 (z_1 z_2^n - z_1^n z_2)} = \frac{a_0 z_1 - a_1}{b_0 z_1 - b_1} = \frac{c (c- y z_1)}{c (y - z_1) + b y z_1}
$$
In the case at hand $x_0 = \infty$ and $x_1 = b$, corresponding to the limit of $y \to \infty$, in which case the value of the continued fraction becomes:
$$
\lim_{n \to \infty} x_n = \lim_{y \to 0} \frac{c (c- y z_1)}{c (y - z_1) + b y z_1} = - \frac{c}{z_1} = z_2
$$
|
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|
Given digits $2,2,3,3,4,4,4,4$ how many distinct $4$ digit numbers greater than $3000$ can be formed? Given digits $2,2,3,3,4,4,4,4$ how many distinct $4$ digit numbers greater than $3000$ can be formed?
one of the digits which can be formed is $4444$
$4$ digit numbers greater than $3000$, which consists of only $2's$ and $4's$ are $4224$, $4242$, $4244$, $4422$, $4424$, $4442$
is there a well defined technique to solve this question.
|
You want to make a $4$ digit number. Now lets do it with Permutation-Combination.We have 4 places to fill different numbers.
First place can have either $3$ or $4$. So we have two choices. Lets analyse it.
If first place is $3$ - so we have to choose 3 digits from $(2,2,3, 3,4,4,4,4)$. So any place can have either 2 or 3 or 4, but $(222,333)$ is not possible, because we have just two $3$'s and two $2$'s
So total = $3 \times 3\times 3 - 2 = 25$
If first place is $4$ - so now we have to choose 3 digits from $(2,2,3, 3,3,4,4,4)$. So any place can have either $2$ or $3$ or $4$, but $(222)$ is not possible, because we have just two $2$'s
So total = $3 \times 3\times 3 - 1 = 26$
So, total choices = $25+26 = 51$, which is your answer.
|
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|
Proving $n^4 + 4 n^2 + 11$ is $16k\,$ for odd $n$
if $n$ is an odd integer, prove that $n^4 + 4 n^2 + 11$ is of the form
$16 k$.
And I went something like:
$$\begin{align*}
n^4 +4 n^2 +11
&= n^4 + 4 n^2 + 16 -5 \\
&= ( n^4 +4 n^2 -5) + 16 \\
&= ( n^2 +5 ) ( n^2-1) +16
\end{align*}$$
So, now we have to prove that the product of $( n^2 +5 )$ and $( n^2-1)$ is a multiple of 16.
But, how can we do this?
If anybody has any idea of how I can improve my solution, please share it here.
Edit updated to include the necessary hypothesis that $n$ is odd.
|
*
*$n=2k$:
$$n^4+4n^2+11\\=(n^2-1)(n^2+5)+16\\=(4k^2-1)(4k^2+5)+16\\=16k'^4+16k''^2+11\\=16k+11$$
Which is not $16k$.
*
*$n=2k+1$:
$$n^4+4n^2+11\\=(n^2-1)(n^2+5)+16\\=(4k^2+4k)(4k^2+4k+6)+16\\=8\underbrace{k(k+1)}
_{2k}(2k^2+2k+3)+16$$
Which is $16k$.
|
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|
Prove that $\frac{2}{1+\tan A}+\frac{2}{1+\tan B}+\frac{2}{1+\tan C} \le 3(\sqrt{3}-1)$ let $ABC$ be an acute triangle with all angles greater than $45^o$
Prove that
$$\frac{2}{1+\tan A}+\frac{2}{1+\tan B}+\frac{2}{1+\tan C} \le 3(\sqrt{3}-1)$$
I let $\tan A=a$, $\tan B=b$, $\tan C=c$ with $a+b+c=abc$ then the inequality equivalent to $\frac{2}{1+a}+\frac{2}{1+b}+\frac{2}{1+c} \le 3(\sqrt{3}-1)$ but it's not work.
I don't known what to do. Algebra or trigonometric?
|
Putting $D=A-45^{\circ}$ etc, $D+E+F=(A+B+C-135^{\circ})=45^{\circ}$
$1+\tan A=1+\tan(D+45^{\circ})=\frac{2}{1-\tan D}$ appyling $\tan(A+B)=\frac{\tan A+ \tan B}{1-\tan A \tan B}$
So, $\frac{2}{1 + \tan A}=1-\tan D$
Now the problem reduces to minimize $\tan D+\tan E+ \tan F$ where $D+E+F=45^{\circ}$ and $D,E,F>0$
If we can show that the minimum value occurs when $D=E=F$, we are done.
(i) Observe that $\tan x $ is convex for $x=D,E$ or $F$, as $\frac{d\tan x}{dx}=sec^2x$ and $\frac{d^2\tan x}{dx^2}=2\sec^2x\tan x>0$ as $D+E+F=45^{\circ}$ and $D,E,F>0$.
Using the Jensen's Inequality, we get that $\sum_{i=1}^n \tan A_i \ge n \cdot \tan \left( \frac{A_1 + A_2 + \cdots+ A_n}{n} \right) $ if $\sum_{i=1}^n A_i$ is constant.
So, the value of $\tan D+\tan E+ \tan F$ is minimized when $D=E=F$.
(ii) Alternatively,
$\tan D+\tan E+ \tan F=\tan D+\tan E+ \tan (45^{\circ} - D - E)=g(D,E)$(say),
Applying partial derivative wrt to D, $\frac{\partial g}{\partial D}=\sec^2D+sec^2(45^{\circ} - D - E)\cdot (-1)$
$\frac{\partial g}{\partial D}=0=>\sec^2D=sec^2(45^{\circ} - D - E)=>D=±(45^{\circ} - D - E)$
If $D=-(45^{\circ} - D - E)$, then $E=45^{\circ}$ which is impossible as $D,E,F>0$,
so, $D=45^{\circ} - D - E=>E=45^{\circ} - 2D$
Putting this is in $D+E+F=45^{\circ}=>F=D$
Applying partial derivative wrt to E, $\frac{\partial g}{\partial E}=\sec^2E+sec^2(45^{\circ} - D - E)\cdot (-1)$
$\frac{\partial g}{\partial E}=0=>\sec^2E=sec^2(45^{\circ} - D - E)$
So, $\sec^2E=\sec^2D=>D=E$ as $D,E,F>0$,
So, $\frac{\partial g}{\partial E}=0=\frac{\partial g}{\partial D}=>D=E=F$
Using Second Derivative Test for $g(D,E)$, we can show that $min(g(D,E))=g(D,E)_{D=E=F}=g(15^{\circ}, 15^{\circ})$
So, the minimum value of $\tan D+\tan E+ \tan F$ is $3 (\tan15^{\circ})=3(2-\sqrt 3)$
|
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|
I know that, $S_{2n}+4S_{n}=n(2n+1)^2$. Is there a way to find $S_{2n}$ or $S_{n}$ by some mathematical process with just this one expression? $S_{2n}+4S_{n}=n(2n+1)^2$, where $S_{2n}$ is the Sum of the squares of the first $2n$ natural numbers, $S_{n}$ is the Sum of the squares of the first $n$ natural numbers.
when, $n=2$
$S_{2n}=S_{4}=1^2+2^2+3^2+4^2=30$
$S_{n}=S_{2}=1^2+2^2=5$
$S_{4}+4S_{2}=2(2*2+1)^2=50$
|
Let $S_n=an^3+bn^2+cn+d$ where $a,b,c,d$ are rational numbers.
So, $S_{2n}+4S_n=n^3 12a+n^2 8b + n 6c+5d$
$\implies n^3 12a+n^2 8b + n 6c+5d= n(2n+1)^2=4n^3+4n^2+n$
Comparing the coefficients of the different powers on $n$,
$12a=4,8b=4,6c=1,d=0$
So, $6S_n=2n^3+3n^2+n=n(n+1)(2n+1)\implies S_n=\frac{n(n+1)(2n+1)}{6}$
Also, $S_n -S_{n-1}=n^2\implies S_n=n^2+S_{n-1}=\sum_{1≤r≤n}r^2+S_0=\sum_{1≤r≤n}r^2$
Observe that, we don't need to know the nature or formula of $S_n$. Solution of any such difference equation of any positive integer degree can be attempted this way.
|
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|
Limit of equation as x tends to -1 I was given the following expression and had to find the limit as: $$ x \rightarrow 1, x \rightarrow - 1, x \rightarrow \infty $$
$$ \lim_{x \to -1} \frac{x^2 +3x +2}{x^2 -1} = \lim_{x \to -1} \frac{\frac{x^2}{x^2} + \frac{3x}{x^2} + \frac{2}{x^2}}{\frac{x^2}{x^2} - \frac{1}{x^2}} = \lim_{x \to -1} \frac{\frac{1}{1} + 0 + 0}{\frac{1}{1} - 0} = \lim_{x \to -1} \frac{1}{1} = 1
$$
So for $-1$, I got 1. However the text book says it's $1/2$. I tried pluging in -1 but I don't get $1/2$, no matter how I shift this.
|
$$
\require{cancel}
\begin{equation*}
\lim \frac{x^2 +3x +2}{x^2 -1}= \lim \frac{\cancel{(x+1)}(x+2)}{\cancel{(x+1)}(x-1)}= \lim \frac{x+2}{x-1}=
\begin{cases} -\frac 12 & \text{if $x \to -1$,}
\\
+\infty &\text{if $x \to 1$.}
\end{cases}
\end{equation*}$$
When $x \to \infty$ you should consider the terms with the biggest power of the main variable in both denominator and numerator, which is $x$ and $x$ here, so the answer would be $1$.
|
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|
$ \sum_{k=1}^{\infty} \ln{\left(1 + \frac{1}{4 k^2}\right)}$ Computing this sum Compute the limit:
$$ \sum_{k=1}^{\infty} \ln{\left(1 + \frac{1}{4 k^2}\right)}$$
My teacher says it can be solved by only using high school knowledge, but I don't see how. What did I try? Well, I thought of Riemann sums but I see no way to connect this sum to it. Thanks.
I'm only interested in a solution at high school level if possible !
UPDATE: Now, I'm my own teacher.
|
As an alternative to @Seiros solution, albeit far less elegant is to use $$\log\left(1+\frac{1}{4k^2}\right) = \sum_{n=0}^\infty \frac{(-1)^n}{n+1} \frac{1}{4^{n+1} k^{2n+2}}$$
Then
$$\begin{eqnarray}
\sum_{k=1}^\infty \log\left(1+\frac{1}{4k^2}\right) &=& \sum_{n=0}^\infty \frac{(-1)^n}{n+1} \frac{\zeta(2n+2)}{4^{n+1}} = \sum_{n=1}^\infty \frac{\pi^{2n}}{2n} \frac{B_{2n}}{(2n)!} \\ &=& \sum_{n=1}^\infty \frac{\pi^{n}}{n} \frac{B_{n}}{(n)!} + \frac{\pi}{2} \\&=& \frac{\pi}{2} + \int_0^1 \left(\sum_{n=1}^\infty \pi^{n} t^{n-1}\frac{B_{n}}{(n)!}\right) \mathrm{d} t \\ &=& \frac{\pi}{2} + \int_0^1 \left(\frac{\pi}{\mathrm{e}^{\pi t}-1} - \frac{1}{t} \right) \mathrm{d} t \\ &=& \frac{\pi}{2} + \int_0^\pi \left(\frac{1}{\mathrm{e}^{u}-1} - \frac{1}{u} \right) \mathrm{d} u \\ &=&
\frac{\pi}{2} + \left. \log\left(\frac{1-\mathrm{e}^{-u}}{u}\right) \right|_{0}^{\pi} = \frac{\pi}{2} + \log\left(\frac{1-\exp(-\pi)}{\pi}\right) = \log\left(\frac{2}{\pi} \cdot \sinh\frac{\pi}{2}\right)
\end{eqnarray} \tag{1}
$$
|
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|
Proving the inequality $a^2+b^2+c^2+ab+bc+ca\ge6$ Given that $a$, $b$, $c$ are non-negative real numbers such that $a+b+c=3$, how can we prove that:
$a^2+b^2+c^2+ab+bc+ca\ge6$
|
$$ a^2 + b^2 + c^2 + ab + bc + ac = (a+b+c)^2 - (ab + bc + ac) = 9 - (ab + bc + ac)$$
Now it remains to show that max value of $(ab + bc + ac)$ is $3$. For that, we know the AM-GM equality ( for $a,b, c >0$ ) that $3(a^2 + b^2 + c^2) \geq (a+ b + c)^2 \geq 3(ab +bc +ac)$. From the last two part we have $(a+ b + c)^2 \geq 3(ab +bc +ac) \implies 9 \geq 3 (ab +bc +ac) \implies 3 \geq ab +bc +ac$
Hence we have
$$ a^2 + b^2 + c^2 + ab + bc + ac = 9 - (ab + bc + ac) \geq 9 - 3 = 6$$
|
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|
What is the right approach to use for factoring a rational inequality Here is an example:
$\frac{x^2 - 3x + 2}{x + 1} -5 > 0$
My approach would be to factor, find the undefined areas and the zeros, and then pick some points in the intervals left to see what I find. I'm not really sure if that's the right way to go, and to find the zeros I would just throw an equals sign in instead of the inequality and find the roots. Is this the right approach, and are there any techniques to improve it?
|
This is not a general approach, but in this example $x=-1$ is 'special', so split the consideration into $I_- = (-\infty,-1)$ and $I_+ = (-1, +\infty)$.
On $I_+$, we have $x^2 - 3x + 2 - 5 (x + 1) = x^2-8x-3 > 0$. The factors are $4 \pm \sqrt{19}$, hence if $x \in I_+$, then $\frac{x^2 - 3x + 2}{x + 1} -5 > 0$ iff $x > 4 + \sqrt{19}$ or $x < 4 - \sqrt{19}$.
On $I_-$, we have $x^2 - 3x + 2 - 5 (x + 1) =x^2-8x-3 < 0$. From the previous reasoning, we have that if $x \in I_-$, then $\frac{x^2 - 3x + 2}{x + 1} -5 \leq 0$.
Hence $\frac{x^2 - 3x + 2}{x + 1} -5 > 0$ iff $ x \in (-1,4 - \sqrt{19}) \cup (4 + \sqrt{19}, +\infty)$.
|
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|
Inequality. $\frac{a}{(b+c)^4}+\frac{b}{(c+a)^4}+\frac{c}{(a+b)^4} \geq \frac{3}{2(a+b)(b+c)(c+a)}$
For $a,b,c >0$ prove that :
$$\frac{a}{(b+c)^4}+\frac{b}{(c+a)^4}+\frac{c}{(a+b)^4} \geq \frac{3}{2(a+b)(b+c)(c+a)}.$$
I don't know how should I start. It is difficult for me because the denominators has the power equals with 4, I think it must decrease the rank of the power.
danke :)
|
The inequality can be written
$$
X := \sum_{cyc} \frac {a(a + b)(a + c)}{(b + c)^3} \geq \frac 3 2
$$
Let's suppose $a \leq b \leq c$. We have
$$
\frac 1 {b + c} \leq \frac 1 {a + c} \leq \frac 1 {a + b} \\
\frac {a(a + b)(a + c)} {(b + c)^2} \leq
\frac {b(b + c)(b + a)} {(c + a)^2} \leq
\frac {c(c + a)(c + b)} {(a + b)^2}
$$
Applying Rearrangement Inequality we get
$$
X \geq \sum_{cyc} \frac {a(a + b)}{(b + c)^2}=:Y
$$
Reapplyng Rearrangement Inequality and Nesbitt's Inequality we arrive to
$$
Y \geq \sum_{cyc} \frac a {b + c} \geq \frac 3 2
$$
|
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|
How to evaluate this limit: $\lim\limits_{x\to 1} \frac{\sqrt{3+x}-2}{\sqrt[3]{7+x}-2}$? I have difficulties in evaluating $$\lim_{x\to 1} \frac{\sqrt{3+x}-2}{\sqrt[3]{7+x}-2}$$
Could you give me a hint how to start solving this? (I know the result is $3$)
Thanks a lot !
|
For convenience, change variables to $t = x - 1$, so you're looking at
$\dfrac{\sqrt{4+t}-2}{(8+t)^{1/3}-2}$.
$$ \sqrt{4+t} = 2 \sqrt{1+\frac{t}{2}} = 2 \left(1 + \frac{t}{4} + \ldots\right) $$
$$ (8+t)^{1/3} = 2 \left(1 + \frac{t}{8}\right)^{1/3} = 2 \left(1 + \frac{t}{24} + \ldots\right) $$
|
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|
Solve $(\sqrt{5+2\sqrt{6}})^{x}+(\sqrt{5-2\sqrt{6}})^{x}=10$.
