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Math Help - Power series (being revised)

  1. #1
    MHF Contributor Mathstud28's Avatar
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    Power series (being revised)

    This is a thread meant to help those who are having trouble with power series. Also I will cover some useful uses for these power series.

    Part 1. What is a power series?

    A power series bascially works of this observation

    "Hey at x=0 e^{x} and x+1 are equal, and also for numbers close to x=0 x+1 is a good approximation for e^{x}. I wonder if I can find a way to find a good approximation for numbers like x=7?"

    Well you do this by first taking one of these cases and saying since e^{x}=x+1,x=0. Next you can say "well if they increase at a similar rate( f'(x) the approximation will be better...and hey if the rate of the rate( f''(x)) increases similarly then the approximation will be even better..and if the rate of the rate of the rate....ad infinitum"

    This is how The classic f(c)=\sum_{x=a}^{\infty}\frac{f^{(n)}(c)(x-c)^n}{n!} arises....

    This series will ensure that for all the values at which the series converges the sereis will follow the pattern of the series's rate being comparable to the rate of the function being approximated.

    Now I am writing this assuming you knew this. So That is all I am going to go into regarding that


    Part 2. Why power series converge only for a certain values?

    To show this make a show of good faith and assume that the following is correct

    \ln(1-x)=-\sum_{n=0}^{\infty}\frac{x^{n+1}}{n+1},\forall{x}\  in[-1,1)

    I will show later how this was gotten...You may ask why there has to be a limiting interval?

    This is called the interval of convergence. This is exactly what it sounds like, the series only converges for x values that are an element of the interval. To show this I will take one of the values contained x=-1.

    If we imput that into our series we have

    -\sum_{n=0}^{\infty}\frac{(-1)^{n+1}}{n+1}=\sum_{n=0}^{\infty}\frac{(-1)^{n}}{n+1}

    Now since a_{n+1}<a_n and \lim_{n\to\infty}a_n=0 we see this converges by the alternating series test

    So we see that this converges for x=-1

    Next let's show why x=1 is excluded. When x=1 the series becomes

    -\sum_{n=0}^{\infty}\frac{1}{n+1}

    which diverges by the integral test

    \int_0^{\infty}\frac{dx}{x+1}=\ln(x+1)\bigg|_0^{\i  nfty}=\infty

    Now lets test another random value...say x=e^{\pi} Gelfond's constant

    Imputting that we get -\sum_{n=0}^{\infty}\frac{e^{n\pi}}{n+1}

    Which obviously diverges due to the n-th term test

    \lim_{n\to\infty}\frac{e^{n\pi}}{n+1}\ne{0}

    This is also true \forall{x}\notin[-1,1)

    Part 3. How do I find power series for unfamilar functions?(I will do this for Maclaurin only)

    First off let me be blunt. You are going to have much difficulty in this topic if you don't memorize the basic power series

    To learn them look here Maclaurin Series -- from Wolfram MathWorld

    The must knows are

    \frac{1}{1-x}=\sum_{n=0}^{\infty}x^{n},\forall{x}\in[-1,1]

    \frac{1}{1+x}=\sum_{n=0}^{\infty}(-1)^nx^{n},\forall{x}\in[-1,1]

    \cos(x)=\sum_{n=0}^{\infty}\frac{(-1)^{n}x^{2n}}{(2n)!},\forall{x}\in\mathbb{R}

    e^{x}=\sum_{n=0}^{\infty}\frac{x^{n}}{n!},\forall{  x}\in\mathbb{R}

    From these you can use tricks to find all the others you'll most likely encounter

    First lets look at the previously stated power series for \ln(1-x)

    Here is how this was derived since

    -\int\frac{dx}{1-x}=\ln(1-x)

    and we know that \frac{1}{1-x}=\sum_{n=0}^{\infty}x^{n}

    we can make the jump that

    \ln(1-x)=-\int\frac{dx}{1-x}=-\int\sum_{n=0}^{\infty}x^{n}dx=-\sum_{n=0}^{\infty}\int{x^{n}}dx=-\sum_{n=0}^{\infty}\frac{x^{n+1}}{n+1}

