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$\lim_{n\to \infty} F(k)=\frac{(1^{k}+2^{k}+3^{k}+.....+n^{k})}{(1^{2}+2^{2}+3^{2}+.....+n^{2})(1^{3}+2^{3}+3^{3}+.....+n^{3})}$ Find F(5) and F(6) Find the value of F(5) & F(6).It is given that
$$F(k)= \lim_{n\to \infty} \frac{(1^{k}+2^{k}+3^{k}+.....+n^{k})}{(1^{2}+2^{2}+3^{2}+.....+n^{2})(1^{3}+2^{3}+3^{3}+.....+n^{3})}$$ where 'k' is a Natural number
My approach Denominator is $\frac{(n)^{3}*(n+1)^{3}*(2n+1)}{24}$
I dont know how to proceed from here as the Numerator has power in k. I also tried definite integral formula but not able to solve it
|
note that $$\sum_{i=1}^ni^2=\frac{1}{6}n(n+1)(2n+1)$$
$$\sum_{i=1}^n i^3=\frac{1}{4}n^2(n+1)^2$$
$$\sum_{i=1}^ni^5=\frac{1}{12}n^2(n+1)^2(2n^2+2n-1)$$
and we get $$\lim_{n\to \infty}\frac{\frac{1}{12}n^2(n+1)^2(2n^2+2n-1)}{\frac{1}{6}n(n+1)(2n+1)\frac{1}{4}n^2(n+1)^2}$$
the searched limit is $0$
note for the next:
$$\sum_{n=1}^n i^6=1/42\,n \left( n+1 \right) \left( 2\,n+1 \right) \left( 3\,{n}^{4}+6
\,{n}^{3}-3\,n+1 \right)
$$
|
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|
closed form solution for $ \displaystyle \sum_{x=1}^{k-1} {\frac{1}{x(k-x)}} $ Is there any closed form solution for this summation?
$$ \sum_{x=1}^{k-1} {\frac{1}{x(k-x)}} $$
$k$ is a finite integer constant.
|
We have that $\frac{1}{x(k-x)} = \frac{1}{k} \left( \frac{1}{x} + \frac{1}{k-x} \right)$. Therefore,
$$\sum_{x=1}^{k-1} \frac{1}{x(k-x)} = \frac{1}{k} \left( \sum_{x=1}^{k-1} \frac{1}{x} + \sum_{x=1}^{k-1} \frac{1}{k-x} \right).$$
By reversing the order of summation in the second sum, we get that this is equal to $\frac{2}{k} H_{n-1}$, where $H_{n-1} = \sum_{x=1}^{k-1} \frac{1}{x}$ is the harmonic sum. (There is no simpler closed-form expression for $H_n$, though a great deal is known about it.)
|
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|
How to show that $\sqrt{x}$ grows faster than $\ln{x}$. So I have the limit $$\lim_{x\rightarrow \infty}\left(\frac{1}{2-\frac{3\ln{x}}{\sqrt{x}}}\right)=\frac{1}2,$$ I now want to motivate why $(3\ln{x}/\sqrt{x})\rightarrow0$ as $x\rightarrow\infty.$ I cam up with two possibilites:
*
*Algebraically it follows that $$\frac{3\ln{x}}{\sqrt{x}}=\frac{3\ln{x}}{\frac{x}{\sqrt{x}}}=\frac{3\sqrt{x}\ln{x}}{x}=3\sqrt{x}\cdot\frac{\ln{x}}{x},$$
Now since the last factor is a standard limit equal to zero as $x$ approaches infinity, the limit of the entire thing should be $0$. However, isn't it a problem because $\sqrt{x}\rightarrow\infty$ as $x\rightarrow \infty$ gives us the indeterminate value $\infty\cdot 0$?
*One can, without having to do the arithmeticabove, directly motivate that the function $f_1:x\rightarrow \sqrt{x}$ increases faster than the function $f_2:x\rightarrow\ln{x}.$ Is this motivation sufficient? And, is the proof below correct?
We have that $D(f_1)=\frac{1}{2\sqrt{x}}$ and $D(f_2)=\frac{1}{x}$. In order to compare these two derivateives, we have to look at the interval $(0,\infty).$ Since $D(f_1)\geq D(f_2)$ for $x\geq4$, it follows that $f_1>f_2, \ x>4.$
|
For any $a > c > 0, x > 1$,
we have
$\begin{array}\\
\ln(x)
&=\int_1^x \frac{dt}{t}\\
&\lt\int_1^x \frac{dt}{t^{1-c}}\\
&=\int_1^x t^{c-1}dt\\
&=\dfrac{t^c}{c}\big|_1^x\\
&=\dfrac{x^c-1}{c}\\
&<\dfrac{x^c}{c}\\
\text{so}\\
\dfrac{\ln(x)}{x^a}
&<\dfrac1{x^a}\dfrac{x^c}{c}\\
&=\dfrac{x^{c-a}}{c}\\
&\to 0
\qquad\text{as } x \to \infty
\text{ since } a > c\\
\end{array}
$
For your case,
where $a = \frac12$,
we can choose
$c = \frac14$
to get
$\dfrac{\ln(x)}{x^{1/2}}
\lt\dfrac{x^{\frac14-\frac12}}{\frac14}
=4x^{-\frac14}
\to 0
$.
In general
we can choose
$c = a/2$
to get
$\dfrac{\ln(x)}{x^{a}}
\lt\dfrac{x^{a/2-a}}{a/2}
=\dfrac{2}{ax^{a/2}}
\to 0
$.
|
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|
Finding common vectors in $R^5$ of both spans
Determine which vectors in $R^5$ belong to both $sp${$(1, 0, 3, 5, 1),
(0, 2, 0, 1, 5)$} and $sp${$(1, 0, 2, 5, 1), (1, 3, 1, 1, 0), (2, 7,
0, 0, 1)$}.
My approach:
I've interpreted this question as a particular vector $v$ $ϵ$ $R^5$, such that
$v$ $ϵ$ $sp${$(1, 0, 3, 5, 1),
(0, 2, 0, 1, 5)$} $∩$ $sp${$(1, 0, 2, 5, 1), (1, 3, 1, 1, 0), (2, 7,
0, 0, 1)$}
which exists real number $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}$.
$\implies$ $v = a_{1} (1, 0, 3, 5, 1) + a_{2} (0, 2, 0, 1, 5) = a_{3} (1, 0, 2, 5, 1) + a_{4} (1, 3, 1, 1, 0) + a_{5} (2, 7, 0, 0, 1)$
$\implies$ $$(a_{1}, 2a_{2}, 3a_{1}, 5a_{1} + a_{2}, a_{1} + 5a_{2}) = (a_{3} + a_{4} + 2a_{5}, 3a_{4} + 7a_{5}, 2a_{3} + a_{4}, 5a_{3} + a_{4}, a_{3} + a_{5})$$
$\implies$
$a_{1} = a_{3} + a_{4} + 2a_{5}$
$2a_{2} = 3a_{4} + 7a_{5}$
$3a_{1} = 2a_{3} + a_{4}$
$5a_{1} + a_{2} = 5a_{3} + a_{4}$
$a_{1} + 5a_{2} = a_{3} + a_{5}$
I understand the need to find a single relationship, so that I can list the linear combinations of the vectors that can span both. I'm stuck here, or am I doing it wrong?
|
Hint: You're having a system of equations with $5$ variables. Bring it back to matrix $\mathbf{Ax}=0$ and solve it. The answer will be all vectors $\alpha \cdot \mathbf{x}$ where $\alpha \in \mathbf{R}$.
|
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|
Limit of $e^{n^{3/4}} ((1- c/n^{1/4})^{n^{1/4}})^{n^{3/4}/c}$
Let $c\ne0$ be a constant. Consider the limit of $$f(n)=e^{n^{3/4}} ((1- c/n^{1/4})^{n^{1/4}})^{n^{3/4}/c}$$ as $n \to \infty$.
I think it is zero because for large $n$,
$$e^{n^{3/4}} ((1- \frac{c}{n^{1/4}})^{n^{1/4}})^{\frac{n^{3/4}}{c}} \approx e^{n^{3/4}} (e^{-c})^{ \frac{n^{3/4}}{c}}$$
But, how do I prove it formally?
|
If we set $ h=\frac{1}{n^{1/4}} $ then
\begin{split}\lim_{n\to \infty}f(n)&=&\lim_{n\to \infty}e^{n^{3/4}} ((1- c/n^{1/4})^{n^{1/4}})^{n^{3/4}/c}\\
&=& \lim_{n\to \infty}e^{n^{3/4}} (1- c/n^{1/4})^{n/c}
\\&=&\lim_{h\to 0}e^{ \frac{1}{ h^3}}\left( 1-ch\right)^{1/ch^4}
\\&=&\lim_{h\to 0}\exp\left( \frac{1}{ h^3}\right)\exp\left( \frac{\ln(1-ch)}{ch^4} \right)\\
&=&\lim_{h\to 0}\exp\left( \frac{1}{ h^3} \left( \frac{\ln(1-ch)}{ch}+1\right) \right)
\\&=& \lim_{h\to 0}\exp\left( \frac{1}{ h^3} \left( -\frac{ch } {2} -\frac{c^2h^2}{3} -\frac{c^3h^3}{4}
+o(h^3)\right) \right)
\\&=&\lim_{h\to 0}\exp\left( -\frac{c } {2 h^2} -\frac{c^2}{3h} -\frac{c^3}{4}
+o(1)\right)=0\end{split}
Given that
$$\ln(1-ch)= -ch-\frac{c^2h^2}{2} -\frac{c^3h^3}{3} -\frac{c^4h^4}{4} +
o(h^4).$$
|
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|
Concrete Mathematics: Clarification on Euclidean algorithm Let $m$ and $n$ be two positive integers, then:
$$m'm + n'n = gcd(m, n)$$
We can rewrite this as:
$$\overline{r}r + \overline{m}m = gcd(r, m)$$
with $r = n - \lfloor n/m \rfloor m$. Inserting the value of $r$ in the above equation, we have:
$$\overline{r}\left(n - \Bigl\lfloor \frac{n}{m} \Bigr\rfloor m\right) + \overline{m}m = gcd(m, n)\\
\left(\overline{m} - \Bigl\lfloor \frac{n}{m} \Bigr\rfloor \overline{r}\right)m + \overline{r}n = gcd(m, n)$$
After that it gives an example.
For $m = 12$ and $n = 18$, this method results in:
$$6 = 0 \times 0 + 1 \times 6 = 1 \times 6 + 0 \times 12 = (-1) \times 12 + 1 \times 18$$
I do not understand how it arrives at $1\times6 + 0\times12$ from $0\times0 + 1\times6$. We know $gcd(12, 18) = 6$, so $6$ can be written as $gcd(0, 6) = 0 \times 0 + 1 \times 6$.
$$n' = 0, m' = 1 - \lfloor n/m \rfloor \times 0, m = 6$$
How can we calculate $n$? How did the text know that $n = 12$?
|
In the previous page it is mentioned that, we can calculate $m'$ and $n'$ recursively until $m \neq 0$. I missed the crucial point as I over read it.
Here's what is written
Euclid's algorithm also gives us more.: We can extend it so that it will compute integers $m'$ and $n'$ satisfying
$$m'm + n'n = gcd(m, n).$$
Here's how. If $m = 0$, we simply take $m' = 0$ and $n' = 1$. Otherwise we let $r = n\ mod\ m$ and apply the method recursively with $r$ and $m$ in place of $m$ and $n$, computing $\overline{r}$ and $\overline{m}$ such that
$$\overline{r}r + \overline{m}m = gcd(r, m).$$
So basically what it did was:
$$
\square \times 12 + \square \times 18 = gcd(12, 18) = 6\\
\square \times 6 + \square \times 12 = 6\\
\square \times 0 + \square \times 6 = 6\\
0 \times 0 + 1 \times 6 = 6\\
1 \times 6 + 0 \times 12 = 6\\
-1 \times 12 + 1 \times 18 = 6\\
$$
|
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|
How to show that $\int_{0}^{\infty}e^{-x}(x^2+3x+3)\cos(x){\mathrm dx\over (1+x)^3}=1?$ How to show that $(1)$ is
$$\int_{0}^{\infty}e^{-x}(x^2+3x+3)\cos(x){\mathrm dx\over (1+x)^3}=1?\tag1$$
$$x^2+3x+3=\left(x+{3\over 2}\right)^2+{3\over 4}$$
$$\int_{0}^{\infty}e^{-x}\cos(x){\mathrm dx\over (1+x)^3}+\int_{0}^{\infty}e^{-x}\left(x+{3\over 2}\right)^2\cos(x){\mathrm dx\over (1+x)^3}=I_1+I_2\tag2$$
$$(1+x)^{-3}=\sum_{n=0}^{\infty}(-1)^n{n+2\choose n}x^n$$
$$\sum_{n=0}^{\infty}(-1)^n{n+2\choose n}\int_{0}^{\infty}x^{n}e^{-x}\cos(x)\mathrm dx=I_1\tag3$$
$$\sum_{n=0}^{\infty}(-1)^n{n+2\choose n}\int_{0}^{\infty}x^{n}e^{-x}\left(x+{3\over 2}\right)^2\cos(x)\mathrm dx=I_2\tag4$$
|
Start with partial fractions:
$$ \frac{x^2+3x+3}{(1+x)^3} = \frac{1}{1+x} + \frac{1}{(1+x)^2} + \frac{1}{(1+x)^3} $$
Now let
$$ J_k = \int_0^\infty e^{-(1+i)x}\frac{dx}{(1+x)^k} $$
Integrating by parts,
$$ J_k = \frac{1-i}{2} - \frac{((1-i)k}{2} J_{k+1} $$
Using this for $k=1$ and $k=2$, $J_1 + J_2 + J_3$ simplifies to $1 - i/2$.
Your integral is the real part of this, namely $1$.
|
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|
Non unique factorization in $\mathbb{Z}_5$
Prove $3X^3+4X^2+3=(X+2)^2(3X+2)=(X+2)(X+4)(3X+1)$ in $\mathbb{Z}_5$.
I see that $2$ is a root (since $-24+16+3=-5$) , then $3X^3+4X^2+3=(X+2)(3X^2)$, and also $-4$ (since -$64\cdot3+16\cdot4+3=-125$) and although $x=-1/3$ is not in $\mathbb{Z}_5$, because it also works as a root then $3x+1$ is a nother factor.
Then it should be $3X^3+4X^2+3=(X+2)(X+4)(3X+1)$ but the product of the right side gives $3X^3+24X^2+30X+8=3X^3+4X^2+3+5(4X^2+6X^2+1)=3X^3+4X^2+3.$
But why is it $3X^3+4X^2+3=(X+2)^2(3X+2)$? It seems $(X+2)^2(3X+2)=(X^2+2X+4)(3X+2)=3X^4+8X^2+10X+8=3X^4+3X^2+3+5(X+2X+1)=3X^4+3X^2+3$ which is not the initial polynomial.
Also, $\mathbb{Z}_5$ is UFC domain, wouldn't the double factorization imply a contradiction?
|
Factorization is only unique up to associates; e.g. over the reals, we can factor $4x^2 - 1$ in many different ways
$$ 4x^2 - 1 = (2x-1)(2x+1) = 4\left(x - \frac{1}{2}\right)\left(x + \frac{1}{2}\right) = (4x-2)\left(x + \frac{1}{2} \right)$$
In $k[x]$, we most commonly normalize by factoring out an overall unit so as to make the nonconstant factors monic polynomials; the individual factors of this form are unique. (the ordering is not)
Doing so, you would further factor
$$ (X+2)^2 (3X+2) = 3 (X+2)^2 (X+4) $$
$$ (X+2)(X+4) (3X+1) = 3 (X+2)(X+4)(X+2) $$
and we can see that, in both cases, the factorizations really do have the same factors.
In other words, what you were overlooking was:
*
*$3X+2$ and $X+4$ are associates
*$3X+1$ and $X+2$ are associates
|
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|
Is it possible to determine which of these numbers is greater: $\sqrt{2}^{\sqrt{3}}$ or $\sqrt{3}^{\sqrt{2}}$ without approximating anything? Is it possible to determine which of these numbers is greater: $\sqrt{2}^{\sqrt{3}}$ or $\sqrt{3}^{\sqrt{2}}$?
I tried to express both in terms of powers of 3, but in the first number I got $log_32 $ as part of the product of the overall power of two, which - again - makes things hard.
Is there a simple way to do this?
|
We compare
$$\sqrt{2}^\sqrt{3} \text{ vs. } \sqrt{3}^\sqrt{2}$$
or (taking natural logarithms)
$$(\sqrt{3}/2) \ln 2 \text{ vs. } (\sqrt{2}/2) \ln 3$$ or (multiplying by $2$) $$\sqrt{3} \ln 2 \text{ vs. } \sqrt{2} \ln 3$$ or (rearranging) $$\sqrt{3/2} \text{ vs. } (\ln3)/(\ln2) = \log_2 3.$$
Now $2^{\sqrt{3/2}} < 2^{3/2} = 2 \sqrt{2} < 3$, because $(2\sqrt{2})^2 = 8$ whereas $3^2 = 9$. Thus $\sqrt{3/2} < \log_2 3$ and we conclude $\sqrt{2}^{\sqrt{3}} < \sqrt{3}^{\sqrt{2}}$, chasing back the comparisons.
|
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|
Factorising $(4 + 3i)z^2 + 26iz + (-4+3i)$? Quadratic formula attempt included. I'm trying to factorise $(4 + 3i)z^2 + 26iz + (-4+3i)$.
I tried to use the quadratic formula to factorise $(4 + 3i)z^2 + 26iz + (-4+3i)$, where $b = 26i$, $a = (4 + 3i)$, $c = (-4+3i)$. This gives us the roots $z = \dfrac{-4i - 12}{25}$ and $z =-4i - 3$.
But $\left( z + \dfrac{4i + 12}{25} \right) \left( z + 4i + 3 \right) \not= (4 + 3i)z^2 + 26iz + (-4+3i)$.
So I did what we do when using the quadratic formula with real numbers and have that $a \not= 0, 1$: $(4 + 3i)\left( z + \dfrac{4i + 12}{25} \right) \left( z + 4i + 3 \right)$. But, as far as I can tell, $(4 + 3i)\left( z + \dfrac{4i + 12}{25} \right) \left( z + 4i + 3 \right) \not= (4 + 3i)z^2 + 26iz + (-4+3i)$.
So what is the correct way to go about this?
I would greatly appreciate it if people could please take the time to explain this.
|
Another way, let $a = 4+3i$, then the equation is $az^2 + (a \overline a+1)iz - \overline{a}=0$.
$$\implies (az)^2 + (a \overline ai+i)(az) - a\overline{a}=0$$
Knowing Vieta, it is easy to observe the roots are $az \in \{ -a\overline ai, -i\}$, which implies $z \in \{-\overline ai, -\dfrac{i}a \} = \{-3-4i, -\frac3{25}-\frac4{25}i\}$.
|
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|
If $a,b$ are positive real numbers, such that $a+b=1$, then prove that $(a+ \dfrac {1}{a})^3 +(b+ \dfrac {1}{b})^3 \ge \dfrac {125}{4}$
If $a,b$ are positive real numbers, such that $a+b=1$, then prove that $$\bigg(a+ \dfrac {1}{a}\bigg)^3 +\bigg(b+ \dfrac {1}{b}\bigg)^3 \ge \dfrac {125}{4}$$
I just learnt to prove If $a$ and $b$ are positive real numbers such that $a+b=1$, then prove that $\big(a+\frac{1}{a}\big)^2 +\big(b+\frac{1}{b}\big)^2 \ge\frac{25}{2}$ a few days ago, and it was just basic application of CS. But I find this one really difficult. I can not apply CS here directly on the numbers $a+\frac{1}{a}$ and $b+\frac{1}{b}$ because CS is for squares. So some manipulation is needed. Anything from hint to full answer will be appreciated.
|
Let $a=\cos^2 t,b=\sin^2 t$.
Note
$$ \cos^6t+\sin^6t=(\cos^2t+\sin^2t)(\cos^4t-\cos^2t\sin^2t+\sin^4t)=1-3\cos^2t\sin^2t. $$
Then
\begin{eqnarray}
&&(a+\frac1a)^3+(b+\frac1b)^3\\
&=&(\cos^2t+\sec^2t)^3+(\sin^2t+\csc^2t)^3\\
&=&\cos^6t+3\cos^4t\sec^2t+3\cos^2t\sec^4t\\
&&+\sin^6t+3\sin^4t\csc^2t+3\sin^2t\csc^4t+\csc^6t\\
&=&(\cos^6t+\sin^6t)+3(\cos^2t+\sin^2t)+3(\sec^2t+\csc^2t)+\frac{\cos^6t+\sin^6t}{\cos^6t\sin^6t}\\
&=&4-3\cos^2t\sin^2t+\frac{3}{\sin^2t\cos^2t}+\frac{1-3\cos^2t\sin^2t}{\cos^6t\sin^6t}
\end{eqnarray}
Let $x=\cos^2t\sin^2t$ and then $x\in(0,\frac14]$. Define
$$ f(x)=4-3x+\frac3x+\frac{1-3x}{x^3}. $$
Then
$$ f'(x)=-3-\frac3{x^2}+\frac{8x-3}{x^4}<0, x\in(0,\frac12]$$
and hence
$$ f(x)\ge f(\frac12)=\frac{125}{4}. $$
So
$$ (a+\frac1a)^3+(b+\frac1b)^3\ge\frac{125}{4}$$
|
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|
Prove that if $p|x^p+y^p$ then $p^2|x^p+y^p$ I can show that $5|x^5+y^5$, by considering $(x+y)^5$ and using binomial expansion. But I am not sure how to show that $25|x^5+y^5$.
More generally, if p is a prime and $p>2$, how do I prove that if $p|x^p+y^p$ then $p^2|x^p+y^p$?
|
Assuming $x,y\in \mathbb{Z}$, if we have $5\mid x^5 + y^5$ then by binomial expansion it follows that $5\mid (x+y)^5$. But by prime factorization, it is obvious that $5^5\mid (x+y)^5$ and hence $25\mid (x+y)^5$. To see that $25\mid x^5 + y^5$, it is enough to show that $5x^4y + 10x^3y^2 + 10x^2y^3 +5xy^4$ is divisible by $25$ or equivalently $(x^3 +2x^2y + 2xy^2 +y^3)$ is divisible by 5.
Doing this by brute force check for pairs $(x,y)= (1,9) \mod 10$, $(x,y)= (2,8) \mod 10$, $(x,y)= (3,7) \mod 10$, $(x,y)= (4,6) \mod 10$, $(x,y)= (5,0) \mod 10$. [$\mod 10$ these are the only possible solutions]
|
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|
Find roots of $x^4 -6x^3 + 12x^2 - 12 x + 4 = 0$ the original equation is:
$$(x^2 + 2)^2 -6x(x^2+2) + 8x^2=0.$$
cannot see how to go solving this. I tried following way to factorise:
$$(x^2+2)(x^2-6x+2) + 8x^2 = 0.$$
But this has no help to solve.
Thank you people, but I need the thinking process, not the answer.
|
One way to find the factorizations others have given is to complete the square. If you recognize $y^2-6y+9=(y-3)^2$ you can do $(x^2+2)^2-6(x^2+2)+8x^2=((x^2+2)-3x)^2-x^2=(x^2+2-4x)(x^2+2-2x)$
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/2494962",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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|
Finding the Inverse of a function with natural logs The function is
$$f(x)= \frac{7 e^{x} +2}{5 e^{x} - 6}$$
and I am suppose to find the inverse, so I switched the $x$'s and $y$'s.
I know I am suppose to multiple the denominator out and I end up with $5 \, e^{y} \, x - 6 x = 7 \, e^{y} + 2$. From there I do not know what to do.
|
Consider:
$$f(x) = \frac{7 \, e^{x} + 2}{5 \, e^{x} - 6}$$
for which
\begin{align}
y &= \frac{7 \, e^{x} + 2}{5 \, e^{x} - 6} \\
5 y \, e^{x} - 6 y &= 7 \, e^{x} + 2 \\
(5 y - 7) \, e^{x} &= 6 y + 2 \\
e^{x} &= \frac{6 y + 2}{5y - 7} \\
x &= \ln\left( \frac{6 y + 2}{5y - 7} \right)
\end{align}
It can now be stated that
$$f^{-1}(x) = \ln\left( \frac{6 x + 2}{5x - 7} \right).$$
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/2495753",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Show that if $n \geq 2$ is even, then: $\frac {2n}{3} \leq \sum_{k=1}^{n} \frac{p(k)}{k} \leq \frac {2(n+1)}{3}$ Show that if $n \geq 2$ is even, then: $$\frac {2n}{3} \leq \sum_{k=1}^{n} \frac{p(k)}{k} \leq \frac {2(n+1)}{3}$$ where $p(k)$ is the greatest odd integer that divides $k$.
I think I'm almost done but cannot complete it.
Well, it's known that there are $\frac {n}{2}$ odd naturals $ < n$. And for these integers, we have $\frac {p(k)}{k}=1$
Hence, we are left to show that: $$\frac {n}{6}\leq \sum_{k={\text {even}}, k\leq n}\frac{p(k)}{k}\leq \frac {n}{6} + \frac {2}{3}$$
Define $v_c(k)=d$ such that $c^d \mid k$ and $c^{d+1}\nmid k$.
If $k=2^x y$, then $\frac {p(k)}{k}=2^{-x}$.
Hence, if we find of $v_2(k)=1, v_{2^2}(k)=1, v_{2^3}(k)=1, \cdots$ for $k=2,4,6, \cdots , n$, and multiply $2^{-r}$ to each $v_{2^r}$, we get the value of the equation.
Then, we can substitute it to show that it $\in [\frac {n}{6}, \frac{n+4}{6}]$
And the equation becomes, upon simplifying:
$$\frac {n}{6}\leq \sum_{x=1}^{e} \left\lfloor\frac{n-2^x}{2^{x+1}}\right\rfloor \leq \frac {n}{6}+\frac {2}{3}$$ where we assume $2^e \leq n < 2^{e+1}$ for some $e\in \mathbb{N}$.
But how to prove this? Maybe it's quite simple but I can't do it unfortuantely.
|
As you found, we have
$$\frac{p(k)}{k} = 2^{-v_2(k)}.$$
Thus we can compute the sum if we know how many $k \leqslant n$ have $v_2(k) = m$ for every $m$. That's the number of multiples of $2^m$ not exceeding $n$ that are not multiples of $2^{m+1}$. The number of multiples of $r$ not exceeding $x$ is $\bigl\lfloor \frac{x}{r}\bigr\rfloor$, hence there are $\bigl\lfloor \frac{n}{2^m}\bigr\rfloor - \bigl\lfloor \frac{n}{2^{m+1}}\bigr\rfloor$ multiples of $2^m$ that aren't multiples of $2^{m+1}$ not exceeding $n$, and
$$\sum_{k = 1}^n \frac{p(k)}{k} = \sum_{m = 0}^{\infty} \frac{1}{2^m}\biggl(\biggl\lfloor \frac{n}{2^m}\biggr\rfloor - \biggl\lfloor \frac{n}{2^{m+1}}\biggr\rfloor\biggr).$$
The last sum is actually finite, since the terms are $0$ when $n < 2^m$, but $\infty$ is quicker to write. We can regroup this last sum,
\begin{align}
\sum_{m = 0}^{\infty} \frac{1}{2^m}\biggl(\biggl\lfloor \frac{n}{2^m}\biggr\rfloor - \biggl\lfloor \frac{n}{2^{m+1}}\biggr\rfloor\biggr)
&= \sum_{m = 0}^{\infty} \frac{1}{2^m}\biggl\lfloor \frac{n}{2^m}\biggr\rfloor - \sum_{m = 0}^{\infty} \frac{2}{2^{m+1}}\biggl\lfloor \frac{n}{2^{m+1}}\biggr\rfloor \\
&= \sum_{m = 0}^{\infty} \frac{1}{2^m}\biggl\lfloor \frac{n}{2^m}\biggr\rfloor - 2 \sum_{k = 1}^{\infty} \frac{1}{2^k} \biggl\lfloor \frac{n}{2^k}\biggr\rfloor \tag{$k = m+1$} \\
&= n - \sum_{k = 1}^{\infty} \frac{1}{2^k}\biggl\lfloor \frac{n}{2^k}\biggr\rfloor.
\end{align}
Now $\lfloor x\rfloor \leqslant x$ yields, together with the fact that all but finitely many terms are $0$, so the inequality is strict for almost all terms,
$$\sum_{k = 1}^{\infty} \frac{1}{2^k} \biggl\lfloor \frac{n}{2^k}\biggr\rfloor < \sum_{k = 1}^{\infty} \frac{n}{4^k} = \frac{n}{3}$$
and hence we have the lower bound
$$\sum_{k = 1}^n \frac{p(k)}{k} > \frac{2n}{3}.$$
This also holds for odd $n$.