Solve $(\sqrt{5+2\sqrt{6}})^{x}+(\sqrt{5-2\sqrt{6}})^{x}=10$
I square the both sides and get $(5+2\sqrt{6})^{x}+(5-2\sqrt{6})^{x}=98$. But I don't know how to carry on. Please help. Thank you.
|
Let $t=( \sqrt{5 + 2\sqrt{6}})^{x}\implies (\sqrt{5-2\sqrt{6}})^{x}=\frac{1}{t} $
Thus the equation becomes,
$ t+\frac{1}{t}= 10\implies t^2-10t+1=0$ which is a quadratic equation and have roots $t=5+2\sqrt 6,5-2\sqrt 6$
If $t=5+2\sqrt 6\implies ( \sqrt{5 + 2\sqrt{6}})^{x}=5+2\sqrt 6\implies (5+2\sqrt 6)^{1-x/2}=1\implies 1-x/2=0\implies x=2$
If $t=5-2\sqrt 6 \implies (\sqrt{5+2\sqrt{6}})^{x}=5-2\sqrt 6=\frac{1}{5+2\sqrt 6}\implies (5+2\sqrt 6)^{1+x/2}=1\implies x=-2$
Thus, $x=2,-2$ are the two solutions.
Verification:
Putting $x=2$ in the original equation gives L.H.S=$5+2\sqrt 6+5-2\sqrt 6=10=$R.H.S
Similarly, putting $x=-2$ also gives L.H.S=$5-2\sqrt 6+5+2\sqrt 6=10=$R.H.S
|
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"timestamp": "2023-03-29T00:00:00",
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|
Finding domains? I need to find what $(f + g)(x)$ would equal here..
$$f(x) = \sqrt{25 − x^2},\quad g(x) = \sqrt{x^2 − 4}$$
Am I supposed to just add $\sqrt{25 − x^2}$ and $\sqrt{x^2 − 4}$? And if so, how? And how would I then determine the domain?
|
If $f(x)=\sqrt{25-x^2}$ and $g(x)=\sqrt{x^2-4}$, then $(f+g)(x)=\sqrt{25-x^2}+\sqrt{x^2-4}$, so yes: to find the value of $f+g$ at any point $x$, just add the values of $f(x)$ and $g(x)$.
In the context of this question the domain of $f+g$ is the set of all real numbers $x$ for which $(f+g)(x)$ can be calculated as a real number. In other words, you need to be able to take both of those square roots. This means that you need to have $25-x^2\ge 0$ and $x^2-4\ge 0$: you have to find those values of $x$ that satisfy both inequalities.
There are several ways to do this, but I’d begin by rewriting them as $x^2\le 25$ and $x^2\ge 4$. $x^2\le 25$ if and only if $|x|\le 5$, which in turn is true if and only if $-5\le x\le 5$. Similarly, $x^2\ge 4$ if and only if $|x|\ge 2$, which is true if and only if $x\le-2$ or $x\ge 2$. Thus, you need to find all those real numbers $x$ such that
$$-5\le x\le 5\quad\mathbf{and}\qquad\Big(x\le -2\quad\mathbf{or}\quad x\ge 2\Big)\;.\tag{1}$$
At this point it’s just a matter of converting $(1)$ into a simpler, more readily understandable form.
|
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|
How to derive the equation for x in a quadratic equation?
Possible Duplicate:
Why can ALL quadratic equations be solved by the quadratic formula?
How to derive this:
$x = \frac{-b + {\sqrt{b^2 + 4ac}}}{2a}$
From this:
$ax^2 + bx + c = 0$
I know this may be a little elementary :)
|
from $ax^2 + bx + c = 0 \Leftrightarrow 4a^2x^2 + 4abx + b^2 = b^2 - 4ac$.
Hence $(2ax + b)^2 = b^2 - 4ac$ then $2ax + b = \pm \sqrt{b^2 - 4ac}$.
from this, we have $ x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
|
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|
Help? I cannot do the integration by parts correctly?? Can someone please show me how to integrate
$$\int_0^\infty\frac4{\pi b^2}x^2e^{-x^2/b^2}dx\;?$$
please show steps how to integrate this problem. This is what i have so far.
$$\frac4{\pi b^2}\int_0^\infty x^2 e^{-x^2/b^2}dx\;.$$
Then i take the property I^2 = $$\int_0^\infty\frac4{\pi b^2}x^2e^{-x^2/b^2}dx\;*\int_0^\infty\frac4{\pi b^2}y^2e^{-y^2/b^2}dx\;?$$
Then i substitue in x=rcos0 and y=rsin0 dxdy=rdrd0
Then I get: $$\int_0^{\pi/2}\int_0^\infty r^5\cos^2\theta\sin^2\theta e^{(-r^2)/b^2}drd\theta=\int_0^{\pi/2}\cos^2\theta\sin^2\theta\int_0^\infty r^5e^{(-r^2)/(b^2)}drd\theta\;.$$
Then I do $$\int_0^\infty r^5e^{(-r^2)/(b^2)}drd\theta\;.$$ using integration by parts
$u=r^4$, $dv=re^{-r^2}dr$, so that $du=4r^3dr$ and $v=-\frac12e^{-r^2/b^2}$. That will leave you with something of the form $r^4(-\frac12e^{-r^2/b^2})(from 0 to infinity)-\int_0^\infty r^3e^{-r^2/b^2}dr$
Then took the limit of r from 0 to $infty$ of $r^4(-\frac12e^{-r^2/b^2})$ I got infinity. So now my problem looks like $\infty-\int_0^\infty r^3e^{-r^2/b^2}dr$
Then I did integration by parts on $\int_0^\infty r^3e^{-r^2/b^2}dr$. I let $w=r^2$, $dz=re^{-r^2/b^2}dr$, so that $dw=2rdr$ and $z=-\frac12e^{-r^2/b^2}$. Then i have $\infty-(wz-\int_0^\infty r^2e^{-r^2/b^2}dr)$.
Then I do integration by parts one more time. But when i find w and z and take the limit i get inifinity again, so i get something that looks like $\infty-(infty-\int_0^\infty r^2e^{-r^2/b^2}dr)$ . Can someone please tell me what I am doing wrong?
|
$$\begin{align*}u=&x&u'=&1\\v'=&xe^{-x^2/b^2}&v=&-\frac{b^2}{2}e^{-x^2/b^2}\end{align*}\;\;\;\;\;\;\Longrightarrow$$
$$\Longrightarrow \int_0^\infty x^2e^{-x^2/b^2}dx=\left.-\frac{b^2x}{2}e^{-x^2/b^2}\right|_0^\infty+\frac{b^2}{2}\int^\infty_0e^{-x^2/b^2}dx=\frac{b^3}{2}\sqrt\frac{\pi}{2}$$
And thus your integral equals
$$\frac{4}{\pi b^2}\frac{b^3}{2}\sqrt\frac{\pi}{2}=\sqrt\frac{2}{\pi}\,b$$
|
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|
How do you find a polynomial with integer coefficients with this sum of radicals as a root? Let $$x=\sqrt{a + \sqrt{b}} + \sqrt{c + \sqrt{d}}$$
How do you find the polynomial with this value as a root? Where a, b, c, and d are integers.
|
For the sake of defending my honor, here's most of the solution I alluded to involving squaring. It is somewhat messy, and the solution alluded to by Patrick Da Silva is nicer.
First square:
$$x^2 = a + \sqrt{b} + 2 \sqrt{(a + \sqrt{b})(c + \sqrt{d})} + c + \sqrt{d}.$$
Second square:
$$(x^2 - a - \sqrt{b} - c - \sqrt{d})^2 = 4 (a + \sqrt{b})(c + \sqrt{d}).$$
Expand:
$$x^4 - 2x^2 (a + \sqrt{b} + c + \sqrt{d}) + (a^2 + 2a \sqrt{b} + b) + 2(ac + c \sqrt{b} + a \sqrt{d} + \sqrt{bd}) + c^2 + 2c \sqrt{d} + d = 4(ac + c \sqrt{b} + a \sqrt{d} + \sqrt{bd})$$
Rearrange:
$$x^4 - 2x^2 (a + c) + (-2x^2 + 2a - 2c) \sqrt{b} + (a^2 + b - 2ac + c^2 + d) = (2x + 2a - 2c) \sqrt{d} + 2 \sqrt{bd}.$$
I won't write everything out from here (I've probably already made a mistake). Squaring a third time removes all of the radicals except those of the form $\sqrt{b}$. Rearranging and squaring a fourth time removes these.
This method does not generalize; putting too many additional square roots into the problem will cause squaring to keep giving you more terms regardless of how cleverly you rearrange.
|
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|
Using generating functions find the sum $1^3 + 2^3 + 3^3 +\dotsb+ n^3$ I am quite new to generating functions concept and I am really finding it difficult to know how to approach problems like this. I need to find the sum of $1^3 + 2^3 + 3^3 +\dotsb+ n^3$ using generating functions. How do I proceed about it?
|
Note that if $A(z) = \sum_{n \ge 0} a_n z^n$, then
$$
z \frac{\mathrm{d}}{\mathrm{d} z} A(z)
= \sum_{n \ge 0} n a_n z^n
$$
and also:
$$
\frac{A(z)}{1 - z}
= \sum_{n \ge 0} \left( \sum_{0 \le k \le n} a_k \right) z^n
$$
Starting with:
$$
\sum_{n \ge 0} z^n
= \frac{1}{1 - z}
$$
the generating function for the sum you want is:
$\begin{align}
\frac{z}{1 - z}
\frac{\mathrm{d}}{\mathrm{d} z}
\left( z \frac{\mathrm{d}}{\mathrm{d} z}
\left( z \frac{\mathrm{d}}{\mathrm{d} z} \frac{1}{1 - z} \right)
\right)
&= \frac{z (1 + 4 z + z^2)}{(1 - z)^5}
\end{align}$
Thus:
$\begin{align}
\sum_{0 \le k \le n} k^3
&= [z^n] \frac{z (1 + 4 z + z^2)}{(1 - z)^5} \\
&= [z^n] (z + 4 z^2 + z^3) \sum_{k \ge 0} (-1)^k \binom{-5}{k} z^k \\
&= [z^n] (z + 4 z^2 + z^3) \sum_{k \ge 0} \binom{k + 5 - 1}{5 - 1} z^k \\
&= \binom{n - 1 + 4}{4}
+ 4 \binom{n - 2 + 4}{4}
+ \binom{n - 3 + 4}{4} \\
&= \frac{n^2 (n + 1)^2}{4}
\end{align}$
Note that:
$$
\sum_{0 \le k \le n} k^3
= \left( \sum_{0 \le k \le n} k \right)^2
$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
how to find the series $x + x^{1 + \frac{1}{2}} + x^{1 + \frac{1}{2}+ \frac{1}{3}} +...$ is convergent. The series $x + x^{1 + \frac{1}{2}} + x^{1 + \frac{1}{2}+ \frac{1}{3}} +...$
is convergent if
(A) $x>e$
(B) $x<e $
(C) $x<1/e$
(D) $x>1/e$
I think the answer is C, but I could not determine the condition... how to solve it. Plz help.
|
We have
\begin{align*}
x^{\sum_{i=1}^n \frac 1i} &\le x^{\log n + 1}\\
&= x \cdot \exp(\log n\cdot \log x)\\
&= x \cdot n^{\log x}
\end{align*}
and
\begin{align*}
x^{\sum_{i=1}^n \frac 1i} &\ge x^{\log n}\\
&= \exp(\log n\cdot \log x)\\
&= n^{\log x}
\end{align*}
and hence
\[ \sum_{n=1}^\infty x^{\sum_{i=1}^n \frac 1i} < \infty \iff \sum_{n=1}^\infty n^{\log x} < \infty \]
which is true exactly for $\log x < -1\iff x < \frac 1e$.
|
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|
Common tangent to two circles
Find the equations of the common tangents to the 2 circles:
$$(x - 2)^2 + y^2 = 9$$
and
$$(x - 5)^2 + (y - 4)^2 = 4.$$
I've tried to set the equation to be $y = ax+b$, substitute this into the 2 equations and set the discriminant to zero, we then get a simultaneous quadratic equations. But they are really difficult to solve. So is there any simpler way to do this? Thank you.
|
Like in my other answer, the equation of any tangent of $$(x-2)^2+y^2=3^2$$ will be $$x\cos t+y\sin t-(2\cos t+3)=0$$
Now if this has to be tangent of $(x-5)^2+(y-4)^2=4$ if the radius $=$ the distance from the center$(5,4)$
$$\dfrac{|5\cos t+4\sin t-(2\cos t+3)|}{\sqrt{\cos^2t+\sin^2t}}=2$$
$$\implies4\sin t+3\cos t-3=\pm2$$
Case $\#1:+\implies4\sin t+3\cos t=5$
Use Find $4\cos\theta-3\sin\theta$, given that $4\sin \theta +3\cos \theta = 5$ to solve for $\sin t,\cos t$
Case $\#2:-\implies4\sin t+3\cos t=1$
Use Weierstrass substitution to find $\sin t,\cos t$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Frequency of a trigonometric function - Where is my mistake? I need to find the frequency of the following trigonometric function.$$y=\sin^4(x)+\cos^4(x)$$
The "answers" section says the answer is: $$F_y=\frac{\pi}{2}$$
This is what i did:
Finding $\sin(x)^4$ frequency (I'll call it F1):
$$\cos(2x)=1-\sin^2(x)$$
$$\sin^2(x)=\frac{1-\cos(2x)}{2}$$
$$\sin^4(x)=\frac{\cos^2(2x)-2\cos(2x)+1}{4}=\frac{cos^2(2x)+4\sin^2(x)-1}{4}$$
Finding $\cos(2x)^2$ frequency:
$$\cos(4x)=2\cos^2(2x)-1$$
$$\cos^2(2x)=\frac{\cos(4x)+1}{2}$$
$$f_1=\frac{2\pi}{4}=\frac{\pi}{2}$$
Finding $\sin(x)^2$ frequency:
$$\cos(2x)=1-2\sin^2(x)$$
$$\sin^2(x)=\frac{1-\cos(2x)}{2}$$
$$f_2=\frac{2\pi}{2}=\pi$$
$$F_1: \frac{f_1}{f_2}=\frac{\frac{\pi}{2}}{\pi}=\frac{1}{2}$$
$$F_1=\frac{\pi}{2}\times2=\pi$$
Finding $cos(x)^4$ frequency (I'll call it F2):
$$\cos(2x)=2\cos^2(x)-1$$
$$\cos^2(x)=\frac{\cos(2x)+1}{2}$$
$$\cos^4(x)=\frac{\cos^2(2x)+2\cos(2x)+1}{4}$$
Finding $\cos(2x)$ frequency (we already have $\cos(2x)^2$ frequency - f1):
$$f_3=\frac{2\pi}{2}=\pi$$
$$F_2: \frac{f_1}{f_3}=\frac{\frac{\pi}{2}}{\pi}=\frac{1}{2}$$
$$F_2=\frac{\pi}{2}\times2=\pi$$
Finding $y$'s frequency:
$$F_y: \frac{F_1}{F_2}=\frac{\pi}{\pi}=\frac{1}{1}$$
$$F_y=\pi\times1=\pi$$
|
You might consider using Euler's formula, which can be used to obtain
$$ \cos(x) = \frac{e^{ix} + e^{-ix}}{2} $$ and
$$ \sin(x) = \frac{e^{ix} - e^{-ix}}{2i}. $$
Put $z = e^{ix}$, so that
$$\cos(x) =\frac{z+1/z}{2}$$
and
$$\sin(x) = \frac{z-1/z}{2i}$$
This gives
$$\cos(x)^4+\sin(x)^4 =
\left(\frac{z+1/z}{2}\right)^4 + \left(\frac{z-1/z}{2i}\right)^4 =
\frac{z^4}{8} + 3/4 + \frac{1/z^4}{8}$$
which is
$$\frac{3}{4} + \frac{1}{4} \cos(4x).$$
So your frequency is $$ \frac{2\pi}{4} = \frac{\pi}{2}.$$
|
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|
When is $1^5 + 2^5 + \ldots + n^5$ a square? When is $1^5 + 2^5 + \ldots + n^5$ a square? I found that this happens sometimes: $n=13$ gives $1001^2$, $n=133$ gives $9712992^2$ and $n=1321$ gives $942162299^2$.
I feel that the identity$$\displaystyle\sum_{i=1}^n i^5 = \tfrac{1}{12}[2n^6+6n^5+5n^4-n^2]$$ will be useful, since it's all square powers except one.. but I see no way to connect that.
|
This is OEIS A031138, which lists some more and says $a(n) =11\cdot(a(n-1)-a(n-2)) + a(n-3) \\ a(n)=-1/2+((3-\sqrt 6)/4)\cdot(5+2\sqrt 6)^n+((3+\sqrt 6)/4)\cdot(5-2\sqrt 6)^n$
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Expected Number of Successes in a Sample
$200$ calculators are ordered and of those $200$, $20$ are broken. $10$ calculators are selected at random. Calculate the expected value of broken calculators in the selection.
Solution:
Chance of broken calculator: $\dfrac{1}{10}$.
Do I need to calculate the odds of $0$ - $10$ calculators being broken, multiply the probabilities with the respective $0$ - $10$ and add those together?
For example:
\begin{align}
{10 \choose 0} \cdot \left(\frac{1}{10}\right)^0 \cdot \left(\frac{9}{10}\right)^{10} &= 0.3487 \tag{0}, \\
{10 \choose 1} \cdot \left(\frac{1}{10}\right)^1 \cdot \left(\frac{9}{10}\right)^9 &= 0.3874 \tag{1}, \\
\vdots & \\
{10 \choose 10} \cdot \left(\frac{1}{10}\right)^{10} \cdot \left(\frac{9}{10}\right)^0 &= 1 \cdot 10^{-10}. \tag{10}
\end{align}
Adding them together:
\begin{align}
E[X] = (0) \cdot 0 + (1) \cdot 1 + \ldots + (10) \cdot 10.