    Note that although integrating a power series does not change its radius of convergence it can change its endpoint behavior

    I will go over three more

    THe first is this: Find the power series for \sin(\sqrt{x})

    Well first we need to find the power series for \sin(x) before we can tackle this...first we make this observation

    \int\cos(x)dx=\sin(x)\therefore\sin(x)=\int\sum_{n  =0}^{\infty}\frac{(-1)^{n}x^{2n}}{(2n)!}=\sum_{n=0}^{\infty}\frac{(-1)^{n}x^{2n+1}}{(2n+1)!}

    You can verify the details missed yourself

    So if

    \sin(x)=\sum_{n=0}^{\infty}\frac{(-1)^{n}x^{2n+1}}{(2n+1)!}\Rightarrow{\sin(\sqrt{x})  =\sum_{n=0}^{\infty}\frac{(-1)^{n}(\sqrt{x})^{2n+1}}{(2n+1)!}}
    -------------------------------------------------------------------------
    The second one is \frac{6x}{5-2x}

    The key for these type is to get it into one of our normal forms \frac{1}{1-x}

    To do this we first need to have a 1-u(x) on the bottom so we rewrite it as this \frac{6x}{5}\frac{}{1-\frac{2}{5}x}

    The second part of which looks like our known formula so we apply the power series replacing

    \frac{1}{1-\frac{2}{5}x} with \sum_{n=0}^{\infty}\bigg(\frac{2}{5}x\bigg)^{n}

    To get \frac{6x}{5-2x}=\frac{6x}{5}\sum_{n=0}^{\infty}\bigg(\frac{2}{  5}\bigg)^{n}x^{n}=\frac{6}{5}\sum_{n=0}^{\infty}\b  igg(\frac{2}{5}\bigg)^{n}x^{n+1}

    Check interval of convergence by using ratio test


    I will be back later for third example and uses.
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Now for the third, fourth, and fifth example of finding a power series( I decided to expand a little =D)

    3. Find the power series for g(x)=\frac{-1}{(x+1)^3}

    This one is one that through experience with recognition becomes easy. The first thing you should see is the \frac{1}{1+x} contatined within

    Next comes the trick if you look closely you will see that \frac{D^2[\frac{1}{1+x}]}{dx^2}=\frac{2}{(x+1)^3}

    \therefore\frac{-1}{(1+x)^3}=\frac{-1}{2}\frac{D^2[\frac{1}{1+x}]}{dx^2}

    So now calling f(x)=\frac{1}{1+x}=\sum_{n=0}^{\infty}(-1)^{n}x^{n}

    we have that f'(x)=\sum_{n=1}^{\infty}n(-1)^{n}(x)^{n-1}

    and f''(x)=\sum_{n=2}^{\infty}n(n-1)(x)^{n-2}

    \therefore\frac{-1}{(1+x)^3}=\frac{-1}{2}\sum_{n=2}^{\infty}n(n-1)(x)^{n-2}

    Since I did not ask to have the interval of convergence checked we won't but to test if you are right take an element of the interval of convergence and test it in the actual function versus the power series. To pick what value to test we must realize that the power series for
    \frac{1}{1+x} is [-1,1],

    so in realizing this we can say that since differentiating only effects endpoint behavior this series we derived converges AT LEAST on (-1,1)

    So we test x=.5

    \frac{-1}{(1+(.5))^3}=-.296=\frac{-1}{2}\sum_{n=2}^{\infty}(-1)^{n}n(n-1)(.5)^{n-2}
    -------------------------------------------------------------------------------
    The fourth example will be this