For the upper bound, the estimate $\lfloor x\rfloor > x-1$ is not enough, that would yield an upper bound of $\frac{2n}{3} + 1$ rather that the desired $\frac{2n+2}{3}$. But for integer $n$ we have $\bigl\lfloor \frac{n}{2^k}\bigr\rfloor \geqslant \frac{n}{2^k} - 1 + \frac{1}{2^k}$, and for even $n$, we have $\bigl\lfloor \frac{n}{2^k}\bigr\rfloor \geqslant \frac{n}{2^k} - 1 + \frac{2}{2^k}$, with a strict inequality for all large enough $k$, hence
\begin{align}
\sum_{k = 1}^{\infty} \frac{1}{2^k}\biggl\lfloor \frac{n}{2^k}\biggr\rfloor
&> \sum_{k = 1}^{\infty} \frac{1}{2^k}\biggl( \frac{n}{2^k} - 1 + \frac{a}{2^k}\biggr) \\
&= \frac{n}{3} - 1 + \frac{a}{3}
\end{align}
where $a = 1$ if $n$ is odd and $a = 2$ if $n$ is even. Thus we obtain
$$\frac{2n}{3} < \sum_{k = 1}^n \frac{p(k)}{k} < \frac{2n + (3-a)}{3},$$
and the upper bound $\frac{2(n+1)}{3}$ also holds for all $n$, and for even $n$ we can reduce it to $\frac{2n+1}{3}$.
|
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|
How to find the Laplace Transform of $\frac{1-\cos(t)}{ t}$? What I tried is to find the transform of $f(t) = \frac{1-\cos(t)}{t}$ with $$\int_{s}^{\infty }\frac{1}{u} du + \int_{s}^{\infty }\frac{u}{u^2+1}du$$
$\int_{s}^{\infty }\frac{1}{u} du = \ln(\infty)-\ln(s)$. Is it a valid integral if I get an infinite value as a result?
$$\int_{s}^{\infty }\frac{u}{u^2+1}du = \frac{1}{2} \ln(\infty) - \frac{1}{2} \ln(s^2+1)$$
If it is right to have infinite values in the result of the integral, $\ln(\infty)$ cancels with $\frac{1}{2} \ln(\infty)$
and $F(s) = \ln(\infty) - \ln(s) - \frac{1}{2} \ln(\infty) + \frac{1}{2} \ln(s^2+1) $
$F(s)= \frac{1}{2} \ln(s^2+1) - \ln(s) = \ln\left(\frac{\sqrt{s^2+1}}{s}\right) $
|
Define the Laplace transform,
$$f(s) = \int_{0}^{\infty} e^{- s t} f(t) \, dt,$$
with the short notation $f(t) \doteqdot f(s)$.
The long method:
\begin{align}
\frac{1 - \cos(a t)}{t} &\doteqdot \int_{0}^{\infty} e^{- s t} \, \frac{1 - \cos(a t)}{t} \, dt \\
&\doteqdot \int_{s}^{\infty} \int_{0}^{\infty} e^{- u t} \, (1 - \cos(at)) \, dt \, du \\
&\doteqdot \int_{s}^{\infty} \left(\frac{1}{u} - \frac{u}{u^{2} + a^{2}} \right)
\, du \\
&\doteqdot \frac{1}{2} \, \left[ \ln(u^{2}) - \ln(u^{2}+a^{2}) \right]_{s}^{\infty} \\
&\doteqdot - \frac{1}{2} \, \ln\left(1 + \frac{a^{2}}{\infty^{2}}\right) + \frac{1}{2} \, \ln\left(\frac{s^{2} + a^{2}}{s^{2}}\right) \\
&\doteqdot \ln\left(\frac{\sqrt{s^{2}+a^{2}}}{s}\right).
\end{align}
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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|
Solving Laplace's equation for PDE problem: Contradiction Between Fourier Analysis and Instructor's Solution? Confusion with How to Proceed. I am told to find a general solution for the following PDE (partial differential equation):
$\dfrac{\partial^2{u}}{\partial{x}^2} + \dfrac{\partial^2{u}}{\partial{y}^2} = 0, u(x, 0) = 0, u(x,2) = 0, u(0, y) = 0$, and $u(3, y) = \begin{cases}
y, & 0 \le y \le 1 \\
2 - y, & 1 \le y \le 2
\end{cases}$
I understand that this type of PDE is Laplace's equation, and I know how to go about solving Laplace's equation.
For the sake of clarity and completeness, I will type out my full solution to the problem. However, if you are only interested in the specific problem that I have encountered, please scroll to the bottom of my solution, where I have indicated in bold text the beginning of my problem.
My Solution
$\\$
Using the separation of variables method for solving PDEs, the solution will be of the form $u(x,t) = X(x)Y(y)$.
$\therefore X''Y + Y''X = 0$
$\therefore \dfrac{X''}{X} = \lambda = -\dfrac{-Y''}{Y}$
$\therefore X'' - \lambda X = 0$ and $Y'' + \lambda Y = 0$ are the two homogeneous differential equations that we are looking for (as part of the separation of variables procedure).
$\\$
We now first use the boundary condition for $Y$:
$u(x, 0) = 0, u(x, 2) = 0$
$u(x, 0) = X(x)Y(y)$
$= 0$
$\therefore Y(0) = 0$
$u(x, 2) = X(x)Y(2)$
$= 0$
$\therefore Y(2) = 0$
$\\$
We can now begin solving $Y'' + \lambda Y = 0$
$r^2 + \lambda = 0$ This is the auxiliary equation
$\implies r = \pm \sqrt{-\lambda}$
The three possibilities are $\lambda = 0, \lambda > 0,$ and $\lambda < 0$.
Skipping to $\lambda > 0$, we let $\lambda = \beta^2$ Since $\beta^2$ is always positive.
$r = \pm i \beta$ From the auxiliary equation
$\therefore Y(y) = A\cos(\beta y) + B \sin(\beta y)$
$Y(0) = A$
$= 0$
$Y(2) = B\sin(2\beta)$
$= 0$
$\therefore \beta = n \dfrac{\pi}{2} \forall n \in \mathbb{N}^+$.
$\therefore \beta^2 = n^2 \dfrac{\pi^2}{4} = \lambda_n$ is the corresponding eigenvalue of the eigenfunction $Y(y) = B \sin \left( n \dfrac{\pi}{2} y \right)$.
$\\$
We now find the solution for $X'' - \lambda X = 0$.
$X'' - \lambda X = 0$
$r^2 - \lambda = 0$ Auxiliary equation
$\implies r = \pm \sqrt{\lambda}$
We had that $\beta^2 = n^2 \dfrac{\pi^2}{4} = \lambda_n \forall n \in \mathbb{N}^+$.
$\therefore r = \pm \sqrt{n^2 \dfrac{\pi^2}{4}}$
$= \pm n \dfrac{\pi}{2}$
$\therefore X(x) = Ae^{n \dfrac{\pi}{2} x} + Be^{-n \dfrac{\pi}{2} x}$
$\\$
Therefore, the general solution to the PDE is
$u(x, t) = \sum_{n = 1}^{\infty} B \left( -e^{n \dfrac{\pi}{2} x} + e^{-n \dfrac{\pi}{2} x} \right) \sin \left( n \dfrac{\pi}{2} y \right)$
$\\$
!!!Here is where I'm having problems!!!
We can now fit the other boundary condition.
$u(3, y) = \begin{cases}
y, & 0 \le y \le 1 \\
2 - y, & 1 \le y \le 2
\end{cases}$
We found the general solution to the PDE to be
$u(x, t) = \sum_{n = 1}^{\infty} B \left( -e^{n \dfrac{\pi}{2} x} + e^{-n \dfrac{\pi}{2} x} \right) \sin \left( n \dfrac{\pi}{2} y \right)$
$\therefore u(3, y) = \sum_{n = 1}^{\infty} B \left( -e^{n \dfrac{3\pi}{2}} + e^{-n \dfrac{3\pi}{2}} \right) \sin \left( n \dfrac{\pi}{2} y \right)$
Note that this is a Fourier sine series, where the Fourier coefficient $b_n = B \left( -e^{n \dfrac{3\pi}{2}} + e^{-n \dfrac{3\pi}{2}} \right)$.
Our goal here is to find $b_n$. We can do this using the formula for the coefficient of a Fourier sine series:
$b_n = \dfrac{2}{P} \int_{x_0}^{x_0 + P} f(x) \sin \left( \dfrac{2 \pi n x}{P} \right) \ dx$,
where $P$ is the periodicity of the series.
Since our general solution for the PDE has the term $\sin \left( n \dfrac{\pi}{2} y \right)$, we must have $P = 4$, right?
However, the instructor has written
$B \left( -e^{n \dfrac{3\pi}{2}} + e^{-n \dfrac{3\pi}{2}} \right) = \int^1_0 y \sin \left( \dfrac{n\pi y}{2} \right) \ dy + \int_1^2 (2-y) \sin \left( \dfrac{n\pi y}{2} \right) \ dy$
The instructor's solution seems to maintain a period of $P = 4$ since it has $\sin \left( \dfrac{n\pi y}{2} \right)$.
Based on my understanding of Fourier series and how it applies to this situation, I am confused as to what is going on here. Recall the formula for the Fourier coefficient $b_n$ (as shown above). According to it, we should have
$B \left( -e^{n \dfrac{3\pi}{2}} + e^{-n \dfrac{3\pi}{2}} \right) = \int^{x_0 + 4}_{x_0} y \sin \left( \dfrac{n\pi y}{2} \right) \ dy + \int^{x_0 + 4}_{x_0} (2-y) \sin \left( \dfrac{n\pi y}{2} \right) \ dy$
But instead, the instructor has used the limits of the boundary condition as the limits of integration, whilst simultaneously using a period of $P = 4$ in $\sin \left( \dfrac{n\pi y}{2} \right)$ ?
In other words, the instructor's solution has $P = 1$ in $x_0 + P$ for the limits of integration and $P = 4$ in $\sin \left( \dfrac{2 \pi n x}{P} \right)$.
I'm awfully confused as to the correct way to proceed here. There seems to be a contradiction between the formula for the Fourier coefficient $b_n$ and the instructor's solution.
I apologise for the long post. I would greatly appreciate it if people could please take the time to clarify my confusion and explain the correct way to proceed.
|
Before the "having problem zone" $u(x,y)$ is
$$u(x,y) = \sum_{n=1}^{\infty} B_{n} \, \sinh\left(\frac{n \pi \, x}{2}\right) \, \sin\left(\frac{n \pi \, y}{2}\right).$$
Now let
$$C_{n} = B_{n} \, \sinh\left(\frac{3 n \pi }{2}\right)$$
to obtain
$$u(3,y) = f(y) = \sum_{n=1}^{\infty} C_{n} \sin\left(\frac{n \pi \, y}{2}\right)$$
which leads to the Fourier coefficients found by use of Paul's Online Math Notes, where, for $0 \leq y \leq L$,
$$C_{n} = \frac{2}{L} \, \int_{0}^{L} f(y) \, \sin\left(\frac{n \pi y}{L}\right) \, dy,$$
and leads to, for this problem with $L=2$, $0 \leq y \leq 2$,
$$C_{n} = \frac{2}{2} \, \int_{0}^{2} f(y) \, \sin\left(\frac{n \pi \, y}{2}\right) \, dy.$$
The evaluation then follows:
\begin{align}
C_{n} &= \int_{0}^{2} f(y) \, \sin\left(\frac{n \pi \, y}{2}\right) \, dy \\
&= \int_{0}^{1} y \, \sin\left(\frac{n \pi \, y}{2}\right) \, dy + \int_{1}^{2} (2 - y) \, \sin\left(\frac{n \pi \, y}{2}\right) \, dy \\
&= \frac{4}{n^{2} \pi^{2}} \, \left[ \sin\left(\frac{n \pi \, y}{2}\right) - \frac{n \pi}{2} \cos\left(\frac{n \pi \, y}{2}\right) \right]_{0}^{1} + \frac{4}{n^{2} \pi^{2}} \, \left[ - \sin\left(\frac{n \pi \, y}{2}\right) + \frac{ n \pi \, y}{2} \cos\left(\frac{n \pi \, y}{2}\right) - n \pi \cos\left(\frac{n \pi \, y}{2}\right) \right]_{1}^{2} \\
&= \frac{2}{n \pi} + \frac{8}{n^{2} \pi^{2}} \, \sin\left(\frac{n \pi}{2}\right)
\end{align}
Put it all together to obtain
$$u(x,y) = \frac{2}{\pi^{2}} \, \sum_{n=1}^{\infty} \frac{n \pi + 4 \, \sin\left(\frac{n \pi}{2}\right)}{n^{2} \, \sinh\left(\frac{3 n \pi}{2}\right)} \, \sinh\left(\frac{n \pi \, x}{2}\right) \, \sin\left(\frac{n \pi \, y}{2}\right).$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Prove That $ \sqrt{\frac xy + \frac yz}+ \sqrt{\frac yz + \frac zx}+ \sqrt{\frac zx + \frac xy}\ge 3 $ for x, y, z $\gt 0$, prove that: $$ \sqrt{\frac xy + \frac yz}+ \sqrt{\frac yz + \frac zx}+ \sqrt{\frac zx + \frac xy}\ge 3 $$
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By AM-GM inequality we have
$$
\begin{align*}
\frac{x}{y}+\frac{y}{z} &\ge 2\sqrt{\frac{x}{z}}\\
\frac{y}{z}+\frac{z}{x} &\ge 2\sqrt{\frac{y}{x}}\\
\frac{z}{x}+\frac{x}{y} &\ge 2\sqrt{\frac{x}{z}}.
\end{align*}
$$
Then again by AM-GM inequality we have
$$
\begin{align*}
\frac{1}{3}\left(\sqrt{\frac{x}{y}+\frac{y}{z}}+\sqrt{\frac{y}{z}+\frac{z}{x}}+\sqrt{\frac{z}{x}+\frac{x}{y}} \right) &\ge \sqrt[3]{\sqrt{\frac{x}{y}+\frac{y}{z}}\sqrt{\frac{y}{z}+\frac{z}{x}}\sqrt{\frac{z}{x}+\frac{x}{y}}}\\
&\ge \sqrt[3]{\sqrt{2\sqrt{\frac{x}{z}}}\sqrt{2\sqrt{\frac{y}{x}}}\sqrt{2\sqrt{\frac{z}{y}}}}\\
&=2^{1/2}\left(\frac{xyz}{zxy}\right)^{1/12}\\
&> 1,
\end{align*}
$$
and our result follows.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Evaluate $2\int_{0}^{\pi\over 2}\ln^2(\tan^2{x})\mathrm dx$ How to show that $(1)$
$$2\int_{0}^{\pi\over 2}\ln^2(\tan^2{x})\mathrm dx=\pi^3?\tag1$$
$u=\tan^2{x}$ then $\mathrm dx={\mathrm du\over \sqrt{u}(u-1)}$
$$\int_{0}^{\infty}\ln^2{u}{\mathrm du\over \sqrt{u}(u-1)}\tag2$$
$$-\sum_{n=0}^{\infty}\int_{0}^{\infty}u^{n-1/2}\ln^2{u}\mathrm du\tag3$$
huhhh? $(3)$ seems to be diverging.
|
Setting $I = \int^{\pi/2}_0 \ln^2 (\tan^2 x) \, dx$, after letting $u = \tan x$ the integral becomes
$$I = 4 \int^\infty_0 \frac{\ln^2 u}{1 + u^2} \, du.$$
Writing this integral as
$$I = 4 \int^1_0 \frac{\ln^2 u}{1 + u^2} \, du + 4 \int^\infty_1 \frac{\ln^2 u}{1 + u^2} \, du,$$
if, in the second of these integrals we set $u = 1/x$, one has
$$I = 4 \int^1_0 \frac{\ln^2 u}{1 + u^2} \, du + 4 \int^1_0 \frac{\ln^2 u}{1 + u^2} \, du = 8 \int^1_0 \frac{\ln^2 u}{1 + u^2} \, du.$$
Writing the term $1/(1 + x^2)$ that appears in the integrand as a geometric series, namely
$$\frac{1}{1 + x^2} = \sum^\infty_{n = 0} (-1)^n x^{2n}, \quad |x| < 1,$$
after interchanging the summation with the integration (by Tonelli's theorem) we have
$$I = 8 \sum^\infty_{n = 0} (-1)^n \int^1_0 x^{2n} \ln^2 x \, dx.$$
Integrating by parts twice, we find
$$I = 16 \sum^\infty_{n = 0} \frac{(-1)^n}{(2n + 1)^3}.$$
To evaluate the sum we write
\begin{align*}
I &= 16 \sum^\infty_{n = 0} \frac{(-1)^n}{(2n + 1)^3} = 2 \sum^{\infty}_{n = 0} \frac{(-1)^n}{(n + \frac{1}{2})^3}\\
&= 2 \sum^{\infty}_{n = 0,n \in \text{even}} \frac{1}{(n + \frac{1}{2})^3} - 2 \sum^{\infty}_{n = 0,n \in \text{odd}} \frac{1}{(n + \frac{1}{2})^3}.
\end{align*}
Reindexing, in the first sum let $n \mapsto 2n$ while in the second sum let $n \mapsto 2n + 1$. Thus
\begin{align*}
I &= 2 \sum^\infty_{n = 0} \frac{1}{(2n + \frac{1}{2})^3} - 2 \sum^\infty_{n = 0} \frac{1}{(2n + \frac{3}{2})^3}\\
&= \frac{1}{4} \sum^\infty_{n = 0} \frac{1}{(n + \frac{1}{4})^3} - \frac{1}{4} \sum^\infty_{n = 0} \frac{1}{(n + \frac{3}{4})^3}\\
&= -\frac{1}{4} \left [\zeta(3,\frac{3}{4}) - \zeta(3, \frac{1}{4}) \right ],
\end{align*}
where $\zeta (s,z)$ is the Hurwitz zeta function.
Now the polygamma function $\psi^{(m)} (z)$ is related to the Hurwitz zeta function by
$$\zeta (1 + m, z) = \frac{(-1)^{m + 1}}{m!} \psi^{(m)}(z).$$
Using this result to rewrite the Hurwitz zeta functions in terms of polygamma functions we have
\begin{align*}
\zeta (3, \frac{1}{4}) &= \zeta (1 + 2, \frac{1}{4}) = -\frac{1}{2} \psi^{(2)} (\frac{1}{4})\\
\zeta (3, \frac{3}{4}) &= \zeta (1 + 2, \frac{3}{4}) = -\frac{1}{2} \psi^{(2)} (\frac{3}{4})
\end{align*}
So our integral becomes
$$I = \frac{1}{8} \left [\psi^{(2)}(\frac{3}{4}) - \psi^{(2)} (\frac{1}{4}) \right ].$$
Now making use of the reflection formula for the polygamma function, namely
$$\psi^{(m)}(1 - z) + (-1)^{m + 1} \psi^{(m)}(z) = (-1)^m \pi \frac{d^m}{dz^m} \cot (\pi z),$$
setting $m = 2$ we have
\begin{align*}
\psi^{(2)}(\frac{3}{4}) - \psi^{(2)} (\frac{1}{4}) &= \psi^{(2)}(1 - \frac{1}{4}) - \psi^{(2)} (\frac{1}{4})\\
&= \pi \left. \frac{d^2}{dz^2} \cot (z \pi) \right |_{z = 1/4}\\
&= \pi [2 \pi^2 \cot (z\pi) \text{cosec}^2 (z \pi)]_{z = 1/4}\\
&= \pi \cdot 4\pi^2 = 4 \pi^3.
\end{align*}
So finally
$$I = \frac{1}{8} \cdot 4 \pi^3 = \frac{\pi^3}{2},$$
as required.
|
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|
Conditional Expected Value of a Joint Probability Density Function I was wondering if anyone could help me with the following question specifically:
The continuous random variables $X_1$ and $X_2$ have the following joint probability density function: $$f(x_1,x_2)=\frac2{27}$$ over $0< x_1< 3$ and $0< x_2< 9-3x_1.$
Find $$E(X_1\mid X_2=5).$$
I tried to use the method of integrating $xh(x_1\mid x_2)$ but I couldn't get it to work. Thank you in advance!
|
From the definition of conditional expectation,
\begin{equation}
\mathbb{E}(X_{1}|X_{2} = x_{2}) = \int_{-\infty}^{\infty}x_{1}f_{X_{1}|X_{2}}(x_{1}|x_{2})\, dx_{1}.
\end{equation}
Furthermore,
\begin{equation}
f_{X_{1}|X_{2}}(x_{1}|x_{2}) = \dfrac{f_{X_{1}, X_{2}}(x_{1}, x_{2})}{f_{X_{2}}(x_{2})},
\end{equation}
and
\begin{equation}
f_{X_{2}}(x_{2}) = \int_{x_{1}} f_{X_{1}, X_{2}}(x_{1}, x_{2}) \, dx_{1} = \int_{0}^{\frac{9-x_{2}}{3}} \dfrac{2}{27} dx_{1} = \dfrac{2}{27} \times \dfrac{9-x_{2}}{3}.
\end{equation}
Therefore,
\begin{align}
\mathbb{E}(X_{1}|X_{2} = 5) & = \int_{-\infty}^{\infty} x_{1}f_{X_{1}|X_{2}}(x_{1}|x_{2} = 5) \, dx_{1} \\
& = \int_{-\infty}^{\infty} x_{1} \dfrac{2/27}{\frac{2}{27} \times \frac{9-5}{3}} \, dx_{1} = \int_{0}^{\frac{9-5}{3}}\dfrac{3x_{1}}{4}\, dx_{1} \\
& = \dfrac{3}{4} \times \left.\dfrac{x_{1}^{2}}{2}\right\vert_{0}^{4/3} = \dfrac{3}{4} \times \dfrac{(4/3)^{2}}{2} = \dfrac{2}{3}
\end{align}
|
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|
Express the rectangles area as a function of $x$.
Translation: 3214 C A rectangle is drawn inside a semicircle as the figure shows.
FIG (See picture)
a) Express the rectangles area as a function of $v$.
b) Express the rectangles area as a function of $x$.
c) Determine the largest value of the rectangles area.
My solution:
a) c)
$x=12\sin v$
$a=12\cos v$
$A=2ax=144\sin 2v$ where $v=\frac{\pi}{4}$
$A''=-576\sin 2v$
$A''\left(\frac{\pi}{4}\right)\lt0:\:max$
$A_{max}\left(\frac{\pi}{4}\right)=144\sin 2\frac{\pi}{4}=144\;\text{cm}^2$
Answer: $A_{max}=144\;\text{cm}^2$.
b)
One correct answer is $A=2x\sqrt{144-x^2}$, but I do not know how to get there. Any hints, tips or solution(s) is appreciated.
|
Let $y$ be the side adjacent to the angle $v$. Then, the area of the rectangle would be equal to $x\cdot 2y$.
a)
$$
\sin{v}=\frac{x}{12}\implies x=12\cdot \sin{v}\\
\cos{v}=\frac{y}{12}\implies y=12\cdot \cos{v}\
$$
Therefore, the area of the rectangle expressed as a function of $v$ is $A(v)=2\cdot 12^2\cdot \sin{v} \cdot \cos{v}$.
Answer: $A(v)=288\cdot \sin{v} \cdot \cos{v}$.
b) Notice that the sides marked $x$, $12$ and $y$ form a right triangle. Therefore:
$$
x^2+y^2=12^2\implies y=\sqrt{12^2-x^2}
$$
So, the area of the rectangle expressed as a function of $x$ is $A(x)=2x\sqrt{12^2-x^2}$.
Answer: $A(x)=2x\sqrt{12^2-x^2}$.
c) Given $A(x)=2x\sqrt{12^2-x^2}$ with the domain $x\in (0,12)$ (that's the interval on which $x$ makes geometrical sense for this problem), the largest area is where the given function reaches a maximum value and that can be found by setting the first derivative to zero and solving for $x$.
$$
A'(x)=\left(2x\sqrt{144-x^2}\right)'=(2x)'\sqrt{144-x^2}+2x\left(\sqrt{144-x^2}\right)'=\\
2\sqrt{144-x^2}-\frac{2x^2}{\sqrt{144-x^2}}
$$
Now, set the derivative to zero and solve for $x$:
$$
2\sqrt{144-x^2}-\frac{2x^2}{\sqrt{144-x^2}}=0\\
2(144-x^2)-2x^2=0\\
x^2=\frac{288}{4}\\
x=\pm\sqrt{72}\\
x=\pm6\sqrt{2}
$$
$x=-6\sqrt{2}$ goes out the window because it's not part of the domain. So, the largest area you can get is $A(6\sqrt{2})=2\cdot 6\sqrt{2}\sqrt{144-(6\sqrt{2})^2}=144$ $cm^2$.
Answer: $144$ $cm^2$.
|
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|
Sum of the multiplication of two geometric progressions I want to calculate the convolution of two geometric progressions by the convolution definition:
$$2^n*3^n=\sum_{k=0}^{n}f_1[k]f_2[n-k]=\sum_{k=0}^{n}2^k3^{n-k}=3^n\sum_{k=0}^n 2^k 3^{-k}=3^n\sum_{k=0}^n \bigg(\frac{2}{3}\bigg)^k=3^n\frac{1-\big(\frac{2}{3}\big)^n}{1-\frac{2}{3}}=3^n\frac{1-\big(\frac{2}{3}\big)^n}{\frac{1}{3}}=3^{n+1}\bigg(1-\Big(\frac{2}{3}\Big)^n\bigg)=3^{n+1}-2^n3^{-n}3^{n+1}=3^{n+1}-3\times 2^n$$
Since the sum is in k, the $3^n$ is out of the sum.
This result must be equivalent to that obtained by the Z transform:
$$2^n*3^n=F_1[z]F_2[z]=\frac{z}{z-2}\times \frac{z}{z-3}=G(z)$$
$$\frac{G(z)}{z}=\frac{z}{(z-2)(z-3)}=\frac{-2}{z-2}+\frac{3}{z-3}$$
Making the Z-transform I get:
$$g(n)=-2\times2^n+3\times 3^n=-2^{n+1}+3^{n+1}$$
I can not get the same result in both expressions. what did I do wrong?
|
Use
\begin{eqnarray*}
\sum_{k=0}^{n} r^k = \frac{1-r^{n+1}}{1-r}
\end{eqnarray*}
with $r=\frac{2}{3}$. So we have
\begin{eqnarray*}
2^n*3^n=\sum_{k=0}^{n}f_1[k]f_2[n-k]=\sum_{k=0}^{n}2^k3^{n-k}=3^n\sum_{k=0}^n 2^k 3^{-k} =3^n \frac{1-(\frac{2}{3})^{n+1}}{1-\frac{2}{3}} =3^{n+1}-2^{n+1}.
\end{eqnarray*}
|
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|
Let $f(x) =|x|+|x+1|+...+|x+99|$ then find: $f'\left(\frac{-9}{2}\right)=?$
Let $f(x) =|x|+|x+1|+...+|x+99|$ then find :
$$f'\left(\frac{-9}{2}\right)=?$$
My Try :
$$\lim_{x\to \frac{-9}{2}}\dfrac{f(x)-f\left(\frac{-9}{2}\right)}{\left(x+\frac{9}{2}\right)}
=
\lim_{x\to \frac{-9}{2}}\dfrac{(90x+95(52)-10)-f\left(\frac{-9}{2}\right)}
{\left(x+\frac{9}{2}\right)}
=?
$$
Now what ?
|
we have $f(x)=\left| x\right| +\left| x+1\right| +\left| x+2\right| +\left| x+3\right| +\left| x+4\right|+\ldots +\left| x+99\right|$
In the interval $\left[-4.6,-4.4\right]$ the function is
$f(x)=-x-x-1-x-2-x-3-x-4+(x+5)+(x+6)+(x+7)+(x+8)+(x+9) +\ldots+x+99$
the first $10$ terms containing $-x$ and $x$ simplify to $25$ while the other $90$ give
$f(x)=90x+M$
differentiating will give $f'(-9/2)=90$
Hope this helps
|
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|
Prove that if $x$ is divisible by $4$, then $x=a^2-b^2$
If $x\in Z$ and divisible by $4$, then there are existing $a,b \in Z$ , so: $x=a^2-b^2$
I had been trying to figure out which way this should be solved.