\end{align}
Or is it enough to calculate
\begin{align}
E[X] &= (1+2+3+4+5+6+7+8+9+10) \cdot \left(\frac{1}{10}\right).
\end{align}
Thanks.
|
Your shorter formula doesn't work. Suppose the question asked you to select two calculators instead of ten. To compute the expected value "by hand", there are three cases to consider (we don't care about the case when no calculators are broken).
First Calculator Broken: This event happens with probability $(1/10)(9/10) = 9/100$.
Second Calculator Broken: This event happens with probability $(9/10)(1/10) = 9/100$.
Both Calculators Broken: This event happens with probability $(1/10)(1/10) = 1/100$.
The expected number of broken calculators (in my modified example) is therefore
$$
(1 \cdot 9/100) + (1 \cdot 9/100) + (2 \cdot 1/100) = 20/100 = 0.2.
$$
The fact that we had two instances of "one calculator broken" would normally be accounted for by the factor of $C(2,1) = 2$.
To address your concern more directly, your shorter formula does not reflect the fact that "exactly one broken calculator" really means "one is broken and nine are functional", which is why that formula is giving a higher expected value than is correct.
|
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|
How to eliminate these extra solutions? (finding the gcd of two expressions)
Prove that for any integer $n$, $\gcd (3n^2+5n+7, n^2+1)=1$ or $41$.
The following answer is convoluted because I've intentionally created excess solutions. However, I can't figure out how to eliminate them! Anyone?
Let $$d=\gcd (3n^2+5n+7, n^2+1).$$
Then $$d|[(3n^2+5n+7)-3(n^2+1)]$$
$$d |(5n+4)$$
And
$$d | [5(3n^2+5n+7)-3n(5n+4)]$$
$$d |(13n+35)$$
And
$$d |[5(13n+35)-13(5n+4)]$$
$$d |123$$
Therefore, $d= 1$ or $3$ or $41$ or $123$.
|
Suppose that
$$
(3n^2+5n+7,n^2+1)=(5n+4,n^2+1)\ne1\tag{1}
$$
then either
$$
(5n+4,n+i)=(4-5i,n+i)\ne1\tag{2}
$$
or
$$
(5n+4,n-i)=(4+5i,n-i)\ne1\tag{3}
$$
Since $4-5i$ is a Gaussian prime, $(2)\Rightarrow4-5i\,|\,n+i$. That is,
$$
\frac{n+i}{4-5i}=\frac{(4n-5)+(5n+4)i}{41}\in\mathbb{Z}[i]\tag{4}
$$
which is true if and only if $n\equiv32\pmod{41}$.
Since $4+5i$ is a Gaussian prime, $(3)\Rightarrow4+5i\,|\,n-i$. That is,
$$
\frac{n-i}{4+5i}=\frac{(4n-5)-(5n+4)i}{41}\in\mathbb{Z}[i]\tag{5}
$$
which is true if and only if $n\equiv32\pmod{41}$.
Thus, $(1)$ implies either
$$
(2)\Rightarrow4-5i\,|\,(5n+4,n^2+1)\text{ iff }n\equiv32\pmod{41}\tag{6}
$$
or
$$
(3)\Rightarrow4+5i\,|\,(5n+4,n^2+1)\text{ iff }n\equiv32\pmod{41}\tag{7}
$$
Therefore,
$$
(1)\Rightarrow n\equiv32\pmod{41}\tag{8}
$$
It is easy to verify that
$$
n\equiv32\pmod{41}\Rightarrow41\,|\,(3n^2+5n+7,n^2+1)\tag{9}
$$
Again, $(1)$ implies either
$$
(2)\Rightarrow4-5i\,|\,(3n^2+5n+7,n^2+1)\Rightarrow4+5i\,|\,(3n^2+5n+7,n^2+1)\tag{10}
$$
or
$$
(3)\Rightarrow4+5i\,|\,(3n^2+5n+7,n^2+1)\Rightarrow4-5i\,|\,(3n^2+5n+7,n^2+1)\tag{11}
$$
Therefore,
$$
(1)\Rightarrow41=(4-5i)(4+5i)\,|\,(3n^2+5n+7,n^2+1)\tag{12}
$$
Finally, as Pambos points out, the Euclidean Algorithm yields
$$
(15n+13)(n^2+1)-(5n-4)(3n^2+5n+7)=41\tag{13}
$$
Therefore,
$$
(3n^2+5n+7,n^2+1)\,|\,41\tag{14}
$$
|
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"url": "https://math.stackexchange.com/questions/218915",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
$\cos^n x-\sin^n x=1$ For $0 < x < 2\pi$ and positive even $n$, the only solution for $\cos^n x-\sin^n x=1$ is $\pi$.
The argument is simple as $0\le\cos^n x, \sin^n x\le1$ and hence $\cos^n x-\sin^n x=1$ iff $\cos^n x=1$ and $\sin^n x=0$.
My question is that any nice argument to show the following statement?
'For $0 < x < 2\pi$ and positive odd $n$, the only solution for $\cos^n x-\sin^n x=1$ is $\frac{3\pi}{2}$.'
|
We leave the case $n = 1$ and $n = 2$ separately, and assume $n \geq 3$ from now on.
Observe that if $|r| \leq 1$, then $|r^n| \leq r^2$ with equality if and only if $r = 0$ or $|r| = 1$. Then it follows that
$$1 = \left|\cos^n x - \sin^n x\right| \leq \left|\cos^n x\right| + \left|\sin^n x\right| \leq \cos^2 x + \sin^2 x = 1. $$
This forces every intermediate inequality to be equality. In particular, we must have
$$ \cos x , \sin x \in \{0, \pm 1\}.$$
Thus $x \in \{ \frac{\pi}{2}, \pi, \frac{3\pi}{2} \}$. Now the rest is clear.
|
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"timestamp": "2023-03-29T00:00:00",
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|
What is the number of trailing zeros in a factorial in base ‘b’? I know the formula to calculate this, but I don't understand the reasoning behind it:
For example, the number of trailing zeros in $100!$ in base $16$:
$16=2^4$,
We have:
$\frac{100}{2}+\frac{100}{4}+\frac{100}{8}+\frac{100}{16}+\frac{100}{32}+\frac{100}{64}=97$
Number of trailing zeros$ =\frac{97}{4} = 24$.
Why do we divide by the power of '$2$' at the end?
|
Suppose that $b=p^m$, where $p$ is prime; then $z_b(n)$, the number of trailing zeroes of $n!$ in base $b$, is
$$z_b(n)=\left\lfloor\frac1m\sum_{k\ge 1}\left\lfloor\frac{n}{p^k}\right\rfloor\right\rfloor\;.\tag{1}$$
That may look like an infinite summation, but once $p^k>n$, $\left\lfloor\frac{n}{p^k}\right\rfloor=0$, so there are really only finitely many non-zero terms.
The summation counts the number of factors of $p$ in $n!$. The set $\{1,2,\dots,n\}$ of integers whose product is $n!$ contains $\left\lfloor\frac{n}p\right\rfloor$ multiples of $p$, $\left\lfloor\frac{n}{p^2}\right\rfloor$ multiples of $p^2$, and so on $-$ in general $\left\lfloor\frac{n}{p^k}\right\rfloor$ multiples of $p^k$. Each multiple of $p$ contributes one factor of $p$ to the product $n!$; each multiple of $p^2$ contributes an additional factor of $p$ beyond the one that was already counted for it as a multiple of $p$; each multiple of $p^3$ contributes an additional factor of $p$ beyond the ones already counted for it as a multiple of $p$ and as a multiple of $p^2$; and so on.
Let $s=\sum_{k\ge 1}\left\lfloor\frac{n}{p^k}\right\rfloor$; then $n!=p^sk$, where $k$ is not divisible by $p$. Divide $s$ by $m$ to get a quotient $q$ and a remainder $r$: $s=mq+r$, where $0\le r<m$. Then $$n!=p^sk=p^{mq+r}k=(p^m)^qp^rk=b^qp^rk\;,$$ where $p^rk$ is not divisible by $b$. Since $p^rk$ isn’t divisible by $b$, in base $b$, it will not end in $0$. Multiplying it by $b$ will simply tack a $0$ on the righthand end of it, just as multiplying $123$ by $10$ in base ten tacks a $0$ on the end to make $1230$. Multiplying by $b^q$ is multiplying by $b$ a total of $q$ times, so it tacks $q$ zeroes onto a number that did not end in $0$; the result is that $n!$ ends up with $q$ zeroes in base $b$. But $$q=\left\lfloor\frac{s}m\right\rfloor=\left\lfloor\frac1m\sum_{k\ge 1}\left\lfloor\frac{n}{p^k}\right\rfloor\right\rfloor\;,$$ showing that $(1)$ is correct.
In your example $b=2^4$, so $p=2$ and $m=4$, and with $n=100$, $(1)$ becomes
$$\begin{align*}
z_{16}(100)&=\left\lfloor\frac14\sum_{k\ge 1}\left\lfloor\frac{100}{2^k}\right\rfloor\right\rfloor\\\\
&=\left\lfloor\frac14\left(\left\lfloor\frac{100}2\right\rfloor+\left\lfloor\frac{100}4\right\rfloor+\left\lfloor\frac{100}8\right\rfloor+\left\lfloor\frac{100}{16}\right\rfloor+\left\lfloor\frac{100}{32}\right\rfloor+\left\lfloor\frac{100}{64}\right\rfloor\right)\right\rfloor\\\\
&=\left\lfloor\frac14(50+25+12+6+3+1)\right\rfloor\\\\
&=\left\lfloor\frac{97}4\right\rfloor\\\\
&=24
\end{align*}$$
The value of the summation is $97$, which tells you that there are $97$ factors of $2$ in $100!$: $100!=2^{97}k$, where $k$ is an odd number. $97=4\cdot24+1$, so $100!=2^{4\cdot24+1}k=(2^4)^{24}2k=16^{24}(2k)$, where $2k$ is not a multiple of $16$ (since it’s just $2$ times an odd number). Thus, the base $16$ representation of $2k$ does not end in $0$. Each of the $24$ multiplications of this number by $16$ tacks another $0$ on the end in base $16$, so you end up with $24$ zeroes on the end.
The original sum counts the factors of $2$ in $100!$, but the number of zeroes on the end isn’t the number of factors of $2$: it’s the number of factors of $2^4$, the base. Every four factors of $2$ give you one factor of $2^4$, so you divide by $4$ (and throw away the remainder) to see how many factors of $2^4$ you can build out of your $97$ factors of $2$.
When the base is not a power of a prime, counting the trailing zeroes is a little harder, but it can be done using exactly the same ideas.
|
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|
Question involving some roots of the equation $31z^{15} - z^{10} + 32 = 0 $ Consider the equation $$31z^{15} - z^{10} + 32 = 0.$$ What would be the sum of all those roots of the equation whose real part is positive? Only trivially trying to solve the equation I find not helpful. Even factorizing we get , by putting $z^5=x $ ,
$31x^3 - x^2 + 32 = (x+1)(31x^2 - 32x + 32) =0 $ which is of no help .
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$31z^{15}-z^{10}+32=0\implies 31(z^{15}+1)-(z^{10}-1)=0$
$31(z^5+1)(z^{10}-z^5+1)-(z^5+1)(z^5-1)=0$
$(z^5+1)\{31(z^{10}-z^5+1)-(z^5-1) \}=0 $
$(z^5+1)\{31z^{10}-32z^5+32 \}=0 $
If $z^5+1=0, z^5=-1=e^{(2m+1)\pi i} $ where $m$ is any integer.
So, $z=e^{\frac{(2m+1)\pi i}5}$ where any $5$ in-congruent values of $m\pmod 5$ will give us essentially the same set of $5$ distinct solutions, find the explanation here, the simplest example can be $0,1,2,3,4$.
Now, $e^\frac{(2m+1)\pi i}5=\cos \frac{(2m+1)\pi}5+i\sin \frac{(2m+1)\pi}5$
For the real part to be positive, $-\frac \pi 2<\frac{(2m+1)\pi}5< \frac \pi 2$
$\implies -\frac74<m<\frac 3 4\implies m=-1,0$
So, those roots are $\cos \frac{\pi}5\pm \sin \frac{\pi}5,$ the sum being $2\cos \frac{\pi}5$
If $31z^{10}-32z^5+32=0,z^5=\frac{16\pm12\sqrt 5 i}{31}$
Let $r\cos A= \frac{16}{31}--->(1), r\sin A= \frac{12\sqrt 5 }{31}--->(2)$ where $r>0$
so that $z^5=r(\cos A\pm i\sin A)=re^{\pm iA}=re^{i(2n\pi\pm A)}$ where $n$ is any integer.
$z=r^\frac 15 e^{\frac{i(2n\pi\pm A)}5}$ where $0\le n<5$
Dividing $(2)$ by $(1),\tan A=\frac{12\sqrt 5}{16}=\frac{3\sqrt5}4$
So, $A=\arctan \frac{3\sqrt5}4$ where $A$ lies $\in(0,\frac \pi 2)$ as $\cos A,\sin A>0$
Squaring and adding $(2)$ and $(1), r^2=\left(\frac{16}{31}\right)^2+\left(\frac{12\sqrt 5 }{31}\right)^2$
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Finding Lebesgue Integral of $\frac{1}{\sqrt{x}}$ over $(0,1]$ How do I rigorously discover what
$$
\int_{(0,1]} \frac{1}{x^{1/2}} = \underset{0 \le \phi \le \frac{1}{\sqrt{x}}}{\sup} \int_{(0,1]} \phi
$$
(for $\phi$ a simple function) is? Note that I have already shown in another exercise that such an integral exists and that it is finite.
|
Here is a brute force approach using the dominated convergence theorem (DCT). The idea is to create a sequence of simple functions $s_n$ such that $s_n(x) \leq \frac{1}{\sqrt{x}}$, and $s_n(x) \to \frac{1}{\sqrt{x}}$. The DCT shows that $\int s_n \to \int \frac{1}{\sqrt{x}}$.
Let $s_n = \sum_{k=n}^{n^2-1} \frac{k}{n} 1_{[(\frac{n}{k+1})^2, (\frac{n}{k})^2)}$. Then $s_n(x) \leq \frac{1}{\sqrt{x}}$, $s_n(x) \to \frac{1}{\sqrt{x}}$, and $s_n$ is supported in $[\frac{1}{n^2},1)$.
Then
\begin{eqnarray}
\int s_n & = & \sum_{k=n}^{n^2-1} \frac kn\left(\left(\frac{n}{k}\right)^2 - \left(\frac{n}{k+1}\right)^2\right) \\
& = & n \sum_{k=n}^{n^2-1}\left(\frac{1}{k} - \frac{k}{(k+1)^2}\right) \\
& = & n \sum_{k=n}^{n^2-1}\left(\frac{1}{k} - \frac{1}{k+1} +\frac{1}{(k+1)^2}\right) \\
& = & n\left(\frac{1}{n}-\frac{1}{n^2}\right) + n \sum_{k=n}^{n^2-1} \frac{1}{(k+1)^2}.
\end{eqnarray}
Noting that $\int_{k+1}^{k+2} \frac{dx}{x^2} \leq \frac{1}{(k+1)^2} \leq \int_k^{k+1} \frac{dx}{x^2}$, we obtain the bounds $\frac{1}{n+1}-\frac{1}{n^2+1} \leq \sum_{k=n}^{n^2-1} \frac{1}{(k+1)^2} \leq \frac{1}{n}-\frac{1}{n^2}$. It follows that $\lim_{n \to \infty} \int s_n = 2$, which coincides with the value obtained from the improper Riemann integral.
In terms of the original question, we note that if $s$ is simple, and $s(x) \leq \frac{1}{\sqrt{x}}$, then $\int s \leq \int \frac{1}{\sqrt{x}}$, and so the above shows that $\int \frac{1}{\sqrt{x}} = \sup_{s \, \mathbb{simple}, s(x) \leq \frac{1}{\sqrt{x}}} \int s = 2$.
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
How to find solutions of $x^2-3y^2=-2$? According to MathWorld,
Pentagonal Triangular Number: A number which is simultaneously a pentagonal number $P_n$ and triangular number $T_m$. Such numbers exist when
$$\frac{1}{2}n(3n-1)=\frac{1}{2}m(m+1).$$
Completing the square gives
$$(6n-1)^2-3(2m+1)^2=-2.$$
Substituting $x=6n-1$ and $y=2m+1$ gives the Pell-like quadratic Diophantine equation
$$x^2-3y^2=-2,$$
which has solutions $(x,y)=(5,3),(19,11),(71,41),(265,153), \ldots$.
However, it does not state how these solutions for $(x,y)$ were obtained.
I know that the solution $(5,3)$ can be obtained by observing that $1$ is both a pentagonal and a triangular number.
