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Math Help - Finding Maclaurin Series By Division

  1. #1
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    Finding Maclaurin Series By Division

    Find the first four nonzero terms of the Maclaurin series for the function by dividing appropriate Maclaurin series

    secx

    I know secx=1/cosx and the Maclaurin series of cosx=1-x^2/2!+x^4/4!-x^6/6!............

    I do not know how one divide those terms

    Please help me

    Thank you
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by soleilion View Post
    Find the first four nonzero terms of the Maclaurin series for the function by dividing appropriate Maclaurin series

    secx

    I know secx=1/cosx and the Maclaurin series of cosx=1-x^2/2!+x^4/4!-x^6/6!............

    I do not know how one divide those terms

    Please help me

    Thank you
     <br />
\sec(x)=\frac{1}{\cos(x)}=\frac{1}{1-x^2/2!+x^4/4!-x^6/6!+...}<br />
<br />
=\frac{1}{\cos(x)}=\frac{1}{1- (x^2/2!-x^4/4!+x^6/6!-...) }<br />

    ............. =1+(x^2/2!-x^4/4!+x^6/6!-...) +(x^2/2!-x^4/4!+x^6/6!-...)^2
    ................... +(x^2/2!-x^4/4!+x^6/6!-...)^3+..

    The first four non-zero terms will be the 0,\ 2,\ 4,\ 6 th powers of x

    So:

     <br />
\sec(x) = 1 +(x^2/2!-x^4/4!+x^6/6!)+[x^4/(2!)^2-2x^6/(2!4!)]+[x^6/(2!)^3]+O(x^8)<br />

    RonL
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  3. #3
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    Quote Originally Posted by CaptainBlack View Post
     <br />
\sec(x)=\frac{1}{\cos(x)}=\frac{1}{1-x^2/2!+x^4/4!-x^6/6!+...}<br />
<br />
=\frac{1}{\cos(x)}=\frac{1}{1- (x^2/2!-x^4/4!+x^6/6!-...) }<br />

    ............. =1+(x^2/2!-x^4/4!+x^6/6!-...) +(x^2/2!-x^4/4!+x^6/6!-...)^2
    ................... +(x^2/2!-x^4/4!+x^6/6!-...)^3+..

    The first four non-zero terms will be the 0,\ 2,\ 4,\ 6 th powers of x

    So:

     <br />
\sec(x) = 1 +(x^2/2!-x^4/4!+x^6/6!)+[x^4/(2!)^2-2x^6/(2!4!)]+[x^6/(2!)^3]+O(x^8)<br />

    RonL

    I misunderstood
    How the series have 1power, 2power and 3power????
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by soleilion View Post
    I misunderstood
    How the series have 1power, 2power and 3power????

    Those terms don't appear, that is they have zero coefficients.

    RonL
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  5. #5
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    Quote Originally Posted by CaptainBlack View Post
     <br />
\sec(x)=\frac{1}{\cos(x)}=\frac{1}{1-x^2/2!+x^4/4!-x^6/6!+...}<br />
<br />
=\frac{1}{\cos(x)}=\frac{1}{1- (x^2/2!-x^4/4!+x^6/6!-...) }<br />

    ............. =1+(x^2/2!-x^4/4!+x^6/6!-...) +(x^2/2!-x^4/4!+x^6/6!-...)^2
    ................... +(x^2/2!-x^4/4!+x^6/6!-...)^3+..

    The first four non-zero terms will be the 0,\ 2,\ 4,\ 6 th powers of x

    So:

     <br />
\sec(x) = 1 +(x^2/2!-x^4/4!+x^6/6!)+[x^4/(2!)^2-2x^6/(2!4!)]+[x^6/(2!)^3]+O(x^8)<br />

    RonL
    ............. =1+(x^2/2!-x^4/4!+x^6/6!-...) +(x^2/2!-x^4/4!+x^6/6!-...)^2
    ................... +(x^2/2!-x^4/4!+x^6/6!-...)^3+..

    How does this come from 1/cosx

    what method did you use?
    Do you have any similar example?
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  6. #6
    Grand Panjandrum
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    Quote Originally Posted by soleilion View Post
    ............. =1+(x^2/2!-x^4/4!+x^6/6!-...) +(x^2/2!-x^4/4!+x^6/6!-...)^2
    ................... +(x^2/2!-x^4/4!+x^6/6!-...)^3+..

    How does this come from 1/cosx

    what method did you use?
    Do you have any similar example?
    For |u|<1:

    \frac{1}{1-u}=1+u+u^2+...

    In our case, for small x:

    u=x^2/2!-x^4/4!+x^6/6!-...

    RonL
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  7. #7
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    Quote Originally Posted by CaptainBlack View Post
    For |u|<1:

    \frac{1}{1-u}=1+u+u^2+...

    In our case, for small x:

    u=x^2/2!-x^4/4!+x^6/6!-...

    RonL
    Thank you very much
    I got it
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