First I found out that if $x=20$, then $a^2 = 6^2$ and $b^2 = 4^2$ (of course $20=36-16$ ).
now I'm not sure if i could place those examples as an answer, or I need to prove it with some "proving tricks".
anyway, I thought about that $a^2 - b^2 = (a+b)(a-b)$ and $4\mid x$ is also $x=4k$ for some $k$ integer.
|
As $a+b\pm(a-b)$ are even,
$a+b, a-b$ have the same parity
Case$\#1:$ If $a+b$ is even, so will be $a-b\implies(a+b)(a-b)$ must be divisible by $4$
Let $(a+b)(a-b)=4m\iff\dfrac{a+b}2\cdot\dfrac{a-b}2=m$
If $m=pq,$ $\dfrac{a+b}2=p,\dfrac{a-b}2=q$
$\implies a=p+q, b=p-q$
Trivially choose $p$ or $q=1$
Case$\#2:$ If $a+b$ is odd, so will be $a-b\implies(a+b)(a-b)$ must be odd.
Let for odd $x(=uv),(a+b)(a-b)=uv$ where $u,v$ are odd
If $a+b=u,a-b=v;$
$a=\dfrac{u+v}2$ which is an integer as $u+v$ is even
$b=?$
Trivially set $u$ or $v=1$
|
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|
Show that $\sum\limits_{n=1}^{\infty}\frac{(H_{n})^2}{n(n+1)}=3\zeta(3)$, where, for every positive $n$, $H_n=\sum\limits_{k=1}^n\frac1k$ The problem I was considering about is the evaluation of the following series:
\begin{align*}
\sum_{n=1}^{\infty}\frac{(H_{n})^2}{n(n+1)}
\end{align*}
The attempt I could make was to change $(H_{n})^2$ to $H_{n}(H_{n+1}-\frac{1}{n+1})$ to see if the denominator and the numerator could match each other. Then the given series becomes
\begin{align*}
\sum_{n=1}^{\infty}\left(\frac{H_{n}H_{n+1}}{n(n+1)}-\frac{H_n}{n(n+1)^2}\right)
\end{align*}
The second term in the above series calculates to $\zeta{(3)}-\zeta{(2)}$. I have no idea how to evaluate the first term.
Wolfram alpha gives 3$\zeta(3)$ for the value of the desired series. So the first term must be somehow evaluated to $2\zeta(3)+\zeta(2)$.
Other ways to calculate the desired series are also appreciated, particularly if integrals were used.
|
using the identity $$ \sum_{n=1}^{\infty}x^n \left(H_n^2-H_n^{(2)}\right)=\frac{\ln^2(1-x)}{1-x}$$
divide both sides by $x$ then integrate w.r.t $x$ from $x=0$ to $y$ , we get
$$ \sum_{n=1}^{\infty}\frac{y^n}{n}\left(H_n^2-H_n^{(2)}\right)=\int_0^y\frac{\ln^2(1-x)}{1-x}\ dx-\frac13 {\ln^3(1-y)}$$
integrate both sides w.r.t $y$ from $y=0$ to $1$ , we get
\begin{align*}
\sum_{n=1}^{\infty}\frac{H_n^2-H_n^{(2)}}{n(n+1)}&=\int_0^1\int_0^y\frac{\ln^2(1-x)}{x}\ dxdy-\frac13\int_0^1{\ln^3(1-y)}\ dy\\
&=\int_0^1\frac{\ln^2(1-x)}{x}\left(\int_x^1dy\right)dx-\frac13\int_0^1{\ln^3(y)}\ dy\\
&=\int_0^1\frac{\ln^2(1-x)}{x}\left(1-x\right)dx-\frac13(-6)\\
&=\int_0^1\frac{\ln^2(1-x)}{x}dx-\int_0^1\ln^2(1-x)\ dx+2\\
&=\int_0^1\frac{\ln^2x}{1-x}dx-(2)+2\\
&=2\zeta(3)\tag{1}
\end{align*}
\begin{align*}
\sum_{n=1}^{N}\frac{H_n^{(2)}}{n(n+1)}&=\sum_{n=1}^{N}\frac{H_n^{(2)}}{n}-\sum_{n=0}^{N}\frac{H_n^{(2)}}{n+1}\\
&=\sum_{n=1}^{N}\frac{H_n^{(2)}}{n}-\sum_{n=1}^{N+1}\frac{H_{n-1}^{(2)}}{n}\\
&=\sum_{n=1}^{N}\frac{H_n^{(2)}}{n}-\sum_{n=1}^{N+1}\frac{H_{n}^{(2)}}{n}+\sum_{n=1}^{N+1}\frac1{n^3}\\
&=\sum_{n=1}^{N}\frac{H_n^{(2)}}{n}-\sum_{n=1}^{N}\frac{H_{n}^{(2)}}{n}-\frac{H_{N+1}^{(2)}}{N+1}+\sum_{n=1}^{N+1}\frac1{n^3}\\
&=-\frac{H_{N+1}^{(2)}}{N+1}+\sum_{n=1}^{N+1}\frac1{n^3}
\end{align*}
letting $N$ approach $\infty$ yields
$$\sum_{n=1}^{\infty}\frac{H_n^{(2)}}{n(n+1)}=0+\sum_{n=1}^\infty\frac1{n^3}=\zeta(3)\tag{2}$$
plugging $(2)$ in $(1)$ we have
$$\sum_{n=1}^{\infty}\frac{H_n^2}{n(n+1)}=3\zeta(3)$$
|
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|
How to solve $e^{2x - 1} = -x^2 + 3x - 0.25$ I know the answer is $x=0.5$, but I only know it because my calculator showed me the intersection point of the two graphs. What I want to know is how to get to this result algebraically. I've tried to use the Lambert-W function, but I ended up with $$W((1-2x)/(-x^2+3x-0.25))=1-2x$$ and this didn't help me either. If smn has an idea on how to solve this, pls let me know.
Obs: I don't know how to use the Newton-raphson's method, so that's why I haven't tried it.
|
For
$$e^{2x-1} + x^2 = 3 x - \frac{1}{4}$$
it can be seen that
\begin{align}
e^{2x-1} + x^2 &= 3 x - \frac{1}{4} \\
e^{2x -1} + x^{2} - x + \frac{1}{4} &= 2 x = 2 \left(x - \frac{1}{2}\right) + 1 \\
e^{2 x -1} + \left(x - \frac{1}{2}\right)^{2} - 2 \left( x - \frac{1}{2} \right) + 1 &= 2 \\
e^{2 x - 1} + \left( \left(x - \frac{1}{2}\right) - 1 \right)^{2} &= 2 \\
e^{2 \left(x - \frac{1}{2}\right)} + \left(x - \frac{3}{2}\right)^{2} &= 2.
\end{align}
The only solution to this equation is
$$x = \frac{1}{2}.$$
|
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|
Proving $(1+a)^n\ge 1+na+\frac{n(n-1)}{2}a^{2}+\frac{n(n-1)(n-2)}{6}a^3$ for all $n\in\mathbb N$ and all $a\ge -1.$ I was asked to prove the the following without induction. Could someone please verify whether my proof is right? Thank you in advance.
For any real number $a\ge -1$ and every natural number n, the statement, $$(1+a)^n\ge 1+na+\frac{n(n-1)}{2}a^{2}+\frac{n(n-1)(n-2)}{6}a^3,$$ holds.
Proof. From the binomial theorem we can see that, $$(1+a)^n\ge 1+na+\frac{n(n-1)}{2}a^{2}+\frac{n(n-1)(n-2)}{6}a^3,$$ becomes, $$\sum^{n}_{k=0}\binom{n}{k}a^{k}\ge\sum^{n}_{k=0}\binom{n}{k}a^{k}-\sum^{n}_{k=4}\binom{n}{k}a^{k}.$$ Also, if $a\lt b$, let us define $\sum^{a}_{i=b}f(i)=0$, by convention. Then for n=1 we have,\begin{align}\sum^{1}_{k=0}\binom{n}{k}a^{k}&\ge\sum^{1}_{k=0}\binom{n}{k}a^{k}-\sum^{1}_{k=4}\binom{n}{k}a^{k}\\ \sum^{1}_{k=0}\binom{n}{k}a^{k}&\ge\sum^{1}_{k=0}\binom{n}{k}a^{k}-0,\end{align} which is true, since both remaining sums are equal to each other. It is similarly the case for $n=2$ and $n=3$. More generally, \begin{align}\sum^{n}_{k=0}\binom{n}{k}a^{k}&\ge\sum^{n}_{k=0}\binom{n}{k}a^{k}-\sum^{n}_{k=4}\binom{n}{k}a^{k}\\ 1+\binom{n}{1}a^{1}+...+\binom{n}{k}a^{k}+...+\binom{n}{n}a^{n}&\ge1+\binom{n}{1}a^{1}+\binom{n}{2}a^{2}+\binom{n}{3}a^{3}\\ \binom{n}{4}a^4+\binom{n}{5}a^5+...+\binom{n}{k}a^{k}+...+\binom{n}{n}a^{n}&\ge0.\end{align}
To determine whether this last statement is true, we first consider $a\ge0$, and we see that it is trivially true. Next we consider the case where $-1\le a\lt0.$ We let $a=-b$, so $0\lt b\le 1$ and notice that, $$\binom{n}{4}a^4+\binom{n}{5}a^5+...+\binom{n}{k}a^{k}+...+\binom{n}{n}a^{n}$$ then becomes, $$\binom{n}{4}b^4-\binom{n}{5}b^5+...+(-1)^{k}\binom{n}{k}b^{k}+...+(-1)^{n}\binom{n}{n}b^{n}.$$
Also from the binomial theorem we can see that,
\begin{align}
(1+(-1))^{n}-(1+(-1))^{3}+(1+(-1))^{1}\\
=\sum^{n}_{k=0}\binom{n}{k}1^{n-k}(-1)^{k}-\sum^{3}_{k=0}\binom{n}{k}1^{n-k}(-1)^{k}+\sum^{1}_{k=0}\binom{n}{k}1^{n-k}(-1)^{k}\\
0=1-\binom{n}{1}+\binom{n}{4}-\binom{n}{5}+\binom{n}{6}+...+(-1)^{k}\binom{n}{k}+...+(-1)^{n}\binom{n}{n}\\
n-1=\binom{n}{4}-\binom{n}{5}+\binom{n}{6}+...+(-1)^{k}\binom{n}{k}+...+(-1)^{n}\binom{n}{n}\ge0,
\end{align}
and by noticing that $b^4\ge b^5\ge b^6\ge ...\ge b^k$, we can see that each binomial term $\binom{n}{k}$ is multiplied by a factor of $b^k$, making each term smaller than the term before.
Thus,
\begin{align}
\binom{n}{4}b^4-\binom{n}{5}b^5+...+(-1)^n\binom{n}{n}b^n\ge0\\
\binom{n}{4}a^4+\binom{n}{5}a^5+...+\binom{n}{n}a^n\ge0,\tag{by substitution}
\end{align} as desired.
Working backward, \begin{align}\binom{n}{4}a^4+\binom{n}{5}a^5+...+\binom{n}{k}a^{k}+...+\binom{n}{n}a^{n}&\ge0\\ 1+\binom{n}{1}a^{1}+...+\binom{n}{k}a^{k}+...+\binom{n}{n}a^{n}&\ge1+\binom{n}{1}a^{1}+\binom{n}{2}a^{2}+\binom{n}{3}a^{3}\\ \sum^{n}_{k=0}\binom{n}{k}a^{k}&\ge\sum^{n}_{k=0}\binom{n}{k}a^{k}-\sum^{n}_{k=4}\binom{n}{k}a^{k}\\ (1+a)^n&\ge 1+na+\frac{n(n-1)}{2}a^{2}+\frac{n(n-1)(n-2)}{6}a^3.\ \ \ \blacksquare\end{align}
|
proof-verification:
Proof. From the binomial theorem we can see that,
$$
(1+a)^n\ge 1+na+\frac{n(n-1)}{2}a^{2}+\frac{n(n-1)(n-2)}{6}a^3,
$$
becomes ("is equivalent to"),
$$
\sum^{n}_{k=0}\binom{n}{k}a^{k}\ge\sum^{n}_{k=0}\binom{n}{k}a^{k}-\sum^{n}_{k=4}\binom{n}{k}a^{k}.
$$
Also, if $a\lt b$ (I would use other notations here so that I would not mess up with the "a" in the inequality), let us define $\sum^{a}_{i=b}f(i)=0$, by convention.
I would not bother with the cases n=1,2,3: one can just claim that one can just do it by direct substitution.
More generally Suppose $n\geqslant 4$. Then it suffices to show that (you don't need to write so much here. I have tailored some of your steps.)
$$
\sum^{n}_{k=4}\binom{n}{k}a^{k}\geq 0
$$
which is
$$
\binom{n}{4}a^4+\binom{n}{5}a^5+...+\binom{n}{k}a^{k}+...+\binom{n}{n}a^{n}\ge0.
$$
To determine whether this last statement is true, We first consider $a\ge0$, and we see that it is trivially true. Next we consider the case where $-1\le a\lt0.$ We let $a=-b$ (I would write b=-a instead. When defining something new, it is clearer to write it on the left.), so $0\lt b\le 1$ and notice that,
$$
\binom{n}{4}a^4+\binom{n}{5}a^5+...+\binom{n}{k}a^{k}+...+\binom{n}{n}a^{n}
\\
=\binom{n}{4}b^4-\binom{n}{5}b^5+...+(-1)^{k}\binom{n}{k}b^{k}+...+(-1)^{n}\binom{n}{n}b^{n}.
$$
Also from the binomial theorem we can see that (A possible big gap here: this step is mysterious to me and totally unmotivated: how do you suddenly get rid of the a and b in the equality above and jump to the following one? I would stop reading here.),
\begin{align}
(1+(-1))^{n}-(1+(-1))^{3}+(1+(-1))^{1}\\
=\sum^{n}_{k=0}\binom{n}{k}1^{n-k}(-1)^{k}-\sum^{3}_{k=0}\binom{n}{k}1^{n-k}(-1)^{k}+\sum^{1}_{k=0}\binom{n}{k}1^{n-k}(-1)^{k}\\
0=1-\binom{n}{1}+\binom{n}{4}-\binom{n}{5}+\binom{n}{6}+...+(-1)^{k}\binom{n}{k}+...+(-1)^{n}\binom{n}{n}\\
n-1=\binom{n}{4}-\binom{n}{5}+\binom{n}{6}+...+(-1)^{k}\binom{n}{k}+...+(-1)^{n}\binom{n}{n}\ge0,
\end{align}
and by noticing that $b^4\ge b^5\ge b^6\ge ...\ge b^k$, we can see that each binomial term $\binom{n}{k}$ is multiplied by a factor of $b^k$, making each term smaller than the term before.
Thus (Assuming the above is correct, I don't understand this step either.),
\begin{align}
\binom{n}{4}b^4-\binom{n}{5}b^5+...+(-1)^n\binom{n}{n}b^n\ge0\\
\binom{n}{4}a^4+\binom{n}{5}a^5+...+\binom{n}{n}a^n\ge0,\tag{by substitution}
\end{align} as desired.
Working backward(the rest is redundant. Writing could be precise without sacrificing any rigorousness.)
|
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|
How to show this function is a metric: $d(x,y) = \frac{2|x-y|}{\sqrt{1+x^2}+\sqrt{1+y^2}}$? Consider $( \mathbb{R}, d )$ where $$d(x,y) = \frac{2|x-y|}{\sqrt{1+x^2}+\sqrt{1+y^2}}.$$
The problem arises in showing triangle inequality. I can deal with the numerator only with the help of triangle inequality for the mod in $\mathbb{R}$ but for the denominator I get no clue how to handle?
|
Let $x\geq y\geq z$.
Hence, $$\frac{|x-z|}{\sqrt{1+x^2}+\sqrt{1+z^2}}\geq\frac{|x-y|}{\sqrt{1+x^2}+\sqrt{1+y^2}}$$ it's
$$\frac{x-z}{\sqrt{1+x^2}+\sqrt{1+z^2}}\geq\frac{x-y}{\sqrt{1+x^2}+\sqrt{1+y^2}}$$ or
$$\frac{y-z}{\sqrt{1+x^2}+\sqrt{1+z^2}}\geq\frac{x-y}{\sqrt{1+x^2}+\sqrt{1+y^2}}-\frac{x-y}{\sqrt{1+x^2}+\sqrt{1+z^2}}$$ or
$$y-z\geq\frac{(x-y)(z^2-y^2)}{\left(\sqrt{1+x^2}+\sqrt{1+y^2}\right)\left(
\sqrt{1+y^2}+\sqrt{1+z^2}\right)}$$ or
$$\left(\sqrt{1+x^2}+\sqrt{1+y^2}\right)\left(
\sqrt{1+y^2}+\sqrt{1+z^2}\right)\geq(y-x)(y+z),$$
which is true because
$$\left(\sqrt{1+x^2}+\sqrt{1+y^2}\right)\left(
\sqrt{1+y^2}+\sqrt{1+z^2}\right)\geq$$
$$\geq(|x|+|y|)(|y|+|z|)\geq|(y-x)(y+z)|\geq(y-x)(y+z).$$
By the same way we can prove that
$$\frac{|x-z|}{\sqrt{1+x^2}+\sqrt{1+z^2}}\geq\frac{|y-z|}{\sqrt{1+y^2}+\sqrt{1+z^2}}.$$
Thus, it's enough to prove that
$$\frac{x-y}{\sqrt{1+x^2}+\sqrt{1+y^2}}+\frac{y-z}{\sqrt{1+y^2}+\sqrt{1+z^2}}\geq\frac{x-z}{\sqrt{1+x^2}+\sqrt{1+z^2}}$$ or
$$\tfrac{x-y}{\sqrt{1+x^2}+\sqrt{1+y^2}}+\tfrac{y-z}{\sqrt{1+y^2}+\sqrt{1+z^2}}\geq\tfrac{x-y}{\sqrt{1+x^2}+\sqrt{1+z^2}}+\tfrac{y-z}{\sqrt{1+x^2}+\sqrt{1+z^2}}$$ or
$$\frac{(x-y)(\sqrt{1+z^2}-\sqrt{1+y^2})}{\sqrt{1+x^2}+\sqrt{1+y^2}}\geq\frac{(y-z)(\sqrt{1+y^2}-\sqrt{1+x^2})}{\sqrt{1+y^2}+\sqrt{1+z^2}}$$ or
$$(x-y)(z^2-y^2)\geq(y-z)(y^2-x^2)$$
$$(x-y)(y-z)(x+y)-(x-y)(y-z)(y+z)\geq0$$ or
$$(x-y)(y-z)(x-z)\geq0,$$ which is obvious.
Done!
|
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|
Prove $ \frac{1\cdot 3 \cdot 5 \cdot 7 \cdot \dots \cdot 2007}{2 \cdot 4 \cdot 6 \cdot \dots \cdot 2008}< \frac{1}{40} $ Prove:
$$
\frac{1\cdot 3 \cdot 5 \cdot 7 \cdot \dots \cdot 2007}{2 \cdot 4 \cdot 6 \cdot \dots \cdot 2008}< \frac{1}{40}
$$
i have tried to write $1/40$ as $(1/40^{1/2007})^{2007}$ and prove $(1/40^{1/2007})^{2007}$ to be greater than $2007/2008$ but i quickly found out this is not true. Is there another way to manipulate this product? I think telescoping is possible but I just don't know how to do it; maybe split the fractions and work it out? Any help will be appreciated thank you.
|
Let
$$
A=\frac{1\cdot 3 \cdot 5 \cdot 7 \cdot \dots \cdot 2007}{2 \cdot 4 \cdot 6 \cdot \dots \cdot 2008}
$$
$$
B=\frac{2 \cdot 4 \cdot 6 \cdot \dots \cdot 2008}{1\cdot 3 \cdot 5 \cdot 7 \cdot \dots \cdot 2009}
$$
Then $$A<B$$
Then $$A^2<AB=\frac1{2009}$$
Then $$A<\frac1{\sqrt{2009}}<\frac1{\sqrt{1600}}=\frac1{40}$$
Generally:
$$ {1\over 2} \cdot {3\over 4} \cdot {5 \over 6} \cdot \ldots \cdot {2n-1 \over 2n} < {1 \over \sqrt{2n+1}}$$
$$A={1\over 2}\cdot{3\over 4}\cdot{5 \over 6}\cdot\ldots\cdot{2n-1\over 2n}, \quad B={2 \over 3} \cdot{4\over 5}\cdot{6\over 7}\cdot\ldots\cdot{2n \over 2n+1}.$$
$$A < B$$
$$A^2 < AB = {1 \over 2n+1}.$$
|
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|
Prove $ ( \frac{\sin ^{2}\theta}{2}+\frac{2}{\cos ^{2}\theta} )^{1/4} + ( \frac{\cos ^{2}\theta}{2}+\frac{2}{\sin ^{2}\theta} )^{1/4}\ge 68^{1/4}$
Let $0<\theta<\pi/2$. Prove that $$\left ( \frac{\sin ^{2}\theta}{2}+\frac{2}{\cos ^{2}\theta} \right )^{1/4}+\left ( \frac{\cos ^{2}\theta}{2}+\frac{2}{\sin ^{2}\theta} \right )^{1/4}\geqslant (68)^{1/4}$$
and find when the equality case holds.
This is a competition math problem. The material used should only cover up to pre-calculus.
So I quickly found out that equality holds when both of the $\sin^2(\theta)$ and $\cos^2(\theta)$ equals to 1/2, but I am not sure how to prove that this equality is true. I tried to substitute for variables and also use trig identities but just can't find out a way to do this. Thank you guys for helping me.
|
By Minkowski (see here: https://en.wikipedia.org/wiki/Minkowski_inequality) we obtain
$$\sqrt[4]{ \frac{\sin ^{2}\theta}{2}+\frac{2}{\cos ^{2}\theta}}+\sqrt[4]{ \frac{\cos ^{2}\theta}{2}+\frac{2}{\sin ^{2}\theta}}\geq\sqrt[4]{\frac{1}{2}(\sqrt{\sin\theta}+\sqrt{\cos\theta})^4+2\left(\frac{1}{\sqrt{\sin\theta}}+\frac{1}{\sqrt{\cos\theta}}\right)^4}.$$
Thus, it remains to prove that
$$\frac{1}{2}(\sqrt{\sin\theta}+\sqrt{\cos\theta})^4+2\left(\frac{1}{\sqrt{\sin\theta}}+\frac{1}{\sqrt{\cos\theta}}\right)^4\geq68.$$
Now, let $\sin\theta+\cos\theta=2k\sqrt{\sin\theta\cos\theta}$.
Thus, $k\geq1$ and we need to prove that
$$\frac{1}{2}(\sqrt{\sin\theta}+\sqrt{\cos\theta})^4+2(\sin^2\theta+\cos^2\theta)^2\left(\frac{1}{\sqrt{\sin\theta}}+\frac{1}{\sqrt{\cos\theta}}\right)^4\geq68(\sin^2\theta+\cos^2\theta)$$ or
$$(k+1)^2+16(2k^2-1)^2(k+1)^2\geq68(2k^2-1)$$ or
$$(k-1)(64k^5+192k^4+192k^3+64k^2-119k-85)\geq0,$$
which is obvious.
Done!
|
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|
Show that the lines joining the origin to the point of intersection of the line Show that the lines joining the origin to the point of intersection of the line $x+y=1$ with the curve $4x^2+4y^2+4x-2y-5=0$ are at right angles to each other.
How do I approach this? Please help.
|
Solving of the system gives
$$4x^2+4(1-x)^2+4x-2(1-x)-5=0$$ or
$$8x^2-2x-3=0,$$ which gives the following points: $\left(\frac{3}{4},\frac{1}{4}\right)$ and $\left(-\frac{1}{2},\frac{3}{2}\right)$.
Now, $$m_1=\frac{\frac{1}{4}}{\frac{3}{4}}=\frac{1}{3}$$ and
$$m_2=\frac{\frac{3}{2}}{-\frac{1}{2}}=-3$$ and since $m_1m_2=-1$, we are done!
|
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|
Real root of $f(x) = 1+2x+3x^2 +4x^3$ Consider the polynomial $f(x) = 1+2x+3x^2 +4x^3$. Let $s$ be the sum of all distinct real roots of $f(x)$ and let $t = |s|$.
The real number $s$ lies in the interval
A) $\ \ (-\frac{1}{4},0)$
B) $ \ \ (−11,−\frac{3}{4}) $
C) $\ \ (−\frac{3}{4},−\frac{1}{2})$
D) $\ \ (0,\frac{1}{4})$
Answer is 'C'
My approach, $f'(x)=2+6x+12x^2$
$D<0$ for $f'(x)$ hence $f(x)$ has one real root. The root of $f(x)$ are $a , c+id$ and $c-id$.
$a+2c=-\frac{3}{4}$ and $a(c^2-d^2)=-1$
I am not able to proceed from here.
|
f(x) is a polynomial function and so it is continuous and differentiable through out. We have by intermediate value theorem, if f is continous in $[a,b]$ and if $f(a)<0, f(b)>0$ then there exists $c \in (a,b)$ such that f(c) = 0
in above options, consider (C):
$f(-1/2) = positve$
$f(-3/4) = negative$
So there should exist by intermediate value theorem, $c \in (-3/4, -1/2)$ such that f(c)=0
|
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|
Prime factorization of integers Find the prime factorization of the following integers:
$e) 2^{30} -1$
Click here for the solutions
I used the formulas:
$a^2-1^2=(a-1)(a+1)$
$a^3+b^3=(a+b)(a^2-ab+b^2)$
$a^3-b^3=(a-b)(a^2+ab+b^2)$
And I became:
$2^{30} -1$
$=(2^{15}-1)(2^{15}+1)$
$=(2^5-1)(2^{10}+2^5+1)(2^5+1)(2^{10}-2^5+1)$
$=31*1057*32*993$
$=2^5*3*7*31*151*331$
What did I do wrong?
|
One good method to obtain some factors of numbers of the form $a^n-1$ is to
look at the factorization of $x^{n}-1$ into cyclotomic polynomials:
$$
x^n - 1 = \prod_{d\mid n}\Phi_d(x)
$$
$
x^6-1
= \Phi_{1} \Phi_{2} \Phi_{3} \Phi_{6}
=(x - 1) (x + 1) (x^2 + x + 1) (x^2 - x + 1)
$.
Therefore, $10^6-1=9 \cdot 11 \cdot 111 \cdot 91$. It remains to factor $111$ and $91$.
$
x^8-1
= \Phi_{1} \Phi_{2} \Phi_{4} \Phi_{8}
=(x - 1) (x + 1) (x^2 + 1) (x^4 + 1)
$.
Therefore, $10^8-1=9 \cdot 11 \cdot 101 \cdot 10001$. It remains to factor $10001$.
You get the idea...
Of course, finding the factorization of $x^{n}-1$ into cyclotomic polynomials is mostly an application of the identity $x^n-1 = (x-1)(x^{n-1}+ \cdots +x + 1)$ to various factors of $n$.
|
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|
The limit of a sum: $\lim_{n\to \infty} \sum_{k=1}^{n} \left(\frac{k}{n^2}-\frac{k^2}{n^3}\right)$
Evaluate the following limit:
$$
\lim_{n\to \infty} \sum_{k=1}^{n} \left(\frac{k}{n^2}-\frac{k^2}{n^3}\right)
$$
I haven't ever taken the limit of the sum... Where do I start?
Do I start taking the sum?
|
With Riemann sums:
We have
$$
\sum_{k=1}^n \left(\frac{k}{n^2}-\frac{k^2}{n^3}\right)
= \frac{1}{n}\sum_{k=1}^n \left(\frac{k}{n}-\left(\frac{k}{n}\right)^2\right)
= \frac{1}{n}\sum_{k=1}^n \frac{k}{n}\left(1-\frac{k}{n}\right)
$$
which is a Riemann sum for $f\colon[0,1]\to\mathbb{R}$ defined by $f(x)=x(1-x)$. Therefore, we have
$$
\lim_{n\to\infty} \sum_{k=1}^n \left(\frac{k}{n^2}-\frac{k^2}{n^3}\right) = \int_0^1 f(x)dx = \left[\frac{x^2}{2}-\frac{x^3}{3}\right]^1_0 =\boxed{ \frac{1}{6}}\,.
$$
Without Riemann sums:
Here, you can directly use the facts that $\sum_{k=1}^n k = \frac{n(n+1)}{2}$ and $\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}$ to compute the sum, and conclude afterwards.
$$
\sum_{k=1}^n \left(\frac{k}{n^2}-\frac{k^2}{n^3}\right)
= \frac{1}{n^2}\sum_{k=1}^n k-\frac{1}{n^3}\sum_{k=1}^n k^2 = \frac{n(n+1)}{2n^2}-\frac{n(n+1)(2n+1)}{6n^3} \xrightarrow[n\to\infty]{} \frac{1}{2} - \frac{1}{3} = \boxed{\frac{1}{6}}\,.
$$
|
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|
Ordinary Generating Function for the number of solution :$ x_1 + x_2 + \cdot\cdot\cdot + x_k = n$ Let $a_n$ be the number of solutions of the equation:
$$x_1 + x_2 + \cdot\cdot\cdot + x_k = n$$
where $x_i$ is the positive odd integer.