    4. Find the power series for \ln(1-x^2)

    First we differentiate \frac{D[\ln(1-x^2)]}{dx}=\frac{-2x}{1-x^2}


    Seeing the \frac{1}{1-x^2} which by replacing x\rightarrow{x^2}

    we can get the power series

    \frac{1}{1-x^2}=\sum_{n=0}^{\infty}(x^2)^{n}=\sum_{n=0}^{\inf  ty}x^{2n}

    So using this we see that

    \frac{-2x}{1-x^2}=-2x\cdot\sum_{n=0}^{\infty}x^{2n}=-2\sum_{n=0}^{\infty}x^{2n+1}

    Therefore using this we make this jump

    \frac{D[\ln(1-x^2)]}{dx}=\frac{-2x}{1-x^2}=-2\sum_{n=0}^{\infty}x^{2n+1}\Rightarrow{\ln(1-x^2)=\int\frac{-2x}{1-x^2}=-2\int\sum_{n=0}^{\infty}x^{2n+1}} =-2\sum_{n=0}^{\infty}\frac{x^{2n+2}}{2n+2}

    To test this we pick thevalue x=.5

    \ln(1-(.5)^2)=-.288=-2\sum_{n=0}^{\infty}\frac{x^{2n+2}}{2n+2}

    Alternatively this could have been done by doing this \ln(1-x^2)=\ln((1-x)(1+x))=\ln(1-x)+\ln(1+x)=-\int\frac{dx}{1-x}+\int{\frac{dx}{1+x}}
    and finding the power series individually
    -------------------------------------------------------------------------------
    The fifth one is find the power series for

    f(x)=arctan(x)

    This can be done by recognizing arctan(x)=\int\frac{dx}{1+x^2}=\int\sum_{n=0}^{\in  fty}(-1)^{n}x^{2n}dx=\sum_{n=0}^{\infty}\frac{(-1)^{n}x^{2n+1}}{2n+1}

    To test let x=.5

    arctan(.5)=.464=\sum_{n=0}^{\infty}\frac{(-1)^{n}(.5)^{2n+1}}{2n+1}



    I will be back later to show some applications
    Last edited by Mathstud28; May 13th 2008 at 01:01 PM.
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  3. #3
    MHF Contributor Mathstud28's Avatar
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    Part 4. Some applications of power series.

    The first application is this. Let me preface it by talkinga bout the almighty L'hopital's rule. L'hopital's states that if a limit yields an indeterminate result most notably of the form \frac{\infty}{\infty}\text{or}\frac{0}{0} then you may take the limit from the form of

    \lim_{x\to{c}}\frac{f(x)}{g(x)}\Rightarrow\lim_{x\  to{c}}\frac{f'(x)}{g'(x)}

    You can keep taking the derivative of f(x) and g(x) provided that each time you do it it is due to an indeterminate form.

    Now although L'hopital's rule is a godsend in most cases it has one downfall

    for it to be useful there must be some point where f^{(n)}(c)=d and g^{(n)}(c)=b so the limit is \frac{d}{b}

    But what about when you have gotten to the 28th derivative and you still have not yet found a function that gives a determinate form?

    Often you can apply power series. Take for example the following limit

    \lim_{x\to{0}}\frac{(e^{2x^2}-1-2x^2)(\cos(x)-1)}{[\sin(3x)-\ln(1+3x)]x^4}

    I am sure using L'hopital's you would cry....so why not put in power series?