Does obtaining the other solutions simply involve trial-and-error? Or is there a way to obtain these solutions?
|
Here's another approach.
$$X^2-AY^2=B\tag1$$
$$x^2-Ay^2=1\tag2$$
If we know fundamental solution $(a,b/A)$ for $(2)$ and “trivial” solutions $(t,v)$ for $(1)$ then:
$$X_n = \sum_{k=0}^{n}\frac{a^{n-k}b^k\displaystyle\binom{n}{k}\left(\left(\left\lceil\frac{k}{2}\right\rceil -\left\lfloor\frac{k}{2}\right\rfloor\right)v + \left(\left\lceil\frac{k+1}{2}\right\rceil -\left\lfloor\frac{k+1}{2}\right\rfloor\right)t\right)}{A^{\left\lfloor\frac{k}{2}\right\rfloor}}$$
$$Y_n = \sum_{k=0}^n \frac{a^{n-k}b^k\displaystyle\binom{n}{k}\left(\left( \left\lceil\frac{k+1}{2}\right\rceil -\left\lfloor\frac{k+1}{2}\right\rfloor\right)v + \left(\left\lceil\frac{k}{2}\right\rceil - \left\lfloor\frac{k}{2}\right\rfloor\right)t\right)}{A^{\left\lceil\frac{k}{2}\right\rceil}}$$
For $X^2 -3Y^2 = -2$; $t = 1, v = 1, a = 2, b/A = 1$
$$X_n = \sum_{k=0}^n 3^{k-\left\lfloor\frac{k}{2}\right\rfloor}2^{n-k}\binom{n}{k}$$
$$Y_n = \sum_{k=0}^n 3^{k-\left\lceil\frac{k}{2}\right\rceil}2^{n-k}\binom{n}{k}$$
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
As shown in the figure: Prove that $a\,+\,b\,+\,c=d$ Geometry: Auxiliary Lines
As shown in the figure: Prove that $a\,+\,b\,+\,c=d$
|
Using sine law of triangle, $$\frac e{\sin(144^{\circ}-x)}=\frac g{\sin 36^{\circ}}$$ and $$\frac e{\sin(138^{\circ}-x)}=\frac {g+f}{\sin 42^{\circ}}$$
So, $$f=e\left(\frac{\sin 42^{\circ}}{\sin(138^{\circ}-x)}-\frac{\sin 36^{\circ}}{\sin(144^{\circ}-x)}\right)$$
$$=e\left(\frac{\sin 42^{\circ}\sin(144^{\circ}-x)-\sin 36^{\circ}\sin(138^{\circ}-x)}{\sin(138^{\circ}-x)\sin(144^{\circ}-x)}\right)$$
$2\sin 42^{\circ}\sin(144^{\circ}-x)$
$=\cos(102^{\circ}-x)-\cos({186^\circ}-x)$ using $2\sin A\sin B$
$=\cos(102^{\circ}-x)+\cos(6^\circ-x)$ as $\cos(180^{\circ}+y)=-\cos y$
$2\sin 36^{\circ}\sin(138^{\circ}-x)=\cos(102^{\circ}-x)-\cos({174^\circ}-x)$
$=\cos(102^{\circ}-x)+\cos(6^\circ+x)$ as $\cos({174^\circ}-x)=\cos\{180^\circ-(6^\circ+x)\}=-\cos(6^\circ+x)$
So, $$f=e\frac{\cos(6^\circ-x)-\cos(6^\circ+x)}{2\sin(138^{\circ}-x)\sin(144^{\circ}-x)}=\frac{e\sin x\sin 6^\circ }{\sin(138^{\circ}-x)\sin(144^{\circ}-x)}$$
For $f=a, x=12^\circ, a=\frac{e\sin 12^\circ\sin 6^\circ }{\sin126^{\circ}\sin132^{\circ}}$
For $f=b, x=60^\circ, b=\frac{e\sin 60^\circ\sin 6^\circ }{\sin78^{\circ}\sin84^{\circ}}$
For $f=c, x=96^\circ, c=\frac{e\sin 96^\circ\sin 6^\circ }{\sin42^{\circ}\sin48^{\circ}}$
$2\sin42^{\circ}\sin48^{\circ}=\cos 6^\circ -\cos 90^\circ=\cos 6^\circ$
and $\sin 96^\circ=\sin(90+6)^\circ=\cos 6^\circ$
So,$ c=2e\sin 6^\circ $
For $f=d, x=108^\circ, d=\frac{e\sin 108^\circ\sin 6^\circ }{\sin30^{\circ}\sin36^{\circ}}=\frac{2e\sin 72^\circ\sin 6^\circ }{\sin36^{\circ}}=4e\cos 36^\circ \sin 6^\circ$
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Understanding simplification of an algebra problem I have the problem $x^3 - 2x^2$. My book tells me that this problem is simplified to $x^3 (1 -(\frac{2}{x}))$. How does that work?
This step of my book I am in about the "end behavior" of trying to graph a polynomial function. Apparently once you get $x^3 (1 -(\frac{2}{x}))$, its equivalent is just $x^3$.
|
This is only valid when $x \neq 0$. It follows from the distributive rule. If $x \neq 0$, then
\begin{align*}
x^3 \left(1 - \frac{2}{x} \right) &= x^3 \cdot 1 - x^3 \cdot \frac{2}{x} \quad \text{(by distributive rule)}\\
&= x^3 - 2x^2.
\end{align*}
As Cameron Buie explained nicely, $x^3$ and $x^3(1 - \frac{2}{x})$ are asymptotically equal as $|x| \to \infty$.
|
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|
How to calculate $3^{45357} \mod 5$? I wrote some code, here is what it gives:
\begin{align*}
3^0 \mod 5 = 1 \\
3^1 \mod 5 = 3 \\
3^2 \mod 5 = 4 \\
3^3 \mod 5 = 2 \\\\
3^4 \mod 5 = 1 \\
3^5 \mod 5 = 3 \\
3^6 \mod 5 = 4 \\
3^7 \mod 5 = 2 \\\\
3^8 \mod 5 = 1 \\
3^9 \mod 5 = 3 \\
3^{10} \mod 5 = 4 \\
3^{11} \mod 5 = 2 \\\\
3^{12} \mod 5 = 1 \\
3^{13} \mod 5 = 3 \\
...
\end{align*}
As I see, there's a period $1, 3, 4, 2$. And intuitively, it's easy to predict that $3^{45357} \mod 5 = 3$. Unfortunately, I'm almost unfamilar with modular math. Could you give me strict analytic explaination of this thing?
|
Using Fermat's Little theorem,
$a^{p-1}\equiv1\pmod p$ if $(a,p)=1$
$\implies (a^{p-1})^d\equiv1^d\pmod p \equiv1 \pmod p$
if $b=c+(p-1)d$ i.e., $b\equiv c\pmod {p-1}$ where $a,b,c,d$ are non-negative integers and $p$ is prime,
$ a^b=a^{c+(p-1)d}=a^c\cdot(a^{p-1})^d\equiv a^c\pmod p$
Here, $3^{5-1}\equiv 1\pmod 5$ as $(3,5)=1$
As $45357\equiv 1\pmod 4\implies 3^{45357}\equiv 3^1\pmod 5\equiv 3$
|
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|
Integration questions on $\int \frac{x^4\left ( 1-x \right )^4}{1+x^2} \, dx$ and $\int \frac{x^4}{x^4+5x^2+4} \, dx$ I would appreciate any hints on how to solve the following integration problems, they are my homework questions btw:
$$\int \frac{x^4\left ( 1-x \right )^4}{1+x^2} \, dx$$
$$\int \frac{x^4}{x^4+5x^2+4} \, dx$$
Thank you very much in advance!
|
Here are a couple of off-hand suggestions. There are undoubtedly more efficient ways.
For the first integral, you could multiply out the top, and use polynomial long division. You will get something that has the shape $P(x)+\frac{Ax}{1+x^2}+\frac{B}{1+x^2}$.
Integrating the polynomial will be easy. For $\frac{Ax}{1+x^2}$, use $u=1+x^2$.
For the second problem, divide. We get $1-\frac{5x^2+4}{x^4+5x^2+4}$. Then use the fact that $x^4+5x^2+4=(x^2+1)(x^2+4)$, and use partial fractions. We end up needing to integrate $\frac{A}{x^2+1}$ and $\frac{B}{x^2+4}$. The first is immediate. For the second, use $x=2u$.
|
{
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|
Interpolation of a function Given the function
$$f (x) = x\bigg(x − {1\over4}\bigg)\bigg(x − {1\over2}\bigg)$$
How can I interpolate $f(x)$ with $p(x) = a_0T_0(x) + a_1T_1(x) + a_2T_2(x) + a_3T_3(x)$ to show that
$$a_0 = -{3\over8},\ \ a_1 = {7\over8},\ \ a_2 = −{3\over8},\ \ a_3 = {1\over4}$$
|
First, note that, the Chebyshev polynomials have the property
$$\int_{-1}^1 T_n(x)T_m(x)\,\frac{dx}{\sqrt{1-x^2}}= \begin{cases} 0 &: n\ne m \\ \pi &: n=m=0\\ \pi/2 &: n=m\ne 0 \end{cases}\rightarrow (*)\,.$$
The first four polynomials are $ T_0(x)=1, T_1(x)=x, T_2=2x^2-1, T_3=4 x^3-3x.$
Now, we have
$$ f(x) \sim a_0T_0(x) + a_1T_1(x) + a_2T_2(x) + a_3T_3(x) \rightarrow (1). $$
To find $a_i, i=0..3$, we exploit the above property. We start by finding $a_0$. Multiply both sides of $(1)$ by $T_0$ and integrate from $-1$ to $1$ w.r.t. the weight function $\frac{1}{\sqrt{1-x^2}}$ gives
$$ \int_{-1}^{1} \frac{f(x)T_0(x)}{\sqrt{1-x^2}}dx = a_0\int_{-1}^{1} \frac{{T_0}^2(x)}{\sqrt{1-x^2}}dx + a_1\int_{-1}^{1} \frac{{T_0}(x)T_1(x)}{\sqrt{1-x^2}}dx+\dots. $$
$$ \implies -\frac{3}{8}\pi = a_0 \pi + a_1.0 + a_2.0+a_3.0 \implies a_0=-\frac{3}{8} .$$
Notice that, we used the orthogonal property $(*)$ to evaluate the integrals on the right hand side in the above equation.
To find $a_1$, we multiply both sides of equation $(1)$ by $T_1$ and integrate from $-1$ to $1$ w.r.t. the weight function $\frac{1}{\sqrt{1-x^2}}$
$$ \int_{-1}^{1} \frac{f(x)T_1(x)}{\sqrt{1-x^2}}dx = a_0\int_{-1}^{1} \frac{{T_0}(x)T_1(x)}{\sqrt{1-x^2}}dx + a_1\int_{-1}^{1} \frac{{T_1}(x)T_1(x)}{\sqrt{1-x^2}}dx+\dots. $$
$$ \implies \frac{7}{16}\pi = a_0.0 + a_1\frac{\pi}{2}+a_2.0+a_3.0 \implies a_1 = \frac{7}{8}. $$
You can find the rest of $a_2$ and $a_3$ the same way.
|
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Parameter values that make function values side lengths of a triangle I have been trying to solve the following problem for more than a week without any success.
Given the function:
$$f(x)=\frac{x^2+mx+4}{x^2+x+4}$$
Find all possible values of the parameter $m$ such that for any three numbers $a,b,c$ the corresponding values of the function $f(a),f(b)$ and $f(c)$ are sides of a triangle.
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If $f(a), f(b), f(c)$ are sides of a triangle if they satisfy the triangle inequalities
$$ f(a) + f(b) > f(c), \quad f(a) + f(c) > f(b), \quad f(b) + f(c) > f(a).$$
If we require that this holds for every $a,b,c$, then by symmetry we only need to prove that, for all $a,b,c$, $f(a) + f(b) > f(c)$. Writing out this inequality, we have
$$ \frac{a^2 + ma + 4}{a^2 + a + 4} + \frac{b^2 + mb + 4}{b^2 + b + 4} > \frac{c^2 + mc + 4}{c^2 + c + 4}$$
If we rewrite
$$ \frac{a^2 + ma + 4}{a^2 + a + 4} = \frac{a^2 + a + 4 + (m-1)a}{a^2 + a + 4} = 1 + \frac{(m-1)a}{a^2 + a + 4},$$
and similarly do the same rearrangement for the terms with $b$ and $c$, then our inequality becomes
$$ 2 + \frac{(m-1)a}{a^2 + a + 4} + \frac{(m-1) b}{b^2 + b + 4} > 1 + \frac{(m-1)c}{c^2 + c + 4}.$$
Rearranging, we obtain
$$1 > (m-1) \left[ G(c) - G(b) - G(a) \right]$$
where we have defined a new function $G(x) = x/(x^2 + x + 4)$. Notice that $G(x) \rightarrow 0$ as $|x| \rightarrow \infty$. Therefore we can find the maximum and minimum of $G(x)$ by calculating where its derivative vanishes. We have
$$ G'(x) = \frac{1}{x^2 + x + 4} - \frac{x(2x + 1)}{x^2 + x + 4} = \frac{4 - x^2}{x^2 + x + 4},$$
which vanishes at $x = \pm 2$. We calculate $G(2) = 1/5$ and $G(-2) = -1/3$; therefore these are the maximum and minimum values of $G$, respectively. Therefore we have
$$ - \frac{11}{15} = - \frac{1}{3} - \frac{1}{5} - \frac{1}{5} \le G(c) - G(b) - G(a) \le \frac{1}{5} + \frac{1}{3} + \frac{1}{3} = \frac{13}{15}.$$
Furthermore, these upper and lower bounds are achieved by particular choices of $a,b,c$; therefore, in order for the triangle inequality to be satisfied, therefore, if $m > 1$, we have $(m-1)^{-1} > 13/15$, and if $m < 1$, we have $(m-1)^{-1} > -11/15$. As $m = 1$ is also a solution, then we obtain the set of all possible values of $m$:
$$ -\frac{4}{11} < m < \frac{28}{13}$$
|
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|
Prove via induction $\frac{(n+1)^2}{2^n}\le\frac{9}{4}$ How do you prove in induction that:
$$\frac{(n+1)^2}{2^n}\le\frac{9}{4}$$
This is what I keep getting:
Checking for $n=1$ we get $2\le\frac{9}{4}$.
Assuming it's true for $n$ and checking for $n+1$ I get this:
$$\frac{(n+2)^2}{2^{n+1}}=\frac{2(n+1)^2-n^2+2}{2\times2^n}\le\frac{9}{4}-\frac{n^2-2}{2\times2^n}\le\frac{9}{4}$$
Which is true only for $n>1$.
|
The simple answer is to check $n=2$ too, where you have
$$\frac{(n+1)^2}{2^n}= \frac{(2+1)^2}{2^2} = \frac{9}{4} \le\frac{9}{4}$$ and then do the induction.
|
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|
Showing $\lim_{n\rightarrow\infty}\sqrt[3]{n^3+n^2}-\sqrt[3]{n^3+1}\rightarrow\frac{1}{3}$ $$\lim_{n\rightarrow\infty}\sqrt[3]{n^3+n^2}-\sqrt[3]{n^3+1}\rightarrow\frac{1}{3}$$
I tried to say we can erase the $1$ from the equation, as it's a constant. But I don't know how to do the rest without running into this mistake: $$\lim_{n\rightarrow\infty}\sqrt[3]{n^3+n^2}-n=\frac{\sqrt[3]{\frac{n^3}{n^3}+\frac{n^2}{n^3}}-\frac{n}{n}}{\frac{1}{n}}=\frac{1-1}{0}$$
|
My answer here
(Evaluation of $\lim\limits_{n\to\infty} (\sqrt{n^2 + n} - \sqrt[3]{n^3 + n^2}) $)
shows that
$\sqrt[a]{n^a+n^{a-c}}
=n+\dfrac{1}{an^{c-1}}+O(n^{-(2c-1)})
$.
If $a=3, c=1$,
$\sqrt[3]{n^3+n^{2}}
=n+\dfrac{1}{3}+O(n^{-1})
$.
If $a=3, c=3$,
$\sqrt[3]{n^3+1}
=n+\dfrac{1}{3n^2}+O(n^{-5})
=n+O(n^{-2})
$.
Their difference is,
therefore,
$\frac1{3}+O(n^{-1})
\to \frac13
$.
|
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|
Determining the number $N$ Let $1 = d_1 < d_2 <\cdots< d_k = N$ be all the divisors of $N$ arranged in increasing order. Given that $N=d_1^2+d_2^2+d_3^2+d_4^2$, determine $N$. The divisors include $N$. It seems that $130$ is an answer. Is there another possible answer for $N$?
|
$N$ is even (if not then all $d_i$ are odd, making $\sum_{i=1}^4 d_i^2$ even). Therefore $d_1=1$ and $d_2=2$, and at exactly one of $d_3$ and $d_4$ is even.
Suppose that $4 \mid n$. Then one of $d_3, d_4$ is $4$ and the other is an odd prime $p$. Since $N=21+p^2$ and $p \mid N$, we have $p \mid 21$. But $4 \nmid 21+3^2$ and $5 \mid 21+7^2$, ruling out both choices of $p$.
