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    Well seen as I'm horrible at LaTeX I thought I could practice by writing out some random things that interested me. Will kick off with some naive set theory.

    Consider a function \ f , mapping a domain \ X to a co-domain \ Y

    Now consider a family of sets \ A_i \ni  i \in I  \ni  \bigcup_{i\in I} A_i \subseteq X (Where \ \ni means 'such that' )

    Then the image \ f[\bigcup_{i\in I} A_i] = \bigcup_{i\in I} f[A_i]

    Proof: Let \ y \in Y

    If \ y \in f[\bigcup_{i\in I} A_i]  \Leftrightarrow y=f(x) for some \ x \in A_i and for some \ i \in I

    \Leftrightarrow y \in f[A_i] for some \ i \in I

    \ \Leftrightarrow y \in \bigcup_{i \in I} f[A_i]  \Box

    However, \  f[\bigcap_{i\in I} A_i] \subseteq \bigcap_{i \in I} f[A_i]

    Proof: Let \ y \in Y

    If \ y \in f[\bigcap_{i\in I} A_i] \Leftrightarrow y=f(x) for some \ x \in A_i  , \forall i \in I

    \ \Rightarrow y \in f[A_i] , \forall i \in I

    \ \Leftrightarrow y \in \bigcap_{i \in I} f[A_i] \Box

    'But why only the implication on line 2?'

    Well, we shall consider a family of sets who's intersection is the empty set, \ \bigcap_{i \in I} A_i = \emptyset (That is they are disjoint)

    Whereas the intersections of their images is a non-empty set \ \bigcap_{i \in I} f[A_i] = L \neq \emptyset

    Then, \ f[\bigcap_{i \in I} A_i] = f[\emptyset] = \emptyset \neq L Thus we can only show \subseteq and not equality.

    A nice result I think. Probably missing a step or an assumption somewhere somewhere

    Next a simple inequality on the Bell numbers \ B_n where \ B_n in the number of partitions of a set with \ n elements.

    Claim: \ B_n \leq 2^{2^n}

    Proof: Let \ Q be a set with \ n elements, then the power set of \ Q contains \ 2^n elements that is \ |P(Q)|=2^n .

    If we then take, the power set of the power set of \ Q , \ P(P(Q)) .

    We see that, the power set of the power set of \ Q contains \ 2^{2^n} elements.

    But not every element of, the power set of the power set of \ Q is a partition of \ Q

    Thus \ B_n \leq 2^{2^n} \Box

    Hopefully I can try some new things each day. (Typo's will probably be rife:nodots:)
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    keep going ...looking good!
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    I got a headache thanks
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    Well for today I thought I would do something a bit more fundamental , here it goes.

    Theorem: If \ p(z) is a non-constant polynomial over \mathbb{C} , then there exists \ z_0 \in \mathbb{C} such that \ p(z_0)=0 . (FToA)

    Proof: So we shall suppose that there exists no \ z_0 that satisfies \ p(z_0)=0 therefore for some \ R>0 the set

    \mathscr{N} = \{z: |p(z)|^2\leq R\}

    is non-empty, the map \psi : \mathbb{C} \rightarrow \mathbb{R}^+ defined by \psi (z) = |p(z)|^2 is continuous, so \N = \psi ^{-1} ([0,R]) is compact.

    But for a subset on \mathbb{C} this is the same as being closed and bounded. It then follows that \ |p(z)|^2 attains its minimum on \ N and by the definition of \ N this is also its minimum value on \ \mathbb{C}

    Assume the minimum is obtained at \ w \in \mathbb{C} . Then

    \ |p(z)|^2 \geq |p(w)|^2  , \forall z \in \mathbb{C} , and by assumption \ p(w) \neq 0 .

    Consider \ |p(z)|^2 as \ z runs around a small circle centred at \ w , from this we will derive a contradiction.

