# Math Help - maclaurin series problem

1. ## maclaurin series problem

Given that sinh(x) = (e^x - e^-x) / 2, find a series representation for arcsinh(x).

So my book did a Maclaurin series expansion of sinh(x) = x + x^3 / 3! + x^5 / 5! + ...

Then it said: the inverse will have some series expansion which we will write as arcsinh(x) = b0 + b1 x + b2 x^2 + b3 x^3 + ... where the "b"s are coefficients.

we label the coeficients in the series expansion of sinh(x) as a_j. we find that,
b0 = a0, b1 = 1 / a1 = 1, b2 = -a2/(a1)^3 = 0, b3 = [1 / (a5)^6] (2(a2)^2 - a1a3) = -1/6

therefore it follows that arcsinh(x) = x - (1/6)x^3 + ...

how did they get the coeffients? i have no idea where they came from.

2. Originally Posted by oblixps
Given that sinh(x) = (e^x - e^-x) / 2, find a series representation for arcsinh(x).

So my book did a Maclaurin series expansion of sinh(x) = x + x^3 / 3! + x^5 / 5! + ...

Then it said: the inverse will have some series expansion which we will write as arcsinh(x) = b0 + b1 x + b2 x^2 + b3 x^3 + ... where the "b"s are coefficients.

we label the coeficients in the series expansion of sinh(x) as a_j. we find that,
b0 = a0, b1 = 1 / a1 = 1, b2 = -a2/(a1)^3 = 0, b3 = [1 / (a5)^6] (2(a2)^2 - a1a3) = -1/6

therefore it follows that arcsinh(x) = x - (1/6)x^3 + ...

how did they get the coeffients? i have no idea where they came from.
I would advise you to convert $\sinh^{-1}{x}$ into its logarithmic equivalent and use a Taylor series expansion of the logarithm.

If $x = \sinh{y}$

Then $y = \sinh^{-1}{x}$.

But $\sinh{y} = \frac{e^y - e^{-y}}{2}$

So $x = \frac{e^y - e^{-y}}{2}$

$2x = e^{y} - e^{-y}$

$2x = e^y - \frac{1}{e^y}$

$0 = e^y - \frac{1}{e^y} - 2x$

$0 = e^{2y} - 1 - 2x\,e^y$

$0 = e^{2y} - 2x\,e^y - 1$

$0 = e^{2y} - 2x\,e^y + (-x)^2 - (-x)^2 - 1$

$0 = \left(e^{y} - x\right)^2 - x^2 - 1$

$x^2 + 1 = \left(e^y - x\right)^2$

$\sqrt{x^2 + 1} = e^y - x$

$x + \sqrt{x^2 + 1} = e^y$

$y = \ln{\left(x + \sqrt{x^2 + 1}\right)}$.

Therefore $\sinh^{-1}{x} = \ln{\left(x + \sqrt{x^2 + 1}\right)}$.

You should now be able to use the Taylor Series expansion of a Logarithm to find an expansion for $\sinh^{-1}{x}$.

3. Originally Posted by Prove It
I would advise you to convert $\sinh^{-1}{x}$ into its logarithmic equivalent and use a Taylor series expansion of the logarithm.

If $x = \sinh{y}$

Then $y = \sinh^{-1}{x}$.

But $\sinh{y} = \frac{e^y - e^{-y}}{2}$

So $x = \frac{e^y - e^{-y}}{2}$

$2x = e^{y} - e^{-y}$

$2x = e^y - \frac{1}{e^y}$

$0 = e^y - \frac{1}{e^y} - 2x$

$0 = e^{2y} - 1 - 2x\,e^y$

$0 = e^{2y} - 2x\,e^y - 1$

$0 = e^{2y} - 2x\,e^y + (-x)^2 - (-x)^2 - 1$

$0 = \left(e^{y} - x\right)^2 - x^2 - 1$

$x^2 + 1 = \left(e^y - x\right)^2$

$\sqrt{x^2 + 1} = e^y - x$

$x + \sqrt{x^2 + 1} = e^y$

$y = \ln{\left(x + \sqrt{x^2 + 1}\right)}$.

Therefore $\sinh^{-1}{x} = \ln{\left(x + \sqrt{x^2 + 1}\right)}$.

You should now be able to use the Taylor Series expansion of a Logarithm to find an expansion for $\sinh^{-1}{x}$.
the maclaurin series for ln(1+x) is (-1)^(n+1) x^n / n so i did ln(1 + (x+ sqrt(x^2+1) -1) so i got (-1)^(n+1) (x+sqrt(x^2+1)-1)^n / n. is that correct? by the way do you know what my book is doing?

4. is this right? i looked up the series representation of arcsinh(x) and it looks nothing like what i got in the post above.