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    I am looking at Stroud Engineering Mathematics Programme 14 Further Question 2, which says

    "Apply Maclaurin's series to establish a series for ln(1+x). If 1+x=b/a, show that (b^2-a^2)/2ab=x-x^/2+x^3/2-... [It does say x^3/2 and I wonder if that's a typo].
    Hence show that if b is nearly equal to a, then (b^2-a^2)/2ab exceeds ln(b/a) by approximately (b-a)^3/6a^3"

    I can get ln(1+x) = x-x^2/2+x^3/3-... but I am stumped as where to go next.

    Can anyone point me in the right direction?

    Thanks
    Ant
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    (Original post by antlee)
    I am looking at Stroud Engineering Mathematics Programme 14 Further Question 2, which says

    "Apply Maclaurin's series to establish a series for ln(1+x). If 1+x=b/a, show that (b^2-a^2)/2ab=x-x^/2+x^3/2-... [It does say x^3/2 and I wonder if that's a typo].

    As long as you realise the "/2" is not part of the exponent, then no, it's not a typo.

    Divide out the left hand side of your equality, and you'll see how it's constructed from b/a.

    Note: This bit has nothing to do with the log part, yet.
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    Yes, yes, yes...

    I get that bit now

    (b^2-a^2)/2ab

    = 1/2(b^2/ab-a^2/ab)

    = 1/2(b/a - a/b)

    = 1/2((1+x)-(1+x)^-1)

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

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

    = x - x^2/2 + x^3/2 - ....

    So not a typo...

    Thanks very much.

    I may come back for further help on the last bit but very happy at the moment.

    Ant
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    Ghostwalker

    So now I get as b is close to a that

    (b^2-a^2)/2ab-ln(b/a)

    = (x-x/2+x^3/2-...)-(x-x/2+x/3-...)

    = x^3(1/2-1/3) = x^3/6

    = (b/a-1)^3/6

    =((b-a)/a)^3/6

    = (b-a)^3/6a^3

    Ta dah!

    Thanks for the hint I'm very pleased as I was then able to finish it off myself.

    Cheers
    Ant
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    (Original post by antlee)
    Ghostwalker

    So now I get as b is close to a that

    (b^2-a^2)/2ab-ln(b/a)

    = (x-x/2+x^3/2-...)-(x-x/2+x/3-...)

    = x^3(1/2-1/3) = x^3/6
    You want to be careful here, as it doesn't equal x^3(1/2-1/3).

    We can say ignoring higher order terms \approx  x^3(1/2-1/3)

    etc.

    Thanks for the hint I'm very pleased as I was then able to finish it off myself.

    Cheers
    Ant
    Always a good sign, when you only need the smallest hint, and can otherwise do it all yourself.
 
 
 
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