    Let \ h \in \mathbb{C} . Expand \ p(w+h) in powers of \ h to get

    \ p(w+h)= p_0 + p_1h + p_2h^2 + ... + p_nh^n where \ \partial p =n \ (\Phi)

    Here \ p_j are specific complex numbers, (will leave this as a 'challenge':giggle: )

    Clearly \ p_0=p(w) , and we are assuming this is non-zero, so \ p_0 \neq 0 . If \ p_1 = p_2 = ...=p_n =0 then \ p(z)=p_0 is constant, contrary to hypothesis. So some \ p_j \neq 0 . Let \ m be the smallest integer \ \geq 1 from which \ p_m \neq 0 . In \ (\Phi) let \ h=\epsilon e^{i\theta} for small \ \epsilon > 0 . Then

    \ p(w + \epsilon e^{i\theta}) = p_0 + p_m\epsilon ^m e^{mi\theta} + O(\epsilon ^{m+1}) Therefore

    \ |p(w + \epsilon e^{i\theta})|^2 = |p_0 + p_m\epsilon ^m e^{mi\theta}|^2 + O(\epsilon ^{m+1})

    \  |p(w + \epsilon e^{i\theta})|^2 = p_0\bar{p}_0 + p_0\bar{p}_m\epsilon ^me^{-mi\theta} + \bar{p}_0p_m\epsilon ^me^{mi\theta} + O(\epsilon ^{m+1})

    Let \ p_0\bar{p}_m = re^{i\phi} for \ r \geq 0 . Since \ p_0 \neq 0 and \ p_m \neq 0 we have \ r>0 . Setting \ h=0 we see that \ p_0\bar{p}_0 =|p(w)|^2 . Now

    \ |p(w + \epsilon e^{i\theta})|^2 = p_0\bar{p}_0 + re^{i\phi}\epsilon ^me^{mi\theta} + re^{-i\phi}\epsilon ^me^{-mi\theta} +O(\epsilon ^{m+1})

    \ |p(w + \epsilon e^{i\theta})|^2 = |p(w)|^2 +2\epsilon ^mr\cos(m\theta + \phi) + O(\epsilon ^{m+1})

    Set \ \theta = \frac{1}{m}(\phi + \pi) , so that \ \phi =( \pi - m \theta) . Then \ \cos(m\theta + \phi) = \cos(\pi) =-1 , and

    \ |p(w + \epsilon e^{i\theta})|^2 = |p(w)|^2 -2\epsilon ^mr + O(\epsilon ^{m+1})

    But \ \epsilon , r >0 , so for suffciently small \ \epsilon we have

    \ |p(w + \epsilon e^{i\theta})|^2 < |p(w)|^2

    contradicting the definition of \ w . Therefore \ \exists z_0 \in \mathbb{C} such that \ p(z_0) = 0.             \Box
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    \displaystyle \Gamma(t) = \int_{0}^{\infty} x^{t-1} e^{-x} dx ,we now sub \ x=su^{n} ,

    to get, \displaystyle \Gamma(t) = \int_{0}^{\infty} s^{t} n u^{nt-1} e^{-su^{n}} du

    ,removing constants we obtain,\displaystyle \frac{\Gamma(t)}{ns^{t}} = \int_{0}^{\infty} u^{nt-1} e^{-su^{n}} du

    we now consider the case that s is complex, which therefore gives us an equation for its conjugate,  \bar{s} , that being,

    \displaystyle \frac{\Gamma(t)}{ n\bar{s}^{t}} = \int_{0}^{\infty} u^{nt-1} e^{-\bar{s}u^{n}} du ,

    we then add/sub the two equations together,

    \displaystyle (\frac{\Gamma(t)}{s^{t}n} \pm \frac{\Gamma(t)}{\bar{s}^{t}n}) = \int_{0}^{\infty} u^{nt-1} (e^{-su^{n}} \pm e^{-\bar{s}u^{n}}) du ,

    then \displaystyle \frac{\Gamma(t)}{n}(\frac{1}{s^{  t}} \pm \frac{1}{\bar{s}^{t}}) = \int_{0}^{\infty} u^{nt-1} (e^{-su^{n}} \pm e^{-\bar{s}u^{n}}) du