Hereto I would like to find the ordinary generating function for the sequence $a_n$
ODG has is a function satisfies the below :
$$G(a_n;n) = \sum_0^\infty a_nx^n$$
But what is the typical algorithm to find the ordinary generating function that uniquely determines its coefficient that corresponds to the regarding combinatorial situation?
Anyone can guide me where to start from?
|
Let $X$ be the set of positive odd numbers.
For any formal power series $f(t) = \sum_{j=0}^\infty f_k t^k$, let $[t^k]f(t)$ be the coefficient $f_k$.
When one expand following product into a series of $t$.
$$\left(\sum_{x_1\in X} t^{x_1}\right)\cdots\left(\sum_{x_k\in X} t^{x_k}\right)
= \sum_{x_1,\ldots,x_k \in X} t^{x_1+\ldots + x_k}
= \sum_{n=1}^\infty \alpha_n t^n$$
One will notice there is a one-one correspondence between terms contributing
to $\alpha_n$ and solution $x_1 + \ldots + x_k = n$ for $(x_1,\ldots,x_k) \in X^k$. For example, in the expansion of
$$(t + t^3 + t^5 + \cdots)(t + t^3 + t^5 + \cdots)(t + t^3 + t^5 + \cdots) =
x^3+3x^5+6x^7+10x^9 + \cdots
$$
There are $6$ terms that contribute to $\alpha_7 = 6$. Namely,
$$
\begin{array}{lcr}
(t + t^3 + \color{blue}{t^5} + \cdots)
(\color{blue}{t} + t^3 + t^5 + \cdots)
(\color{blue}{t} + t^3 + t^5 + \cdots) &\leftrightarrow & 5+1+1 = 7\\
(\color{blue}{t} + t^3 + t^5 + \cdots)
(t + t^3 + \color{blue}{t^5} + \cdots)
(\color{blue}{t} + t^3 + t^5 + \cdots) &\leftrightarrow & 1+5+1 = 7\\
(\color{blue}{t} + t^3 + t^5 + \cdots)
(\color{blue}{t} + t^3 + t^5 + \cdots)
(t + t^3 + \color{blue}{t^5} + \cdots) &\leftrightarrow & 1+1+5 = 7\\
(\color{blue}{t} + t^3 + t^5 + \cdots)
(t + \color{blue}{t^3} + t^5 + \cdots)
(t + \color{blue}{t^3} + t^5 + \cdots)&\leftrightarrow & 1+3+3 = 7\\
(t + \color{blue}{t^3} + t^5 + \cdots)
(\color{blue}{t} + t^3 + t^5 + \cdots)
(t + \color{blue}{t^3} + t^5 + \cdots)&\leftrightarrow & 3+1+3 = 7\\
(t + \color{blue}{t^3} + t^5 + \cdots)
(t + \color{blue}{t^3} + t^5 + \cdots)
(\color{blue}{t} + t^3 + t^5 + \cdots)&\leftrightarrow & 3+3+1 = 7\\
\end{array}
$$
In general, this leads to
$$\alpha_n = \sum_{n = \sum_{j=1}^k x_j} 1 = a_n
\quad\leftarrow \text{ the number of solutions we seek. }
$$
Since $\displaystyle\;\sum_{x\in X}t^x = \sum_{j=0}^\infty t^{2j+1} =
t \sum_{j=0}^\infty (t^2)^j = \frac{t}{1-t^2},$ the OGF at hand is simply
$$\sum_{n=0}^\infty a_n t^n = \frac{t^k}{(1-t^2)^k}
\quad\implies\quad a_n = [t^n] \frac{t^k}{(1-t^2)^k} = [t^{n-k}]\frac{1}{(1-t^2)^k}
$$
In order for $a_n \ne 0$, we need $n \ge k$ and $n$ and $k$ has same parity.
When this happens, we have
$$a_n = [t^{(n-k)/2}] \frac{1}{(1-t)^k} = \binom{\frac{n-k}{2} + k - 1}{\frac{n-k}{2}} = \binom{\frac{n+k}{2}-1}{\frac{n-k}{2}}$$
Note
Above answer is under the assumption we want the number of solutions for a fixed $k$. If one want the number of solutions for a given $n$ but the number of $k$ can vary, the corresponding OGF will be
$$1 + \frac{t}{1-t^2} + \frac{t^2}{(1-t^2)^2} + \cdots = \sum_{k=0}^\infty \left(\frac{t}{1-t^2}\right)^k
= \frac{1}{1 - \frac{t}{1-t^2}} = 1 + \frac{t}{1 - t - t^2}
$$
The rightmost expression is the famous OGF for
Fibonacci numbers.
$$\frac{t}{1 - t - t^2} = \sum_{n=1}^\infty F_n t^n$$
This means for any $n > 0$, the number of ways of breaking
$n$ into an ordered list of odd numbers $(x_1, \ldots, x_k)$ which sum to $n$ is simply $F_n$.
|
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|
Coefficient of $x^{n-2}$ in $(x-1)(x-2)(x-3)\dotsm(x-n)$ Question
Find the coefficient of $x^{n-2}$ in the expression $$(x-1)(x-2)(x-3)\dots(x-n)~~.$$
My approach
The coefficient of $x^n$ is $1$. The coefficient of $x^{n-1}$ is $- \frac{n(n+1)}{2}$
But I cannot proceed from here.
I would appreciate any help.
|
For $~x^{n-1}~$ the coefficient would be sum of the roots that is $~1+2+3+4+5+\cdots+n~$ for $~x^{n-2}~$ coefficient would be sum of roots taken two at a time that is $~1.2+2.3+3.4+4.5+\cdots+(n-1)n~$ so let it be $~S~$ them $~S~$ can be find using steps below
$$(1+2+3+4+5+\cdots+n)^2=(1^2+2^2+3^3+\cdots+n^2)+2(1.2+2.3+3.4+\cdots+(n-1)n)$$
So
$$(1+2+3+4+5+\cdots+n)^2=(1^2+2^2+3^3+\cdots+n^2)+2(S)$$
So this can be find as we know sum of $~n~$ natural no
That is $~\frac{1}{2}\{n(n+1)\}~$ and sum of squares of first $~n~$ natural number is $~\frac{1}{6}\{n(n+1)(2n+1)\}~$ so this value can be found
Thanks for opportunity to solve the problem
|
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|
Help with proving the surjectivity of a function satisfying $ f ( x + y ) \big( f ( x ) + f ( y ) \big) = f ( x y ) $ I have the following functional equation for injective $ f : \mathbb R ^ * \to \mathbb R ^ * $
$$ x + y \ne 0 \implies f ( x + y ) \big( f ( x ) + f ( y ) \big) = f ( x y ) $$
Found $ f ( 2 ) = \frac 1 2 $ and I am trying to find $ f ( 1 ) $, so I let
$$ x + y = x y \implies y = \frac x { x - 1 } $$
$$ f \left( \frac x { x - 1 } \right) = 1 - f ( x ) $$
then declared $ g : \mathbb R ^ * \setminus \{ 1 \} \to \mathbb R ^ * \setminus \{ 1 \} $
$$ g(x) = \frac{x}{x-1} $$
and was able to prove $ g $'s bijectivity. So, if I can prove $ f $ is surjetive (or alternatively that it is multiplicative or that, at least $ \exists x \, f ( x ) = 1 $) it follows that $ f ( 1 ) = 1 $, $ \forall q \in \mathbb Q \, f ( q ) = q ^ { - 1 } $ and from this point it should be easy to finish the problem.
Am I going the wrong path? Is there an easier way to deal with this problem? Can someone give me hints?
Update:
I was able to make a system of equations with $ f ( - 1 ) $, $ f ( 1 ) $, $ f ( 2 ) $ and $ f ( - 2 ) $, and found $ f ( - 1 ) = - 1 $, $ f ( - 2 ) = \frac { - 1 } 2 $, $ f ( 1 ) = 1$ and $f \left( \frac 3 2 \right) = \frac 2 3 $. Not as elegant as the method I wanted to use but it works.
Using the identity with $ f \circ g ( x ) $ got the values for $ f \left( \frac k { k - 1 } \right)$ plugging positive integers and $ f \left( \frac { k - 1 } k \right) $ with negative ones. Now I am doing some manipulation so I can get from this to $ \frac 1 k $ and from there it should be easy to induce for the rationals. I will post my calculations soon.
The problem is, having no info about continuity, how could we expand this to the reals? That's harder than I thought it would be. We do know, however, the function is decreasing for positive rational numbers. This may be useful.
Second Update:
I got it, the secret is to look at $ \frac 1 { f ( x ) } $, will probably answer this question myself next week.
|
You can show that the only functions satisfying (I will assume that $ x , y \ne 0 $ everywhere I use them)
$$ x + y \ne 0 \Longrightarrow f ( x + y ) \big( f ( x ) + f ( y ) \big) = f ( x y ) \tag 0 \label 0 $$
where $ f : \mathbb R ^ * \to \mathbb R ^ * $, are $ f ( x ) = \frac 1 x $ and $ f ( x ) = \frac 1 2 $. It's obvious that these two are solutions. To show that these are the only solutions, we define $ g : \mathbb R ^ * \to \mathbb R ^ * $ such that $ g( x ) = \frac 1 { f ( x ) } $. Then \eqref{0} yields
$$ x + y \ne 0 \Longrightarrow \frac { g ( x ) + g ( y ) } { g ( x + y ) } = \frac { g ( x ) g ( y ) } { g ( x y ) } \tag 1 \label 1 $$
It's worth mentioning that \eqref{1} tells us that if $ x + y \ne 0 $ then $ g ( x ) + g ( y ) \ne 0 $. Now, letting $ x = y = 2 $ in \eqref{1} we get $ g ( 2 ) = 2 $. Putting $ x = 2 $ and $ y = - 1 $ gives us $ \frac { g ( - 1 ) + 2 } { g ( 1 ) } = \frac { 2 g ( - 1 ) } { g ( - 2 ) } $, and letting $ x = y = -1 $ we'll have $ \frac { 2 g ( - 1 ) } { g ( - 2 ) } = \frac { g ( - 1 ) ^ 2 } { g ( 1 ) } $. Combining these two equations, we get $ g ( - 1 ) = - 1 $ or $ g ( - 1 ) = 2 $.
Let's suppose that $ g ( - 1 ) = 2 $. Then by the above equations $ g ( - 2 ) = g ( 1 ) $. Now, letting $ x = 1 $ and $ y = -3 $ in \eqref{1} we have $ g ( 1 ) + g ( - 3 ) = g ( 1 ) ^ 2 $, while $ x = - 1 $ and $ y = - 2 $ gives us $ 2 + g ( 1 ) = g ( 1 ) g ( - 3 ) $. Combining these two and rearranging the terms, we get $ \big( g ( 1 ) - 2 \big) \big( g ( 1 ) ^ 2 + g ( 1 ) + 1 \big) = 0 $, and hence $ g ( 1 ) = 2 $. Therefore, if $ x \ne - 1 $ then letting $ y = 1 $ in \eqref{1} we have $ g ( x + 1 ) = 1 + \frac { g ( x ) } 2 $, or equivalently, we have $ g ( x - 1 ) = 2 \big( g ( x ) - 1 \big) $, for $ x \ne 1 $. Now, if $ x \ne 1 $ then letting $ y = - 1 $ in \eqref{1} we'll have $ g ( - x ) = \frac { 4 g ( x ) ( g ( x ) - 1 ) } { g ( x ) + 2 } $. Substituting $ - x $ for $ x $ in the last equation, and using the equation itself, one can get
$$ g ( x ) = \frac { 4 \frac { 4 g ( x ) ( g ( x ) - 1 ) } { g ( x ) + 2 } \left( \frac { 4 g ( x ) ( g ( x ) - 1 ) } { g ( x ) + 2 } - 1 \right) } { \frac { 4 g ( x ) ( g ( x ) - 1 ) } { g ( x ) + 2 } + 2 } $$
for $ x \ne -1 , 1 $. Straightforward algebraic manipulation yields $ 3 g ( x ) \big( g ( x ) - 2 \big) \big( 10 g ( x ) ^ 2 - 5 g ( x ) - 2 \big) = 0 $.
$ g ( x ) $ can't be equal to $ \frac { 5 \pm \sqrt {105} } { 20 } $ since in that case the value of $ g ( x + 1 ) = 1 + \frac { g ( x ) } 2 $ or $ g ( x - 1 ) = 2 \big( g ( x ) - 1 \big) $ would be unacceptable. Thus we have $ g ( x ) = 2 $ for $ x \ne -1 , 1 $, which easily shows that $ f ( x ) = \frac 1 2 $ for every $ x $.
Now, suppose $ g ( - 1 ) = - 1 $. This shows that $ g ( - 2 ) = - 2 g ( 1 ) $. Using this fact, if we let $ x = 1 $ and $ y = - 3 $ in \eqref{1} we get $ g ( 1 ) + g ( - 3 ) = - 2 g ( 1 ) ^ 2 $, and for $ x = - 1 $ and $ y = - 2 $ we have $ - 1 - 2 g ( 1 ) = g ( 1 ) g ( - 3 ) $. Combining these equations we have $ \big( 2 g ( 1 ) + 1 \big) \big( g ( 1 ) + 1 \big) \big( g ( 1 ) - 1 \big) = 0 $. If $ g ( 1 ) = - \frac 1 2 $ then by the above equation $ g ( - 3 ) = 0 $ which can't happen. We can also rule out the case $ g ( 1 ) = - 1 $ by considering different values of $ x $ and $ y $ in \eqref{1}, which I omit. Thus we have $ g ( 1 ) = 1 $.
Now, if $ x \ne - 1 $, letting $ y = 1 $ in \eqref{1} we have $ g ( x + 1 ) = g ( x ) + 1 $, or equivalently $ g ( x - 1 ) = g ( x ) - 1 $ for $ x \ne 1 $.
Thus letting $ x = - 1 $ in \eqref{1}, we get $ g ( - x ) = - g ( x ) $ for $ x \ne 1 $, which can easily be generalized to include every $ x $, and thus $ g $ is an odd function. Now, using a simple induction, we can get $ g ( x + n ) = g ( x ) + n $, for every nonnegative integer $ n $ and $ x \ne - n , - n + 1 , \dots , - 1 $. Letting $ y = n $ in \eqref{1}, we get $ g ( n x ) = n g ( x ) $ for $ x \ne - n , - n + 1 , \dots , - 1 $, since by a simple induction we have $ g ( n ) = n $. This together with $ g $ being odd, shows that in fact for every $ x $ and every nonzero integer $ n $, $ g ( n x ) = n g ( x ) $. Now letting $ y = x $ in \eqref{1}, we get $ g \big( x ^ 2 \big) = g ( x ) ^ 2 $. This shows that if $ x > 0 $ then $ g ( x ) > 0 $, and since $ g $ is odd, the converse is also true. Now, for every two integers $ m $ and $ n $, if $ m \ne 0 $ and $ m x + n \ne 0 $, then we have $ m x + n > 0 $ iff $ g ( m x + n ) > 0 $. Thus $ m x + n > 0 $ iff $ m g ( x ) + n > 0 $. This means that $ x $ and $ g ( x ) $ must be equal, since otherwise we can find a rational number between them, which will give a counterexample to what we've just proven. Thus in this case we have $ f ( x ) = \frac 1 x $ for every $ x $.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Sum of odd terms of a binomial expansion: $\sum\limits_{k \text{ odd}} {n\choose k} a^k b^{n-k}$ Is it possible to find a closed form expression for the sum
$$\sum_{k \text{ odd}} {n\choose k} a^k b^{n-k}$$
in terms of $a$ and $b$ ?
|
I found a rather interesting alternate method to solve this problem (from a problem in communication theory). Please share your thoughts.
Let
$
A =
\left[ {\begin{array}{cc}
a & b \\
b & a \\
\end{array} } \right]
$
Now, $$[A^n]_{1,1} = \sum_{k \text{ even}} {n \choose k} a^k b^{n-k}$$
and $$[A^n]_{1,2} = \sum_{k \text{ odd}} {n \choose k} a^k b^{n-k}$$
$A$ can be decomposed as
$$ A = T^{-1}
\left[ {\begin{array}{cc}
a+b & 0 \\
0 & a-b \\
\end{array} } \right]T $$
where $ T = \left[ {\begin{array}{cc}
1 & 1 \\
1 & -1 \\
\end{array} } \right] $
Now, $$ A^n = T^{-1}
\left[ {\begin{array}{cc}
(a+b)^n & 0 \\
0 & (a-b)^n \\
\end{array} } \right]T $$
$$ A^n =
\left[ {\begin{array}{cc}
\frac{1}{2}[(a+b)^n + (a-b)^n] & \frac{1}{2}[(a+b)^n - (a-b)^n] \\
\frac{1}{2}[(a+b)^n - (a-b)^n] & \frac{1}{2}[(a+b)^n + (a-b)^n] \\
\end{array} } \right] $$
which gives the odd and even terms of the binomial expression.
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/2528974",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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|
Integration by parts of $x^5\ln(x)$ I have to integrate the function: $x^5\ln(x)$
My Attempt
$$\int(x^5\ln(x))dx$$
$u=\ln(x)$ and $du=\frac{1}{x}dx$
$dv=x^5dx$ and
$v=\frac{x^6}{6}$
Using this I can then integrate the function using the method
$$uv-\int v du$$
Which then substituting I get:
$$\frac{x^6\ln(x)}{6}-\int\frac{x^6}{6}\frac{1}{x}dx$$
$$\frac{x^6\ln(x)}{6}-\int\frac{x^5}{6}dx$$
$$\frac{x^6\ln(x)}{6}-\frac{x^6}{36}+C$$
My question is if I used a different u instead of the first u which I used my first attempt how will getting the same answer look like? I tried setting $u=x^5$ and it just sort of repeated itself when integrating, I know the answer is the same but how does one get there?
What I mean
$u=x^5$ and $du=5x^4$
$dv=\ln(x)dx$ and $v=x\ln(x)-x$
Then using the method previously stated I get:
$$x^6\ln(x)-x^6-\int(x\ln(x)-x)(5x^4)dx$$
|
Setting $u = x^{5}$, $v' = \ln x$ and noting that $$\int \ln x = x \ln x - x + C$$ yields
\begin{align}
I &= \int x^{5} \ln x dx \\
&= x^{6} \ln x - x^{6} - 5 \int (x^{5} \ln x - x^{5}) dx \\
&= x^{6} \ln x - x^{6} - 5 I + \frac{5 x^{6}}{6} + C \\
&= x^{6} \ln x - \frac{x^{6}}{6} - 5 I + C \\
\implies 6I &= x^{6} \ln x - \frac{x^{6}}{6} + C \\
\implies I &= \frac{ x^{6} \ln x }{6} - \frac{x^{6}}{36} + C
\end{align}
|
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"question_score": "3",
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|
Calculate the length of the portion of the hypocycloid $x^{\frac{2}{3}}+y^{\frac{2}{3}}=1$ Calculate the length of the portion of the hypocycloid $x^{\frac{2}{3}}+y^{\frac{2}{3}}=1$ in the first quadrant from the point $(\frac{1}{8},\frac{3\sqrt{3}}{8})$ to the point (1.0)
$x^{\frac{2}{3}}+y^{\frac{2}{3}}=1$
$y^{\frac{2}{3}}=1-x^{\frac{2}{3}}$
$y=(1-x^{\frac{2}{3}})^{\frac{3}{2}}$
$\frac{dy}{dx}=(1-x^{\frac{2}{3}})^{\frac{3}{2}}$
$\frac{dy}{dx}= \frac {3}{2}(1-x^{\frac{2}{3}})^{\frac{1}{2}}(0-\frac{2}{3}x)$
i am not sure where to go from there
|
Formula for arc length is
$$L=\int_{\frac18}^1\sqrt{1+\left(y'\right)^2}dx$$
It is easy to find the derivative:
$$y'=-\frac{\sqrt{1-x^{\frac23}}}{x^\frac13}$$
The length is now:
$$\begin{align}L&=\int_{\frac18}^1\sqrt{1+\left(-\frac{\sqrt{1-x^{\frac23}}}{x^\frac13}\right)^2}dx\\&=\int_{\frac18}^1x^{-\frac13}dx\\&=\left.\frac32x^\frac23\right|_\frac18^1\\&=\frac98\end{align}$$
|
{
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"url": "https://math.stackexchange.com/questions/2531156",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Find limit $\lim_{x\to 1} \dfrac{4x^2\sqrt{x+3}-17x+9}{(x-1)^2}$
Find the following limit without using L'Hopital's rule:
$$\lim_{x\to 1} \dfrac{4x^2\sqrt{x+3}-17x+9}{(x-1)^2}$$
My attempt:
$$x-1=u\implies x=u+1$$
So we have:
$$\lim_{u\to 0} \dfrac{4(u+1)^2\sqrt{u+4}-17(u+1)+9}{(u)^2}$$
What now?
|
$$\lim_{x\to 1} \dfrac{4x^2\sqrt{x+3}-17x+9}{(x-1)^2}=\lim_{x\to 1} \dfrac{16x^4(x+3)-(17x-9)^2}{(x-1)^2(4x^2\sqrt{x+3}+17x-9)}=$$
$$=\lim_{x\to 1} \dfrac{16x^3+80x^2+144x-81}{4x^2\sqrt{x+3}+17x-9}=\frac{159}{16}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2532738",
"timestamp": "2023-03-29T00:00:00",
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|
Prove $\lim_{n\to\infty} 2^n \sqrt{2-x_n} = \pi$ Given the sequence $x_1 = 0$, $x_{n+1} = \sqrt{2+x_n}$, proove:
$$\lim_{n\to\infty} 2^n \sqrt{2-x_n} = \pi$$
I have used the relations
$$2\cos({\frac x 2}) = \sqrt{2 + 2\cos(x)}$$
and
$$2\sin({\frac x 2}) = \sqrt{2 - 2\cos(x)}$$
Observe:
$x_1 = 0 = 2\cos(\alpha) \iff \alpha = \frac \pi 2$
I then hoped we can set $x_n = 2\cos(\frac{\alpha}{2^{n-1}})$ but if we can do this, why is that?
I assummed we can, so:
$\lim_{n\to\infty} 2^n \sqrt{2-x_n} = \lim_{n\to\infty} 2^n \sqrt{2-2\cos(\frac{\alpha}{2^{n-1}})} = \lim_{n\to\infty} 2^n \sqrt{4 \sin^2(\frac{\alpha}{2^n})}$
We know $$\alpha = \frac \pi 2$$
By substituting:
$ = \lim_{n\to\infty} 2^{n+1} \sin(\frac{\pi}{2^{n+1}})$
Again, by substituting
$$\frac{\pi}{2^{n+1}} = \frac 1 m$$
$= \lim_{m\to\infty} \pi m \sin(\frac 1 m)= \pi$
Is this proof correct, and if not, how to prove it?
The even more important question is the why is that? part mentioned above.
|
$\cos(x) = \cos(2\cdot \frac{x}{2})= \cos^2(\frac{x}{2}) - \sin^2(\frac{x}{2}) = -1+ 2\cos^2(\frac{x}{2})$ and hence $\cos(\frac{x}{2})= \sqrt{\frac{1}{2}+\frac{1}{2}\cos(x)} = \frac{1}{2}\sqrt{2+2\cos(x)}$. That is the inductive step for the calculation.
Geometrically, you can check that $2^n \sqrt{2-x_n}$ is half the perimeter of a regular $2^{n+1}$-gon which is inscribed in the unit circle. Which for $n \to \infty$ approaches half the circumference of a unit circle, which is $\pi$.
Added: About your question what happens when you choose a different starting value $x_1 \in [-2,2]$ (note that outside this interval, at least one of the square roots in the calculation becomes unreal, and accordingly, the geometric interpretation breaks down).
*
*In the calculation: You now get $\alpha = \displaystyle \arccos\left(\frac{x_1}{2}\right)$. The rest of the computation stays literally the same, just at the end you get the limit $\lim_{n \to \infty} 2^{n+1}\cdot \sin\displaystyle\left(\frac{\alpha}{2^n}\right) = 2\alpha$.
*Geometrically: You get a polygonal chain of $2^n$ pieces, each of which is a chord within a circular segment of arc length $\displaystyle\frac{\alpha}{2^{n-1}}$. So the total arc in which this polygonal chain is inscribed has arc length $2\alpha$. As $n$ grows, the polygonal chain approximates the arc which gives the geometric reason for the limit being just $2\alpha$.
(Archimedes would be so happy with this.)
|
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|
Lie group representation, exponential, and $\theta$-periodicty
SU(2)
I know we can view the group element in the SU(2) Lie group as
$$ g = \exp\left(\theta\sum_{k=1}^{3} i t_k \frac{\sigma_k}{2}\right) $$
where $(t_1,t_2,t_3)$ forms a unit vector [effectively pointing in some direction on a unit 2-sphere $S^2$], and $\sigma_k$ are Pauli matrices:
\begin{align}
\sigma_1 = \sigma_x &=
\begin{pmatrix}
0&1\\
1&0
\end{pmatrix} \\
\sigma_2 = \sigma_y &=
\begin{pmatrix}
0&-i\\
i&0
\end{pmatrix} \\
\sigma_3 = \sigma_z &=
\begin{pmatrix}
1&0\\
0&-1
\end{pmatrix} \,.
\end{align}
Notice that any group element on $SU(2)$ can be parametrized by some $\theta$ and $(t_1,t_2,t_3)$. Also $\theta$ has a periodicity $[0,4 \pi)$, instead of $2 \pi$.
Notice that in this case we also have
$$ g = \exp\left(\theta\sum_{k=1}^{3} i t_k \frac{\sigma_k}{2}\right)
=\cos(\frac{\theta}{2})+i \sum_{k=1}^{3} t_k \sigma_k\sin(\frac{\theta}{2})$$
question 1: SU(3)
(1) Is this true that all $SU(3)$ group elements can be written as:
$$ g = \exp\left(\theta\sum_{k=1}^{8} i t_k \frac{\lambda_k}{2}\right)=\cos(\frac{\theta}{2})+i \sum_{k=1}^{8} t_k \lambda_k\sin(\frac{\theta}{2}) $$
where $\lambda_k$ are Gell-Mann_matrices? And does the second equality still hold? Here Tr$(\lambda_k^2)=2.$
Also $\theta$ has a periodicity $[0,4 \pi)$?
question 2: SU(n), for $n=4, ...$
(2) Is this true that all $SU(4)$ group elements can be written as:
$$ g = \exp\left(\theta\sum_{k=1}^{4^2-1} i t_k \frac{\lambda_k}{2}\right)=\cos(\frac{\theta}{2})+i \sum_{k=1}^{4^2-1} t_k \lambda_k\sin(\frac{\theta}{2}) $$
where $\lambda_k$ are generalized rank-4 Gell-Mann_matrices in eq.(3)? And does the second equality still hold? Here Tr$(\lambda_k^2)=2.$
Also $\theta$ has a periodicity $[0,4 \pi)$?
(3) How to determine the Right Hand side equation and $\theta$ periodicity?
|
At least for SU(3), the exponential of arbitrary Lie-algebra elements in the fundamental representation is a bit more complicated, and, as a consequence of the Cayley-Hamilton theory, it also has a term bilinear in the Lie algebra elements, unlike the simple SU(2) expression you may be hankering for. See Curtright & Zachos 2015, and arXiv.