    using e^{2x^2}=\sum_{n=0}^{\infty}\frac{(2x^2)^{n}}{n!}

    \cos(x)=\sum_{n=0}^{\infty}\frac{(-1)^{n}x^{2n}}{(2n)!}

    \sin(3x)=\sum_{n=0}^{\infty}\frac{(-1)^{n}(3x)^{2n+1}}{(2n+1)!}

    and \ln(1+3x)=\sum_{n=0}^{\infty}\frac{(-1)^{n}3^{n+1}x^{n+1}}{n+1}

    By imputting the series in ... form we get

    \lim_{x\to{0}}\frac{\bigg[\bigg(1+2x^2+\frac{(2x^2)^2}{2!}+...\bigg)-1-2x^2\bigg)\bigg]\bigg[(1-\frac{x^2}{2!}+\frac{x^4}{4!}+...)-1\bigg]}{\bigg[\bigg(3x-\frac{(3x)^3}{3!}+\frac{(3x)^5}{5!}-...\bigg)-\bigg(3x-\frac{9x^2}{2}+\frac{27x^3}{3}\bigg)\bigg]x^4}

    Cancelling like terms we get

    \lim_{x\to{0}}\frac{\bigg[\frac{(2x^2)^2}{2}+...\bigg]\bigg[\frac{x^2}{2!}+\frac{x^4}{4!}+...\bigg]}{\bigg[\frac{9x^2}{2}-\frac{27x^3}{2}+...\bigg]x^4}=\lim_{x\to{0}}\frac{\bigg[\frac{2x^4}{2}+...\bigg]\bigg[\frac{-x^2}{2}+\frac{x^4}{24}-...\bigg]}{\frac{9x^6}{2}-\frac{27x^7}{2}+...}

    Since as x\rightarrow{0} the only terms that matter are the lowest exponents we rewrite this as

    \lim_{x\to{0}}\frac{\bigg(2x^4\bigg)\bigg(\frac{x^  2}{2}\bigg)}{\frac{-9x^6}{2}}=\lim_{x\to{0}}\frac{-2}{9}=\frac{-2}{9}

    I will be back later for some more examples and other uses
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    Another limit would be this one

    \lim_{x\to{0}}\frac{e^{x}-e^{-x}-2x}{x^2-x\ln(1+x)}

    First you establish your power series(all Maclaurin of course =D)

    e^{x}=\sum_{n=0}^{\infty}\frac{x^n}{n!}

    e^{-x}=\sum_{n=0}^{\infty}\frac{(-x)^{n}}{n!}=\sum_{n=0}^{\infty}\frac{(-1)^{n}x^{n}}{n!}

    x\ln(1+x)=x\cdot\int\frac{dx}{1+x}=x\cdot\int\sum_  {n=0}^{\infty}(-1)^{n}x^{n}dx=x\cdot\sum_{n=0}^{\infty}\frac{(-1)^{n}x^{n+1}}{n+1}=\sum_{n=0}^{\infty}\frac{(-1)^{n}x^{n+2}}{n+1}

    So now writing the limit in expanded ... form we get

    \lim_{x\to{0}}\frac{\bigg(1+x+\frac{x^2}{2!}+\frac  {x^3}{3!}\bigg)-\bigg(1-x+\frac{x^2}{2!}-\frac{x^3}{3!}\bigg)-2x}{x^2-\bigg(x^2-\frac{x^3}{2}+\frac{x^4}{3}\bigg)}

    Cancelling terms we get

    \lim_{x\to{0}}\frac{2x+\frac{x^3}{3}+...-2x}{\frac{x^3}{2}-\frac{x^4}{3}+...}\sim\lim_{x\to{0}}\frac{\frac{x^  3}{3}}{\frac{x^3}{2}}=\lim_{x\to{0}}\frac{2}{3}=\f  rac{2}{3}

    Where \sim denotes the relevance of only the lowest exponents of x as x\rightarrow{x}...alternatively you could just divide by x^3....
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  5. #5
    Rhymes with Orange Chris L T521's Avatar
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    [Power] Series is one of my weak points in math. However, it's all starting to make sense. Thanks man!!!
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  6. #6
    MHF Contributor Mathstud28's Avatar
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    Part 5. Using power series for integration

    This part of the thread will discuss the use of power series in integration and how you can obviate much confusion when integrating a function not integrable by normal means

