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Show arsinh(x) is approx ln2 + lnx + 1/(4x2)
By Anonymous on Wednesday, February 24, 1999 - 01:25 pm:

I can't find a way of showing

arsinh(x) is approx= ln2 + lnx + 1/(4x2)

given that x is large and positive.
None of my class or even my teacher were able to do it
We'd be grateful if anyone could help.

By Alex Barnard (Agb21) on Wednesday, February 24, 1999 - 01:51 pm:

Okay... The definition of sinh(y) is as follows:

sinh(y) = (ey - e-y) / 2.

So to work out arsinh(x) we need to be able to reverse the above equation to get y back out. So, let me call sinh(y) = x. Now we have to solve the following for y:

x = 1/2 (ey - e-y).

This might look horrible, but actually it is just a quadratic equation. Multiply up by 2ey and we have:

2xey = e2y - 1 which is a quadratic in ey.

Solving:

ey = (1/2)(2x +/- sqrt(4x2 + 4)) = x +/- sqrt(x2+1)

Obviously we want ey to be positive so we take the + sign rather than the - sign. Hence we have the following:

arsinh(x) = ln[x + sqrt(x2 + 1)]

Now to get the approximation you want we just have to approximate the exact formula above.

sqrt(x2+1) = x.sqrt(1+1/x2) = x(1+1/(2x2)+...)
= x + 1/(2x) + ...

so arsinh(x) ~ ln[2x + 1/(2x)]

Now ln[2x + 1/(2x)] = ln(2x) + ln[1 + 1/(4x2)]

And ln[1 + 1/(4x2)] ~ 1/(4x2).

So combining all of this we get:

arsinh(x) ~ ln(x) + ln(2) + 1/(4x2).

Please write back if you don't understand anything I've done here.

AlexB.