Specifically, the generic SU(3) group element generated by a traceless 3×3 Hermitian matrix H, normalized as trH2 = 2, can be expressed as a second order matrix polynomial in H,
$$\begin{align}
\exp(i\theta H) ={}
&\left[-\frac{1}{3} I\sin\left(\varphi + \frac{2\pi}{3}\right) \sin\left(\varphi - \frac{2\pi}{3}\right) - \frac{1}{2\sqrt{3}}~H\sin(\varphi) - \frac{1}{4}~H^2\right]
\frac{\exp\left(\frac{2}{\sqrt{3}}~i\theta\sin(\varphi)\right)}
{\cos\left(\varphi + \frac{2\pi}{3}\right) \cos\left(\varphi - \frac{2\pi}{3}\right)} \\[4pt]
{}+{} &\left[-\frac{1}{3}~I\sin(\varphi) \sin\left(\varphi - \frac{2\pi}{3}\right) - \frac{1}{2\sqrt{3}}~H\sin\left(\varphi + \frac{2\pi}{3}\right) - \frac{1}{4}~H^{2}\right]
\frac{\exp\left(\frac{2}{\sqrt{3}}~i\theta \sin\left(\varphi + \frac{2\pi}{3}\right)\right)}
{\cos(\varphi) \cos\left(\varphi - \frac{2\pi}{3}\right)} \\[4pt]
{}+{} &\left[-\frac{1}{3}~I\sin(\varphi) \sin\left(\varphi + \frac{2\pi}{3}\right) - \frac{1}{2\sqrt{3}}~H \sin\left(\varphi - \frac{2\pi}{3}\right) - \frac{1}{4}~H^2\right]
\frac{\exp\left(\frac{2}{\sqrt{3}}~i\theta \sin\left(\varphi - \frac{2\pi}{3}\right)\right)}
{\cos(\varphi)\cos\left(\varphi + \frac{2\pi}{3}\right)}
\end{align} $$
where
$$\varphi \equiv \frac{1}{3}\left[\arccos\left(\frac{3\sqrt{3}}{2}\det H\right) - \frac{\pi}{2}\right]. $$
|
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|
Compute $\int_0^x \sqrt{s(2-s)}\, ds $ Could anyone shed some light on how to compute the following two integrals?
$$ \int_0^x \sqrt{s(2-s)}\, ds \ \ \textrm{ for $0<x<1$}, $$
and
$$\int_x^0 \sqrt{s(s-2)}\, ds \ \ \textrm{ for $-1<x<0$}. $$
|
Solution without Finding Antiderivative: Observe
\begin{align}
\int^x_0\sqrt{2s-s^2}\ ds = \int^x_0\sqrt{1-(1-s)^2}\ ds = \int^{1}_{1-x} \sqrt{1-s^2}\ ds
\end{align}
which is just the area underneath the unit semi-circle from $1-x$ to $1$. But that area is given by sector of the disk minus a right triangle, i.e. we have
\begin{align}
\int^1_{1-x} \sqrt{1-s^2}\ ds =&\ \text{ area of sector} - \text{ area of triangle}\\
=&\frac{1}{2}\theta -\frac{1}{2}(1-x)\sqrt{1-(1-x)^2} \\
=&\ \frac{1}{2}\arctan \frac{\sqrt{1-(1-x)^2}}{1-x} - \frac{1}{2}(1-x)\sqrt{1-(1-x)^2}
\end{align}
where $\theta = \arctan(y/x)$ is the angle of the sector.
Additional Remark: Notice the area underneath the unit semi-circle from $1-x$ to $1$ is also equal to the area of quarter disk in the first quadrant minus the area underneath the unit semi-circle from $0$ to $1-x$, let us call that $A$. Observe $A$ is actually the sum of a sector and a right triangle, i.e.
\begin{align}
A =&\ \text{ right triangle } + \text{ sector of disk}\\
=&\ \frac{1}{2} (1-x)\sqrt{1-(1-x)^2} + \frac{1}{2}\left(\frac{\pi}{2}-\theta\right).
\end{align}
Using the trig identity
\begin{align}
\sin\left(\frac{\pi}{2}-\theta \right) = \cos\theta
\end{align}
and the fact that
\begin{align}
\tan\theta = \frac{\sqrt{1-(1-x)^2}}{1-x}
\end{align}
then we arrive at the conclusion that
\begin{align}
\cos\theta = 1-x.
\end{align}
Hence it follows
\begin{align}
\frac{\pi}{2}-\theta = \arcsin\cos\theta = \arcsin(1-x)
\end{align}
which also means
\begin{align}
A = \frac{1}{2}(1-x)\sqrt{1-(1-x)^2} + \frac{1}{2}\arcsin(1-x)
\end{align}
and
\begin{align}
\int^1_{1-x}\sqrt{1-s^2}\ ds =&\ \frac{\pi}{4}- A\\
=&\ \frac{\pi}{4}-\frac{1}{2}(1-x)\sqrt{1-(1-x)^2} - \frac{1}{2}\arcsin(1-x).
\end{align}
|
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|
Factorization of polynomial $2x^4 - 2x^2 + 4$ I'm struggling with the polynomial factorization of the polynomial function $2x^4 - 2x^2 + 4$ in fields $\mathbb{R}, \mathbb{Q}, \mathbb{Z}, \mathbb{Z}_5, \mathbb{Z}_7$.
So far I have determined that it cannot be factored into something*function of degree 1 since it has no roots (used Newton calculus to find that it's "suspicious of extreme" points are positive as is an $f(0)=4>0$ making it a positive function).
That means it either is of irreducible factors already OR it is factored into multiplication of two polynomial functions of degree 2.
I belove its the first case but don't know how to prove it since it ends in $5$ equations of $6$ variables (aka our function = $(ax^2 + bx + c)(ux^2 + vx + w)$).
I'm not sure how to prove it/continue,
thanks for reply C:
|
You can get disregard, for a while, the common factor $2$ and look for the factorization of $x^4-x^2+2$.
First let's look at $\mathbb{Q}$ and $\mathbb{R}$ that are related. Since the polynomial $y^2-y+2$ has no real roots, we can complete the square in a different way than usual:
$$
x^4-x^2+2=x^4+2\sqrt{2}x^2+2-(1+2\sqrt{2})x^2=
(x^2+\sqrt{2})^2-(\sqrt{1+2\sqrt{2}}x)^2
$$
from which you obtain the factorization over the reals and also that the polynomial is irreducible over $\mathbb{Q}$ (why?). However, the given polynomial $2x^4-2x^2+4$ is reducible over $\mathbb{Z}$ (why?).
Over the five element field you can observe that $y^2-y+2$ has no root by direct inspection, so $x^4-x^2+2$ has no root either. Hence its factorization can only be as the product of two monic degree $2$ polynomials:
$$
(x^2+ax+b)(x^2+cx+d)=x^4+(a+c)x^3+(b+d+ac)x^2+(ad+bc)x+bd
$$
so
$$
\begin{cases}
a+c=0\\
b+d+ac=-1\\
ad+bc=0\\
bd=2
\end{cases}
$$
The first equation yields $c=-a$ and the third becomes $a(d-b)=0$. If $a=0$, we obtain $b+d=-1$ and $bd=2$, so $b$ and $d$ should be roots of $y^2+y+2$ that has none. If $d=b$ we get $b^2=2$, which is impossible.
Can you work out the case of the seven element field? Note that $y^2-y+2=y^2+6y+9$ here.
|
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How can I calculate the limit $\lim\limits_{x \to 7} \frac{\sqrt{x+2} - \sqrt[3]{x+20}}{\sqrt[4]{x+9} - 2}$ without L'Hospital's rule? I have a problem with calculation of the limit:
$$\lim\limits_{x \to 7} \frac{\sqrt{x+2} - \sqrt[3]{x+20}}{\sqrt[4]{x+9} - 2}$$
Is there a way to calculate it? How can I do it?
|
With Taylor's formula at order $1$:
Set $x=7+h$. You obtain
\begin{align}
\frac{\sqrt{x+2} - \sqrt[3]{x+20}}{\sqrt[4]{x+9} - 2}&=\frac{\sqrt{9+h} - \sqrt[3]{27+h}}{\sqrt[4]{16+h} - 2}=\frac{3\sqrt{1+\frac h9} - 3\sqrt[3]{1+\frac h{27}}}{2\sqrt[4]{1+\frac h{16}} - 2}\\
&=\frac{\bigl(3+\frac h6+o(h)\bigr)-\bigl(3+\frac h{27}+o(h)\bigr)}{2+\frac h{32}+o(h)-2}=\frac{\frac{7h}{54}+o(h)}{\frac h{32}+o(h)}=\frac{\frac{7}{54}+o(1)}{\frac 1{32}+o(1)}\to \frac{112}{27}.
\end{align}
|
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|
Error in $\int\limits_0^{\infty}dx\,\frac {\log^2 x}{a^2+x^2}$ This is my work for solving the improper integral$$I=\int\limits_0^{\infty}dx\,\frac {\log^2x}{x^2+a^2}$$I feel like I did everything write, but when I substitute values into $a$, it doesn’t match up with Wolfram Alpha.
First substitute $x=\frac {a^2}u$ so$$\begin{align*}I & =-\int\limits_{\infty}^0du\,\frac {2\log^2a-\log^2 u}{x^2+a^2}\\ & =\int\limits_0^{\infty}du\,\frac {2\log^2a}{a^2+x^2}-\int\limits_0^{\infty}du\,\frac {\log^2 u}{a^2+x^2}\end{align*}$$Hence$$\begin{align*}I & =\int\limits_0^{\infty}du\,\frac {\log^2a}{a^2+x^2}\\ & =\frac {\log^2a}{a^2}\int\limits_0^{\pi/2}dt\,\frac {a\sec^2t}{1+\tan^2t}\\ & =\frac {\pi\log^2a}{2a}\end{align*}$$However, when $a=e$ Wolfram Alpha evaluates the integral numerically as$$I\approx 2.00369$$however the input that I arrived at evaluates numerically$$\frac {\pi}{2e}\approx0.5778$$Where did I go wrong? And how would you go about solving this integral?
|
It appears that you wrote
$$
\log\left(\frac{a^2}u\right)^2=2\log(a)^2-\log(u)^2
$$
when it should be
$$
4\log(a)^2-4\log(a)\log(u)+\log(u)^2
$$
Using the Dirichlet beta function as evaluated in this answer
$$
\begin{align}
&\int_0^\infty\frac{\log(x)^2}{a^2+x^2}\,\mathrm{d}x\tag1\\
&=\frac1a\int_0^\infty\frac{\log(x)^2+2\log(a)\log(x)+\log(a)^2}{1+x^2}\,\mathrm{d}x\tag2\\
&=\frac1a\int_{-\infty}^\infty\frac{x^2+2\log(a)x+\log(a)^2}{2\cosh(x)}\,\mathrm{d}x\tag3\\
&=\frac1a\int_0^\infty\frac{x^2+\log(a)^2}{\cosh(x)}\,\mathrm{d}x\tag4\\
&=\frac2a\int_0^\infty\left(x^2+\log(a)^2\right)\sum_{k=0}^\infty (-1)^ke^{-(2k+1)x}\,\mathrm{d}x\tag5\\
&=\frac2a\sum_{k=0}^\infty\int_0^\infty\frac{(-1)^k}{(2k+1)^3} x^2e^{-x}\,\mathrm{d}x+\frac{2\log(a)^2}a\sum_{k=0}^\infty\int_0^\infty \frac{(-1)^k}{2k+1}e^{-x}\,\mathrm{d}x\tag6\\
&=\frac{2\Gamma(3)}a\beta(3)+\frac{2\Gamma(1)\log(a)^2}a\beta(1)\tag7\\[6pt]
&=\frac{\pi^3}{8a}+\frac{\pi\log(a)^2}{2a}\tag8
\end{align}
$$
Explanation:
$(2)$: substitute $x\mapsto ax$
$(3)$: substitute $x\mapsto e^x$
$(4)$: remove the integral of the odd function and apply symmetry
$(5)$: write $\frac1{\cosh(x)}$ for $x\gt0$ as a series
$(6)$: split the sum and substitute $x\mapsto\frac{x}{2k+1}$
$(7)$: apply the beta sum and Gamma integral
$(8)$: evaluate the beta and Gamma functions
|
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|
prove that $\sum_{n=1}^\infty \frac{(-1)^n}{n}$ convergies using Cauchy criterion Prove using Cauchy criterion that the sum of the series $$\sum_{n=1}^\infty \frac{(-1)^n}{n}$$
converges.
The cauchy criterion says that $\sum_{i=0}^\infty a_{i}$ converges if and only if for every $\epsilon>0$ exists $N$ such that $\forall m>n>N$ implies $\vert\sum_{k=n+1}^m a_{k}\vert<\epsilon$
Thats what I got so far $\vert\sum_{k=n+1}^m \frac{(-1)^k}{k}\vert=\vert\frac{(-1)^{n+1}}{n+1}+\frac{(-1)^{n+2}}{n+2}+...+\frac{(-1)^{m}}{m}\vert\le\vert\frac{(-1)^{n+1}}{n+1}\vert+\vert\frac{(-1)^{n+2}}{n+2}\vert+\vert\frac{(-1)^{n+3}}{n+3}\vert...+\vert\frac{(-1)^{m}}{m}\vert$ but this is the harmonic series and we know that she is divergent so what am I missing?
|
Hint:
$$\frac{1}{n+1} - \frac{1}{n+2} + \frac{1}{n+3} - \frac{1}{n+4} \pm \ldots \\= \frac{1}{n+1} - \left(\frac{1}{n+2} - \frac{1}{n+3}\right) - \left(\frac{1}{n+4} - \frac{1}{n+5}\right) - \ldots \\\leqslant \frac{1}{n+1}$$
|
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|
Find $x$ such that $2^x+3^x-4^x+6^x-9^x=1$ The question:
Find values of $x$ such that $2^x+3^x-4^x+6^x-9^x=1$, $\forall x \in \mathbb R$.
Notice the numbers $4$, $6$ and $9$ can be expressed as powers of $2$ and/or $3$. Hence let $a = 2^x$ and $b=3^x$.
\begin{align}
1 & = 2^x+3^x-4^x+6^x-9^x \\
& = 2^x + 3^x - (2^2)^x + (2\cdot3)^x-(3^2)^x\\
& = 2^x + 3^x - (2^x)^2 + 2^x\cdot3^x-(3^x)^2 \\
& = a+b-a^2+ab-b^2 \\
0 & = a^2-ab+b^2-a-b+1
\end{align}
\begin{align}
0 & = a^2-ab+b^2-a-b+1 \\
& = 2a^2-2ab+2b^2-2a-2b+2 \\
& = (a^2-2ab+b^2)+(a^2-2a+1)+(b^2-2b+1) \\
& = (a-b)^2 + (a-1)^2 + (b-1)^2
\end{align}
This is where I am stuck. I am convinced that this factorisation could help solve the question, but I don't know how. Also, once we find values for $x$, we must prove that there are no further values of $x$. Could someone complete the question?
|
A sum of squares equals zero if and only if each of the squares equals zero. So you get $a=b=1$.
|
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|
Find locus of point vertex of right angle triangle whose hypotenuse is chord of circle. Let $AB$ be a variable chord of length 5 to the circle $$x^2+y^2=\frac{25}{2}$$. A triangle $ABC$ is constructed such that $BC=4$ and $CA=3$. If the locus of C is $x^2+y^2=a$ find all possible values of $a$
I tried using the parametric form to solve this question but it resulted in a very cumbersome task with no avail because of so much of variables and their eliminations. Can anyone explain any easier method to solve it.
|
Consider the particular position of $A\left(-\frac52,\frac52\right);\;B\left(\frac52,\frac52\right)$
The line $AC_2$ has equation $y-\frac{5}{2}=\frac{4}{3} \left(x+\frac{5}{2}\right)$
And the circle with center $A$ and radius $3$ has equation $\left(x+\frac{5}{2}\right)^2+\left(y-\frac{5}{2}\right)^2=9$
The intersection point is $C_2(-0.7,\;4.9)$
Similarly for $AC_1$. The equation is $y-\frac{5}{2}=-\frac{4}{3} \left(x+\frac{5}{2}\right)$
Intersecting with the same circle center $A$ and radius $3$ we get
$C_1(-0.7,0.1)$
The radii of the two loci are
$OC_1=\sqrt{0.7^2+0.1^2}=\frac{1}{\sqrt{2}}$
so the first value for $a_1=\frac12$
And $OC_2=\sqrt{0.7^2+4.9^2}=\frac{7}{\sqrt{2}}$
and we get $a_2=\frac{49}{2}$
Hope this helps
$$...$$
|
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Finding $\lim_{x \to \infty} x(\ln(1+x) - \ln(x))$ without l'Hopital I solved the limit
$$\lim_{x \to \infty} x(\ln(1+x) - \ln(x))$$
by writing it as $\lim_{x \to \infty} \frac{\ln(\frac{1+x}{x})}{\frac{1}{x}}$ and applying l'Hopital rule but is it possible to solve it without using l'Hopital rule?
|
$$\log(x+1)-\log(x)=\log\left(\frac{x+1}{x}\right)=\log\left(1+\frac{1}{x}\right)$$
We know that $\log(1+t)=t-\frac{t^2}{2}+\frac{t^3}{3}+O(t^4)$ (for $t \in (-1,1)$); substituting $t=\frac{1}{x}$ we get: $\log\left(1+\frac{1}{x}\right)=\frac{1}{x}-\frac{1}{2} \frac{1}{x^2}+\frac{1}{3} \frac{1}{x^3}+O\left(\frac{1}{x^4}\right)$, so:
$$x(\log(x+1)-\log(x))=x\left(\frac{1}{x}-\frac{1}{2} \frac{1}{x^2}+\frac{1}{3} \frac{1}{x^3}+O\left(\frac{1}{x^4}\right)\right)$$
$$x(\log(x+1)-\log(x))=1-\frac{1}{2} \frac{1}{x}+\frac{1}{3} \frac{1}{x^2}+O\left(\frac{1}{x^3}\right) \to 1$$
|
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|
Prove that the roots of $ax^2+bx+c=0$, $a\neq 0$ and $a,b,c\in R$ will be real if $a(a+7b+49c)+c(a-b+c)<0$
Prove that the roots of $ax^2+bx+c=0$, $a\neq 0$ and $a,b,c\in \mathbb{R}$ will be real if $$a(a+7b+49c)+c(a-b+c)<0$$
My Attempt:
Given
\begin{align}
a(a+7b+49c)+c(a-b+c) &< 0 \\
49a \left( \dfrac {a}{49} + \dfrac {b}{7} + c \right)+c(a-b+c) &< 0 \\
49a \cdot f\left( \dfrac {1}{7} \right) + f(0)\cdot f(-1) &< 0
\end{align}
|
a) Suppose that $a$ and $c$ have opposite sign. Then $b^2-4ac\geq 0$, and we are done.
b) Suppose that $f(1/7)$ and $f(-1)$ have opposite sign. Then by the intermediate value theorem, $ax^2+bx+c=0$ have a real root, hence two real roots, and we are done.
c) Suppose now that $a,c$ have the same sign, say $e_1$, and that $f(1/7)$ and $f(-1)$ have also the same sign, say $e_2$.
By the condition, we cannot have $e_1=e_2$. Hence $e_2=-e_1$; for example, if $e_1=1$, then $f(x)\to +\infty$ if $x\to +\infty$, and $f(1/7)\leq 0$, hence $f(x)=0$ has one, hence two real roots. The same for the case $e_1=-1$.
|
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|
Find the limit of a series of fractions starting with $\frac{1}{2}, \, \frac{1/2}{3/4}, \, \frac{\frac{1}{2}/\frac{3}{4}}{\frac{5}{6}/\frac{7}{8}}$ Problem
Let $a_{0}(n) = \frac{2n-1}{2n}$ and $a_{k+1}(n) = \frac{a_{k}(n)}{a_{k}(n+2^k)}$ for $k \geq 0.$
The first several terms in the series $a_k(1)$ for $k \geq 0$ are:
$$\frac{1}{2}, \, \frac{1/2}{3/4}, \, \frac{\frac{1}{2}/\frac{3}{4}}{\frac{5}{6}/\frac{7}{8}}, \, \frac{\frac{1/2}{3/4}/\frac{5/6}{7/8}}{\frac{9/10}{11/12}/\frac{13/14}{15/16}}, \, \ldots$$
What limit do the values of these fractions approach?
My idea
I have calculated the series using recursion in C programming, and it turns out that for $k \geq 8$, the first several digits of $a_k(1)$ are $ 0.7071067811 \ldots,$ so I guess that the limit exists and would be $\frac{1}{\sqrt{2}}$.
|
Only a remark, not a complete answer.
For $q\in \mathbb{N}$, put $s_2(q)=$ the sum of the digits of the base two expansion of $q$, ie $s_2(3)=2$, $s_2(4)=1$, etc. The following formula can be proven by induction :
$$a_m(n)=\prod_{0\leq q<2^m}\left(1-\frac{1}{2n+2q}\right)^{(-1)^{s_2(q)}}$$
For $m=0$, we have only $q=0$ to consider, and the formula gives $\displaystyle \frac{2n-1}{2n}$; if the formula is true for $m$, we have
$$a_{m+1}(n)=\prod_{0\leq q<2^m}\left(1-\frac{1}{2n+2q}\right)^{(-1)^{s_2(q)}}\prod_{0\leq q<2^m}\left(1-\frac{1}{2n+2q+2^{m+1}}\right)^{(-1)^{s_2(q)+1}}$$
and as $\{q; 0\leq q<2^{m+1}\}$ is the disjoint union of $\{q; 0\leq q<2^{m}\}$ and $\{q+2^m; 0\leq q<2^{m}\}$, and that if $0\leq q<2^m$, we have $s_2(q+2^m)=s_2(q)+1$, the formula is true for $m+1$.
Now taking the log, using that $-\log(1-x)=x+\dfrac{x^2}2+o(x^2)$, the convergence of the whole infinite product $\displaystyle \prod_{0\leq q}\left(1-\frac{1}{2n+2q}\right)^{(-1)^{s_2(q)}}$ is equivalent to the convergence of the series $\displaystyle \sum_{q\geq 0}\frac{(-1)^{s_2(q)}}{n+q}$, and I do not know if this series is convergent...
|
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Number of ways to pick 12 elements from 3 different kinds of different no of elements Question:
A florist has 5 aspidistras, 6 buttercups, and 7 chrysanthemums. How many different kinds of bouquets of a dozen flowers (it is not required to use all types of flowers) can she make from these?
My Solution:
The solution I came up with is as follows:
Since it seems that the order of flowers doesn't matter, thus we can just add them all up and choose a dozen flowers out of them.
5 aspidistras + 6 buttercups + 7 chrysanthemums = 18 total flowers
answer = 18 choose 12 = 18564
Is the above answer correct or if not can anyone tell me what I am doing wrong? Any help is greatly appreciated.
|
Let $a$ be the number of aspidistras, $b$ be the number of buttercups, and $c$ be the number of chrysanthemums. The number of ways of forming a bouquet of a dozen flowers is the number of solutions of the equation
$$a + b + c = 12 \tag{1}$$
in the nonnegative integers subject to the constraints $a \leq 5$, $b \leq 6$, and $c \leq 7$.
A particular solution of equation 1 corresponds to the placement of two addition signs in a row of twelve ones. For instance,
$$1 1 1 1 + 1 1 1 1 1 + 1 1 1$$
corresponds to the solution $a = 4$, $b = 5$, and $c = 3$, while
$$+ 1 1 1 1 1 1 + 1 1 1 1 1 1$$
corresponds to the solution $a = 0$, $b = c = 6$. The number of such solutions is
$$\binom{12 + 2}{2} = \binom{14}{2}$$
since we must choose which two of the fourteen positions required for twelve ones and two addition signs will be filled by addition signs.
In general, the equation
$$x_1 + x_2 + \ldots + x_k = n$$
has
$$\binom{n + k - 1}{k - 1}$$
solutions in the nonnegative integers since a particular solution corresponds to the placement of $k - 1$ addition signs in a row of $n$ ones, and we must choose which $k - 1$ of the $n + k - 1$ positions required for $n$ ones and $k - 1$ addition signs will be filled with addition signs.
Now, we must subtract the number of solutions that violate one or more of the constraints. Notice that it is not possible to violate two constraints simultaneously since $6 + 7 = 13 > 12$.
Suppose $a > 5$. Then $a' = a - 6$ is a nonnegative integer. Substituting $a' + 6$ for $a$ in equation 1 yields
\begin{align*}
a' + 6 + b + c & = 12\\
a' + b + c & = 6 \tag{2}
\end{align*}
Equation 2 is an equation in the nonnegative integers with
$$\binom{6 + 2}{2} = \binom{8}{2}$$
solutions.
Suppose $b > 6$. Then $b' = b - 7$ is a nonnegative integer. Substituting $b' + 7$ for $b$ in equation 1 yields
\begin{align*}
a + b' + 7 + c & = 12\\
a + b' + c & = 5 \tag{3}
\end{align*}
Equation 3 is an equation in the nonnegative integers with
$$\binom{5 + 2}{2} = \binom{7}{2}$$
solutions.
Suppose $c > 7$. Then $c' = c - 8$ is a nonnegative integer. Substituting $c' + 8$ for $c$ in equation 1 yields
\begin{align*}
a + b + c' + 8 & = 12\\
a + b + c' & = 4 \tag{4}
\end{align*}
Equation 4 is an equation in the nonnegative integers with
$$\binom{4 + 2}{2} = \binom{6}{2}$$
solutions.
Hence, the number of bouquets that can be formed is
$$\binom{14}{2} - \binom{8}{2} - \binom{7}{2} - \binom{6}{2} = 27$$
|
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|
Set of values of $x$
If the inequality
$$(1-a^2)x^2+(2a-3)x+1<0$$
is true for all values of $a$ then the set of values of $x$ is?
I took two cases (the parabola opens upwards and the parabola opens downwards). For the first case the value of $x$ is all real except the interval containing roots. For the second case the value of $x$ is the interval containing the roots. I couldn't proceed with this as I couldn't eliminate $a$. Any ideas? Thanks
|
$$\left(1-a^2\right)x^2+(2 a-3) x+1<0$$
Expand and collect wrt $a$
$$-a^2 x^2+2 a x+\left(x^2-3 x+1\right)<0$$
Multiply both sides by $-1$ and change the $<$ into $>$
$$a^2x^2-2ax-\left(x^2-3 x+1\right)>0$$
This is true for any $a$ if and only if the discriminant $\Delta$ of the polynomial
$P(a)=a^2x^2-2ax-\left(x^2-3 x+1\right)$ is negative
$\Delta=4x^2+4x^2(x^2-3x+1)=4x^2(x-2) (x-1)<0$
That is $\color{red}{1<x<2}$
Hope this is useful
|
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|
Find : $\lim\limits_{x \rightarrow 0}\left(\frac{1}{\sin(x)} - \frac{1}{x}\right)$ with l'Hôpital's rules I want to calculate :
$$\lim\limits_{x \rightarrow 0}\left(\frac{1}{\sin(x)} - \frac{1}{x}\right)$$
with l'Hôpital's rules.
I get
$$\lim\limits_{x \rightarrow 0}\left(\frac{1}{\sin(x)} - \frac{1}{x}\right) = \lim\limits_{x \rightarrow 0}\left(\frac{x-\sin(x)}{x\sin(x)}\right) = \lim\limits_{x \rightarrow 0}\left(\frac{1-\cos(x)}{\sin(x)+x\cos(x)}\right)$$
but I don't see useful next steps.
|
You can also use Taylor series:
$$\sin(x)=x-\frac{x^3}{6}+O(x^4)$$
So your function:
$$\frac{1}{\sin(x)}-\frac{1}{x}=\frac{x-\sin(x)}{x\sin(x)}=\frac{\frac{x^3}{6}+O(x^4)}{x^2-\frac{x^4}{6}+O(x^5)}=\frac{\frac{x}{6}+O(x^2)}{1-\frac{x^2}{6}+O(x^3)} \to \frac{0}{1}=0$$
|
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|
Convergence of $\sum_{n=0}^{+ \infty} \frac{n^3 - 5n^2 + \pi}{2n^5 - \sqrt{3}}$
Does $$\sum_{n=0}^{+\infty} \frac{n^3 - 5n^2 + \pi}{2n^5 - \sqrt{3}}$$
converge?
My attempt:
$$\forall n \in \mathbb{N} \setminus\{0\}:\frac{n^3-5n^2 + \pi }{2n^5 -\sqrt{3}} = \frac{1}{n^2}\frac{1-5/n + \pi/n^2}{2- \sqrt{3}/n^5}$$
And because
$$\frac{1-5/n + \pi/n^2}{2- \sqrt{3}/n^5} \to \frac{1}{2}$$
there exists $n_0$ such that $\forall n > n_0: \left\vert \dfrac{1-5/n + \pi/n^2}{2- \sqrt{3}/n^5} - \dfrac{1}{2}\right\vert < 1$
Hence, for $n \geq n_0$:
$$ \frac{1-5/n + \pi/n^2}{2- \sqrt{3}/n^5} < \frac{3}{2}$$
Thus: $$\frac{n^3-5n^2 + \pi }{2n^5 -\sqrt{3}} < \frac{3}{2}\frac{1}{n^2}$$
and for $n_1$ sufficiently large, the sequence of terms is positive (as both the numerator and the denumerator converge to $+ \infty$, so for $n$ large enough the numerator and denumerator are positive, and hence the quotient is positive).