    ,we then define \ s ,and. \bar{s} in polar and Cartesian form, so, \ s=a + ib , \bar{s}=a -ib , s=|s|e^{i \alpha} , \bar{s}=|s|e^{-i \alpha} ,applying these we get,

    \displaystyle \frac{\Gamma(t)}{n|s|^{t}}(e^{-i \alpha t} \pm e^{i \alpha t}) = \int_{0}^{\infty} u^{nt-1} e^{-au^{n}} (e^{-ibu^n} \pm e^{ibu^n}) du ,

    these reduce to, \displaystyle \frac{\Gamma(t)}{n|s|^{t}} \cos( \alpha t) = \int_{0}^{\infty} u^{nt-1} e^{-au^{n}}\cos(bu^{n}) du and \displaystyle \frac{\Gamma(t)}{n|s|^{t}} \sin( \alpha t) = \int_{0}^{\infty} u^{nt-1} e^{-au^{n}}\sin(bu^{n}) du ,

    only the second of these we shall use.

    That being the second one. \displaystyle \frac{\Gamma(t)}{n|s|^{t}} \sin( \alpha t) = \int_{0}^{\infty} u^{nt-1} e^{-au^{n}}\sin(bu^{n}) du we now set \ a=0, b=1 ,and, n=1 thus meaning \ |s|=1 ,and, \alpha = \frac{\pi}{2} ,

    we must also use the Euler reflection formula,  \Gamma(t) \Gamma(1-t) = \frac{\pi}{\sin(\pi t)} ,using what we know, now we get a much more simple formula,

    \displaystyle \sin\left(\frac{\pi t}{2}\right) \frac{\pi t}{\sin(\pi t) \Gamma(1-t)} = \int_{0}^{\infty} u^{t-1}\sin(u) du ,

    now we take the limit as \ t \rightarrow 0 ,so we have \displaystyle \int_{0}^{\infty} u^{-1}\sin(u) du = \lim_{t\to 0} \frac{\pi}{2} \frac{\sin(\frac{\pi}{2})}{\frac  {\pi t}{2}} \frac{\pi t}{\sin(\pi t) \Gamma(1-t)} ,

    the right hand side having had some \ t's , and, \pi 's added, but all mathematically consistent.

    Evaluating the RHS we get \displaystlye \frac{\pi}{2} \frac{1}{\Gamma(1)}

    as we know \displaystlye \lim_{t\to 0} \frac{\sin(t)}{t} =1,from this we have,

    \displaystyle \int_{0}^{\infty} \frac{\sin(u)}{u} du = \frac{\pi}{2} \Box
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    Claim: \ e is transcendental.

    Proof: Assume \ e is not transcendental.

    Then\ a_me^m +...+a_1e + a_0 =0

    Without the loss of generality we can assume \ a_j \in \mathbb{Z}, \forall j and \ a_0 \neq 0

    Define \ f(x)=\frac{x^{p-1}(x-1)^p(x-2)^p...(x-m)^p}{(p-1)!}

    Where p is an arbitrary prime.

    Then \ f is a polynomial in \ x, \partial =mp + (p-1) .

    Put\ F(x)=f(x) + f^{\prime}(x) + ... + f^{(mp + p -1)}(x) and note that \ f^{(mp + p -1)}(x) =0

    \ \frac{d}{dx}(e^{-x}F(x)) = e^{-x}(F^{\prime}(x) - F(x)) = -e^{-x}F(x)

    Hence for any \ j \displaystyle a_j \int_{0}^{j} e^{-x}f(x)dx = a_j [\left -e^{-x}F(x)\right]_{0}^{j}

    \ = a_j F(0) - a_j e^{-j}F(j)