Thus $4 \nmid n$. $d_3$ is an odd prime $p$ and $d_4$ is even. Since $d_4/2$ is a smaller divisor, it could only be $d_3$. Therefore $N = 1+4+p^2+4p^2 = 5(1+p^2)$. $3$ cannot divide a number of this form, and clearly $5$ does. Therefore $d_3=5$ and $d_4=10$, uniquely determining $N=130$.
|
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|
solving the equation $x^4-5x^3+11x^2-13x+6=0$ by given condition 1.(a) solve the equation $x^4-5x^3+11x^2-13x+6=0$ , given that two of its roots $p$ & $q$ are connected by the relation $3p+2q=7$
(b) solve the equation $x^4-5x^3+11x^2-13x+6=0$ which has two roots whose difference is $1$
did I need to solve these problems by taking roots $x_1,x_2,x_3,x_4$ and using the relation between roots and coefficients and the given facts. it will be then a very lengthy process.is there any alternative short process
|
Let $$f(x) = x^4 - 5x^3 + 11x^2 - 13x + 6$$ We get that $$f(1) = 1 - 5 + 11 - 13 + 6 = 0$$ and $$f(2) = 16 - 5 \times 8 + 11 \times 4 - 13 \times 2 + 6 = 16 - 40 + 44 - 26 + 6 = 0$$
Hence, we have that $$f(x) = (x-1)(x-2)(x^2 + ax + b)$$ Plugging in $x=0$, we get that
$$f(0) = 2b = 6 \implies b = 3.$$ Plugging in $x=3$, we get that $$f(3) = 2(9 +3a+3) = 81-5 \times 27 + 11 \times 9 - 13 \times 3 + 6$$
This gives us that
$$6a + 24 = 12 \implies a = -2$$
Hence, we have that
$$f(x) = (x-1)(x-2)(x^2 - 2x+3)$$
Hence, $$f(x) = 0 \implies x = 1 \text{ or } x= 2 \text{ or }x^2 -2x+3 = 0$$
$$x^2 -2x+3 = 0 \implies (x-1)^2 + 2 =0 \implies x = 1 \pm i\sqrt{2}$$
Hence, the roots are $$x=1,2,1 \pm i \sqrt{2}$$
|
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|
Find the maxium $\frac{bc}{a^{2}b+a^{2}c}+\frac{ac}{b^{2}a+b^2c}+\frac{ab}{c^2a+c^{2}b}$ 1) $a, b, c$ are triangle edges's length such that $abc = 1$. Find max:
$$\frac{bc}{a^{2}b+a^{2}c}+\frac{ac}{b^{2}a+b^2c}+\frac{ab}{c^2a+c^{2}b}$$
My idea: $$\frac{bc}{a^{2}b+a^{2}c}+\frac{ac}{b^{2}a+b^2c}+\frac{ab}{c^2a+c^{2}b}=\frac{abc}{a^3(b+c)}+\frac{abc}{b^{3}(c+a)}+\frac{abc}{c^3(a+b)}$$
Then use AM-GM ? I just can find min: $$\frac{1}{a^{3}\left ( b+c \right )}+\frac{1}{b^{3}\left ( c+a \right )}+\frac{1}{c^{3}\left ( a+b \right )}\geq \frac{3}{2}$$
2) Find for $x$, $y$, $z$ such that $\left\{\begin{matrix}
xy + 2(x+y)=0\\
\ yz + 2(y+z)=-3\\
zx + 2(z+x)=5
\end{matrix}\right.$
(Some one should edit my post: correct grammar...)
|
Hint: For your second question, subtracting the equations in pairs suggests the substitution $$a=x+2,\quad b=y+2, \quad c=z+2.$$
Try this in the original set of equations.
|
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|
Is this series convergent or divergent? Kindly asking, what can I do about series
$$
\left(\frac{1}{3}\right)^2+\left(\frac{1\times 4}{3\times 6}\right)^2+\left(\frac{1\times 4\times 7}{3\times 6\times 9}\right)^2+...+\left(\frac{1\times 4\times 7\times...\times (3n-2)}{3\times 6\times 9\times...\times3n}\right)^2+...$$
Indeed, the ratio test fails. Thank you.
|
$$\frac{1\times 4\times 7\times\cdots\times (3n-2)}{3\times 6\times 9\times\cdots\times3n}\leqslant\frac1{(n+1)^{2/3}}$$
|
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|
An algebra problem related to trigonometry Let $\alpha $ be an acute angle such that $\sin \alpha = \dfrac{3 \sqrt 3}{14}$.
Prove that
$$\frac{2\cdot7^{n}}{\sqrt 3}\sin \left(n\alpha + \dfrac{\pi}{3} \right) \in \mathbb{Z} \qquad\forall n>0. $$
|
This is equivalent to saying that for all $n>0$
$$
\frac{2\cdot7^{n}}{\sqrt{3}}\sin\left(n\alpha+\frac\pi3\right)\in\mathbb{Z}
$$
Note that $\cos(\alpha)=\frac{13}{14}$, so that
$$
e^{i\alpha}=\frac{13}{14}+i\frac{3\sqrt3}{14}
$$
therefore
$$
\begin{align}
e^{i(n\alpha+\pi/3)}
&=\left(\frac{13}{14}+i\frac{3\sqrt3}{14}\right)^{\large\!n}\left(\frac12+i\frac{\sqrt{3}}{2}\right)\\[6pt]
&=\frac1{7^n}\left(\frac{13}{2}+i\frac{3\sqrt3}{2}\right)^{\large\!n}\left(\frac12+i\frac{\sqrt{3}}{2}\right)
\end{align}
$$
Note that
$$
a_n=\left(\frac{13}{2}+i\frac{3\sqrt3}{2}\right)^{\large\!n}\left(\frac12+i\frac{\sqrt{3}}{2}\right)
$$
and $\frac2{\sqrt{3}}$ times its imaginary part are solutions to
$$
a_n=13a_{n-1}-49a_{n-2}
$$
$\dfrac2{\sqrt{3}}\mathrm{Im}(a_0)=1$ and $\dfrac2{\sqrt{3}}\mathrm{Im}(a_1)=8$. Thus, $\dfrac2{\sqrt{3}}\mathrm{Im}(a_n)\in\mathbb{Z}$. Substituting back, we get the desired result.
|
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|
How to solve second degree recurrence relation? For first degree recurence relation it is as simple as $f(n)=a^n\cdot f(0)+b\dfrac{a^n-1}{a-1}$.
But how do you solve second degree?
For example
$$f(n)=\begin{cases}
1,&\text{for }n=1\\
2,&\text{for }n=2\\
-3f(n-1)+4f(n-2),&\text{for }n>2\;.
\end{cases}$$
I tried googling "How to solve second degree recurrence relation?" but it gave me solutions only for first degree and some other random stuff.
|
Or use generating functions. Rewite the recurrence as:
$$
f(n + 2) = - 3 f(n + 1) + 4 f(n) \qquad f(1) = 1, f(2) = 2
$$
Define:
$$
F(z) = \sum_{n \ge 0} f(n + 1) z^n
$$
By the properties of ordinary generating functions (see e.g. Wilf's "generatingfunctionology"):
$$
\frac{F(z) - f(1) - f(2) z}{z^2} = - 3 \frac{F(z) - f(1)}{z} + 4 F(z)
$$
My own tame computer algebra system gives:
$$
F(z) = \frac{6}{5} \cdot \frac{1}{1 - z} + \frac{1}{5} \cdot \frac{1}{1 + 4 z}
$$
This expands as two geometric series:
$$
\begin{align*}
f(n) &= \frac{6}{5} + \frac{1}{5} (-4)^{n - 1} \\
&= \frac{6}{5} - \frac{(-4)^n}{20}
\end{align*}
$$
|
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|
Confused about an integral from MIT integration bee 2012 One of the integrals is:
$$\int \frac{\mathrm{d}x}{2+2\sin x + \cos x}\, \mathrm{d}x $$
How can there be two $\mathrm{d}x$?
MIT's Integration Bee
|
The extra $dx$ is definitely a typo. To solve this integral we have to substitute $u=\tan\frac{x}{2}$. Then, $\sin x=\frac{2u}{1+u^2}$, $\cos x=\frac{1-u^2}{1+u^2}$ and $dx=\frac{du}{1+u^2}$. Thus,
$$\int\frac{dx}{2+2\sin x+\cos x}=\int\frac{du}{2+2u^2+4u+1-u^2}=\int\frac{du}{u^2+4u+3}=\int\frac{du}{(u-1)(u-3)}=\frac{1}{2}\int\frac{du}{1-u}+\frac{1}{2}\int\frac{du}{u-3}=\frac{1}{2}\ln\left|1-u\right|+\frac{1}{2}\ln\left|u-3\right|+c=\\
\frac{1}{2}\ln\left|1-\tan\frac{ x}{2}\right|+\frac{1}{2}\ln\left|\tan\frac{ x}{2}-3\right|+c$$
|
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|
Exercise about MacLaurin's polynomial and small-o In class the professor wrote the following limit:
$\lim_{x\to 0} \frac{\sinh^2 (x) -x^2}{x^4}$
So he "expanded" (sorry for my English) the MacLaurin's formula for $\sinh x$ up to the 3rd power, and got: $x + \frac{x^3}{3!} + o(x^4)$
When he squared the MacLaurin's polynomial, he wrote the following steps:
$(\sinh x)^2 = (x + \frac{x^3}{3!} + o(x^4))^2 = x^2 + (\frac{x^3}{3!})^2 + (o(x^4))^2 + 2x\frac{x^3}{3!} + 2xo(x^4) + 2\frac{x^3}{3!}o(x^4) = x^2 + \frac{x^4}{3} + o(x^5)$
Now, my question is: why he used $o(x^5)$? As far as I know, when $x\to0$, you have to consider the "$o$" with the highest power, because there is a $o(x^7)$ and an $o(x^8)$
Thank you!
|
The $o(x^7)$ and $o(x^8)$ are "absorbed" into the $o(x^5)$.
Suppose for example that $f(x)=o(x^7)$. Formally, what that means is that
$$\lim_{x\to 0} \frac{f(x)}{x^7}=0.$$
We show that $f(x)=o(x^5)$. Suppose that $f(x)=o(x^7)$. Then
$$\lim_{x\to 0}\frac{f(x)}{x^5}=\lim_{x\to 0}x^2\frac{f(x)}{x^7}=(0)(0)=0.$$
Informally, if $f(x)$ goes to $0$ faster than $x^7$, then it certainly goes to $0$ faster than $x^5$.
|
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|
Find a matrix $B$ such that $B^2=A$ $A= \begin{pmatrix} 1 & 2+3i \\ 2-3i & -1 \end{pmatrix}$
What is matrix $B$ such that $B^2=A$?
Its eigenvalues are $\sqrt{14}, -\sqrt{14}$
and I tried to use formula $B=U\sqrt{\lambda}U^{H}$ where $U$ is unitary matrix.
But then $\sqrt{ }$ of $-\sqrt{14}$ is not possible.
How can I solve this?
|
The characteristic polynomial of $A$ is $\lambda^2 - 14$, so $A^2 - 14 I = 0$.
Thus $(a A + bI)^2 = (14 a^2 + b^2) I + 2 a b A$. To make this equal to $A$, we want
$2ab = 1$ and $14 a^2 + b^2 = 0$. Thus $b = 1/(2a)$ and $14 a^2 = -1/(4a^2)$, or
$a^4 = -1/56$. Take $a$ to be any of the four fourth roots of $-1/56$, and you have a solution.
|
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|
If $\gcd(a,b)=1$, then $\gcd(a+b,a^2 -ab+b^2)=1$ or $3$. Hint: $a^2 -ab +b^2 = (a+b)^2 -3ab.$
I know we can say that there exists an $x,y$ such that $ax + by = 1$. So in this case,
$(a+b)x + ((a+b)^2 -3ab)y =1.$
I thought setting $x = (a+b)$ and $y = -1$ would help but that gives me $3ab =1.$ Any suggestions?
|
Hint $\,\ c\mid a\!+\!b,\overbrace{a^2\!-\!ab\!+\!b^2}^{\large f(b)}\Rightarrow\,c\mid 3a^2,3ab,3b^2\Rightarrow\,c\mid(3a^2,3ab,3b^2)=3(a,b)^2 = 3 $
by $\,\ {\rm mod}\ \ a\!+\!b\!:\ \ \color{#c00}{b\equiv -a}\,\Rightarrow\, f(\color{#c00}b)\equiv f(\color{#c00}{-a})\equiv 3a^2\equiv 3a(-b)\equiv 3(-b)(-b) $
|
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|
Prove that if $a^x=b^y=(ab)^{xy}$, then $x+y=1$ using logarithms
Prove that if $a^x=b^y=(ab)^{xy}$, then $x+y=1$.
How do I use logarithms to approach this problem?
|
Using logarithms:
Since $a^x = b^y$,
$$
\log a^x = \log b^y \quad \Rightarrow \quad x \log a = y \log b
\quad \Rightarrow \quad \log a = \frac{y}{x} \log b
$$
Then, since $b^y = (ab)^{xy}$,
$$
\log b^y = \log (ab)^{xy}
\quad \Rightarrow \quad
y \log b = xy \log (ab) = xy \left( \log a + \log b\right)
$$
Let's assume $y \neq 0$ (since if $y=0$, then we must also have $x=0$ and get the required conditions without having $x+y=1$ -- meaning the original question must have had some restriction such as $x, y \neq 0$). Cancel $y$ on both sides of the last equation:
$$
\log b = x\left(\log a + \log b\right)
= x\left( \frac{y}{x} \log b + \log b \right)
= y \log b + x \log b = (y + x)\log b
$$
Then as long as $b \neq 1$ (which is guaranteed if $x \neq 0$), we know that $\log b \neq 0$, hence we can cancel in the equation:
$$
\log b = (x+y)\log b
\quad \Rightarrow \quad
1 = x + y.
$$
|
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|
Prove that $\int_0^{+\infty} \frac{\ln x}{a^2+x^2} dx = \frac{\pi\ln a}{2a}$ Is that true $$\int_0^{+\infty} \cfrac{\ln x}{a^2+x^2} dx = \cfrac{\pi\ln a}{2a},$$ where $a>0$ ? And how to compute it?
|
Let
$$
\mathcal{I}(a)=\int_0^\infty\frac{\ln x}{x^2+a^2}\ dx.
$$
Using substitution $u=\dfrac{a^2}{x}\;\Rightarrow\;x=\dfrac{a^2}{u}\;\Rightarrow\;dx=-\dfrac{a^2}{u^2}\ du$ yields
\begin{align}
\mathcal{I}(a)&=\int_0^\infty\frac{\ln \left(\dfrac{a^2}{u}\right)}{\left(\dfrac{a^2}{u}\right)^2+a^2}\cdot \dfrac{a^2}{u^2}\ du\\
&=\int_0^\infty\frac{2\ln a-\ln u}{a^2+u^2}\ du\\
&=2\ln a\int_0^\infty\frac{1}{a^2+u^2}\ du-\int_0^\infty\frac{\ln u}{u^2+a^2}\ du\\
\color{red}{\mathcal{I}(a)}&=2\ln a\int_0^\infty\frac{1}{a^2+u^2}\ du-\color{red}{\mathcal{I}(a)}\\
\mathcal{I}(a)&=\ln a\int_0^\infty\frac{1}{a^2+u^2}\ du.
\end{align}
The last integral can easily be evaluated since it is a common integral. Using substitution $u=\tan\theta$, the integral turns out to be
\begin{align}
\mathcal{I}(a)&=\frac{\ln a}{a}\int_0^{\Large\frac\pi2} \ d\theta\\
&=\large\color{blue}{\frac{\pi\ln a}{2a}}.
\end{align}
|
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|
Finding the number of points on the straight line joining $(-4,11)$ and $(16,-1)$ Find the number of points on the straight line which joins $(-4,11)$ and $(16,-1)$ whose coordinates are positive integers.
a) $1$
b) $2$
c) $3$
d) $4$
|
The slope of this line is $$\frac{-1-11}{16-(-4)}=\frac{-12}{20}=-\frac35\;.$$ Thus, the equation of the line is $y-11=-\frac35(x-(-4))=-\frac35x-\frac{12}5=-\frac15(3x+12)$. Clearly $y$ is an integer precisely when $3x+12$ is a multiple of $5$, which is the case precisely when $x+4$ is a multiple of $5$. These values of $x$ occur $5$ units apart, so between $-4$ and $16$ they are $-4,1,6,11$, and $16$, and the corresponding $y$-coordinates are $11,8,5,2$, and $-1$.
You don’t actually need the equation of the line. The slope of $-\frac35$ means that every $1$-unit increase in $x$ causes a decrease in $y$ of $\frac35$. Thus, it takes a $5$-unit increase in $x$ to decrease $y$ by an integer (namely, $3$), and the values of $x$ between $-4$ and $16$ that give integral values of $y$ must be $-4,1,6,11$, and $16$. The corresponding $y$-coordinates are $11,8,5,2$, and $-1$.
By either approach we see that the line segment contains exactly $3$ points whose coordinates are both positive integers.
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|
Help with Inequality Given that $x, y, z$ are nonnegative real numbers such that :
$$x^2 + y^2 + z^2 + xyz = 4$$
Prove that
$0 ≤ xy + yz + zx − xyz ≤ 2$
|
Let $x=\frac{2a}{\sqrt{(a+b)(a+c)}}$ and $y=\frac{2b}{\sqrt{(a+b)(b+c)}}$, where $a$, $b$ and $c$ be positives.