Let $N := \max\{n_0,n_1\}$
Then, because $\sum\dfrac{1}{n^2}$ converges, we conclude using the comparison test that the series
$$\sum_{n > N}^{+ \infty} \frac{n^3-5n^2 + \pi }{2n^5 -\sqrt{3}}$$
converges, and hence, the given series converges.
Is this correct? (I tried to explain every single step). Do you have any comments? Is there an easier approach?
|
A rational expression $\dfrac{P(n)}{Q(n)}$ is asymptotic to $\dfrac{p_p}{q_q}n^{p-q}$ where $p,q$ are the respective degrees and $p_p,q_q$ the leading coefficients. (You can check this by factoring out the powers $n^p$ and $n^q$ from the fraction.)
And as you know, the summation of $n^{p-q}$ converges for $p-q<-1$. The other terms can be neglected.
|
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Can someone explain this proof to me for $2n < 2^n - 1? \,\, n >= 3$ Prove $2n < 2^n - 1;n \geq 3$
Prove for $n = 3$
$2\cdot 3 < 2^6 - 1$
$6 < 7$
Prove for $n \mapsto n + 1$
$$2(n + 1) = 2n + 2$$
$$< 2^n - 1) + 2$$
$$= 2^n + 1$$
$$< 2^n + 2^n - 1$$
$$ = 2^{n + 1} - 1$$
Can someone explain this proof to me? I get lost on the third step: $2^n + 1$ and I don't understand how this shows that $2n < 2^n − 1$
|
The inductive step: assuming that $2n < 2^n - 1$, show that $2(n+1) < 2^{n+1} - 1$.
*
*$2(n+1) = 2n + 2\qquad$distribution
*$2n + 2 < (2^n -1) + 2\qquad$ by the inductive hypothesis that $2n < 2^n -1$. (Just add two to both sides of inductive hypothesis to see that this is true.)
*$(2^n -1) + 2 = 2^n -1 + 2 = 2^n + 1\qquad$ Remove parentheses
*$2^n + 1 < 2^n + 2^n + 1\qquad$ Adding $2^n$ makes it larger.
*$2^n + 2^n + 1 = 2^{n+1} + 1\qquad$ Because $2^n + 2^n = 2(2^n) = 2^{n+1}$.
|
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|
Convergence and divergence of an infinite series The series is
$$1 + \frac{1}{2}.\frac{x^2}{4} + \frac{1\cdot3\cdot5}{2\cdot4\cdot6}.\frac{x^4}{8} + \frac{1\cdot3\cdot5\cdot7\cdot9}{2\cdot4\cdot6\cdot8\cdot10}.\frac{x^6}{12}+... , x\gt0$$
I just stuck over the nth term finding and once I get nth term than I can do different series test but here I am unable to find the nth term of the given series please help me out of this.
The question is different as it contains x terms and its nth term will be totally different from marked as duplicate question
|
Hint:
\begin{align}
\frac{1\cdot 3 \cdot 5 \cdot 7 \cdot 9}{2\cdot 4 \cdot 6 \cdot8\cdot10}.\frac{x^6}{12} &=\frac{(1\cdot 3 \cdot 5 \cdot 7 \cdot 9)(2\cdot 4 \cdot 6 \cdot8\cdot10)}{(2\cdot 4 \cdot 6 \cdot8\cdot10)^2}.\frac{x^6}{12} \\
&=\frac{10!}{2^{10}(1\cdot 2 \cdot 3 \cdot4\cdot 5)^2} \frac{x^6}{12}\\
&=\frac{1}{2^{10}} \cdot \binom{10}{5} \frac{x^6}{2(6)}\\
&= \frac{1}{2^{2\cdot (6-1)}} \cdot \binom{2\cdot(6-1)}{6-1} \frac{x^6}{2(6)}
\end{align}
Edit:
$$ \frac{1}{2^{2(n-1)}}\frac{(2n-2)!}{((n-1)!)^2}\frac{x^{2n}}{2n}$$
|
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|
Is the curve $ z = e^{i\theta}\left(\frac{7}{8} + \frac{1}{4} e^{6i\theta}\right) $ algebraic? Is this spirograph curve algebraic? I an only write it in polar coordinates:
$$ z = e^{i\theta}\left(\frac{7}{8} + \frac{1}{4} e^{6i\theta}\right) $$
and here is a picture. It is a six-sided rose-shaped curve, a hypotrochoid.
I read somewhere that all spirograph curves are algebraic, so this must be the solution of some polynomial equation $p(x,y) = 0$. Then I could ask questions about this curve as a Riemann surface.
|
Given
$$ x=\frac{7}{8}\cos\theta+\frac{1}{4}\cos(7\theta),\qquad y=\frac{7}{8}\sin\theta+\frac{1}{4}\sin(7\theta) $$
we have $x^2+y^2=\frac{1}{64}\left(53+28\cos(6\theta)\right)$ and the given hypotrochoid is an algebraic curve since we may eliminate the $t$ variable in
$$ \left\{\begin{array}{ccc} x& =& \frac{7}{8} T_1(t) + \frac{1}{4}T_7(t)\\x^2+y^2&=&\frac{53}{64}+\frac{7}{16}T_6(t).\end{array}\right.$$
|
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|
Integral involving $\phi$
$$\int_{0}^{\pi/2}\arctan\left({2\over \cos^2{x}}\right)\mathrm dx=\pi\arctan\left({1\over \sqrt{\phi}}\right)\tag1$$
$\phi$ is the golden ratio
$2\sec^2{x}=2\tan^2{x}+2$
$u=\sec^2{x}$ then $\mathrm du=2{\tan{x}\over \cos^2{x}}\mathrm dx$
$$\int{\cos^2{x}\over \tan{x}}\arctan\left({2\over u}\right)\mathrm du\tag2$$
$$\int{1\over u\sqrt{u-1}}\arctan\left({2\over u}\right)\mathrm du\tag3$$
Not so sure what is next step...
How do we show that $(1)=\pi\arctan\left({1\over \sqrt{\phi}}\right)$
|
$$\begin{eqnarray*}\int_{0}^{\pi/2}\arctan\left(\frac{2}{\cos^2\theta}\right)\,d\theta&=&\int_{0}^{\pi/2}\arctan\left(\frac{2}{\sin^2\theta}\right)\,d\theta\\&=&\int_{0}^{1}\frac{\arctan\frac{2}{u^2}}{\sqrt{1-u^2}}\,du\\&=&\int_{0}^{1}\frac{\frac{\pi}{2}-\arctan\frac{u^2}{2}}{\sqrt{1-u^2}}\,du\\&=&\frac{\pi^2}{4}-\sum_{n\geq 0}\frac{(-1)^n}{(2n+1)\,2^{2n+1}}\int_{0}^{1}\frac{u^{4n+2}}{\sqrt{1-u^2}}\,du\\&=&\frac{\pi^2}{4}-\sum_{n\geq 0}\frac{\pi(-1)^n (4n+1)}{(2n+1)^22^{6n+3}}\binom{4n}{2n}\end{eqnarray*}\tag{A}$$
The last hypergeometric series can be computed from
$$ \sum_{n\geq 0}\frac{2n+1}{(n+1)^2}\binom{2n}{n}z^n =\frac{\log 2-\log\left(1+\sqrt{1-4z}\right)}{z}\tag{B} $$
leading to
$$ \sum_{n\geq 0}\frac{4n+1}{(2n+1)^2}\binom{4n}{2n}z^n =\frac{1}{2\sqrt{z}}\,\log\left(\frac{1+\sqrt{1+4\sqrt{z}}}{1+\sqrt{1-4\sqrt{z}}}\right) \tag{C}$$
$$ \sum_{n\geq 0}\frac{z^n\,\Gamma\left(2n+\tfrac{3}{2}\right)}{(2n+1)\,\Gamma(2n+2)}=\sqrt{\frac{\pi}{z}}\,\text{arctanh}\sqrt{\frac{1-\sqrt{1-z}}{2}}\tag{D}$$
and finally to
$$ \int_{0}^{\pi/2}\arctan\left(\frac{2}{\cos^2\theta}\right)\,d\theta = \pi\arctan\sqrt{\frac{2}{1+\sqrt{5}}}.$$
As a corollary, $z=\frac{1-\sqrt{5}}{2}$ is a special value for the hypergeometric function ${}_2 F_1\left(\tfrac{3}{4},\tfrac{3}{4};\tfrac{7}{4};z\right)$.
This can be done also by differentiation under the integral sign, since $\int_{0}^{\pi/2}\frac{\partial}{\partial a}\arctan\left(\frac{a}{\cos^2\theta}\right)\,d\theta$ is not that difficult to compute.
|
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|
How to prove that $\sqrt{2+\sqrt3}-\sqrt{2-\sqrt3}=\sqrt2$ without squaring both sides I have been asked to prove:
$$\sqrt{2+\sqrt3}-\sqrt{2-\sqrt3}=\sqrt2$$
Which I can easily do by converting the LHS to index form, then squaring it and simplifying it down to get 2, which is equal to the RHS squared, hence proved.
However I know you can't square a side during proof because it generates an extraneous solution. So: how do you go about this proof without squaring both sides? Or can my method be made valid if I do this:
$$\sqrt{2+\sqrt3}-\sqrt{2-\sqrt3}=\sqrt2$$
$$...=...$$
$$2=2$$
$$\lvert\sqrt2\rvert=\lvert\sqrt2\rvert$$
$$\sqrt2=\sqrt2\text{ hence proved.}$$
Cheers in advance :)
|
You can square both sides in a proof if you note the extraneous solutions are added.
Example $\sqrt{2 + \sqrt{3}} -\sqrt{2 - \sqrt{3}} = k$.
First $2 > \sqrt 3$ so $\sqrt{2 - \sqrt{3}} > 0$ so $k > 0$. !!!TAKE NOTE OF THAT!!!
$(\sqrt{2 + \sqrt{3}} -\sqrt{2 - \sqrt{3}})^2 = k^2$
$2 + \sqrt3 + 2-\sqrt 3 - 2(\sqrt{2 + \sqrt{3}}\sqrt{2 - \sqrt{3}}) = k^2$
$-2(\sqrt{2 + \sqrt{3}}\sqrt{2 - \sqrt{3}}) = k^2 - 4$
$\sqrt {4 -3} = \frac {4-k^2}{2}$
$4-3 = (\frac {4-k^2}{2})^2$ !!!NOTE!!! $\frac {4-k^2}{2} \ge 0$.
$1 = (\frac {4-k^2}{2})^2$ so
$\frac {4-k^2}{2} = \pm 1$. !!!BUT we took note that $\frac {4-k^2}{2}\ge 0$.!!!
So $\frac {4-k^2}{2} = 1$
So $4-k^2 = 2$
and $k^2 = 2$ so
$k = \pm \sqrt 2$ !!!But we took note that $k>0$ so $k = \sqrt 2$.
That is valid.
|
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|
Solving a congruence using primitive roots Suppose we know that $3$ is a primitive root of $17$.
How can that help us solving $7^x \equiv 6 \pmod {17}$?
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Use Discrete Logarithm wrt primitive root $3\pmod{17},$
$x$ind$_37\equiv$ind$_3(2\cdot3)\pmod{\phi(17)}\equiv$ind$_32+1$
Now $3^3\equiv10\pmod{17},3^5\equiv9\cdot10\equiv5\implies7\equiv5^{-1}\equiv3^{-5}\equiv3^{16-5}$
$2\equiv(-1)3\cdot5\pmod{17}\equiv3^{8+1+5}$ as $3$ is a primitive root, $3^{(17-1)/2}\equiv-1\pmod{17}$
$\implies11x\equiv14+1\pmod{16}\equiv-1$
$\implies3\cdot11x\equiv-1\cdot3\pmod{16}$
$\iff x\equiv-3\equiv13\pmod{16}$
|
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|
Prove: $(\frac{1+i\sqrt{7}}{2})^4+(\frac{1-i\sqrt{7}}{2})^4=1$ $$\left(\frac{1+i\sqrt{7}}{2}\right)^4+\left(\frac{1-i\sqrt{7}}{2}\right)^4=1$$
I tried moving the left exponent to the RHS to then make difference of squares exp. $(x^2)^2$. Didn't get the same on both sides though. Any help?
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We have $\dfrac {1+i \sqrt 7}{2} = \sqrt 2 \left(\cos \theta + i \sin \theta \right)$ where $ \cos \theta = \dfrac{1}{2 \sqrt 2}$
We need to calculate $2 \Re \left(\dfrac {1+i \sqrt 7}{2} \right)^4 = 2 \times (\sqrt 2)^4 \times \cos 4\theta = 8 \cos 4 \theta$
We have $\cos 2 \theta = 2 \cos^2 \theta -1 = -\dfrac{3}{4}$ and use that to get $\cos 4 \theta = \dfrac{1}{8}$
Hence we see that the sum evaluates to $8 \cos 4 \theta = 1$
|
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|
Argue that $f$ is not uniformly continuous Let $f(x)=x^3$. We want to show that $f$ is not uniformly continuous.
Can someone please explain from $|f(x_n)-f(y_n)=\big|(n+\frac{1}{n})^3 - n^3)\big| = 3n+\displaystyle\frac{3}{n}$? How does $\big|(n+\frac{1}{n})^3-n^3\big| $ yield $3n+\displaystyle\frac{3}{n}$? ${}{}{}{}$
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You're missing the $\frac{1}{n^3}$ term. Indeed, $$\left(n+\frac{1}{n}\right)^3-n^3=n^3+3n^2\frac{1}{n}+3n\frac{1}{n^2}+\frac{1}{n^3}-n^3=3n+3\frac{1}{n}+\frac{1}{n^3}$$
|
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|
$\lim_{x\to 0} \left(\frac{1}{1-\cos x}-\frac{2}{x^2}\right)$ Find the limits :
$$\lim_{x\to 0} \left(\frac{1}{1-\cos x}-\frac{2}{x^2}\right)$$
My Try :
$$\lim_{x\to 0} \left(\frac{1}{1-\cos x}-\frac{2}{x^2}\right)=\lim_{x\to 0}\frac{x^2-2(1-\cos x)}{x^2(1-\cos x)}$$
Now what do I do ?
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$$\left(\frac{1}{1-\cos x}-\frac{2}{x^2}\right)=\frac{x^2-2(1-\cos x)}{x^2(1-\cos x)}=\frac{x^2(1+\cos x)-2\sin^2 x}{x^2\sin^2 x}=$$
from here by Taylor series:
$\begin{cases}\cos x=1-\frac{x^2}{2}+o(x^2)\\\sin x=x-\frac{x^3}{6}+o(x^3)\implies \sin^2 x=x^2-\frac{x^4}{3}+o(x^4)\end{cases}$
thus:
$$=\frac{x^2(1+1-\frac{x^2}{2}+o(x^2))-2\left( x^2-\frac{x^4}{3}+o(x^4) \right)}{x^2\left(x^2-\frac{x^4}{3}+o(x^4)\right)}=\frac{2x^2-\frac{x^4}{2}-2x^2+\frac{2x^4}{3}+o(x^4)}{x^4+o(x^4)}=\frac{\frac{x^4}{6}+o(x^4)}{x^4+o(x^4)}=\frac{\frac{1}{6}+o(1)}{1+o(1)}\to \frac16 $$
|
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|
Finding a base and dimension to a polynomial space under conditions I need to find a base and dimension for the polynomial space $V$ while the factor of $x$ is 58: $$V=\left \{ \right.p(x)\in P_{4}[x]: p(1)=p''(1)=0\left. \right \}$$
What I did is to create a general polynome: $p(x)=e+dx+cx^2+bx^3+ax^4$
Then I calculated:
$$p(1)=e+d+c+b+a=0$$
$$p''(1)=2c+6b+12a=0$$
I've expressed $p(x)$ as bellow: $$p(x)=(2/3c+e+58)x^4+(-3/5c-2e-116)x^3+cx^2+58x+e$$
Here I'm stuck because I don't know how to continue.
Help will be more than appreciated!
|
Solution:
Let's write a general polynomial: $p(x)=ax^4+bx^3+cx^2+dx+e$
Then: $p''(x)=12ax^2+6bx+2c$
Under the condition $p''(1)=p(1)=0$: $$a+b+c+d+e=0$$ $$12a+6b+c=0$$
Let's express $a$ and $b$ using $c,d=58,e$: $$a=2/3c+58+e$$ $$b=-5/3c-116-2e$$
$$\Rightarrow p(x)=(2/3c+58+e)x^4+(-5/3c-116-2e)x^3+cx^2+58x+e$$
$$\Rightarrow p(x)=sp\left \{ 58(x^4-2x^3+x)+c(2/3x^4-5/3x^3+x^2)+e(x^4-2x^3+1) \right \}$$
We'll check that our polynomials are l.i.: $$\begin{pmatrix}
1 & -2 & 0 & 1 & 0 \\
2/3 & -5/3 & 1 & 0 & 0 \\
1 & -2 & 0 & 0 & 1
\end{pmatrix} \xrightarrow[r2\rightarrow r2-2/3r1]{r3\rightarrow r3-r1}\begin{pmatrix}
1 & -2 & 0 & 1 & 0\\
0 & -1/3 & 1 & -2/3 & 0\\
0 & 0 & 0 & -1 & 1
\end{pmatrix}
$$
$\Rightarrow$The polynomials are l.i.
$\Rightarrow$A base for V while the coefficient of $x$ is 58: $\left \{ (58x^4-116x^3+58x),(29x^3-87x^2+58x),(-58x+58) \right \}$
|
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|
find the range of the function : $y=(3\sin 2x-4\cos 2x)^2-5$ find the range of the function :
$$y=(3\sin 2x-4\cos 2x)^2-5$$
My try :
$$y=9\sin^22x+16\cos^22x-24\sin 2x\cos 2x-5\\y=9+7\cos^22x-12\sin4x-5$$
now what do I do؟
|
Hints:
Every real number $t$ can be expressed as $2x$ for some $x$, so the range of the function
$$f(x) = (3\sin 2x-4\cos 2x)^2-5$$
is the same as the range of
$$g(t)=(3\sin t-4\cos t)^2-5$$
But the expression
$$3\sin t-4\cos t$$
can be expressed as
$$5\left({\small{\frac{3}{5}}}\sin t-{\small{\frac{4}{5}\cos t}}\right)$$
which suggests what formula?
|
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|
Proving whether the series $\sum_{n=1}^\infty \frac{(-1)^n}{n-(-1)^n}$ converges. I've updated my proof to be complete now, edited for proof-verification!
We know that for the partial sums with even an uneven terms, the following holds:
$S_{2N}=\sum_{n=1}^{2N} \frac{(-1)^n}{n-(-1)^n} = -\frac{1}{2} + \frac{1}{1} -\frac{1}{4} +\frac{1}{3} -\frac{1}{6} +\frac{1}{5}-\dots+\frac{1}{2N-1}$
$= \frac{1}{2\times1} +\frac{1}{4\times3} + \frac{1}{6\times5}+\dots+\frac{1}{2N(2N-1)} = \sum_{n=1}^{2N} \frac{1}{2n(2n-1)}$
We may rewrite the series in pairs as we know it will have an even amount of terms.
$S_{2N+1} = \sum_{n=1}^{2N+1} \frac{(-1)^n}{n-(-1)^n} = -\frac{1}{2} + \frac{1}{1} -\frac{1}{4} +\frac{1}{3}-\dots+\frac{1}{2N-1} -\frac{1}{2N+2}$
$=\frac{1}{2\times1} +\frac{1}{4\times3} + \frac{1}{6\times5}+\dots+\frac{1}{2N(2N-1)} - \frac{1}{2N+2} = \sum_{n=1}^{2N} \frac{1}{2n(2n-1)}-\frac{1}{2N+2}$
As $n\in\mathbb{N}$, we know that $n\geq1$ so:
$n\geq1 \iff 3n\geq3 \iff 3n^2\geq3n \iff 3n^2-3n\geq0 \iff 4n^2-2n \geq n^2+n$
So: $2n(2n-1)\geq n(n+1) \iff \frac{1}{2n(2n-1)}\leq \frac{1}{n(n+1)}$ for all $n\geq1$.
As the series of the latter sequence converges, we can conclude, by the comparison, test that the series $\sum_{n=1}^{2N} \frac{1}{2n(2n-1)}$ converges.
Suppose it converges to $s$, then we know $^{\lim S_{2N}}_{N\to\infty} = s$ and thus $\lim_{N\to\infty}[S_{2N+1}] = s - (\lim_{N\to\infty}[\frac{1}{2N+2}]) = s-0 = s.$ As the partial sums ending with even and uneven terms both converge to the same limit, the series $\sum_{n=1}^\infty \frac{(-1)^n}{n-(-1)^n}$ converges. $\tag*{$\Box$}$
|
As you computed, it's the alternating harmonic series in a different order. In fact the first N terms are just the first N terms of the alternating harmonic series permuted (edit: with an extra term which decays like 1/N in case N is odd), so the sequence of partial sums is the same.
|
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|
How to I approach this problem? Trigonometric or Algebraic computations? The radicals as denominator is not all that encouraging too and I don't see any promising method to evaluate it.
Solve for $n$ in the equation below:
$$\frac{\sin\left(\frac{90^\circ}{2^n}\right)}{\underbrace{\sqrt{2+\sqrt{2+\sqrt{2+\ldots\sqrt{2}}}}}_{n \text{ radicals}}}=\frac{-1+\sqrt2}{2}$$
|
Use the half-angle formula:
$$\cos{x}=2\cos^2\frac{x}{2}-1$$
So
$$2+2\cos{x}=4\cos^2\frac{x}{2}$$
Then, for $0\leq x\leq\frac{\pi}{2}$,
$$2\cos\frac{x}{2}=\sqrt{2+2\cos x}$$
So
$$2\cos\frac{x}{2^{k+1}}=\sqrt{2+2\cos \frac{x}{2^k}}=\sqrt{2+\sqrt{2+2\cos \frac{x}{2^{k-1}}}}=\ldots$$
We have the denominator
$${\underbrace{\sqrt{2+\sqrt{2+\sqrt{2+\ldots\sqrt{2}}}}}_{n \text{ radicals}}}$$
and we know that $\cos \pi =0$, so
$$2\cos\frac{\pi}{2^{n+1}}={\underbrace{\sqrt{2+\sqrt{2+\sqrt{2+\ldots\sqrt{2}}}}}_{n \text{ radicals}}}$$
Hence, the LHS of the expression becomes
$$\frac{\sin\left(\frac{90^\circ}{2^n}\right)}{\underbrace{\sqrt{2+\sqrt{2+\sqrt{2+\ldots\sqrt{2}}}}}_{n \text{ radicals}}}=\frac{\sin\left(\frac{\pi}{2^{n+1}}\right)}{2\cos\left(\frac{\pi}{2^{n+1}}\right)}=\frac12\tan{\frac{\pi}{2^{n+1}}}$$
So we evaluate
$$\frac12\tan{\frac{\pi}{2^{n+1}}}=\frac{-1+\sqrt2}{2}$$
which simplifies to
$$\tan{\frac{\pi}{2^{n+1}}}=\sqrt2-1$$
$n$ can be easily found from here.
|
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|
Derivative of $f\left(x\right)=\arctan\left(\sqrt{\frac{1+x}{1-x}}\right)$ I am trying to find the derivative of $f\left(x\right)=\arctan\left(\sqrt{\frac{1+x}{1-x}}\right)$ by only using the formula $\arctan\left(u\left(x\right)\right)'=\frac{u'\left(x\right)}{u\left(x\right)^2+1}$. I don't honestly understand this formula, here are however my calculations. Is this the right way to find the derivative? And how do I proceed?
$$\LARGE
\frac{\frac{1}{2\sqrt{\frac{1+x}{1-x}}}\cdot \frac{\left(1-x\right)-\left(1+x\right)}{x^2-2x+1}}{\frac{x+1}{1-x}+1}
$$
I also appreciate if you can explain what that "formula" exactly is.
|
We have
\begin{eqnarray*}
\frac{d}{dx} \tan^{-1}(x) = \frac{1}{1+x^2}
\end{eqnarray*}
and
\begin{eqnarray*}
\frac{d}{dx} \sqrt{x} = \frac{1}{2\sqrt{x}}
\end{eqnarray*}
and
\begin{eqnarray*}
\frac{d}{dx} \frac{1+x}{1-x} = \frac{ (1-x)--(1+x)}{(1-x)^2}= \frac{2 }{(1-x)^2}.
\end{eqnarray*}
Now we need to do a $3-$ fold chain rule,
\begin{eqnarray*}
\frac{d}{dx} f(g(h(x))) = h'(x) g'(h(x)) f'(g(h(x))).
\end{eqnarray*}
So in your case we get
\begin{eqnarray*}
\frac{d}{dx} \tan^{-1} \left( \sqrt{\frac{1+x}{1-x}} \right) = \frac{\frac{2 }{(1-x)^2} \frac{1}{2\sqrt{\frac{1+x}{1-x}} }}{1+ \frac{1+x}{1-x}} = \frac{1}{2 \sqrt{1-x^2}}.
\end{eqnarray*}
|
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|
2nd solution of $\cos x \cos 2x\cos 3x= \frac 1 4 $
$\cos x \cos 2x\cos 3x= \dfrac 1 4 $
Attempt explained:
$(2\cos x \cos 3x)\cos 2x = \frac1 2 $
$(\cos 4x +\cos 2x )\cos 2x = \frac 1 2 \\\cos ^2y + \cos y (2\cos^2y- 1)= \frac1 2 \\ $
(Let, y = 2x)
$\implies 4\cos^3 y+2\cos^2y- 2\cos y-1=0$
I solved this equation using Rational Root Theorem and got $y = \frac 1 2$
$\implies x= m\pi \pm \dfrac\pi3 \forall x\in \mathbb {Z}$
Using Remainder theorem, the other solution is $\cos^2 y = \dfrac 1 2 $
$\implies x = \dfrac n2\pi \pm\dfrac \pi 8 $
But answer key states: $ x= m\pi \pm \dfrac\pi3or x =(2n+1)\dfrac \pi 8$
Why don't I get the second solution correct?
|
Remember that a cubic equation can have up to three real solutions. You did find that $\cos y = -\dfrac{1}{2}$ is one of the solutions, but you still need to find the other two.
Since $\cos y = -\dfrac{1}{2}$ is a root of $4\cos^3 y+2\cos^2y- 2\cos y-1=0$, we can factor the equation as:
\begin{align*}
4\cos^3 y+2\cos^2y- 2\cos y-1 &= 0
\\
(2\cos y + 1)(2\cos^2 y - 1) &= 0
\\
(2\cos y + 1)(\sqrt{2}\cos y + 1)(\sqrt{2}\cos y - 1) &= 0
\end{align*}
So the solutions are $$\cos y = -\dfrac{1}{2}, -\dfrac{1}{\sqrt{2}}, \dfrac{1}{\sqrt{2}}$$
Now, solve these for $y$, and then divide by $2$ to get $x$.
|
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|
$\sin^2 x \cos^2 x + \sin x \cos x -1 = 0$
$\sin^2 x \cos^2 x + \sin x \cos x -1 = 0$
Attempt:
$\sin^2 2x + 2\sin 2x -4 = 0 \\ \implies \sin 2x= \sqrt 5 - 1$
Now, using the formula $\sin 2 x = \dfrac {2\tan x}{1+\tan^2x}$, I couldn't express the answer in terms of $\tan()$
The answer given is in terms of tan. How do I express it in that way?
|
Hint:
Observe the roots of $t^2+2t-4=0$ are $-1\pm\sqrt 5$ (no denominator), and their absolute values are greater than $1$.
|
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|
Prove that $\sqrt{s(s-a)(s-b)(s-c)}=A$ In 50 AD, the Heron of Alexandria came up with the well-known formula, that, given the three side lengths of a triangle (or even two and an angle, thanks to trigonometry) you can get the area of said triangle by using this formula:
$$
\text{if } s=\frac{a+b+c}{2},\\
\text{then} A=\sqrt{s(s-a)(s-b)(s-c)}
$$
But the book it came with (mathematics 1001) did not give a mathematical proof to this. So I tried and failed to get a proof to this. This is how my train wreck went…
$$\begin{align}
\text{equation }&\sqrt{s(s-a)(s-b)(s-c)}\\
\text{substitute }&\sqrt{\frac{a+b+c}{2}(\frac{a+b+c}{2}-a)(\frac{a+b+c}{2}-b)(\frac{a+b+c}{2}-c)}\\
\text{eliminate fractions}&\sqrt{a+b+c(a+b+c-2a)(a+b+c-2b)(a+b+c-2c)}\\
\text{comb. Like terms }&\sqrt{a+b+c(b+c-a)(a-b+c)(a+b-c)}\\
\text{associative property }&\sqrt{(a+(b+c)((b+c)-a))((a-(b+c))(a+(b-c)))}\\
\text{difference of 2 squares }&\sqrt{((b+c)^2-a^2)(a^2-b^2-c^2)}\\
\text{square of two numbers }& \sqrt{((b^2+bc+c^2)-a^2)(a^2-b^2-c^2)}\\
\text{polynomial multiplication}& \sqrt{2(a^2b^2-b^2c^2+a^2c^2)+a^2(b-c+a^2)-(b+c)(b^3+c^3)}\\
\end{align}$$
The issue is, I don't know how to get to $A$, the area. Does anyone know how to continue the proof train?? Thanks in advance.
|
Consider a triangle $ABC$.
