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# Hyperbolic trig...? watch

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1. 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
2. When x is so small that powers of x above the fifth can be neglected,

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

ln(x + sqrt(1 + x^2))
ln(1 + x + (1/2)x^2 - (1/8)x^4)
[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
[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
3. If we let then you're simply deriving the Maclaurin series for f(x).

That is,

If x is small then the high powers of x are very small and so we can approximate

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:

and so
4. Ta muchly
5. 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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Updated: September 11, 2005
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