Hence, $z=\frac{2c}{\sqrt{(a+c)(b+c)}}$ and the left inequality it's
$$4\sum_{cyc}\frac{ab}{(a+b)\sqrt{(a+c)(b+c)}}\geq\frac{8abc}{(a+b)(a+c)(b+c)}$$ or
$$\sum_{cyc}ab\sqrt{(a+c)(b+c)}\geq2abc,$$
which is true because
$$\sum_{cyc}ab\sqrt{(a+c)(b+c)}\geq\sum_{cyc}abc\geq2abc.$$
The right inequality.
We need to prove that
$$4\sum_{cyc}\frac{ab}{(a+b)\sqrt{(a+c)(b+c)}}-\frac{8abc}{(a+b)(a+c)(b+c)}\leq2$$ or
$$2\sum_{cyc}ab\sqrt{(a+c)(b+c)}\leq(a+b)(a+c)(b+c)+4abc,$$
which is AM-GM:
$$2\sum_{cyc}ab\sqrt{(a+c)(b+c)}\leq\sum_{cyc}ab(a+c+b+c)=$$
$$=\sum_{cyc}(a^2b+a^2c+2abc)=(a+b)(a+c)(b+c)+4abc.$$
Done!
|
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|
limits calculus I am having trouble understanding part of the solution to this simple problem.
$\lim_{x \to 2} (x^2 + 3x) = 10$
Solution:
Let $\epsilon > 0$
$| x - 2 | < \delta$ and $| x^2 +3x -10 | < \epsilon$
since $x^2 +3x -10 = (x - 2)^2 + 7x -14 = (x - 2)^2 + 7x -14 = ( x -2 )^2 +7(x-2)$
$|(x-2)^2 +7(x-2)| \leq |(x-2)|^2 +7|(x-2)|$
$\delta^2 + 7\delta < \epsilon$
let $\delta$ be the minimum of $1$ and $\epsilon/8$, $\delta^2 \leq \delta$.
then $8\delta < \epsilon$
$\delta < \epsilon/8$.
My Question:
I worked my way through the question down to $\delta^2 + 7\delta < \epsilon$
I then got confused by the end of this statement
Let $\delta$ be the minimum of $1$ and $\epsilon/8$, $\delta^2 \leq \delta$.
and in particular $\delta^2 \leq \delta$. I see how this allows me to prove the limit but I cannot make sense out of $\delta^2 \leq \delta$.
Could anyone explain this to me?
|
We have been challenged with an $\epsilon$, perhaps $\epsilon=1/1000$. We want to come up with a $\delta$ such that if $|x-2|\lt \delta$, then for sure $|x^2+3x-10|\lt \epsilon$.
Suppose that after some calculation, we announce that $\delta=\epsilon/8$ does the job. Then triumphantly the challenger could say that she had $\epsilon=47$ in mind, and that in that case taking $\delta=47/8$ is insufficient. Of course, that is not playing fair. But we might as well come up with a $\delta=\delta(\epsilon)$ that always works.
Some algebra shows that $x^2+3x-10=(x-2)^2+7(x-2)$. So we want to make sure that $|(x-2)^2+7(x-2)|\lt \epsilon$. Note that
$$|(x-2)^2+7(x-2)|\le (x-2)^2+7|x-2|.$$
We want to make the right-hand side "small," by choosing $x$ appropriately close to $2$. Suppose we had been given a ridiculous $\epsilon$, like $47$. If $x$ is within $47$ of $2$, the number $(x-2)^2$ could be very large. So the first task is to make sure $\delta$ is small enough not to allow $(x-2)^2$ to be large.
So we say first of all, let $\delta\le 1$. Then if $|x-2|\lt \delta$, it follows that $(x-2)^2\lt \delta$. For $(x-2)^2=(x-2)(x-2)$. The "first" $x-2$ has absolute value $\le 1$, and the second has absolute value $|x-2|$, so the product has absolute value $\le |x-2|$.
It follows that as long as $\delta\le 1$, we have $(x-2)^2+7|x-2|\le 8|x-2|$. To make sure this is $\lt \epsilon$, it is enough to make $\delta=\frac{1}{8}\epsilon$.
However, in deriving our simplified inequality, we assumed that $\delta\le 1$. So we know that everything will work if $\delta=\min(1,\epsilon/8)$.
It might have been better to observe that
$$(x-2)^2+7|x-2|=|x-2|\left(|x-2|+7\right).$$
Then it is clear that if $|x-2|\le 1$, then $(x-2)^2+7|x-2|\le 8|x-2|$.
|
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|
If $m,n\in\mathbb{N}$ and $f(m) \mid f(n)$ for the $f$ given below, must $m=n$? If for two natural number $m$ and $n$, $(2^{2m+1}-1)2^{4m-2}(2^{2m+1}+2^{m+1}+1)\mid(2^{2n+1}-1)2^{4n-2}(2^{2n+1}+2^{n+1}+1)$, then $m=n$?
|
False as stated. The divisibility also holds for example, when $n=1$ and $m=13$.
When $m=1$, the left hand side is
$$
(2^3-1)2^2(2^3+2^2+1)=7\cdot4\cdot13.
$$
When $n=13$ we obviously have that
$$
(2^3-1)\mid(2^{27}-1)\qquad\text{and}\qquad2^2\mid2^{4\cdot13-2}.
$$
The last factor is a bit trickier, but from Little Fermat we see that modulo $13$ we have the congruences $2^{13+1}\equiv2^{1+1}$ and $2^{2\cdot13+1}\equiv2^{2\cdot1+1}$, so also
$$
2^{27}+2^{14}+1\equiv2^3+2^2+1\equiv0\pmod{13}.
$$
It seems to me that is should be easy to find several other examples of `factor wise' divisibility.
|
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|
Help me solve this olympiad challenge? Given:
$$p(x) = x^4 - 5773x^3 - 46464x^2 - 5773x + 46$$
What is the sum of all arctan of all the roots of $p(x)$?
|
Let $p_1,p_2,p_3,p_4$ be the roots of $p(x)=0$
So, we need $\arctan p_1+\arctan p_2+\arctan p_3+\arctan p_4$
Using Vieta's Formulae,
$\sum p_i=p_1+p_2+p_3+p_4=\frac{5773}1$
$\sum p_ip_j=p_1p_2+p_1p_3+p_1p_4+p_2p_3+p_2p_4+p_3p_4=\frac{46464}1$
$\sum p_ip_jp_k=p_1p_2p_3+p_1p_2p_4+p_1p_3p_4+p_2p_3p_4=\frac{5773}1$
$p_1p_2p_3p_4=\frac{46}1$
We know, $$\tan(A+B+C+D)$$
$$=\frac{\tan A+\tan B+\tan C+\tan D -(\tan A\tan B\tan C+\tan A\tan B\tan D+\tan A\tan C\tan D+\tan B\tan C\tan D)}{1- (\tan A\tan B+\tan A\tan C+\tan A\tan D+\tan B\tan C+\tan B\tan D+\tan C\tan D)+ \tan A\tan B\tan C\tan D}$$
If we put $p_1=\tan A,p_2=\tan B$ etc.,
$$\tan(A+B+C+D)=\frac{p_1+p_2+p_3+p_4 -(p_1p_2p_3+p_1p_2p_4+p_1p_3p_4+p_2p_3p_4)}{1- (p_1p_2+p_1p_3+p_1p_4+p_2p_3+p_2p_4+p_3p_4)+ p_1p_2p_3p_4}$$
$$=\frac{5773-5773}{1-46464+46}=0$$
So, $\tan(A+B+C+D)=0\implies A+B+C+D=\arctan (0)$
$\implies \arctan p_1+\arctan p_2+\arctan p_3+\arctan p_4=\arctan (0)$
The general value of $\arctan (0)$ is $n\pi$ where $n$ is any integer, the special value being $0$(putting $n=0$)
|
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|
A relationship between matrices, bernoulli polynomials, and binomial coefficients We define the following polynomials, for $n≥0$:
$$P_n(x)=(x+1)^{n+1}-x^{n+1}=\sum_{k=0}^{n}{\binom{n+1}{k}x^k}$$
For $n=0,1,2,3$ this gives us,
$$P_0(x)=1\enspace P_1(x)=2x+1\enspace P_2(x)=3x^2+3x+1\enspace P_3(x)=4x^3+6x^2+4x+1$$
We then define the set $P_{(3)}=\{P_0,P_1,P_2,P_3\}$. It can be easily shown that this set is a basis over the vector space of polynomials of degree $3$ and lower. We take $3$ for the sake of brevity.
Taking the coefficients of these polynomials and turning them into column vectors, we can construct the matrix (coefficients from the lowest term to the highest term)
$$\large{M_{P_{(3)}}}=\begin{pmatrix}
1 & 1 & 1 & 1 \\
0 & 2 & 3 & 4 \\
0 & 0 & 3 & 6 \\
0 & 0 & 0 & 4
\end{pmatrix}$$
We'll call this matrix the pascal in the context of this post, and the above polynomials as pascal polynomials.
The inverse of this matrix is the matrix,
$$M_{P_{(3)}}^{-1}=\begin{pmatrix}
1 & -\frac{1}{2} & \frac{1}{6} & 0 \\
0 & \frac{1}{2} & -\frac{1}{2} & \frac{1}{4} \\
0 & 0 & \frac{1}{3} & -\frac{1}{2} \\
0 & 0 & 0 & \frac{1}{4}
\end{pmatrix}$$
We'll factor this matrix into two matrices as follows:
$$M_{P_{(3)}}^{-1}=\begin{pmatrix}
1 & 0 & 0 & 0 \\
0 & \frac{1}{2} & 0 & 0 \\
0 & 0 & \frac{1}{3} & 0 \\
0 & 0 & 0 & \frac{1}{4}
\end{pmatrix}×\begin{pmatrix}
1 & -\frac{1}{2} & \frac{1}{6} & 0 \\
0 & 1 & -1 & \frac{1}{2} \\
0 & 0 & 1 & -\frac{3}{2} \\
0 & 0 & 0 & 1
\end{pmatrix}$$
We can see the Bernoulli numbers in the first row of the matrix. Every column is a coefficient vector of a Bernoulli polynomial
The following are extended versions of these matrices:
*
*Pascal Matrix
*Inverse Pascal Matrix
*Last, Bernoulli Matrix
Are there accepted names for these matrices and polynomials? What is the meaning of these relationships?
In particular, is there some treatment of using these matrices as change of basis transformations between representations of polynomials? E.g. from a linear combination of pascal polynomials to a linear combination of monomial terms.
|
Here's another way to look at this. The Bernoulli polynomials can be defined by the property
$$\int_x^{x+1} B_n(u) \, du = x^n.$$
So if we let $T$ be the operator from the set of polynomials to itself given by $(Tf)(x) = \int_x^{x+1} f(u) \, du$, then we have $(TB_n)(x) = x^n$. The operator $T$ sends $x^n$ to $$\int_x^{x+1} u^n \, du = \frac{1}{n+1}\left((x+1)^{n+1} - x^{n+1}\right) = \sum_{k=0}^{n} \frac{1}{n+1} {n+1 \choose k}x^k.$$
Writing $T$ as an infinite matrix with respect to the basis $1,x,x^2, \ldots$, gives
$$T =
\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rrrr}
1 & \frac{1}{2} & \frac{1}{3} & \frac{1}{4} & \ldots\\
0 & 1 & 1 & 1 & \ldots\\
0 & 0 & 1 & \frac{3}{2} & \ldots\\
0 & 0 & 0 & 1& \ldots\\
\vdots & \vdots & \vdots & \vdots & \ddots
\end{array}\right).
$$
We can approximate the inverse by taking the inverse of this truncation, giving the matrix form of $T^{-1}$:
$$T^{-1} = \newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rrrr}
1 & -\frac{1}{2} & \frac{1}{6} & 0 & \ldots\\
0 & 1 & -1 & \frac{1}{2} & \ldots\\
0 & 0 & 1 & -\frac{3}{2} & \ldots\\
0 & 0 & 0 & 1& \ldots\\
\vdots & \vdots & \vdots & \vdots & \ddots
\end{array}\right)$$
This operator sends $x^n$ to $B_n(x)$, so the columns are the coefficients of Bernoulli polynomials.
To see how this fits with your equations, note that we may factor the first matrix $T$ to remove the fraction $\frac{1}{n+1}$ in the formula for its coefficients, since multiplying by a diagonal matrix on the left scales the columns of a matrix by the diagonal entries. This gives $$ T = \newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rrrr}
1 & 1 & 1 & 1 & \ldots\\
0 & 2 & 3 & 4 & \ldots\\
0 & 0 & 3 & 6 & \ldots\\
0 & 0 & 0 & 4 & \ldots\\
\vdots & \vdots & \vdots & \vdots & \ddots
\end{array}\right)\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rrrr}
1 & 0 & 0 & 0 & \ldots\\
0 & 1/2 & 0 & 0 & \ldots\\
0 & 0 & 1/3 & 0 & \ldots\\
0 & 0 & 0 & 1/4 & \ldots\\
\vdots & \vdots & \vdots & \vdots & \ddots
\end{array}\right)$$
The matrix on the left is your $M_P$ (infinitely extended). Calling the diagonal matrix on the right $D$, we have $T = M_pD$, or $DT^{-1} = M_P^{-1}$ which is your last equation.
To my mind, this is the most natural way to compute Bernoulli polynomials. A few years ago I was playing around with this idea before I knew what a Bernoulli polynomial was. I was slightly disappointed, although not too surprised, to hear that someone else had discovered this first. I was beaten by some three hundred years, no less. It ended up as part of my undergraduate thesis. :)
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|
Find the acute angle $x$ for $\tan x = \tan(x+10^\circ)\tan(x+20^\circ)\tan(x+30^\circ)$. How to solve the following equation?
$$\tan x= \tan(x+10^\circ)\tan(x+20^\circ)\tan(x+30^\circ)$$
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$$\frac{\sin x\cos(x+10)}{\cos x\sin(x+10)}=\frac{\sin(x+20)\sin(x+30)}{\cos(x+20)\cos(x+30)}$$
Applying componendo and dividendo,
$$\frac{\cos x\sin(x+10)+\sin x\cos(x+10)}{\cos x\sin(x+10)-\sin x\cos(x+10)}
=\frac{\cos(x+20)\cos(x+30)+\sin(x+20)\cos(x+30)}{\cos(x+20)\cos(x+30)-\sin(x+20)\cos(x+30)}$$
$$\frac{\sin(2x+10)}{\sin10}=\frac{\cos 10}{\cos(2x+50)}$$ applying $\sin(A\pm B)$ and $\cos(A\pm B)$
So, $$\sin(2x+10)\cos(2x+50)=\sin10 \cos 10$$
$$\sin(4x+60)-\sin40=\sin20$$ (applying $2\sin A\cos A=\sin2A$ and $2\sin A\cos B==\sin(A+B)+\sin(A-B)$)
$$\sin(4x+60)=\sin40+\sin20=2\sin\frac{20+40}2\cos\frac{40-20}2=\cos10$$
(applying $\sin 2C+\sin 2D =2\sin(C+D)\cos(C-D)$ and $\sin(90\pm A)=\cos A$)
Now, $\sin(4x+60)=\cos\{90-(4x+60)\}=\cos(30-4x)=\cos(4x-30)$ as $\cos(-A)=\cos A$
So, $\cos(4x-30)=\cos10$
or, $4x-30=360n\pm10$ where $n$ is any integer.
Find suitable $n$ to keep $x\in[0,90]$
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|
please solve a 2013 th derivative question? $ f(x) = 6x^7\sin^2(x^{1000}) e^{x^2} $
Find $ f^{(2013)}(0) $
A math forum friend suggest me to use big O symbol, however have no idea what that is, so how does that helping?
|
Note that,
$$ 6\,x^{7} \sin\left(x^{1000}\right)\sin\left(x^{1000}\right)e^{x^2} $$
$$ = 6\,x^{7} \left( x^{1000}-\frac{x^{3000}}{3!}+\dots \right)\left( x^{1000}-\frac{x^{3000}}{3!}+\dots \right)\left(1+\frac{x^2}{1!}+\frac{x^4}{4!}+\dots\right) $$
$$ = 6x^7x^{2000}\left( 1-\frac{x^{2000}}{3!} +\dots\right)^2\left(1+\frac{x^2}{1!}+\frac{x^4}{2!}+\dots\right) $$
$$ = 6x^{2007}\left(1+\frac{x^2}{1!}+\frac{x^4}{2!}+\frac{x^6}{3!}+\dots\right)\left( 1-\frac{x^{2000}}{3!} +\dots\right)^2 $$
Now, it is clear that the coefficient of $x^{2013}$ is $1$, which implies that
$$ \frac{f^{(2013)}(0)}{(2013)!} = 1 \implies f^{(2013)}(0)=(2013)!. $$
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|
$\lim_{x \to 0}\frac{|x|\sin \left(\frac{1}{3 \sqrt{x}}\right)}{\sqrt{x^4+4x^2+7}}$
Find $$\lim_{x \to 0}\frac{|x|\sin \left(\frac{1}{3 \sqrt{x}}\right)}{\sqrt{x^4+4x^2+7}}$$
I know that $\lim_{x \to 0} \frac{\sin x}{x}=1$ But here $\sin \left(\frac{1}{3 \sqrt{x}}\right)$ is given when $x \to 0$. Need help.