    Example one: The oh so classic

    \int{e^{-x^2}dx}

    How would one go about doing this? There is no derivative of the quantity? There is no useful substitution? Short of some really advanced inegration techniques this is hard to do...or is it. Using power series this becomes simple

    e^{x}=\sum_{n=0}^{\infty}\frac{x^{n}}{n!}\Rightarr  ow{e^{-x^2}=\sum_{n=0}^{\infty}\frac{(-x^2)^{n}}{n!}=\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n}}{n!}}

    Thus \int{e^{-x^2}dx}=\int\sum_{n=0}^{\infty}\frac{(-1)^{n}x^{2n}}{n!}=\sum_{n=0}^{\infty}\int\frac{(-1)^{n}x^{2n}}{n!}dx=\sum_{n=0}^{\infty}\frac{(-1)^{n}x^{2n+1}}{(2n+1)n!}+C

    Try \int_0^{1}e^{-x^2}dx=F(1)-F(0)

    where as we just showed F(x)=\sum_{n=0}^{\infty}\frac{(-1)^{n}x^{2n+1}}{(2n+1)n!}

    So \int_0^{1}e^{-x^2}dx=\sum_{n=0}^{\infty}\frac{(-1)^{n}}{(2n+1)n!}-0=.747
    ---------------------------------------------------------------------------
    Before I continue let me note that you may only use this method on definite integrals whose limits of integration are contained within the interval of convergence

    for example

    \int_0^{\frac{1}{2}}\frac{dx}{1-x}=\int_0^{\frac{1}{2}}\sum_{n=0}^{\infty}x^{n}dx

    But \int_0^{2}\frac{dx}{1-x}\ne\int_0^2\sum_{n=0}^{\infty}x^{n}dx
    ---------------------------------------------------------------------------------
    Lastly I am going to consider using integration with power series to solve a general case integral that is irksome.

    \int\sin(\sqrt[y]{x})dx for n>0

    You have two options...the first is cry. The second is use power series. Now this gets a little ugly but hey..Who said math always is Euler's Reflection Formula.

    \sin(x)=\sum_{n=0}^{\infty}\frac{(-1)^{n}x^{2n+1}}{(2n+1)!}\Rightarrow\sin(\sqrt[y]{x})=\sum_{n=0}^{\infty}\frac{(-1)^{n}(\sqrt[y]{x})^{2n+1}}{(2n+1)!}=\sum_{n=0}^{\infty}\frac{(-1)^{n}x^{\frac{2n}{y}+\frac{1}{y}}}{(2n+1)!}


    \therefore\int\sin(\sqrt[y]{x})dx=\int\sum_{n=0}^{\infty}\frac{(-1)^nx^{\frac{2n}{y}+\frac{1}{y}}}{(2n+1)!}dx=\sum_  {n=0}^{\infty}\int\frac{(-1)^{n}x^{\frac{2n}{y}+\frac{1}{y}}}{(2n+1)!}dx =\sum_{n=0}^{\infty}\frac{(-1)^{n}x^{\frac{2n}{y}+\frac{1}{y}+1}}{(\frac{2n}{y  }+\frac{1}{y}+1)(2n+1)!}+C


    Ain't pretty is it? But it works where nothign else I am aware of will.

    To test it we try this

    \int_0^{28}\sin(\sqrt[28]{x})dx=F(28)-F(0)

    and as we just showed F(x)=\sum_{n=0}^{\infty}\frac{(-1)^{n}x^{\frac{2n}{y}+\frac{1}{y}+1}}{(\frac{2n}{y  }+\frac{1}{y}+1)(2n+1)!}

    \therefore\int_0^{28}\sin(\sqrt[28]{x})dx=\sum_{n=0}^{\infty}\frac{(-1)^{n}(28)^{\frac{2n}{28}+\frac{1}{28}+1}}{(\frac{  2n}{28}+\frac{1}{28}+1)(2n+1)!}

    Imputting into mathcad(maple)

    I get \int_0^{28}\sin(\sqrt[28]{x})dx=24.776

    and the moment of truth...I put in F(28)

    and I get.......

