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Math Help - Maclaurin expansion of e^x

  1. #1
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    Maclaurin expansion of e^x

    Here is the question Use the Maclaurin expansion of e^x to find the value e of correct to four decimal places
    here is the series
    e^x=1+x+(x^2)/2!+...
    here is the answer

    Answer is 2.7183
    Problem
    Whilst I obviously have the answer, and I have the sequence to follow, I cannot see how the answer is derived from the series, nor can I reproduce the answer on my calculator (ti89 using solve)
    Can someone please point out thru the progression how the answer is achieved. Thanks
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  2. #2
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    Hi!

    To find 'e' to four decimal places you have to take the first terms of the series for x=1 , i.e.:

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

    for x=1 we have

    e^1= \displaystyle{\sum_{n=0}^{\infty}}\frac{1^n}{n!}

    \displaystyle{\sum_{n=0}^{8}}\frac{1}{n!}=1+1+\fra  c{1}{2}+\frac{1}{6}+\frac{1}{24}+\frac{1}{120}+\fr  ac{1}{720}+\frac{1}{5040}=2,718253968\approx2,7183
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  3. #3
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by dazza67 View Post
    Here is the question Use the Maclaurin expansion of e^x to find the value e of correct to four decimal places
    here is the series
    e^x=1+x+(x^2)/2!+...
    here is the answer

    Answer is 2.7183
    Problem
    Whilst I obviously have the answer, and I have the sequence to follow, I cannot see how the answer is derived from the series, nor can I reproduce the answer on my calculator (ti89 using solve)
    Can someone please point out thru the progression how the answer is achieved. Thanks
    you can use the Remainder Estimation Theorem:

    \left|R_n(x)\right| \leq \frac{M}{(n+1)!}\left|x-x_0\right|^{n+1}
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  4. #4
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by Marine View Post
    Hi!

    To find 'e' to four decimal places you have to take the first terms of the series for x=1 , i.e.:

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

    for x=1 we have

    e^1= \displaystyle{\sum_{n=0}^{\infty}}\frac{1^n}{n!}

    \displaystyle{\sum_{n=0}^{8}}\frac{1}{n!}=1+1+\fra  c{1}{2}+\frac{1}{6}+\frac{1}{24}+\frac{1}{120}+\fr  ac{1}{720}+\frac{1}{5040}=2,718253968\approx2,7183
    true! but that is if you already know that the value of n should be 8..
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  5. #5
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    Quote Originally Posted by Marine View Post
    Hi!

    To find 'e' to four decimal places you have to take the first terms of the series for x=1 , i.e.:

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

    for x=1 we have

    e^1= \displaystyle{\sum_{n=0}^{\infty}}\frac{1^n}{n!}

    \displaystyle{\sum_{n=0}^{8}}\frac{1}{n!}=1+1+\fra  c{1}{2}+\frac{1}{6}+\frac{1}{24}+\frac{1}{120}+\fr  ac{1}{720}+\frac{1}{5040}=2,718253968\approx2,7183
    I see that now, it really is frusrating being so close, but yet so far away. Thanks
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  6. #6
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    Quote Originally Posted by kalagota View Post
    you can use the Remainder Estimation Theorem:

    \left|R_n(x)\right| \leq \frac{M}{(n+1)!}\left|x-x_0\right|^{n+1}
    Can you explain this to me please, I haven't come up against this yet. Thanks
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  7. #7
    Grand Panjandrum
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    Quote Originally Posted by dazza67 View Post
    Here is the question Use the Maclaurin expansion of e^x to find the value e of correct to four decimal places
    here is the series
    e^x=1+x+(x^2)/2!+...
    here is the answer

    Answer is 2.7183
    Problem
    Whilst I obviously have the answer, and I have the sequence to follow, I cannot see how the answer is derived from the series, nor can I reproduce the answer on my calculator (ti89 using solve)
    Can someone please point out thru the progression how the answer is achieved. Thanks
    You are told that:

     <br />
e^x=1+x+(x^2)/2!+...<br />

    and you want e, so we are interested in:

     <br />
e=e^1=1+1+1/2!+...<br />

    Put:

    S_n=1+1+1/2!+ .. + 1/(n-1)!

    S_1=1
    S_2=2
    S_3=S_2+1/2!=2.5
    S_4=S_3+1/3!=2.666667
    S_5=S_4+1/4!=2.708333
    S_6=S_5+1/5!=2.716667
    S_7=S_6+1/6!=2.718056

    and so on.

    RonL
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  8. #8
    MHF Contributor kalagota's Avatar
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    ok.
    Remainder Estimation Theorem: If f can be differentiated n+1 times on an interval I containing x_0 and if M is an upper bound for \left|f^{(n+1)}(x)\right| on I, that is \left|f^{(n+1)}(x)\right| \leq M \, \forall \, x \in I, then
    \left|R_n(x)\right| \leq \frac{M}{(n+1)!}\left|x-x_0\right|^{n+1}

    as an example, you can read this..
    Error Estimation
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  9. #9
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    Brilliant thanks Guys, I am going over the tread from kalagota hoping to find and edge in the next lot of exams, Thanks again for you time..
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