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If we let $1/x=y$
$$\lim_{y \to \infty}\frac{\sin \left(\frac{\sqrt{y}}{3 }\right)}{|y|\sqrt{7}}=0$$
|
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|
Check convergence - look complicated $$\sum_{n=2}^\infty(\cos((n^{3}+\sqrt{n}+7)^{\frac{1}{3}})-\cos((n^{3}-2\sqrt{n}+3)^{\frac{1}{3}}))$$
Check convergence.
Please verify my answer below.
$$\sum_{n=2}^\infty(\cos((n^{3}+\sqrt{n}+7)^{\frac{1}{3}})-\cos((n^{3}-2\sqrt{n}+3)^{\frac{1}{3}}))=$$
$$=\sum_{n=2}^\infty-2\sin\left(\frac{\sqrt[3]{n^{3}+\sqrt{n}+7}-\sqrt[3]{n^{3}-2\sqrt{n}+3}}{2}\right)\cdot\sin\left(\frac{\sqrt[3]{n^{3}+\sqrt{n}+7}+\sqrt[3]{n^{3}-2\sqrt{n}+3}}{2}\right)$$
Checking arbirary convergence
$\sin\left(\frac{\sqrt[3]{n^{3}+\sqrt{n}+7}+\sqrt[3]{n^{3}-2\sqrt{n}+3}}{2}\right)\in\langle-1,1\rangle$
$=\sum_{n=2}^\infty|-2\sin\left(\frac{\sqrt[3]{n^{3}+\sqrt{n}+7}-\sqrt[3]{n^{3}-2\sqrt{n}+3}}{2}\right)\cdot\sin\left(\frac{\sqrt[3]{n^{3}+\sqrt{n}+7}+\sqrt[3]{n^{3}-2\sqrt{n}+3}}{2}\right)|\leq2\sum_{n=1}^\infty\left|\sin\left(\frac{\sqrt[3]{n^{3}+\sqrt{n}+7}-\sqrt[3]{n^{3}-2\sqrt{n}+3}}{2}\right)\right|$
[1]
Checking convergence of
[2] $\frac{1}{2}\sum_{n=2}^\infty\sqrt[3]{n^{3}+\sqrt{n}+7}-\sqrt[3]{n^{3}-2\sqrt{n}+3}=\frac{1}{2}\sum_{n=2}^\infty\frac{4+3\sqrt{n}}{\sqrt[3]{(n^{3}+\sqrt{n}+7)^{2}}+\sqrt[3]{n^{3}+\sqrt{n}+7} \sqrt[3]{n^{3}-2\sqrt{n}+3}+\sqrt[3]{(n^{3}-2\sqrt{n}+3)^{2}}}$
$$\frac{1}{2}\sum_{n=2}^\infty\frac{4+3\sqrt{n}}{\sqrt[3]{(n^{3}+\sqrt{n}+7)^{2}}+\sqrt[3]{n^{3}+\sqrt{n}+7}\cdot\sqrt[3]{n^{3}-2\sqrt{n}+3}+\sqrt[3]{(n^{3}-2\sqrt{n}+3)^{2}}}<\frac{4\sqrt{n}}{n^{2}}=\frac{4}{n^{1.5}}$$
From the comparison test, it converge
Checking product test
$$\frac{\sin\left(\frac{\sqrt[3]{n^{3}+\sqrt{n}+7}-\sqrt[3]{n^{3}-2\sqrt{n}+3}}{2}\right)}{\frac{\sqrt[3]{n^{3}+\sqrt{n}+7}-\sqrt[3]{n^{3}-2\sqrt{n}+3}}{2}}\rightarrow1$$
Limit exist and is in $\mathbb{R}$ so from [2], [1] converge. If our series arbitrary converge, it also converge.
|
Let me suggest some simplifications. The $n$th term is the difference of two cosines and the function cosine is $1$-Lipschitz hence each term is at most the difference of the two cube roots. Furthermore, the derivative of the cube root function is decreasing hence for every $a\gt b\gt0$,
$$
a^{1/3}-b^{1/3}=\int_b^a\frac{\mathrm dx}{3x^{2/3}}\leqslant\frac{a-b}{3b^{2/3}}.
$$
Applying this to $a_n=n^3+\sqrt{n}+7$ and $b_n=n^3-2\sqrt{n}+3$ yields
$$
|\cos(a_n^{1/3})-\cos(b_n^{1/3})|\leqslant a_n^{1/3}-b_n^{1/3}\leqslant\frac{a_n-b_n}{3b_n^{2/3}}.
$$
Now, $a_n-b_n\sim3\sqrt{n}$ and $b_n\sim n^3$ hence the RHS is equivalent to
$$
\frac{3\sqrt{n}}{3(n^3)^{2/3}}=\frac1{n\sqrt{n}}.
$$
The series $\sum\limits_n\frac1{n\sqrt{n}}$ converges hence the series $\sum\limits_n\left[\cos(a_n^{1/3})-\cos(b_n^{1/3})\right]$ converges absolutely.
|
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|
If $r = \min \left\{ a , \frac{b}{ a+ a^2 b^3} \right\}$ , find $ r_{\max}$ Studying differential equations I came cross through this:
Let $ \displaystyle{ r = \min \left\{ a , \frac{b}{ a+ a^2 b^3} \right\} } $, where $ a,b >0$. Find $ r_{ \max} $.
Here is what I did:
Fix $ a> 0$ and define $ \displaystyle{ g(b) = \frac{b}{ a+ a^2 b^3} ,\quad b>0 }$. Then we have that $ \displaystyle{ g'(b) = \frac{ a(1-2ab^3)}{ (a+ a^2 b^3)^2} }$.
Thus, $g$ has maximum at $ \displaystyle{b_0= \frac{1}{ \sqrt[3]{2a} }}$ which gives $ \displaystyle{ g(b) \leq g\left( \frac{1}{ \sqrt[3]{2a}} \right) = \frac{2}{ 3a \sqrt[3]{2a} }}$
So, $ \displaystyle{ r = \min \{ a , \frac{2}{ 3a \sqrt[3]{2a} } \}}$. I think that to find $a$ which gives $ r_{\max}$ I have to solve the equation $ \displaystyle{ a=\frac{2}{ 3a \sqrt[3]{2a}} }$.
Is this correct and why....?
Any ideas?
Thank's in advance!
edit: I edit the title.
|
It looks correct to me. The reason that is correct is because you're finding when the two values are equal. If they aren't equal, then since the real numbers are totally ordered, we know that one will be less than the other.
Solving for $a$, we have $$a^{7/3}=\frac{2^{2/3}}{3}.$$ Taking the $3/7$ root of each side yields $$a=\left(\frac{2^{2/3}}{3}\right)^{3/7}=\left(\frac{4}{27}\right)^{1/7}.$$
|
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|
Use Riemann Sums to Find Area bounded by Curve How can I find the area bound by $\;x=0,\, x=1,\;$ the $\;x$-axis ($y = 0$) and $\;y=x^2+2x\;$ using Riemann sums?
I want to use the right-hand sum. Haven't really found any good resources online to explain the estimation of areas bounded by curves, hoping anyone here can help?
By the way, I would like there to be 100 intervals.
|
Right Riemann Sums place the right corner of the rectangles on the curve.
Right Riemann Sums are an overestimation of area because of all the extra space that is not under the curve that is still calculated in the area because it is inside the rectangles.
For your problem, we have:
$f(x) = x^{2} + 2x$
With: $a = 0, b = 1$ and $100$ rectangles, we have $\frac{b-a}{n} = \frac{1-0}{100} = \frac{1}{100}$
Our formula now becomes:
$\frac{b-a}{n} [f(0) + f(\frac{1}{100}) + f(\frac{2}{100})+ \cdots + f(\frac{100}{100})]$
$\bullet$ For Left Riemann, plug in all numbers except last.
$\bullet$ For Right Riemann, plug in all numbers except first.
$\bullet$ For Midpoint Riemann, average the numbers and plug those in.
So, using Right Riemann, we have:
$f(\frac{1}{100}) = \frac{1^{2}}{100^{2}} + \frac{2}{100} = \frac{201}{10000}$
$f(\frac{2}{100}) = \frac{2^2}{100^{2}} + \frac{4}{100} = \frac{101}{2500}$
$\cdots$
$f(\frac{100}{100}) = \frac{100^2}{100^{2}} + \frac{200}{100} = 3$
You'll get this set of values using WA.
Now, using the formula above, we have:
$\frac{1}{100}[f(\frac{1}{100}) + f(\frac{2}{100})+ \cdots + f(\frac{100}{100})] = (\frac{1}{100})(\frac{26967}{200}) = 1.34835$
Comparing that with the actual $\int_0^1 (x^{2} + 2x) dx = \frac{4}{3} = 1.33333$, we see, as expected, that we got an overestimate.
You can compare these two with the nice pointer by amWhy at Wolfram Math World.
You should repeat this for left and midpoint and compare so you understand!
Regards
|
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|
Number of strings of length n formed by $\{0,1,2\}$ such that $1$ and $2$ do not occur successively. What is the number of ways of forming a string of length $n$ from the set $\{0,1,2\}$ such that $1$ and $2$ do not occur successively.
|
There is an interesting related problem that has a solution using inclusion-exclusion. This is the restriction of the inadmissible words to words where $1$ and $2$ occur in order. This requires that we count the number of $n$-strings having $k$ instances of $1$ followed by $2$ ocurring successively with $1\le k \le \lfloor n/2 \rfloor$. This count can be done by considering the length of the possible gaps between these pairs, which are given by the generating function
$$\left(\frac{1}{1-z}\right)^{k+1}.$$
The total of these gaps must equal $n-2k$, giving
$$[z^{n-2k}] \left(\frac{1}{1-z}\right)^{k+1}
= {n-2k+k\choose n-2k} = {n-2k+k\choose k} = {n-k\choose k}.$$
Therefore by inclusion-exclusion the number of strings of length $n$ containing at least one occurrence of $1$ and $2$ in order is given by
$$\sum_{k=1}^{\lfloor n/2 \rfloor} (-1)^{k+1} {n-k\choose k} 3^{n-2k}$$
and the count where $1$ and $2$ do not occur successively is
$$3^n + \sum_{k=1}^{\lfloor n/2 \rfloor} (-1)^k {n-k\choose k} 3^{n-2k}
= \sum_{k=0}^{\lfloor n/2 \rfloor} (-1)^k {n-k\choose k} 3^{n-2k}.$$
If we compute this for small values we obtain the sequence
$$1, 3, 8, 21, 55, 144, 377, 987, 2584, 6765, 17711, 46368, 121393, 317811, 832040,\ldots$$
which is A001906 from the OEIS.
Call this sequence $\{a_n\}$. The ordinary generating function of this sequence is given by
$$f(z) =
\sum_{n\ge 0} \sum_{k=0}^{\lfloor n/2 \rfloor} (-1)^k {n-k\choose k} 3^{n-2k} z^n.$$
Rearranging the sum we find that
$$f(z) = \sum_{k\ge 0} \sum_{\lfloor n/2 \rfloor\ge k} (-1)^k {n-k\choose k} 3^{n-2k} z^n = \sum_{k\ge 0} (-1)^k \times 3^{-2k}
\sum_{\lfloor n/2 \rfloor\ge k}{n-k\choose k} 3^n z^n .$$
The inner sum can be manipulated as follows:
$$\sum_{n\ge 2k} {n-k\choose k} 3^n z^n
= \sum_{n\ge 0} {n+2k-k\choose k} 3^{n+2k} z^{n+2k}
= 3^{2k} z^{2k} \sum_{n\ge 0} {n+k\choose k} 3^n z^n\\
= 3^{2k} z^{2k} \sum_{n\ge 0} {n+k\choose n} 3^n z^n
= 3^{2k} z^{2k} \left(\frac{1}{1-3z}\right)^{k+1}.$$
This finally gives for the generating function that
$$f(z) = \sum_{k\ge 0} (-1)^k z^{2k} \left(\frac{1}{1-3z}\right)^{k+1}
= \frac{1}{1-3z} \sum_{k\ge 0} (-1)^k z^{2k} \left(\frac{1}{1-3z}\right)^k
= \frac{1}{1-3z} \frac{1}{1+z^2/(1-3z)} = \frac{1}{1-3z+z^2}.$$
The inverses of the poles of $f(z)$ are given by
$$\rho_{0,1} = \frac{3\pm\sqrt{5}}{2}$$
leading to the closed form
$$a_n = \frac{1}{\sqrt{5}}
\left(
\left(\frac{3+\sqrt{5}}{2}\right)^{n+1}-\left(\frac{3-\sqrt{5}}{2}\right)^{n+1}
\right).$$
|
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|
Inequality: $(a^3+a+1)(b^3+b+1)(c^3+c+1) \leq 27$
Let be $a,b,c \geq 0$ such that: $a^2+b^2+c^2=3$.
Prove that:
$$(a^3+a+1)(b^3+b+1)(c^3+c+1) \leq 27.$$
I try to apply $GM \leq AM$ for $x=a^3+a+1$, $y=b^3+b+1,z=c^3+c+1$ and
$$\displaystyle \sqrt[3]{xyz} \leq \frac{x+y+z}{3}$$ but still nothing.
Thanks :-)
|
Let $u:=a^2, v:=b^2, w:=c^2$, we have $u+v+w=3$.
Consider the function
$$f(x)=\ln (1+x^{\frac{1}{2}}+x^{\frac{3}{2}}),\ 0<x\leq 3$$
it's easy to compute that $f''(x)<0$.
by Jensen's inequality, we have
$$\sum\ln (1+u^{\frac{1}{2}}+u^{\frac{3}{2}})\leq3f(\dfrac{\sum u}{3})=3\ln 3$$
that is
$$\prod(a^3+a+1) \leq 27$$
When the max occurs, we have $u=v=w\Rightarrow a=b=c$
Q.E.D.
By the way, I draw a graph of $f''(x)$ on $(0,3]$ by mathematica to show it more directly...
|
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|
Prove $6 \nmid [\left( \sqrt[3]{28} - 3 \right)^{-n}]$ Prove that: $$6 \not\left|\ \left\lfloor\frac 1 {(\sqrt[3]{28} - 3)^{n}}\right\rfloor \ (n \in Z^+)\right.$$
($\lfloor x\rfloor$ = largest integer not exceeding $x$)
I am very bad as English and number theory, please help me
|
If we set $\eta=\sqrt[3]{28}$ and $\omega=\dfrac1{\eta-3}=\dfrac{\eta^3-27}{\eta-3}=\eta^2+3\eta+9$, then, working $\bmod\ \eta^3-28$:
$$
\begin{align}
\omega^0&=1\\
\omega^1&=9+3\eta+\eta^2\\
\omega^2&=249+82\eta+27\eta^2\\
\omega^3&=6805+2241\eta+738\eta^2
\end{align}\tag{1}
$$
Solving the linear equations involved yields
$$
\omega^3-27\omega^2-9\omega-1=0\tag{2}
$$
Looking at the critical points of $x^3-27x^2-9x-1$, we see that it has one real root and two complex conjugate roots. The real root is $\omega\stackrel.=27.3306395$, and since the product of all the roots is $1$, the absolute value of the two conjugate roots is less than $\frac15$.
Let $\omega_0=\omega$ and $\omega_1$ and $\omega_2=\overline{\omega}_1$ be the roots of $x^3-27x^2-9x-1=0$. Symmetric functions and the coefficients of $(2)$ yield
$$
\begin{align}
a_0=\omega_0^0+\omega_1^0+\omega_2^0&=3\\
a_1=\omega_0^1+\omega_1^1+\omega_2^1&=27\\
a_2=\omega_0^2+\omega_1^2+\omega_2^2&=747\quad=27^2-2(-9)
\end{align}\tag{3}
$$
and, because each $\omega_k$ satisfies $(2)$,
$$
a_n=27a_{n-1}+9a_{n-2}+a_{n-3}\tag{4}
$$
Because $|\omega_1|=|\omega_2|<\frac15$, $|\,a_n-\omega^n\,|\le\frac2{5^n}$. Also, $(3)$ and $(4)$ show that $a_n\equiv3\pmod{6}$.