    Multiply by \ e^j and sum over \ j to get

    \displaystyle \sum_{j=0}^{m} \left( a_je^j \int_{0}^{j} e^{-x}f(x)dx \right) = F(0)\sum_{j=0}^{m} a_je^j - \sum_{j=0}^{m} a_jF(j)

    \displaystyle = -\sum_{j=0}^{m} \sum_{i=0}^{mp+p-1} a_j f^{(i)}(j) \ (\Phi)

    from the equation supposedly satisfied by by \ e

    We claim that for each \ f^{(j)}(j) is an integer, and that this integer is divisible by \ p unless \ j=0 and \ i=p-1 . To establish the claim we just apply Liebniz's rule; the only non-zero term arising when \ j \neq 0 come from the factor \ (x-j)^p being differentiated exactly p times. Since \ \frac{p!}{(p-1)}=p , all such terms are integers divisible by \ p .

    In the exceptional case \ j=0 , the frist non-zero term occurs when \ i=p-1 , and then\ f^(p-1)(0)=(-1)^p...(-m)^p The next non-zero terms are all multiples of \ p .

    The value of equation \ (\Phi) is therefore\ K_p +a_0(-1)^p...(-m)^p for some \ K \in \mathbb{Z} . If \ p>max(m,|a_0|) , then the integer \ a_0(-1)^p...(-m)^p is not divisible by \ p . So fot sufficiently large primes \ p the value of the equation \ (\Phi) is an integer not divisible by \ p , hence not zero.

    Now estimating the integral. If \ 0 \leq x \leq m then\ |f(x)| \leq \frac{m^{mp+p-1}}{(p-1)!} 



so\displaystyle \left|\sum_{j=0}^{m} a_je^j \int_{0}^{j} e^{-x}f(x)dx \right| \leq \sum_{j=0}^{m} |a_je^j| \int_{0}^{j} \frac{m^{mp+p-1}}{(p-1)!}dx

    \displaystyle \leq \sum_{j=0}^{m}|a_je^j|j\frac{m^{  mp+p-1}}{(p-1)!} which tends to zero as \ p \rightarrow +\infty , thus we have a contradiction .

    Therefore \ e is transcendental which implies it is also irrational.
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    (Original post by zetamcfc)
    \displaystyle \Gamma(t) = \int_{0}^{\infty} x^{t-1} e^{-x} dx
    Is that the Gamma function I see :awesome: my personal favourite!
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    (Original post by zetamcfc)
    \ \Leftrightarrow y \in \bigcap_{i \in I} f[A_i] \Box
    Lovely. :ahee:

    Personally, I'd prefer \iff over \Leftrightarrow though.
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    Not done a post in a few days, here goes.

    Claim: The cardinality of \ \mathbb{C} is the cardinality of the continuum.

    Proof: We have, \ f(a,b) = a+bi defines a bijection \ \mathbb{R}^2 \rightarrow \mathbb{C}

    So, \ \mathbb{R}^2 \sim \mathbb{C}

    But \ \mathbb{R} \sim \mathbb{R}^2 and \ \mathbb{R}^2 \sim \mathbb{C}

    \ \Rightarrow \mathbb{R} \sim \mathbb{C} \, \Box

    Where \ (A \sim B) means \ A has the same cardinality as\ B

    Some thing to think of is, what is the cardinality of \ \mathbb{R}^n and how would one show that ?
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    Not done this in a while but always good to keep practicing

    Find \displaystyle \int_{-\infty}^{\infty} \frac{dx}{x^6 +1}

    First consider \displaystyle \int_{\gamma} \frac{dz}{z^6 +1} Where \frac{1}{z^6+1} is a complex valued function. And \gamma is the semicircle \ (-\infty,\infty) \cup \{z : |z| < \infty \ \ \ and \ \ \ Im(z)>0 \} traced in the anti-clockwise direction. (Probably should use R then take the limit to infinity, but hey ho)

    We can see that the integrand has 3 simple poles at \ e^{\frac{i \pi}{2}} \ \ \ e^{\frac{i \pi}{6}} \ \ \ e^{\frac{ 5i \pi}{6}}