$[ABC]=\frac{ab}{2}\sin C$
$=\frac{ab}{2}\sqrt{1-\cos^2 C}$
$=\frac{ab}{2}\sqrt{1-\left(\frac{a^2+b^2-c^2}{2ab}\right)^2}$
$=\sqrt{\frac{a^2b^2}{4}\left[1-\frac{(a^2+b^2-c^2)^2}{4a^2b^2}\right]}$
$=\sqrt{\frac{4a^2b^2-(a^2+b^2-c^2)^2}{16}}$
$=\sqrt{\frac{(2ab+a^2+b^2-c^2)(2ab-a^2-b^2+c^2)}{16}}$
$=\sqrt{\frac{((a+b)^2-c^2)(c^2-(a-b)^2)}{16}}$
$=\sqrt{\frac{(a+b+c)(a+b-c)(b+c-a)(a+c-b)}{16}}$
$=\sqrt{s(s-a)(s-b)(s-c)}$
Hence proved. Hope it helps
|
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|
Determine the Eigenvalues and Eigen vectors of the operator $ \ A \ $ Consider the vector space of real polynomials of degree not greater than $ \ n \ $ given by $ \large \mathbb{R}_n[x]=\{\sum_{i=0}^{n} a_i x^i \ | a_i \in \mathbb{R}, \ \forall i \} \ $ .
Define an operator $ \ A \in End (\mathbb{R}_n[x] ) \ $ by $ \ (Ap) (x)=P(2x+1) \ $ .
Determine the Eigenvalues and Eigen vectors of the operator $ \ A \ $.
Here $ \ End (\mathbb{R}_n[x] ) \ $ is an Endomorphism from $ \mathbb{R}_n[x] \to \mathbb{R}_n[x] \ $
Answer:
The basis of $ \mathbb{R}_n[x] \ $ is given by $ \{1,x,x^2, x^3,......, x^n \} \ $
Now,
$ (Ap)(x)=P(2x+1) \ $ gives the following system
$ For \ p(x)=1 , \ \ A(1) =1=1 \cdot 1+ x \cdot 0+............+x^n \cdot 0 \\ For \ p(x)=x , \ \ A(x)=2x+1=1 \cdot 1+2 \cdot x+0 \cdot x^2+........+0 \cdot x^n \\ For \ p(x)=x^2 , \ \ A(x^2)=(2x+1)^2=1+4x+4x^2=1 \cdot 1+4 \cdot x+4 \cdot x^2+......+0 \cdot x^n \\ ..............\\ p(x)=x^n \Rightarrow
A (x^n)=(2x+1)^n=1 \cdot 1+2 \binom{n}{1} \cdot x+2^2 \binom {n}{2} \cdot x^2+...........+2^n \binom{n}{n} \cdot x^n $
Thus the coefficient matrix is
$ A=\begin{bmatrix} 1 & 1 & 1 & 1 & ............ & 1 \\ 0 & 2 & 4 & 8 & ...... ........ & 2n \\ 0 & 0 & 4 & ..... \\ .... & ...... & ..... & .. & . ............ & ... \\ 0 & 0 & 0 & 0 & ............ & 2^n \end{bmatrix} \ $
Am I right So far ?
If the above approach is correct , then how to find the Eigen values and Eigen vectors of the matrix $ \ A \ $ ?
Help me out
|
I like what you have so far.
The eigenvalues of an upper-triangle matrix are the values of the main diagonal
The first few eigenvectors
$\pmatrix{1&1&1&2\\&1&2&3\\&&1&3\\&&&1}$
Well that looks like Pascals triangle.
or $(x+1)^n$ appears to describe the set of eigenvectors.
$A((x+1)^n) = (2x+2)^n = 2^n(x+1)^n$ and that is the characteristic of an eigenvector.
I notice an error in your work above, in the $A_{2,4}$ place, you show an $8$
$A= \pmatrix{1&1&1&1\\&2&4&6&\cdots &2{n\choose 1}\\&&4&12&\cdots& 2^2{n\choose 2}\\&&&8&\cdots &2^3{n\choose 3}\\&&&&2^m{n\choose m}&\vdots\\&&&&&2^n}$
|
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|
product of terms taken $3$ at a time in polynomial expression
Finding product of terms taken $3$ at a time in $\displaystyle \prod^{100}_{r=1}(x+r)$
Try:
$$\displaystyle \prod^{100}_{r=1}(x+r)=x^{100}+(1+2+3+\cdots +100)x^{99}+(1\cdot 2+1\cdot 3+\cdots+100\cdot 99)x^{98}+(1\cdot 2\cdot 3+2\cdot 3 \cdot 4+\cdot\cdots+98\cdot 99\cdot 100)x^{98}+\cdots$$
for $1$ at a time (Coefficient of $x^{99}$) is $\displaystyle \sum^{100}_{i=1}i = 50\cdot 101$
for $2$ at a time (Coefficient of $x^{98}$) is $\displaystyle \sum^{100}_{i=1}\sum^{100}_{j=1\;, (1\leq i<j \leq 100)}i \cdot j= \frac{1}{2}\bigg[\bigg(\sum^{100}_{i=1}i\bigg)^2-\sum^{100}_{i=1}i^2\bigg]$
But could some help me how to find coefficient of $x^{97},$ thanks
|
The coefficient of $x^{97}$ is
$$
\begin{align}
\sum_{i=1}^{100}\sum_{j=1}^{i-1}\sum_{k=1}^{j-1}ijk
&=\sum_{i=1}^{100}\sum_{j=1}^{i-1}\sum_{k=1}^{j-1}ij\binom{k}{1}\\
&=\sum_{i=1}^{100}\sum_{j=1}^{i-1}i\,((j-2)+2)\binom{j}{2}\\
&=\sum_{i=1}^{100}\sum_{j=1}^{i-1}i\left[3\binom{j}{3}+2\binom{j}{2}\right]\\
&=\sum_{i=1}^{100}i\left[3\binom{i}{4}+2\binom{i}{3}\right]\\
&=\sum_{i=1}^{100}\left[3((i-4)+4)\binom{i}{4}+2((i-3)+3)\binom{i}{3}\right]\\
&=\sum_{i=1}^{100}\left[15\binom{i}{5}+12\binom{i}{4}+8\binom{i}{4}+6\binom{i}{3}\right]\\
&=15\binom{101}{6}+20\binom{101}{5}+6\binom{101}{4}\\[9pt]
&=20618771250
\end{align}
$$
|
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|
Having trouble of finding this integral in the standard integral list I am facing difficult problem to deal with this integral
$$\int_{0}^{\pi/2}\ln(9-4\cos^2\theta)\,\mathrm{d}\theta \tag*{(1)}$$
note $(3)^2-(2\cos{\theta})^2=(3-2\cos{\theta})(3+2\cos{\theta}) \tag*{(2)}$
also note that, $$\log(AB)=\log A+\log B \tag*{(3)}$$ using $(2)$ and $(3)$ we have,
$$\int_{0}^{\pi/2}\ln(3-2\cos\theta)\,d\theta+\int_{0}^{\pi/2}\ln(3+2\cos\theta)\,\mathrm{d}\theta$$
Looking through my table of standard integrals I couldn't find the result for this integral$\int \ln(a+b\cos{\theta})$
Standard table of integral
Integral (44) (in the table) has the closest of resemblance
$$\int \ln(a\theta+b)\,d\theta=(\theta+{b\over a})\ln(a\theta+b)-\theta$$
In the original problem is $\cos{\theta}$ and not $\theta$
I am facing quite a problem here, I need some help.
Very much appreciated if you can help.
|
Here a slightly different approach will be considered. We will make use of the generating function for the central binomial coefficients of
\begin{align*}
\sum_{n=0}^{\infty}\binom{2n}{n}x^n=\frac{1}{\sqrt{1-4x}}, \qquad |x|<\frac{1}{4}, \tag1
\end{align*}
a result that follows directly from the binomial series.
Let
$$I = \int^{\pi/2}_0 \ln (9 - 4 \cos^2 \theta) \, d\theta.$$
After rearranging the argument for the log function we can write
\begin{align*}
I &= \int_0^{\pi/2} \ln \left [9 \left (1 - \frac{4}{9} \cos^2 \theta \right ) \right ] \, d\theta\\
&= \pi \ln (3) + \int_0^{\pi/2} \ln \left (1 - \frac{4}{9} \cos^2 \theta \right ) \, d\theta\\
&= \pi \ln (3) + I_\alpha.
\end{align*}
Making use of the following Maclaurin series expansion for $\ln (1 - x)$, namely
$$\ln (1 - x) = -\sum_{n = 1}^\infty \frac{x^n}{n}, \qquad |x| < 1,$$
we have, after interchanging the infinite sum with the integration
$$I_\alpha = -\sum_{n = 1}^\infty \frac{4^n}{n 9^n} \int^{\pi/2}_0 \cos^{2n} \theta \, d\theta.$$
The integral that appears here can be evaluated using a Beta function. Here
\begin{align*}
\int_0^{\pi/2} \cos^{2n} \theta \, d\theta &= \frac{1}{2} \cdot 2 \int_0^{\pi/2} \cos^{2(n + \frac{1}{2}) - 1}\theta \sin^{2(\frac{1}{2}) - 1} \theta \, d\theta\\
&= \frac{1}{2} \text{B} \left (n + \frac{1}{2}, \frac{1}{2} \right )\\
&= \frac{1}{2} \frac{\Gamma (n + 1/2) \Gamma (1/2)}{\Gamma (n + 1)}.
\end{align*}
Since $n \in \mathbb{N}$, we have
$$\Gamma (n + 1) = n! \quad {\rm and} \quad \Gamma \left (n + \frac{1}{2} \right ) = \frac{\sqrt{\pi}}{2^{2n}} \frac{(2n)!}{n!},$$
giving
$$\int_0^{\pi/2} \cos^{2n} \theta \, d\theta = \frac{\pi}{2} \cdot \frac{1}{2^{2n}} \frac{(2n)!}{(n!)^2},$$
Thus
$$I_\alpha = - \frac{\pi}{2} \sum_{n = 1}^\infty \binom{2n}{n} \frac{1}{n 9^n},$$
and yields
$$I = \pi \ln (3) - \frac{\pi}{2} \sum_{n = 1}^\infty \binom{2n}{n} \frac{1}{n 9^n}. \tag2$$
To find the sum in (2) we write the generating function for the central binomial coefficients given by (1) as
$$1 + \sum_{n = 1}^\infty \binom{2n}{n} x^n = \frac{1}{\sqrt{1 - 4x}}.$$
Dividing the above expression by $x$ before integrating it from zero to $x$ we have
\begin{align*}
\sum_{n = 1}^\infty \binom{2n}{n} \int_0^x t^{n - 1} \, dt &=
\int_0^x \frac{1}{t} \left (\frac{1}{\sqrt{1 - 4t}} - 1 \right ) \, dt\\
\sum_{n = 1}^\infty \binom{2n}{n} \left [\frac{t^n}{n} \right ]^x_0 &= \int_0^x \frac{1 - \sqrt{1 - 4t}}{t \sqrt{1 - 4t}} \, dt\\
\sum_{n = 1}^\infty \binom{2n}{n} \frac{x^n}{n} &= \int_0^x \frac{1 - \sqrt{1 - 4t}}{t \sqrt{1 - 4t}} \, dt.
\end{align*}
The remaining integral can be found by enforcing a substitution of $x \mapsto \frac{1}{4} \cos^2 x$. When this is done we find
$$\sum_{n = 1}^\infty \binom{2n}{n} \frac{x^n}{n} = 2 \ln \left (\frac{2}{1 + \sqrt{1 - 4x}} \right ), \quad |x| < \frac{1}{4}.$$
If we set $x = 1/9$ in the above generating function we find
$$\sum_{n = 1}^\infty \binom{2n}{n} \frac{1}{n 9^n} = 2 \ln \left (\frac{6}{3 + \sqrt{5}} \right ),$$
and is the required sum we were after.
Returning to (2) we find
$$I = \pi \ln (3) - \pi \ln \left (\frac{6}{3 + \sqrt{5}} \right ),$$
or
$$\int_0^{\pi/2} \ln (9 - 4 \cos^2 \theta) \, d\theta = \pi \ln \left (\frac{3 + \sqrt{5}}{2} \right ).$$
|
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|
Prove by induction that for every $n \ge 0$ the term $2^{2n+1} - 9n^2 + 3n -2$ is divisible by $54$ What I get is $4 \cdot 54k + 27(n(n-1)) - 2$.
I understand first two terms but $-2$ is giving me a headache. I suppose to be true every term should be divisible by $54$. Or did I make wrong conclusion?
P. S: Is there mathjax tutorial?
|
Let $f(n) = 2^{2n+1}-9n^2+3n-2$; then
\begin{align*}
f(n+1)-4f(n) &= 2^{2n+3}-9(n+1)^2+3(n+1)-2-4(2^{2n+1}-9n^2+3n-2) \\
&= 4\cdot 2^{2n+1} - 9n^2 - 18n - 9 + 3n+3-2-4(2^{2n+1}-9n^2+3n-2) \\
&= 27n^2-27n = 27n(n-1).
\end{align*}
Since either $n$ or $n-1$ is even, this expression is divisible by $54$. Since $4f(n)$ is divisible by $54$ (inductive assumption), so is $f(n+1)$.
|
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|
Inequality : $\sum_{cyc}\frac{\sqrt{2}a^2b}{2a+b} \leq \sum_{cyc} \frac{\sqrt{a^2+b^2}}{2ab+1}$ Let $a, b, c$ be positive real number such that $a+b+c = 3$.
Prove that
$$\displaystyle\sum_{cyc}\frac{\sqrt{2}a^2b}{2a+b} \leq \displaystyle\sum_{cyc} \frac{\sqrt{a^2+b^2}}{2ab+1}$$
My attempt :
By AM-GM,
$\displaystyle\sum_{cyc}\frac{a^2b}{2a+b} \leq \displaystyle\sum_{cyc} \frac{a^2b}{3 \sqrt[3]{a^2b}} \leq \displaystyle\sum_{cyc} \frac{\sqrt[3]{a^4b^2}}{3} $ ---[1]
By AM-GM,
$a^2+2ab \geq 3\sqrt[3]{a^4b^2}$
$b^2+2bc \geq 3\sqrt[3]{b^4c^2}$
$c^2+2ca \geq 3\sqrt[3]{c^4a^2}$
$(a+b+c)^2 = 9 \geq 3 \displaystyle\sum_{cyc}\sqrt[3]{a^4b^2}$
$3 \geq \displaystyle\sum_{cyc}\sqrt[3]{a^4b^2}$
$\displaystyle\sum_{cyc} \frac{\sqrt[3]{a^4b^2}}{3} \leq 1$ ---[2]
From [1], [2], we have
$\displaystyle\sum_{cyc}\frac{\sqrt{2}a^2b}{2a+b} \leq \sqrt{2}$
Please suggest how to show that
$ \sqrt{2} \leq \displaystyle\sum_{cyc} \frac{\sqrt{a^2+b^2}}{2ab+1}$
|
By C-S $$\sqrt2\sum_{cyc}\frac{2a^2b}{2a+b}\leq\frac{\sqrt2}{(2+1)^2}\sum_{cyc}a^2b\left(\frac{2^2}{2a}+\frac{1^2}{b}\right)=\frac{\sqrt2}{9}\sum_{cyc}(2ab+a^2)=\sqrt2.$$
Thus, it's enough to prove that
$$\sum_{cyc}\frac{\sqrt{a^2+b^2}}{1+2ab}\geq\sqrt2.$$
Now, by C-S again
$$\sum_{cyc}\frac{\sqrt{a^2+b^2}}{1+2ab}=\frac{1}{\sqrt2}\sum_{cyc}\frac{\sqrt{(1^2+1^2)(a^2+b^2)}}{1+2ab}\geq\frac{1}{\sqrt2}\sum_{cyc}\frac{a+b}{1+2ab}.$$
Thus, it remains to prove that $$\sum_{cyc}\frac{a+b}{1+2ab}\geq2$$ or
$$\sum_{cyc}\left(\frac{(a+b)(a+b+c)}{(a+b+c)^2+18ab}-\frac{2}{9}\right)\geq0$$ or
$$\sum_{cyc}\frac{7a^2+7b^2-2c^2-22ab+5ac+5bc}{1+2ab}\geq0$$ or
$$\sum_{cyc}\frac{(c-a)(-7a+11b-c)-(b-c)(-7b+11a-c)}{1+2ab}\geq0$$ or
$$\sum_{cyc}(a-b)\left(\frac{-7b+11c-a}{1+2bc}-\frac{-7a+11c-b}{1+2ac}\right)\geq0$$ or
$$\sum_{cyc}\frac{(a-b)((-7b+11c-a)(1+2ac)-(-7a+11c-b)(1+2bc))}{(1+2ac)(1+2bc)}\geq0$$ or
$$\sum_{cyc}\frac{(a-b)^2(3+11c^2-ac-bc)}{(1+2ac)(1+2bc)}\geq0$$ or
$$\sum_{cyc}(a-b)^2(3+11c^2-ac-bc)(1+2ab)\geq0$$ or
$$\sum_{cyc}(a-b)^2(3+11c^2-c(3-c))(1+2ab)\geq0$$ or
$$\sum_{cyc}(a-b)^2(4c^2-c+1)(1+2ab)\geq0.$$
Done!
|
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|
Double integration including e? I would like to calculate the following expression:
$\iint_D x^2e^\sqrt{x^2+y^2} dxdy$
for D = ${ a^2\le x^2+y^2 \le b^2} $ and $ 0 \le a \le b$
How would you do it, in a step-by-step, please?
|
$$\iint_D x^2e^\sqrt{x^2+y^2} dxdy=\iint_D (rcos\phi)^2e^r\cdot r drd\phi= \iint_D r^3cos^2\phi e^r drd\phi=$$
$$=\int^{2\pi}_0d\phi \int^b_a r^3cos^2\phi e^r dr=\int^{2\pi}_0cos^2\phi d\phi \int^b_a r^3 e^r dr=$$
$$=\int^{2\pi}_0cos^2\phi d\phi ( r^3-3r^2+6r-6) e^r|^b_a =\int^{2\pi}_0(( b^3-3b^2+6b-6) e^b-(a^3-3a^2+6a-6) e^a) cos^2\phi d\phi= $$
$$=(( b^3-3b^2+6b-6) e^b-(a^3-3a^2+6a-6) e^a)\int^{2\pi}_0 cos^2\phi d\phi $$
$$=(( b^3-3b^2+6b-6) e^b-(a^3-3a^2+6a-6) e^a)(\frac{1}{2}\phi +\frac{1}{4}sin2\phi)|^{2\pi}_0= $$
$$=(( b^3-3b^2+6b-6) e^b-(a^3-3a^2+6a-6) e^a)\pi $$
|
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|
How to evaluate $1 - \frac{\binom{n^2}{1}}{\binom{n+1}{1}} + \frac{\binom{n^2}{2}}{\binom{n+2}{2}} - \frac{\binom{n^2}{3}}{\binom{n+3}{3}} + ..$ How to evaluate $1 - \frac{\binom{n^2}{1}}{\binom{n+1}{1}} + \frac{\binom{n^2}{2}}{\binom{n+2}{2} } - \frac{\binom{n^2}{3}}{\binom{n+3}{3}} + \frac{\binom{n^2}{4}}{\binom{n+4}{4}} - ......$
I really have no idea how to proceed in this question. Expanding it by the formula isn't helping as much as I can see. And the options are extremely sofisticated as well like 1/n, 1/(n+1), 1. How to proceed?
|
Below is a quite brutal approach : whenever I see inverses of binomial coefficients, I try to use the following relation between the Gamma and Beta functions (I can provide a link if needed) :
$$\mbox{With } 0 < m \le n,\quad \quad \ \frac{1}{\binom{m+n}{m}} = \frac{mn}{m+n} \cdot \displaystyle{\int_0^1} t^{m-1}(1-t)^{n-1} dt$$
Your sum becomes :
$$S_n = 1 + \sum \limits_{k=1}^{n^2} (-1)^k \binom{n^2}{k} \frac{kn}{k+n} \displaystyle{\int_0^1} t^{k-1}(1-t)^{n-1} dt$$
$$S_n = 1 + n \displaystyle{\int_0^1} (1-t)^{n-1} \cdot \Big( \sum \limits_{k=1}^{n^2} \frac{(-1)^k k}{k+n} \binom{n^2}{k} t^{k-1} \Big) dt$$
Short interlude : we can simplify this sum. Denote $f : t \mapsto t^{n+1} \sum \limits_{k=1}^{n^2} \frac{(-1)^k k}{k+n} \binom{n^2}{k} t^{k-1}$. Then $f'(t) = \sum \limits_{k=1}^{n^2} (-1)^k k \binom{n^2}{k} t^{k+n-1} = t^n \frac{d}{dt} \big( \sum \limits_{k=0}^{n^2} \binom{n^2}{k} (-t)^k \Big) = t^n \frac{d}{dt} \Big( (1-t)^{n^2} \Big) = -n^2 t^n (1-t)^{n^2-1}$.
Hence $$S_n = 1 + n \displaystyle{\int_0^1} (1-t)^{n-1} \cdot \frac{1}{t^{n+1}} \displaystyle{\int_0^{t}}-n^2 s^n (1-s)^{n^2-1} ds dt$$
and by exchanging the integrals :
$$S_n = 1 - n^3 \displaystyle{\int_0^1} s^n (1-s)^{n^2-1} \cdot \displaystyle{\int_s^1} \frac{(1-t)^{n-1}}{t^{n+1}} dtds$$
$$S_n = 1 - n^3 \displaystyle{\int_0^1} s^n (1-s)^{n^2-1} \cdot \frac{(1-s)^n}{n s^n}ds$$
$$S_n = 1 -n^2 \displaystyle{\int_0^1} (1-s)^{n^2+n-1} = 1 - \frac{n^2}{n^2+n}$$
There might (must) be a more elegant way, but with this we can conclude that $S_n = \frac{1}{n+1}$.
|
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|
Find $P(XY+Z-XZ=2)$ when $Z$~$N(0,1)$, $Y$~$B(3, \frac{1}{2})$, $X$~$B(1, \frac{1}{3})$ Let $X,Y,Z$ be independent variables such that: \begin{gather*}
Z \sim N(0,1) \,\text{(normal distribution)},\\
Y \sim B(3, \frac{1}{2}) \,\text{(binomial distribution)},\\
X \sim B(1, \frac{1}{3}) \,\text{(binomial distribution)}.
\end{gather*}
We define $W=XY+(1-X)Z$. I need to calculate $P(X=1|W=2)$ , so\begin{align*}
P(X=1|W=2)&=P(X=1|XY+Z-XZ=2)\\
&=\frac{P(XY+Z-XZ=2 \cap X=1)}{P(XY+Z-XZ=2)}\\
&= \frac{P(Y=2)}{P(XY+Z-XZ=2)}.
\end{align*}
And I got stuck in calculating $P(XY+Z-XZ=2)$. Any help would be appreciated.
|
\begin{align*}
P(XY + Z - XZ = 2) &= P(XY + Z - XZ = 2 \mid X = 0) + P(XY + Z - XZ = 2 \mid X = 1)\\
&= P(Z = 2 \mid X = 0) + P(Y = 2 \mid X = 1)\\
&= P(Z = 2) P(X = 0) + P(Y = 2)P(X = 1)\\
&= 0 \cdot \frac{2}{3} + \frac{3}{8} \cdot \frac{1}{3} = \frac{1}{8}.
\end{align*}
|
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|
The Jacobi-Madden equation $a^4+b^4+c^4+d^4 = (a+b+c+d)^4$ and disguised Pythagorean triples I. The Jacobi-Madden equation,
$$a^4+b^4+c^4+d^4 = (a+b+c+d)^4$$
is equivalent to a disguised Pythagorean triple,
$$(a^2+ab+b^2)^2+(c^2+cd+d^2)^2 = \big((a+b)^2+(a+b)(c+d)+(c+d)^2\big)^2$$
II. A special case of the Descartes' circle theorem,
$$2(a^4+b^4+c^4+d^4)=(a^2+b^2+c^2+d^2)^2$$
and Euler showed this is just,
$$(2ab)^2+(2cd)^2 = (a^2+b^2-c^2-d^2)^2$$
III. The Fermat quartic,
$$a^4+b^4 = c^4$$
becomes,
$$(a b - a c + b c + c^2)^2 + (a b + a c - b c + c^2)^2 = (a^ 2 + b^2 + c^2)^2$$
IV. The Pell equation,
$$x^2-2y^2 = -1$$
is also,
$$x^2 + (y^2 - 1)^2 = y^4$$
as well as,
$$\Big(\frac{x-1}{2}\Big)^2+\Big(\frac{x+1}{2}\Big)^2 = y^2$$
with the latter showing there are infinitely many triples where the legs differ by just $1$.
Q: Are there any other examples of simple quadratic or quartic equations that can be expressed as a Pythagorean triple?
|
Not sure if this is what you had in mind, but suppose we have two distinct triangular numbers whose product is a square. Say $[u(u+1)/2][v(v+1)/2] = t^2$, where $u > v > 0$. Define $a = 4t$, $b_1 = u - v$, $b_2 = u + v + 1$, $c = 2uv + u + v$. Then a routine calculation gives
$$a^2 + b_1^2 = c^2,\quad a^2 + b_2^2 = (c + 1)^2.$$
Thus we get two Pythagorean triangles with a common side and hypotenuses differing by $1$. For instance $u = 8$, $v = 1$ gives $(24, 7, 25)$ and $(24, 10, 26)$. Alternatively you can start with two such triangles and reverse the procedure, as is easily proved.
|
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|
Find $f(f(f(f(f(f(\cdots f(x)))))))$ $2018$ times
I was given a problem to calculate $f(f(f(\dots f(x))))$ $2018$ times if $f(x)=\frac{x}{3} + 3$ and $x=4$.
Is it even possible to be done without a computer?
|
Let $x_{n+1} =f(x_n)$ that is $x_{n+1} =\frac{x_n}{3}+3 $
Letting $$u_n =x_n -\frac{3}{2}\implies u_{n+1}=\frac13 u_n \implies u_n = \frac{u_1}{3^{n-1}} \implies x_n = \frac{3}{2}+\frac{x_1 -\frac{3}{2}}{3^{n-1}}$$
Hence, we have $$ \color{blue}{ \underbrace{f(f(f(f(f(f(\cdots f(x)))))))}_{n~~times} = x_n = \frac{3}{2}+\frac{2x -3}{2\cdot3^{n-1}}}$$
|
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|
How to evaluate $\tan^{-1}\left(\frac4x\right)-\tan^{-1}\left(\frac3x\right)$ This is what I did but please correct me if I am wrong.
Let $y=\tan^{-1}\Big(\dfrac{4}{x}\Big)-\tan^{-1}\Big(\dfrac{3}{x}\Big)$.