Therefore, $\omega^0=1$ and for $n\ge1$,
$$
\lfloor\omega^n\rfloor\in\{2,3\}\pmod{6}\tag{5}
$$
|
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|
Improper integral with log and absolute value: $\int^{\infty}_{0} \frac{\log |\tan x|}{1+x^{2}} \, dx$ How do you show that $$ \int^{\infty}_{0} \frac{\log |\tan x|}{1+x^{2}} \, dx = \frac{\pi}{2} \log (\tanh 1)\, ?$$
I know that the integral converges since $\log |\tan x| = \frac{1}{2} \log(\tan^{2}x)$, and $\log (\tan^{2} x)$ behaves like $2 (-1)^{n} \log \left(x-\frac{n \pi}{2} \right)$ near $x= \frac{n \pi}{2}$. Thus the singularities at $x= \frac{n \pi}{2}$ are integrable.
|
How about:
$$
\int_0^\infty \frac{\log | \tan x |}{1+x^2} \mathrm{d} x = \frac{1}{2} \int_0^\infty \frac{\log \tan^2 x }{1+x^2} \mathrm{d} x = \frac{1}{4} \int_{-\infty}^\infty \frac{\log \tan^2 x }{1+x^2} \mathrm{d} x \tag{$\ast$}
$$
Writing the integral over $\mathbb{R}$ as the average of integrals over $\mathbb{R} + i \epsilon$ and $\mathbb{R} - i \epsilon$ and using $\vert \tan(z) \vert^2 = \frac{\sin^2(2 x) + \sinh^2(2y)}{(\cos(2x) + \cosh(2y))^2} \to_{y \to \pm \infty} 1$ we can complete the integration contours and apply the residue theorem:
$$ \begin{eqnarray}
\int_{-\infty}^\infty \frac{\log \tan^2 x }{1+x^2} \mathrm{d} x &=& \frac{1}{2} \int_{-\infty +i 0}^{\infty +i 0} \frac{\log \tan^2 x }{1+x^2} \mathrm{d} x + \frac{1}{2} \int_{-\infty-i 0}^{\infty - i 0} \frac{\log \tan^2 x }{1+x^2} \mathrm{d} x \\
&=& \frac{1}{2} \cdot 2 \pi i \operatorname{Res}_{x=i}\frac{\log \tan^2 x }{1+x^2} - \frac{1}{2} \cdot 2 \pi i \operatorname{Res}_{x=-i}\frac{\log \tan^2 x }{1+x^2} \\
&=& 2 \pi \log \tanh(1)
\end{eqnarray}
$$
Combining with eq. $(\ast)$:
$$
\int_0^\infty \frac{\log | \tan x |}{1+x^2} \mathrm{d} x = \frac{\pi}{2} \log \tanh(1)
$$
|
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|
Evaluate integral with trigonometric functions in denominator $$ \int \frac {1}{\sin x + \cos x} dx$$
How would I go about solving this?
|
The easiest way is to proceed as @AymanHourieh has suggested. Another approach is using Weierstrass substitution. Let $t = \tan(x/2)$. Note that
$$\sin(x) = \dfrac{2 \tan(x/2)}{1+\tan^2(x/2)} = \dfrac{2t}{1+t^2}; \,\,\,\,\, \cos(x) = \dfrac{1-\tan^2(x/2)}{1+\tan^2(x/2)} = \dfrac{1-t^2}{1+t^2}$$
$$dt = \dfrac{\sec^2(x/2) dx}2 \implies dx = \dfrac{2dt}{1+t^2}$$
Hence,
$$I = \int \dfrac{dx}{\sin(x) + \cos(x)} = \int \dfrac{2dt}{1-t^2 + 2t} = \int\dfrac{2dt}{2-(t-1)^2}$$
Now use partial fractions to write $\dfrac{2}{2-(t-1)^2}$ as
$$\dfrac2{2-(t-1)^2} = \dfrac1{\sqrt{2}(t+\sqrt{2}-1)} - \dfrac1{\sqrt{2}(t-\sqrt{2}-1)}$$
Now integrate this out by, recalling that
$$\int \dfrac{dt}{t+c} = \log(\vert t+c \vert) + \text{ constant}$$
|
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|
to compute $\int_{2}^{3}\sqrt{1+\frac{1}{x^{2}}} \: \: dx$ Help me please to compute: $\int_{2}^{3}\sqrt{1+\frac{1}{x^{2}}} \: \: dx$
Thanks a lot!
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Hint: let $x=\tan{\theta}$, $dx=\sec^2{\theta} d \theta$:
$$\begin{align} \int_{2}^{3} dx \: \sqrt{1+\frac{1}{x^{2}}} &= \int_{\arctan{2}}^{\arctan{3}} \frac{d \theta}{\sin{\theta} \cos^2{\theta}} \\ &= \int_{\arctan{2}}^{\arctan{3}} \frac{d \theta \, \sin{\theta}}{\sin^2{\theta} \cos^2{\theta}} \end{align} $$
Now let $y=\cos{\theta}$, $dy=-\sin{\theta} d \theta$:
$$\begin{align} \int_{2}^{3} dx \: \sqrt{1+\frac{1}{x^{2}}} &= \int_{\frac{1}{\sqrt{10}}}^{\frac{1}{\sqrt{5}}} \frac{dy}{y^2 (1-y^2)} \\ &= \int_{\frac{1}{\sqrt{10}}}^{\frac{1}{\sqrt{5}}} dy \: \left ( \frac{1}{y^2} + \frac{1}{1-y^2} \right )\\ &= \int_{\frac{1}{\sqrt{10}}}^{\frac{1}{\sqrt{5}}} dy \: \left [ \frac{1}{y^2} + \frac{1}{2} \left ( \frac{1}{1-y} + \frac{1}{1+y} \right ) \right ] \\ \end{align} $$
I think you can take it from here.
|
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|
Solving $\;5^{2x}-4\cdot 5^x=12$ I need to solve $\quad\displaystyle 5^{2x}-4\cdot 5^x=12$.
I've only gotten this far: $\quad \displaystyle 5^{2x}-20^x=12.$
I don't know what to do next.
Thanks in advance!
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Please note that $4\cdot5^x = 4(5^x)\neq 20^x$
If we let $y=5^x$, then $$(5^x)^2-4(5^x) =12 \;\implies \;y^2 - 4y = 12 \;\implies y^2 - 4y - 12 = (y-6)(y+2) = 0$$
$\implies \; y = 6\;$ or $\;y = -2$
So $y = 5^x = 6\; $ or $\; y = 5^x = -2$
Can you take it from here?
You can omit $\;y= 5^x = -2\;$ as a solution if we are constraining ourselves to real solutions...
So it suffices to solve for $x$ given $\;5^x = 6$.
|
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|
Diff eq. transformation polar coordinates I have $(x',y')=(x-y-x(x^2+y^2)+\frac{xy}{\sqrt{x^2+y^2}},x+y-y(x^2+y^2)-\frac{x^2}{\sqrt{x^2+y^2}} )$
Now I want to use polar coordinates $(x,y)=(r\cos(t),r\sin(t))$ to get $(r',t')=(r(1-r^2),2\sin(\frac{t}{2})^2)$
I do not see this relation. When I put $x=\cos t$, $y=\sin t$ into the system of differential equations, I only get $(r\cos(t)-r\sin(t)-r^3\cos(t)+r\cos(t)\sin(t),r\cos(t)+r\sin(t)-r^3\cos(t)-r\cos(t)^2)$.
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Using $x=r \cos{t}$, $y=r \sin{t}$:
$$x'=(\cos{t}) r' - r (\sin{t}) \, t'$$
$$y'=(\sin{t}) r' + r (\cos{t}) \, t'$$
So we get
$$\left ( \begin{array}\\ \cos{t} & -r \sin{t} \\ \sin{t} & r \cos{t} \end{array} \right ) \left ( \begin{array}\\ r' \\ t' \end{array} \right ) = \left ( \begin{array}\\ r \cos{t} - r \sin{t} - r^3 \cos{t} + \sin{t} \cos{t} \\ r \cos{t} + r \sin{t} - r^3 \sin{t} - \cos^2{t} \end{array} \right ) $$
Multiply both sides by the matrix inverse to get
$$\left ( \begin{array}\\ r' \\ t' \end{array} \right ) = \frac{1}{r} \left ( \begin{array}\\ r\cos{t} & r\sin{t} \\ - \sin{t} & \cos{t} \end{array} \right ) \left ( \begin{array}\\ r \cos{t} - r \sin{t} - r^3 \cos{t} + \sin{t} \cos{t} \\ r \cos{t} + r \sin{t} - r^3 \sin{t} - \cos^2{t} \end{array} \right ) $$
Just do out the multiplication. It is messy, but there is a lot of cancellation and we get
$$\left ( \begin{array}\\ r' \\ t' \end{array} \right ) = \left ( \begin{array}\\ r-r^3 \\ 1 - \cos{t} \end{array} \right ) $$
|
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|
First week in Linear Algebra, need some help on this simple problem If you have
$5x + 2y + z = 0$
$2x + y = 0$ and you're asked to solve using back-substitution how would you go about doing it?
Initially I thought just simply the following:
$x + \frac{2}{5} y + \frac{1}{5} z = 0$ (divide by $5$)
$2x + y = 0$
$\displaystyle x + \frac{2}{5} y + \frac{1}{5} z = 0$
$\displaystyle 2(x + \frac{2}{5} y + \frac{1}{5} z) + \frac{1}{5} y - \frac{2}{5} z = 0$ (sub method I was taught in class)
But after simplifying I realized I basically just made the problem worse because I ended with:
$\displaystyle x + \frac{2}{5} y + \frac{1}{5} z = 0$
$\displaystyle \frac{1}{5} y + \frac{2}{5} z = 0$
So I pretty much still have 3 unknowns. Any suggestions or hints?
|
Of course the accepted answer is completing one, but look at this one. Maybe it looks fine either. We have: $$
\left\{
\begin{array}{ll}
5x+2y+z=0 \\
2x+y=0 &
\end{array}
\right.$$
Now multiply the second equation by $-2$:
$$ \left\{
\begin{array}{l1}
5x+2y+z=0 \\
-4x-2y=0 &
\end{array}
\right.$$ Add two equations above to eliminate $y$: $$(5x-4x)+z=0$$ so $x=-z$ Now, put $x=-z$ into the first equation: $$5x+2y-x=0$$ so $4x+2y=0$ or $y=-2x$. This is what you see in another answer as: $(x,-2x,-x)$ which is a 3D line.
|
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|
How is this linear 2nd-order ODE solved? In this article, the authors present the inhomogeneous equation
$$\ddot{\phi}_2 + \phi_2 + g_2\phi_1^2 + \omega_1\ddot{\phi}_1 = 0,\tag{11}$$
where
$$ \phi_1 = p_1 \cos(\tau + \alpha), \tag{13}$$
followed by its solution
$$\phi_2 = p_2\cos(\tau + \alpha) + q_2\sin(\tau + \alpha) + \frac{g_2}{6}p_1^2[\cos(2\tau + 2\alpha) - 3] $$
$$+ \frac{\omega_1}{4}p_1[2\tau\sin(\tau + \alpha) + \cos(\tau + \alpha)]. \tag{14}$$
Check my solution .
|
Particular solution can be written in the form\begin{align}
\phi_2 = p_2 \cos(\tau + \alpha) + q_2 \sin(\tau + \alpha) + C \cos(2(\tau + \alpha)) + D\sin(2(\tau + \alpha)) + E.
\end{align}
Now after doing derivative, we get,
$$\dot{\phi_2}= -p_2 sin(\tau +\alpha)+q_2cos(\tau+ \alpha)-2Csin(2\tau+2\alpha)+2Dcos(2\tau+2\alpha)$$
$$\ddot{\phi_2}= -p_2 cos(\tau +\alpha)-q_2sin(\tau+ \alpha)-4Csin(2\tau+2\alpha)-4Dsin(2\tau+2\alpha)$$
putting these into equation (11), we get,
$$E-3Ccos(2\tau+2\alpha)-3Dsin(2\tau+2\alpha)+\frac{1}{2}g_2p_1^2+\frac{1}{2}g_2p_1^2cos(2\tau+2\alpha)-\omega_1p_1cos(\tau+\alpha)=0$$
$$-3Dsin(2\tau+2\alpha)+cos(2\tau+2\alpha)(-3C-\frac{1}{2}g_2p_1^2)+[E+\frac{1}{2}g_2p_1^2-\omega_1p_1cos(\tau+\alpha)]=0$$
Now, we get,
$$D=0$$, $$-3C-\frac{1}{2}g_2p_1^2=0$$ $$C=\frac{g_2p_1^2}{6}$$ then E becomes
$$E=\omega_1p_1cos(\tau + \alpha)- \frac{1}{2}g_2p_1^2$$
Therefore we get, the require solution,
\begin{align}
\phi_2 = p_2 \cos(\tau + \alpha) + q_2 \sin(\tau + \alpha) + \frac{g_2p_1^2}{6} \cos(2(\tau + \alpha)) + \omega_1p_1cos(\tau+\alpha)-\frac{1}{2}g_2p_1^2
\end{align}
\begin{align}
\phi_2 = p_2 \cos(\tau + \alpha) + q_2 \sin(\tau + \alpha) + \frac{g_2p_1^2}{6} [\cos(2(\tau + \alpha)) -3]+ \omega_1p_1cos(\tau+\alpha)
\end{align}
check the solution of the last term, whats wrong in my calculations?
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/292406",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
}
|
Is this function convex or non-convex? Let
$$f(a,b,c,d) = \frac{(a-b) \cdot (c-d)}{\sqrt{(a-b)^2+(c-d)^2}}$$
where $a,b,c,d$ are variables. Is this function convex or non-convex?
|
We have
$$\frac{\partial f}{\partial a} = \frac{c-d}{\sqrt{(a-b)^2+(c-d)^2}}-\frac{(a-b)^2(c-d)}{\left[(a-b)^2+(c-d)^2\right]^{3/2}}=\frac{(c-d)^3}{\left[(a-b)^2+(c-d)^2\right]^{3/2}}$$
$$\frac{\partial^2f}{\partial a^2} = -\frac{3(a-b)(c-d)^3}{\left[(a-b)^2+(c-d)^2\right]^{5/2}},$$
which is negative when $a>b$ and $c>d$. Therefore the Hessian of $f$ cannot be positive semi-definite, and so $f$ is not convex.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/293837",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
}
|
A simplification regarding matrices Let $$A = \left( \begin{array}{cc}
0 & -1 \\
1 & 0 \\
\end{array} \right )$$
What can I say about the matrix $A^k$ for arbitrary natural $k$. Is there some way to express $A^k$ in terms of $A$. I have observed that $A^4 = I$.
EDIT: I want a better form of $A^k$ in order to find a closed form expression in terms of elementary functions for $\exp(A)$
|
We can diagonalize $A$ by calculating the Jordan Normal Form, using it's eigenvalues, $\lambda_{1, 2} = \pm i$, where $i$ is the imaginary unit for this matrix as:
$$A = \left( \begin{array}{cc}
0 & -1 \\
1 & 0 \\
\end{array} \right ) = S \cdot J \cdot S^{-1} = \left( \begin{array}{cc}
-i & \ i
\\ 1 & 1 \\
\end{array} \right ) \left( \begin{array}{cc}
-i & \ 0
\\ 0 & i \\
\end{array} \right ) \left( \begin{array}{cc}
\frac{i}{2} & \ \frac{1}{2}
\\ \frac{-i}{2} & \frac{1}{2} \\
\end{array} \right )$$
We can now calculate the exponential of $A$ using
$$\exp(A)=S \cdot \exp(J) \cdot S^{-1}$$
Regards
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/295486",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
}
|
Sum of combinations of n taken k where k is from n to (n/2)+1 I wonder if there's a formula for obtaining the sum of $n\choose k$'s where $k$ is from $n$ to $\frac{n}{2}+1$. I found out that in odd numbers, it is $2^{n-1}$ (powerset divided by $2$).
*
*1 = 1
*3 = 4
*5 = 16
*7 = 64
However, It is not the case for even number. I can't come up with the pattern. Here's my observation
*
*2 = 1
*4 = 5
*6 = 22
*8 = 93
|
Recall that
$$\sum_{k=0}^n \dbinom{n}k = 2^n$$
Also, recall that
$$\dbinom{n}k = \dbinom{n}{n-k}$$
Hence, for odd $n$, we have
\begin{align}
2^n & = \sum_{k=0}^n \dbinom{n}k\\
& = \sum_{k=0}^{(n-1)/2} \dbinom{n}k + \sum_{k=(n+1)/2}^n \dbinom{n}k\\
& = \sum_{k=0}^{(n-1)/2} \dbinom{n}{n-k} + \sum_{k=(n+1)/2}^n \dbinom{n}k\\
& = \sum_{k=(n+1)/2}^n \dbinom{n}k + \sum_{k=(n+1)/2}^n \dbinom{n}k\\
& = 2\sum_{k=(n+1)/2}^n \dbinom{n}k
\end{align}
Hence, if $n$ is odd, we have
$$\sum_{k=(n+1)/2}^n \dbinom{n}k = 2^{n-1}$$
If $n$ is even, we have
\begin{align}
2^n & = \sum_{k=0}^n \dbinom{n}k\\
& = \sum_{k=0}^{n/2-1} \dbinom{n}k + \dbinom{n}{n/2} + \sum_{k=n/2+1}^n \dbinom{n}k\\
& = \sum_{k=0}^{n/2-1} \dbinom{n}{n-k} + \dbinom{n}{n/2} + \sum_{k=n/2+1}^n \dbinom{n}k\\
& = \sum_{k=n/2+1}^n \dbinom{n}k + \sum_{k=n/2+1}^n \dbinom{n}k + \dbinom{n}{n/2}\\
& = 2\sum_{k=n/2+1}^n \dbinom{n}k + \dbinom{n}{n/2}
\end{align}
Hence, if $n$ is even, we have
$$\sum_{k=n/2+1}^n \dbinom{n}k = 2^{n-1} - \dfrac12 \dbinom{n}{n/2}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/299273",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
}
|
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.