    Then we know by the residue theorem that:

    \displaystyle \int_{\gamma} \frac{dz}{z^6 +1} = 2 \pi i (\text{Res}(f,e^{\frac{i \pi}{2}} )+ \text{Res}(f,e^{\frac{i \pi}{6}} )+\text{Res}(f, e^{\frac{ 5i \pi}{6}}) )

    To work out the Residues we will prove the following,

    Lemma: Let \ f(z)=\frac{g(z)}{h(z)} , where g and h are both holomorphic in a neighborhood of c, and where \ h(c)=0 \  , \ \ h'(c) \not = 0 . Then, \ \text{Res}(f,c) = \frac{g(c)}{h'(c)}

    Proof:

    \displaystyle \text{Res}(f,c) = \lim_{z \rightarrow c} g(z) \frac{z-c}{h(z)}



= g(c) \lim_{z \rightarrow c} \frac{z-c}{h(z) -h(c)} 



= \frac{g(c)}{h'(c)}

    So we can continue the solution, to find that,

    \displaystyle \text{Res}(f,e^{\frac{i \pi}{2}}) = \frac{1}{6e^{\frac{5i \pi}{2}}}

    \displaystyle \text{Res}(f,e^{\frac{i \pi}{6}}) = \frac{1}{6e^{\frac{5i \pi}{6}}}

    \displaystyle \text{Res}(f,e^{\frac{5i \pi}{6}}) = \frac{1}{6e^{\frac{25i \pi}{6}}}

    Thus by the Residue theorem:

    \displaystyle \int_{\gamma} \frac{dz}{z^6 +1} = 2 \pi i \frac{1}{6} (e^{\frac{-i \pi}{2}} + e^{\frac{-5i \pi}{6}} + e^{\frac{-i \pi}{6}} ) = \frac{2 \pi}{3}

    Therefore:

    \displaystyle \int_{-\infty}^{\infty} \frac{dx}{x^6 +1} = \frac{2\pi}{3}
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    Lovely! Remember that Latex has a built in imaginary part function: \Im (it uses Frankfurt characters, weird...) and you can always write \text{Res}(...) to make it not look like the variables R, e and s bunched together.

    But looks really good! (and really scary, is that complex analysis? :eek3: )
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    (Original post by Zacken)
    Lovely! Remember that Latex has a built in imaginary part function: \Im (it uses Frankfurt characters, weird...) and you can always write \text{Res}(...) to make it not look like the variables R, e and s bunched together.

    But looks really good! (and really scary, is that complex analysis? :eek3: )
    Yup, but not too challenging. Ty for the ideas.
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    Claim: \displaystyle \int_0^{\infty} \frac{1}{1 + x^n} \ dx  = \frac{\pi}{n} \csc \left(\frac{\pi}{n} \right) when n>1. (I posed the question on TPIT but I don't think anyone answered it.)

    I'm interested to see if there is a Real Analysis proof for this question if anyone could give such a proof i'd be very happy

    Sketch proof (Not that rigorous, but saves time ):

    Consider, \displaystyle \int_\gamma \frac{dz}{1+z^n} for n>1

    So we have \ f(z)= \frac{1}{1+z^n}

    \ \gamma shall be made up of 3 parts \ \gamma_{1} the line extending length \ R along the real line, \ \gamma_{2} will be the sector-arc from \ R to \ R e^{\frac{2 \pi i }{n}}}} ,\ \gamma_{3} the line from \ R e^{\frac{2 \pi i }{n}}} to the origin. From the way I have described we will obviously be going anti-clockwise around.

    For this \ \gamma there will only be one pole inside, which we will call \ c .