$\tan(y)\\
=\dfrac{\tan\Big(\tan^{-1}\Big(\dfrac{4}{x}\Big)\Big)-\tan\Big(\tan^{-1}\Big(\dfrac{3}{x}\Big)\Big)}{1-\tan\Big(\tan^{-1}\Big(\dfrac{4}{x}\Big)\Big)\tan\Big(\tan^{-1}\Big(\dfrac{3}{x}\Big)\Big)}\\
=\dfrac{\dfrac{4}{x}-\dfrac{3}{x}}{1-\dfrac{4}{x}\times\dfrac{3}{x}}\\
=\dfrac{\dfrac{1}{x}}{1-\dfrac{12}{x^2}}\\
=\dfrac{x}{x^2-12}
$
|
Yes, your approach is correct but it must be completed. First of all, what you found is $\tan(y)$ in the end so with the correct formula $$\tan(a-b)=\frac{\tan a-\tan b}{1+\tan a\tan b}$$
the answer becomes $$y = \tan^{-1}\bigg(\frac{x}{x^2+12}\bigg)$$
And here is another approach by using geometry:
Here, notice that $\tan(a) = \frac{3}{x}$ and $\tan(b) = \frac{4}{x}$ and what we are seeking is $\angle ACD = c$. Then, if we use Law of Sines for the triangle $ADC$, we have an equality:
$$\frac{|AD|}{\sin(\angle DCA)} = \frac{|DC|}{\sin(\angle DAC)}$$
which is
$$\frac{1}{\sin(c)} = \frac{\sqrt{x^2+9}}{\frac{x}{\sqrt{x^2+16}}}$$
therefore we have $\sin(c) = \frac{x}{\sqrt{x^4+25x^2+144}}$. Now we can construct another right angle triangle with one of its angle is $c$, also satisfying this condition:
Then notice that by Pythagorean Theorem,
$$|LM|^2 = x^4+25x^2+144-x^2 = x^4+24x^2+144 = (x^2+12x)^2 \implies |LM| = x^2+12x$$
Finally, $$\tan(c) = \frac{|KL|}{|LM|} = \frac{x}{x^2+12} \implies c = \tan^{-1}\bigg(\frac{x}{x^2+12}\bigg)$$
|
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|
Minimise $|z_az_b + z_cz_d|$ where $\{z\}$ are the roots of $x^4 + 14x^3 + 52x^2 + 56x + 16$ Let $f(x) = x^4 + 14x^3 + 52x^2 + 56x + 16.$ Let $z_1, z_2, z_3, z_4$ be the four roots of $f$. Find the smallest possible value of $|z_az_b + z_cz_d|$ where ${a, b, c, d} = {1, 2, 3, 4}.$
So I have tried Vieta, but it is try too complex and plugging in a simple value to try to find a root failed, so I think this is going to be much harder than finding the four roots-- I wonder if there are more involved techniques to solve this problem.
|
I claim the minimum is $8$. I will use dxiv suggestion. Let $g(y) = y^4+7y^3+13y^2+7y+1$ Obviously all roots are real and negative since we have
\begin{array}{cccccc}
y & -5 & -2 & -1 & -0.5 & 0 \\
g(y) & + & - & + & - & + \\
\end{array}
Let $a,b,{1\over a},{1\over b}$ be all roots for $g$ where $a,b,$ are negative.
So we have next possible expresions
\begin{eqnarray}
E_1 &=&ab+{1\over ab} >2\\
E_2 &=&{a\over b}+{b\over a} >2\\
E_3 &=&a{1\over a}+b{1\over b} =2\\
\end{eqnarray}
and does are all. So $E_{\min}=2$. So the minimum value for the given expression is $4\cdot 2 =8$
|
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|
How to prove that $\frac{a+b}{1+a+b} \leq \frac{a}{1+a} + \frac{b}{1+b}$ for non-negative $a,b$? If $a, b$ are non-negative real numbers, prove that
$$
\frac{a+b}{1+a+b} \leq \frac{a}{1+a} + \frac{b}{1+b}
$$
I am trying to prove this result. To that end I added $ab$ to both denominator and numerator as we know
$$
\frac{a+b}{1+a+b} \leq \frac{a+b+ab}{1+a+b+ab}
$$
which gives me
$$
\frac{a}{1+a} + \frac{b}{(1+a)(1+b)}
$$
How can I reduce the second term further, and get the required result?
|
In order to prove that $f(x)=\frac{x}{1+x}$ is sublinear on $\mathbb{R}^+$ it is enough to notice that $f'(x)=\frac{1}{(1+x)^2}$ leads to
$$ \frac{f(a+b)-f(a)}{f(b)-f(0)} = \frac{\int_{a}^{a+b}\frac{dx}{(1+x)^2}}{\int_{0}^{b}\frac{dx}{(1+x)^2}}\leq 1 $$
since $f'(x)$ is decreasing. In other terms, the sublinearity is a consequence of the concavity.
|
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|
How to compute $\lim_{n\to \infty}\prod_{k=1}^{2n} \left(1+ \frac{ k}{n} \right)^{1/n} $
Find
$$ \lim_{n\to \infty}\prod_{k=1}^{2n} \left(1+ \frac{ k}{n} \right)^{1/n}.$$
I tried with ln composition but ineffectively. Any idea?
|
Following the suggestion by Paramanand Singh let indicate
$$a_n=\prod_{k=1}^{2n} \left(1+ \frac{ k}{n} \right)^{1/n}=\sqrt[n] b_n \quad b_n=\prod_{k=1}^{2n} \left(1+ \frac{ k}{n} \right)$$
then
$$\frac{b_{n+1}}{b_n}=\frac{\prod_{k=1}^{2n+2} \left(1+ \frac{ k}{n+1} \right)}{\prod_{k=1}^{2n} \left(1+ \frac{ k}{n} \right)}=\left(1+\frac{2n+2}{n+1}\right)\left(1+\frac{2n+1}{n+1}\right)\prod_{k=1}^{2n} \frac{n(n+k+1)}{(n+1)(n+k)} $$
$$=\left(1+\frac{2n+2}{n+1}\right)\left(1+\frac{2n+1}{n+1}\right)\frac{n^{2n}(3n+1)}{(n+1)^{2n}(n+1)} \to 3\cdot 3\cdot \frac{3}{e^2}=\frac{27}{e^2}$$
and therefore (see Limit of ${a_n}^{1/n}$ is equal to $\lim_{n\to\infty} a_{n+1}/a_n$)
$$\lim_{n\to \infty} a_n = \lim_{n\to \infty} \sqrt[n] b_n =\lim_{n\to \infty} \frac{b_{n+1}}{b_n}=\frac{27}{e^2}$$
|
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|
Find $f(1)+f(3)+f(5)+\dots+ f(999)$ where $f$ is a given function
Given $$f(x)= \frac {1}
{\sqrt[3] {x^2+2x+1} + \sqrt[3] {x^2-1} + \sqrt[3] {x^2-2x+1}}$$
and
$$E= f(1)+f(3)+f(5)+\dots+ f(999).$$ Then find the value of $E$.
My work :-
Let $\sqrt[3] {x+1}= a$, $\sqrt[3] {x-1}=b $
Then the equation reduces to
$$f(x)= \frac {1}{a^2+b^2+ab}.$$
Now for this expression I tried breaking into partial fractions to that the sum telescopes to some finite quantity but the cube root terms restrict this straightforward way. Thanks in advance for the help. Also if someone finds a way to rationalize the denominator, please share.
|
Hint. Since $(a^3-b^3)=(a-b)(a^2+ab+b^2)$, it follows that
$$f(x)=\frac{1}{a^2+ab+b^2}=\frac{a-b}{a^3-b^3}=\frac{\sqrt[3] {x+1}-\sqrt[3] {x-1}}{2}.$$
Therefore the given sum is telescopic:
$$\sum_{k=0}^{n-1}f(2k+1)=\frac{1}{2}\left(\sum_{k=0}^{n-1}\sqrt[3] {2(k+1)}-\sum_{k=0}^{n-1}\sqrt[3] {2k}\right)=\frac{\sqrt[3] {2n}}{2}.$$
|
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|
Definite integral of a rational fraction Can I find the value of $$\int_3^{\infty}\frac{x-1}{(x^2-2x-3)^2}dx$$ by just factoring the fraction?
I tried to wrote:
$$\frac{x-1}{(x^2-2x-3)^2}=\frac{x-1}{(x^2-2x+1-4)^2}=\frac{x-1}{[(x-1)^2-2^2]^2}=\frac{x-1}{(x+1)^2\cdot(x-3)^2}$$ but didn't work out. Any ideas?
|
With $u=x^2-2x-3$ we have $du = 2(x-1)\,dx$, so that
$$\int \frac{x-1}{(x^2-2x-3)^2}\,dx = \frac{1}{2}\int\frac{1}{u^2}\,du
= -\frac{1}{2u} + c= -\frac{1}{2(x^2-2x-3)}+c.$$
|
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|
If $a_{1}>2$and $a_{n+1}=a_{n}^{2}-2$ then Find $\sum_{n=1}^{\infty}$ $\frac{1}{a_{1}a_{2}......a_{n}}$ Question If $a_{1}>2$and $\left\{ a_{n}\right\} be$ a recurrsive
sequence defined by setting $a_{n+1}=a_{n}^{2}-2$ then Find
$\sum_{n=1}^{\infty}$$\frac{1}{a_{1}a_{2}......a_{n}}$
Book's Answer
I have mentioned my problem in the image.Any and all help will be appreciated
|
You have proven that
$$a_n^2-4 = a_1^2 \ldots a_{n-1}^2(a_1^2-4)$$
Divide both sides by $a_1^2 \ldots a_{n-1}^2$ and then take square root on both sides,
We have
$$\sqrt{\frac{a_n^2-4}{a_1^2 \ldots a_{n-1}^2}} = \sqrt{a_1^2-4}$$
Now take limit $n \to \infty$,
\begin{align}\lim_{n \to \infty}\sqrt{\frac{a_n^2-4}{a_1^2 \ldots a_{n-1}^2}} = \sqrt{a_1^2-4 }& \text{, note that RHS is independent of $n$.}\\
\sqrt{\lim_{n \to \infty}\frac{a_n^2-4}{a_1^2 \ldots a_{n-1}^2}} = \sqrt{a_1^2-4} & \text{ since square root is continuous}\\
\sqrt{\lim_{n \to \infty}\frac{a_n^2}{a_1^2 \ldots a_{n-1}^2}- \lim_{n \to \infty}\frac{4}{a_1^2 \ldots a_{n-1}^2}} = \sqrt{a_1^2-4}\\
\sqrt{\lim_{n \to \infty}\frac{a_n^2}{a_1^2 \ldots a_{n-1}^2}} = \sqrt{a_1^2-4} & \text{, the $a_n$ increases to $\infty$, the second term vanishes}\\
\lim_{n \to \infty}\frac{a_n}{a_1 \ldots a_{n-1}} = \sqrt{a_1^2-4}
\end{align}
|
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|
A congruence with the Euler's totient function and number of divisors function Can you provide a proof or a counterexample to the claim given below ?
Inspired by the congruence $1.3$ in this paper I have formulated the following claim :
Let $n$ be a natural number , let $\tau(n)$ be number of divisors function and let $\varphi(n)$ be Euler's totient function , then :
$$(2^{\varphi(n)}-1)(2^{\tau(n)}-1) +2 \equiv 2^{n-1} \pmod {2^{n}-1}$$
for all primes and no composite with the exception of $4$ .
I have tested this claim up to $5 \cdot 10^6$ .
I was searching for a counterexample using the following two PARI/GP codes :
NumdivTotient1(lb,ub)={
forprime(n=lb,ub,
if(!(Mod((2^eulerphi(n)-1)*(2^numdiv(n)-1)+2,2^n-1)==2^(n-1)),print(n)))
}
NumdivTotient2(lb,ub)={
forcomposite(n=lb,ub,
if((Mod((2^eulerphi(n)-1)*(2^numdiv(n)-1)+2,2^n-1)==2^(n-1)),print(n)))
}
Remark
More generally we can formulate the following claim :
Let $b$ and $n$ be a natural numbers , $b \ge 2$ , then
$$\frac{b^{\varphi(n)}-1}{b-1}(b^{\tau(n)}-1) + b \equiv b^{n-1} \pmod {\frac{b^{n}-1}{b-1}}$$
for all primes and no composite with the exception of $4$ .
|
We have
$$n = \sum_{d\mid n} \varphi(d)\,,$$
so
$$\varphi(n) + \tau(n) = n - \sum_{\substack{d\mid n \\ d < n}} \varphi(d) + \sum_{d \mid n} 1 = (n+1) - \sum_{\substack{d \mid n \\ 1 < d < n}}\bigl(\varphi(d) - 1\bigr)\,.$$
If $n$ is composite and has a prime factor $\geqslant 5$, it follows that $\varphi(n) + \tau(n) < n-1$. If $n > 9$ is divisible by $8$ or by $9$, it also follows that $\varphi(n) + \tau(n) < n-1$. If $\varphi(n) + \tau(n) < n-1$, it follows that
$$(b^{\varphi(n)} - 1)(b^{\tau(n)} - 1) + b = b^{\varphi(n) + \tau(n)} - (b^{\varphi(n)} - b) - (b^{\tau(n)} - 1) \leqslant b^{\varphi(n) + \tau(n)} < b^{n-1}\,,$$
whence the congruence cannot hold for such $n$.
The remaining cases of composite $n$ - that is, $n \in \{ 4, 6, 8, 9, 12\}$ - are easily checked by hand, and that the congruence holds for prime $n$ follows from $\varphi(n) = n-1$ and $\tau(n) = 2$:
\begin{align}
\frac{(b^{n-1}-1)(b^2-1)}{b-1} + b &= \frac{b^{n+1} - b^{n-1} - b^2 + 1 + b^2 - b}{b-1} \\
&= \frac{b^{n+1} - b^{n-1} - b + 1}{b-1} \\
&= b^{n-1} + \frac{b^{n+1}-b^n-b+1}{b-1} \\
&= b^{n-1} +(b-1)\frac{b^n-1}{b-1}\,.
\end{align}
|
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|
What can be said about the series $\sum_{n=1}^{\infty} \left[ \frac{1}{n} - \frac{1}{\sqrt{ n^2 + x^2 }} \right]$ This is a sequel to this question.
I recently was browsing through Hansen's "A Table of Series and Products", and I miraculously found the sum that I was looking for:
$$
\sum_{n=1}^\infty K_{0}\left( n z \right) \ = \ \frac{\pi}{2 z} + \frac{1}{2} \log\left( \frac{z}{4 \pi} \right) + \frac{\gamma}{2} - \frac{1}{2} \sum_{n=1}^{\infty} \left[ \frac{1}{n} - \frac{1}{\sqrt{ n^2 + \frac{z^2}{4\pi^2} }} \right]
$$
What a beautiful sum (the entire book is full of such amazing results). Here, $\gamma$ is the Euler-Mascheroni constant and $K$ is the modified Bessel function of the second kind (of order 0).
I'm of course curious about the latter sum now:
$$
F(x) := \sum_{n=1}^{\infty} \left[ \frac{1}{n} - \frac{1}{\sqrt{ n^2 + x^2 }} \right]
$$
I'm assuming this sum cannot be evaluated exactly (although, the book is old from 1975, maybe it's out of date?)...
I am most curious, how does this function $F(x)$ look like in the limit $x \to 0$? Does it diverge? My guess would be yes since $x$ is in a denominator. My only thought of how to attack this series in the limit $x\to 0$ is to take the following series expansion in this limit:
$$
\frac{1}{\sqrt{n^2 + x^2}} = \frac{1}{n} + \frac{x^2}{2n^3} + \mathcal{O}\left( x^4 \right)
$$
Then in this limit we have something like;
$$
F[x] \approx \sum_{n=1}^{\infty} \left[ \frac{1}{n} - \frac{1}{n} - \frac{x^2}{2 n^3} \right] = \frac{1}{2} x^2 \zeta(3)
$$
...in terms of the Riemann-zeta function. This seems like an answer, but I am worried there is something wrong with taking an expansion in the argument of the sum. Is this okay? Is there a better approximation?
|
The approximation you mentioned is just fine. In a very simple (or elementary) way we have
$$\sqrt{n^2+x^2}\geq n \Rightarrow 0\leq \frac{1}{n} - \frac{1}{\sqrt{n^2+x^2}}$$
so (from partials sums an taking limits)
$$F(x) \geq 0, \forall x \tag{1}$$
Next thing to observe is
$$\frac{1}{n} - \frac{1}{\sqrt{n^2+x^2}}=\frac{\sqrt{n^2+x^2}-n}{n\sqrt{n^2+x^2}}\leq\frac{\sqrt{n^2+x^2}-n}{n^2}=\\
\frac{n^2+x^2-n^2}{n^2(\sqrt{n^2+x^2}+n)}\leq \frac{x^2}{2n^3}$$
and this leads to
$$0 \leq F(x) \leq \frac{x^2}{2}\sum\frac{1}{n^3}=\frac{x^2 \zeta(3)}{2} \tag{2}$$
applying squeeze theorem to $(2)$
$$\lim\limits_{x\rightarrow 0+}F(x)=0$$
|
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|
An algebraic inequality involving $\sum_{cyc} \frac1{(a+2b+3c)^2}$ I was reading through the proof of an inequality posted on a different website and the following was mentioned as being easily proven by AM-GM:
Let $a,\ b,\ c>0$, then $$\frac{1}{(a+2b+3c)^2} + \frac{1}{(b+2c+3a)^2} + \frac{1}{(c+2a+3b)^2} \le \frac{1}{4(ab + bc+ca)}$$
I checked that the inequality holds numerically, but I can't find a solution (definitely nothing obvious using AM-GM). I played around with a simpler version of the inequality:
$$\frac{1}{(a+2b)^2} + \frac{1}{(b+2a)^2} \le \frac{2}{9}\frac{1}{ab}$$
which I can prove by brute force algebra and a little AM-GM. However, brute force seems impractical for the three variable inequality. Any suggestions?
|
Hint: Using AM-GM,
$$(a+c+2(b+c))^2=(a+c)^2+4(a+c)(b+c)+4(b+c)^2\geqslant 6(a+c)(b+c)+3(b+c)^2=3(b+c)(2a+b+3c)$$
Hence it is enough to show:
$$\sum_{cyc} \frac{ab+bc+ca}{(b+c)(2a+b+3c)} \leqslant \frac34$$
$$\iff \sum_{cyc}\frac{(b+c)^2+2c^2}{(b+c)(2a+b+3c)}\geqslant\frac32$$
which follows almost directly from CS inequality, as its:
$$LHS \geqslant \frac{\left(\sum_{cyc} (b+c) \right)^2+2(\sum_{cyc} c)^2 }{\sum_{cyc} (b+c)(2a+b+3c)} = \frac{4(a+b+c)^2+2(a+b+c)^2 }{4(a+b+c)^2}=\frac32$$
|
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|
Calculate limit with L'Hopital's rule I want to calculate the limit $\displaystyle{\lim_{x\rightarrow 0}\frac{x^2\cos \left (\frac{1}{x}\right )}{\sin x}}$.
I have done the following:
It holds that $\lim_{x\rightarrow 0}\frac{x^2\cos \left (\frac{1}{x}\right )}{\sin x}=\frac{0}{0}$.
So, we can use L'Hopital's rule:
\begin{align*}\lim_{x\rightarrow 0}\frac{x^2\cos \left (\frac{1}{x}\right )}{\sin x}&=\lim_{x\rightarrow 0}\frac{x^2\cos \left (\frac{1}{x}\right )}{\sin x} \\ &=\lim_{x\rightarrow 0}\frac{\left (x^2\cos \left (\frac{1}{x}\right )\right )'}{\left (\sin x\right )'} =\lim_{x\rightarrow 0}\frac{2x\cdot \cos \left (\frac{1}{x}\right )+x^2\cdot \left (-\sin \left (\frac{1}{x}\right )\right )\cdot \left (\frac{1}{x}\right )'}{\cos x} \\ &=\lim_{x\rightarrow 0}\frac{2x\cdot \cos \left (\frac{1}{x}\right )-x^2\cdot \sin \left (\frac{1}{x}\right )\cdot \left (-\frac{1}{x^2}\right )}{\cos x} \\ & =\lim_{x\rightarrow 0}\frac{2x\cdot \cos \left (\frac{1}{x}\right )+\sin \left (\frac{1}{x}\right )}{\cos x}=\lim_{x\rightarrow 0}\left (2x\cdot \cos \left (\frac{1}{x}\right )+\sin \left (\frac{1}{x}\right )\right ) \\ & =\lim_{x\rightarrow 0}\left (2x\cdot \cos \left (\frac{1}{x}\right )\right )+\lim_{x\rightarrow 0}\left (\sin \left (\frac{1}{x}\right )\right )\end{align*}
We calculate the two limits separately
*
*$\lim_{x\rightarrow 0}\left (2x\cdot \cos \left (\frac{1}{x}\right )\right )$ :
\begin{equation*}\left |\cos \left (\frac{1}{x}\right )\right |\leq 1 \Rightarrow -1\leq \cos \left (\frac{1}{x}\right )\leq 1 \Rightarrow -2x\leq 2x\cdot \cos \left (\frac{1}{x}\right )\leq 2x\end{equation*} we consider the limit $x\rightarrow 0$ and we get \begin{equation*}\lim_{x\rightarrow 0} \left (2x\cdot \cos \left (\frac{1}{x}\right ) \right )=0\end{equation*}
*
*How can we calculate the limit $\lim_{x\rightarrow 0}\left (\sin \left (\frac{1}{x}\right )\right )$ ?
|
L'Hospital's rule is not the alpha and omega of limits computation! When it works, Taylor's formula at order $1$ also works, and it is less dangerous.
This being said, in the present case, doing some asymptotic analysis gives you a fast answer:
Near $0$, $\;\sin x \sim x$ and $ \cos \frac1x$ is bounded, so
$$\frac{x^2\cos \left(\frac{1}{x}\right)}{\sin x}\sim_0\frac{x^2}x\cos\frac1x=x\,O(1)=O(x),$$
and the latter tends to $0$ as $x$ tends to $0$.
|
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|
Show that $Q=X^2+5X+7$ divides $P=(X+2)^m+(X+3)^{2m+3}$ for any $m\in\Bbb N$ Let $$P=(X+2)^m+(X+3)^{2m+3}$$ and $$Q=X^2+5X+7.$$ I need to show that $Q$ divides $P$ for any $m$ natural.
I said like this: let $a$ be a root of $X^2+5X+7=0$. Then $a^2+5a+7=0$.
Now, I know I need to show that $P(a)=0$, but I do not know if it is the right path since I have not found any way to do it.
|
Note that
$$(X+3)^3=X^3+9X^2+27X+27=(X^2+5X+7)(X+4)-1 $$
and
$$(X+3)^2 = X^2+6X+9=(X^2+5X+7)+(X+2).$$
|
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|
Real ordered pair $(a,b)$ in equation
If $$a^2+5b^2+2b=6a+2ab-10$$
then all real ordered pair of $(a,b)$ is?
Try: I am trying to convert it into $$(..)^2+(..)^2=0$$ or $$(..)^2+(..)^2+(..)^2=0.$$
So $$(a-b)^2+(2a-0.75)^2+10-(0.75)^2-6a=0.$$
Could some help me to solve it? Thanks.
|
$a^2+5b^2+2b-6a-2ab+10=0$
Let's suppose we had a $(a+kb+j)^2$ term and that were the only term with $a $. We'd have $(a+kb+j)^2=a^2+k^2b^2+j^2+2kab+2aj+2kjb $. So $k=-1;j=-3$.
So
$a^2+5b^2+2b-6a-2ab+10=0$
$a^2+b^2+9-2ab-6a+6b+4b^2-4b+1=0$
$(a-b-3)^2+(2b-1)^2=0$
So for real solutions
$2b-1=0$ and $b=\frac 12$
And $a-b-3=0$ so $a=3\frac 12$.
|
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|
u-substitution for a definite integral I want to solve the following integral:
$$\int_{-2}^2 \sqrt{4- x^2} dx$$
I am thinking of doing a u-substitution of the whole term inside the root. If I do this, I need to change the limits of the integral. Doing so, the new limits would be equal (0=0). Doesn't this mean that the value of the integral would be 0?
|
The integral can be found with the substitution $x=\sin \theta$.
If we let $u=4-x^2$. Then $du=-2xdx$.
Note that $x=\sqrt{4-u}$ if $x\ge 0$ and $x=-\sqrt{4-u}$ if $x<0$.
So, $\displaystyle dx=\frac{du}{-2\sqrt{4-u}}$ if $x\ge 0$ and $\displaystyle dx=\frac{du}{2\sqrt{4-u}}$ if $x\le 0$.
So we actually do not have $\displaystyle \int_0^0 \frac{\sqrt{u}du}{-2\sqrt{4-u}}$.
\begin{align*}
\int_{-2}^2\sqrt{4-x^2}dx&=\int_{-2}^0\sqrt{4-x^2}dx+\int_0^2\sqrt{4-x^2}dx\\
&=\int_0^4 \frac{\sqrt{u}du}{2\sqrt{4-u}}+\int_4^0 \frac{\sqrt{u}du}{-2\sqrt{4-u}}\\
&=\int_0^4 \frac{\sqrt{u}du}{2\sqrt{4-u}}+\int_0^4 \frac{\sqrt{u}du}{2\sqrt{4-u}}\\
&=\int_0^4 \frac{\sqrt{u}du}{\sqrt{4-u}}
\end{align*}
This does not help us to find the integral. But at least the integral is not $0$.
|
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|
Evaluate the integral $ I=\frac{1}{2\pi i}\int_{\vert z \vert =R}(z-3)\sin \left(\frac{1}{z+2}\right)dz$, Evaluate the integral
$$ I=\frac{1}{2\pi i}\int_{\vert z \vert =R}(z-3)\sin\left(\frac{1}{z+2}\right)dz$$
where $R \geq 4$
My work: Here the zeros of $\sin\left(\frac{1}{z+2}\right)$ is $\frac{1}{n \pi}-2$, so $-2$ is non-isolated singularity. But after I am stuck, Please help. Thanks.
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The integrand function has just ONE singularity at $z=-2$. For any non-zero integer $n$ then by letting $z=\frac{1}{n \pi}-2$, we have that
$$\sin\left(\frac{1}{z+2}\right)=\sin(n\pi)=0.$$
So, for $R>2$, the singularity $z=-2$ is inside the circle $|z|=R$, and, by the Residue Theorem,
$$\begin{align}\frac{1}{2\pi i}\int_{\vert z \vert =R}(z-3)\sin\left(\frac{1}{z+2}\right)dz&=\text{Res}\left((z-3)\sin\left(\frac{1}{z+2}\right), -2\right)\quad(w:=z+2)\\
&=
\text{Res}\left((w-5)\left(\frac{1}{w}-\frac{1}{3! w^3}+\frac{1}{5! w^5}+\dots\right), 0\right)\\
&=
\text{Res}\left(1-\frac{5}{w}-\frac{1}{3! w^2}+\frac{5}{3! w^3}+\dots, 0\right)\\
&=-5.
\end{align}$$
P.S. If $0<R<2$ then the above integral is zero.
|
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|
Trigonometry problem, Evaluate: $\frac {1}{\sin 18°}$ My problem is,
Evaluate: $$\frac {1}{\sin 18°}$$
I tried to do something myself.
It is obvious,
$$\cos 18°= \sin 72°$$
I accept $\left\{18°=x \right\}$ for convenience and here, $\sin (x)>0$
$$\cos (x)=\sin (4x)$$
$$\cos (x)=2× \sin(2x) \cos (2x)$$
$$\cos (x)=2× 2\sin(x) \cos (x)×(1-2\sin^2(x)), \cos(x)>0$$
$$8\sin^3(x)-4\sin(x)+1=0$$
$$(2\sin(x)-1)(4\sin^2(x)+2\sin (x)-1)=0$$
$$4\sin^2(x)+2\sin (x)-1=0, \sin(x)≠\frac 12$$
$$4t^2+2t-1=0$$
$$t_{1,2}=\frac {-1±\sqrt 5}{4}$$
$$t=\frac {\sqrt5-1}{4} ,t>0$$
$$\sin 18°=\frac {\sqrt5-1}{4} .$$
Finally, $$\frac {1}{\sin 18°}=\frac {4}{\sqrt5-1}=\sqrt5+1$$
Is this way correct and is there a better/elegant way to do it? As always, it was an ugly solution.
Thank you!
|
Not an ugly solution at all, but one can do better. Here's an analytical solution based on complex numbers.
Let $\alpha=72^\circ=2\pi/5$, so we can consider $z=e^{\alpha}=\cos\alpha+i\sin\alpha$ that satisfies $z^5-1=0$. Since $z\ne1$, we can deduce
$$
z^4+z^3+z^2+z+1=0
$$
and also, dividing by $z^2$,
$$
z^2+\frac{1}{z^2}+z+\frac{1}{z}+1=0
$$
On the other hand, $z^2+\frac{1}{z^2}=(z+\frac{1}{z})^2-2$ and we can observe that
$$
z+\frac{1}{z}=2\cos\alpha
$$
Therefore we see that $2\cos\alpha$ satisfies the equation $t^2+t-1=0$. Since it is positive, we conclude
$$
2\cos\alpha=\frac{-1+\sqrt{5}}{2}
$$
and therefore
$$
\cos\alpha=\frac{\sqrt{5}-1}{4}
$$
Thus
$$
\frac{1}{\sin18^\circ}=\frac{1}{\cos72^\circ}=\frac{4}{\sqrt{5}-1}=\sqrt{5}+1
$$
|
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|
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.