    F(28)=24.776

    Success! Thus I have once again found a way to apply power series to a completely impractical thing! Yay me!

    I might add more to this some other time

    NOTE: for the last example so it isnt left ambiguous if f(x,y)=\sin(\sqrt[y]{x})

    then \int{f(x,y)dx}=F(x,y)+C

    Where F(x,y)=\sum_{n=0}^{\infty}\frac{(-1)^{n}x^{\frac{2n}{y}+\frac{1}{y}+1}}{(\frac{2n}{y  }+\frac{1}{y}+1)(2n+1)!}

    So technically

    \int_0^{28}\sin(\sqrt[28]{x})dx=F(28,28)-F(0,28)

    Not F(28) but I assumed you knew that the y value in F(x,y) was 28
    --------------------------------------------------------------------------------------------------------------------
    Noting Chris LT's signature if you evaluate \int{e^{-x^2}dx} for \int_0^{\infty}e^{-x^2}dx=\frac{\sqrt{\pi}}{2}

    An easy way you can prove this to yourself is to do this...if you evaluate f(x)=\sum_{n=0}^{\infty}\frac{(-1)^{n}x^{2n+1}}{(2n+1)n!} for

    f(3)
    f(4)
    f(5)
    ..etc

    You notice a pattern

    f(3)=0.88620
    f(4)=0.86622
    f(5)=0.86622

    It seems that as x\to\infty f(x)\to{0.86622}

    and looking at your signature \frac{\sqrt{\pi}}{2}\approx{0.86622}
    Last edited by Mathstud28; May 13th 2008 at 01:09 PM.
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  7. #7
    Moo
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    Ok, so first of all, I'd like you not to think that I'm taking pleasure in criticising what you do. I have a strong faith in criticising, you already know it, but here, you can trust me, it's not pleasant.
    Secondly, I know how much you love power series (although I would never act the way you do by using power series as soon as you can).
    Thirdly, as a summary : I don't think your tutorial is really a tutorial. I'll try to explain (it'll take more than a post).


    Note : I'm not able to provide full corrections to the problems, I'm not able to do as much as what has been done, I'm just trying to be on a newbie's side.

    ~~~~~~~~~~~~~~~~

    Quote Originally Posted by Post #1
    Part 1. What is a power series?
    I have learnt this way, and I checked in Wikipedia :

    Quote Originally Posted by Wikipedia
    A power series is an infinite series of the form \sum_{n=0}^\infty a_n (x-c)^n
    So basically, the way you have presented this is already a problem.

    Quote Originally Posted by Post #1
    A power series bascially works of this observation

    "Hey at x=0 e^{x} and x+1 are equal, and also for numbers close to x=0 x+1 is a good approximation for e^{x}. I wonder if I can find a way to find a good approximation for numbers like x=7?"
    What ? Is it the way you observe that e^x has a power series ? This is certainly not the best introduction I've seen. You didn't even explain what a power series is.
    Oh I know, I'm in University, you're only in grade 11, so I must have learnt things you didn't already learn. Yeah, maybe... But you're making a tutorial, which is due to help people understanding it !
    By the way... When x=0, e^x=3^x.


    Well you do this by first taking one of these cases and saying since e^{x}=x+1,x=0.
    "since e^x=3^x, \ x=0."
    I think this sentence is even not mathematically correct... But that's another problem (or not - forget it).

    Next you can say "well if they increase at a similar rate( f'(x) the approximation will be better... and hey if the rate of the rate( f''(x)) increases similarly then the approximation will be even better..and if the rate of the rate of the rate....ad infinitum"
    Maybe I'm stupid (which could be a direct consequence of my post...), but I really and sincerely don't understand your sentence (if I were a newbie).


