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Hyperbolic trig...? watch

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    Can anyone help with this question? It's not for me and I understand nothing of this but I'm sure I've copied it correctly

    (sinh^-1)x = ln[x+√(x²+1)]
    Show for small values of x, (sinh^-1)xx - (x³/6) + (3x^5/40)

    Much appreciated
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    When x is so small that powers of x above the fifth can be neglected,

    sqrt(1 + x^2) \approx 1 + (1/2)x^2 - (1/8)x^4

    ln(x + sqrt(1 + x^2))
    \approx ln(1 + x + (1/2)x^2 - (1/8)x^4)
    \approx [x + (1/2)x^2 - (1/8)x^4]
    . . . - (1/2)[x + (1/2)x^2 - (1/8)x^4]^2
    . . . + (1/3)[x + (1/2)x^2 - (1/8)x^4]^3
    . . . - (1/4)[x + (1/2)x^2 - (1/8)x^4]^4
    . . . + (1/5)[x + (1/2)x^2 - (1/8)x^4]^5
    \approx [x + (1/2)x^2 - (1/8)x^4]
    . . . - (1/2)[x^2 + x^3 + (1/4)x^4 - (1/4)x^5]
    . . . + (1/3)[x^3 + (3/2)x^4 + (3/4)x^5]
    . . . - (1/4)[x^4 + 2x^5]
    . . . + (1/5)x^5
    = x - (1/6)x^3 + (3/40)x^5
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    If we let f(x)=\sinh^{-1} x=\ln [x+\sqrt{x^2+1}] then you're simply deriving the Maclaurin series for f(x).

    That is,  f(x)=\sinh^{-1} x=f(0)+\frac{f'(0)}{1!}x+\frac{f  ''(0)}{2!}x^2+ \frac{f'''(0)}{3!} x^3 +\frac{f''''(0)}{4!}x^4+\frac{f'  ''''(0)}{5!}x^5+...

    If x is small then the high powers of x are very small and so we can approximate  f(x)=\sinh^{-1} x\approx f(0)+\frac{f'(0)}{1!}x+\frac{f''  (0)}{2!}x^2+\frac{f'''(0)}{3!}x^  3+\frac{f''''(0)}{4!}x^4+\frac{f  '''''(0)}{5!}x^5

    You need to proceed by finding the value of the derivatives (up to the fifth) of f(x) at x=0.

    For example, to get the first two terms:
    f(0)=ln(1)=0
    f'(x)=\frac{1}{x+\sqrt{x^2+1}}(1  +2x(x^2+1)^{-\frac{1}{2}}) and so f'(0)=\frac{1}{1}(1)=1
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    Ta muchly
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    You can do the first part of the question like so:

    x = sinh y (so y = (sinh^-1)x)

    x = ((e^y) - (e^-y))/2

    x = ((e^y)^2 - 1)/2e^y (dividing through the top by e^-y)

    2xe^y = (e^y)^2 - 1

    (e^y)^2 - 2xe^y - 1

    Say a = (e^y)^2 and use the quadratic formula.

    a = (2x ± √(4x^2 + 4))/2

    a = (2x ± √(4)√(x^2 + 1))/2

    a = x ± √(x^2 + 1)

    e^y = x ± √(x^2 + 1)

    y = ln (x ± √(x^2 + 1))

    (sinh^-1) x = ln (x ± √(x^2 + 1))

    Not sure about the ± sign but you could probably show that it had to be + in order for the expression (x ± √(x^2 + 1)) to be positive.
 
 
 
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