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    Hi all, I've some work on Maclaurins series to do, but im not sure how exactly to proceed with this particular example question in my book.

    The example being (e^x + e^-x) / (e^x)

    Am i right in thinking this has to be simplified into one line instead of a fraction and then calculate the series from then?

    Cheers RamJam
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    certainly it would be simpler if you cancel down
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    (Original post by Ramjam)
    Hi all, I've some work on Maclaurins series to do, but im not sure how exactly to proceed with this particular example question in my book.

    The example being (e^x + e^-x) / (e^x)

    Am i right in thinking this has to be simplified into one line instead of a fraction and then calculate the series from then?

    Cheers RamJam
    write it as 1 +e-2x
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    (Original post by Ramjam)
    Hi all, I've some work on Maclaurins series to do, but im not sure how exactly to proceed with this particular example question in my book.

    The example being (e^x + e^-x) / (e^x)

    Am i right in thinking this has to be simplified into one line instead of a fraction and then calculate the series from then?

    Cheers RamJam
    This simplifies to 1 + e^{-2x}

    You know the maclaurin expansion for e^x already (I hope), simply subbing in the -2x into the series, and adding 1.

    Spoiler:
    Show


     e^x = 1 + x + \dfrac{x^2}{2} + \dfrac{x^3}{6} ....

    Replace x with -2x, add 1.

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    Ahh brilliant that's exactly what i thought i had to do, well i better crack on and write the working out!!!

    Thankyou for the extremely speedy replies!
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    Hey guys and gals, I've worked it work from scratch as i like to have my works and to help me understand it better, just wondering if you could give it a quick check to make sure ive done this right

    Using equation:

    f(x)=f(0)+f'(0)x+\frac{f''(0)}{2  !} x^2+\frac{f'''(0)}{3!}x^3+ \frac{fiv(0)}{4!}x^4+\frac{fv](0)}{5!}x^5+.........

    I then worked out each differential for f', f'', f''', f'''' ect.

    I end up with the final expansion of:

    {1+e}^{-2x}= 2-2x+2x^2-(4/3)x^3+(2/3)x^4-(4/15)x^5

    How does that look?
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    (Original post by Ramjam)
    Hey guys and gals, I've worked it work from scratch as i like to have my works and to help me understand it better, just wondering if you could give it a quick check to make sure ive done this right

    Using equation:

    f(x)=f(0)+f'(0)x+\frac{f''(0)}{2  !} x^2+\frac{f'''(0)}{3!}x^3+ \frac{fiv(0)}{4!}x^4+\frac{fv](0)}{5!}x^5+.........

    I then worked out each differential for f', f'', f''', f'''' ect.

    I end up with the final expansion of:

    {1+e}^{-2x}= 2-2x+2x^2-(4/3)x^3+(2/3)x^4-(4/15)x^5

    How does that look?
    I think you are all correct ...
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    (Original post by TeeEm)
    I think you are all correct ...
    Cheers bud, glad ive finally got this understood
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    (Original post by Ramjam)
    Cheers bud, glad ive finally got this understood
    pleased for you.
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    dude maclaurin series is so stupid just learn general taylor series and put f(x+0) as your f(a+h) terms for your approximation far more useful
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    (Original post by HLN_Radium)
    dude maclaurin series is so stupid just learn general taylor series and put f(x+0) as your f(a+h) terms for your approximation far more useful
    Im not doing this for fun bud, Ill never need it in my line of work, im only doing it as if it does come up in an exam im gonna need to know it
 
 
 
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