    This is how The classic f(c)=\sum_{x=a}^{\infty}\frac{f^{(n)}(c)(x-c)^n}{n!} arises....
    Does "The classic" have a name ? "Taylor series" would help people remember it, huh ?
    Was your previous sentence a demonstration of it ? o.O


    This series will ensure that for all the values at which the series converges the sereis will follow the pattern of the series's rate being comparable to the rate of the function being approximated.
    If I were looking for something helpful about power series, I'm afraid I'd skip this...
    I'm not normal, I knew it...

    Now I am writing this assuming you knew this. So That is all I am going to go into regarding that
    This is a thread meant to help those who are having trouble with power series. Also I will cover some useful uses for these power series.
    I'm not sure (hey, I'm only supposing what others would think ! I can't go into everybody's mind) that people having trouble with power series would understand better just by reading this first part


    ~~~~~~~~~~~~~~~~~~~~

    Conclusion of this part :
    - Is this a real introduction ?
    - What is the purpose of your tutorial ?
    - Is it your own interpretation of power series ? Because in case you don't know, you're not everybody.



    Others will follow but later, because my bed doesn't like me being awaken too long.
    Last edited by Moo; May 13th 2008 at 02:35 PM.
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  8. #8
    Moo
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    I'll just take one of your examples.

    Quote Originally Posted by Mathstud28 View Post
    --------------------------------------------------------------------------------------------------------------------
    Noting Chris LT's signature if you evaluate \int{e^{-x^2}dx} for \int_0^{\infty}e^{-x^2}dx=\frac{\sqrt{\pi}}{2}

    An easy way you can prove this to yourself is to do this...if you evaluate f(x)=\sum_{n=0}^{\infty}\frac{(-1)^{n}x^{2n+1}}{(2n+1)n!} for

    f(3)
    f(4)
    f(5)
    ..etc

    You notice a pattern

    f(3)=0.88620
    f(4)=0.86622
    f(5)=0.86622

    It seems that as x\to\infty f(x)\to{0.86622}

    and looking at your signature \frac{\sqrt{\pi}}{2}\approx{0.86622}
    Are you kidding ? Apart from the fact that you used a mathematical software (I'll go back to this tomorrow if I have time...), there is a problem :
    What if we don't know that it's equal to \frac{\sqrt{\pi}}{2} ?
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  9. #9
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Moo View Post
    I'll just take one of your examples.



    Are you kidding ? Apart from the fact that you used a mathematical software (I'll go back to this tomorrow if I have time...), there is a problem :
    What if we don't know that it's equal to \frac{\sqrt{\pi}}{2} ?
    Actually...I did not find your advice all to helpful...for although it is useful to point this out I do not see you making any of these...now I am sure that you could spend two days making one and it would be impeccable...well congratulations...but I do these to help those who have a base knowledge of the subject(any "newbie" who learns a thing or two is just an added benefit), I also do these to help me learn things better, since teaching is the best way of learning a subject thoroughly. And I would appreciate your imput if it wasnt so caustic( despite your disclaimer). Please for future reference personal message me with yoru imput instead of cluttering space actually devoted to helping at least one person...thanks very much Moo

    Mathstud
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  10. #10
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Moo View Post
    I'll just take one of your examples.



    Are you kidding ? Apart from the fact that you used a mathematical software (I'll go back to this tomorrow if I have time...), there is a problem :
    What if we don't know that it's equal to \frac{\sqrt{\pi}}{2} ?
    I did not in any ways say I am offering proof that this equality is true...I merely wanted to relate something, that a supportive poster finds interesting, to the matter at hand.
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  11. #11
    MHF Contributor Mathstud28's Avatar
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    You know what...I have been thinking...maybe Moo is right...if anyone out there would supply their imput I would greatly appreciate it...I will not begrudge any who side with Moo and I am sure she wouldnt with you...I would just like honest raw feeback
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