    Now as our situation is clearer we have an equation collecting everything together,

    \displaystyle \lim_{R \rightarrow \infty} \frac{1}{2 \pi i} \left[ \int_{\gamma_{1}} + \int_{\gamma_{2}} + \int_{\gamma_{3}}  \frac{dz}{1+z^n} \right] = res(f,c)

    First we see what \ res(f,c) is,

    \ res(f,c) = \frac{1}{nc^{(n-1)}}

    \ = \frac{1}{ne^{\frac{i \pi (n-1)  }{n}} }

    \ = \frac{-e^{\pi i}}{n}

    Now we can try to see what the integrals are.

    First we must note that \ \int on \ \gamma_{2} \rightarrow 0 as \ R \rightarrow \infty .

    We know \ \int on \ \gamma_{1} is the expression we want, so we will work on, \ \int on \ \gamma_{3}

    \ \int_{\gamma_{3}} \frac{dz}{1+z^n}

    \ z=r e^{\frac{2 \pi i}{n}} where \ r \in [R,0] we write \ [R,0] because of the direction of integration.

    \ dz= e^{\frac{2 \pi i}{n}} dr

    So the integral now becomes,

    \displaystyle \int_{r=R}^{0} \frac{e^{\frac{2 \pi i}{n}} dr}{(1 + (r e^{\frac{2 \pi i}{n}})^n}

    \displaystyle = -e^{\frac{2 \pi i}{n}}  \int_{0}^{R} \frac{dr}{1+r^n}

    Combining our results to the limit we have earlier, we get,

    \displaystyle \lim_{R \rightarrow \infty } \frac{1}{2 \pi i} \left[ -e^{\frac{2 \pi i}{n}} \int_{0}^{R} \frac{dr}{1+r^n}  + \int_{0}^{R} \frac{dx}{1+x^n}  \right] = \frac{-e^{\frac{i \pi}{n}}}{n}

    As r is a dummy variable we get the equation,

    \displaystyle \int_{0}^{\infty} \frac{dx}{1+x^n} = \frac{-2 \pi i e^{\frac{i \pi}{n}}}{n(1-e^{\frac{2 \pi i}{n}})}

    \ = \frac{-2 \pi i}{n(e^{\frac{-i \pi}{n}} - e^{\frac{i \pi}{n}} )}

    \ = \frac{\pi}{n} \frac{1}{\sin(\frac{\pi}{n})} = \frac{\pi}{n} \csc(\frac{\pi}{n})

    And we are done .

    Zacken I'm sure you can find a Real Analysis proof for this.
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    What the actual f*** I don't understand any of this...
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    (Original post by zetamcfc)
    Zacken I'm sure you can find a Real Analysis proof for this.
    Dunno how real-analysis-y this is, but the first time you posted this, I did something akin to:

    Use the substitution t = \frac{1}{1 +x^n} then

    \displaystyle 

\begin{align*}\int_0^{\infty} \frac{\mathrm{d}x}{1+x^n} & = \frac{1}{n} \int_0^{\infty} t^{-1/n} (1-t)^{1/n - 1} \, \mathrm{d}t \\ & = \frac{1}{n}\Gamma\left(\frac{1}{  n}\right) \Gamma \left(1 - \frac{1}{n}\right) \end{align*}

    which I think you can then mangle into required form with some suitable Gama tricks; Euler's reflection formula should come in handy.
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    (Original post by Zacken)
    Dunno how real-analysis-y this is, but the first time you posted this, I did something akin to:

    Use the substitution t = \frac{1}{1 +x^n} then

    \displaystyle 

\begin{align*}\int_0^{\infty} \frac{\mathrm{d}x}{1+x^n} & = \frac{1}{n} \int_0^{\infty} t^{-1/n} (1-t)^{1/n - 1} \, \mathrm{d}t \\ & = \frac{1}{n}\Gamma\left(\frac{1}{  n}\right) \Gamma \left(1 - \frac{1}{n}\right) \end{align*}

    which I think you can then mangle into required form with some suitable Gama tricks; Euler's reflection formula should come in handy.
    nice substitution. (Remember to change limits for your 2nd integral )
 
 
 
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