# Thread: question on confusing maclaurin series problem

1. ## question on confusing maclaurin series problem

the problem is the following:
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 + ...

i understood how they got the maclaurin series for sinh(x), but how did they get the coefficients for the maclaurin series for arcsinh(x)? i have no idea how my book got them.

2. Given an analytic complex function $f(z)$ that has a zero of order 1 in z=0...

$\displaystyle w= f(z)= a_{1}\ z + a_{2}\ z^{2} + a_{3}\ z^{3} + ...$ , $a_{1} \ne 0$ (1)

... it exists the inverse function that is also analytic with a zero of order 1 in w=0...

$\displaystyle z= f^{-1} (w) = b_{1}\ w + b_{2}\ w^{2} + b_{3}\ w^{3} + ...$ (2)

... and the $b_{n}$ are...

$\displaystyle b_{n} = \frac{1}{n!}\ \lim_{z \rightarrow 0} \frac{d^{n-1}}{d z^{n-1}} (\frac{z}{f(z)})^{n}$ (3)

In Your case is $w= \sinh z$ , so that the coefficients of the inverse function $z= \sinh^{-1} w$ are...

$\displaystyle b_{1}= \lim_{z \rightarrow 0} \frac{z}{\sinh z} = 1$

$\displaystyle b_{2}= \frac{1}{2}\ \lim_{z \rightarrow 0} \frac{d}{dz} \frac{z^{2}}{\sinh^{2} z} = 0$

$\displaystyle b_{3}= \frac{1}{6}\ \lim_{z \rightarrow 0} \frac{d^{2}}{dz^{2}} \frac{z^{3}}{\sinh^{3} z} = - \frac{1}{6}$

...

Further computation is a little time expensive of course... fortunately somebody has done the job in the past centuries and the results are in Your book! ...

Kind regards

$\chi$ $\sigma$

3. Originally Posted by oblixps
the problem is the following:
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 + ...

i understood how they got the maclaurin series for sinh(x), but how did they get the coefficients for the maclaurin series for arcsinh(x)? i have no idea how my book got them.
This is an example of reversion of series

CB

4. Originally Posted by CaptainBlack
This is an example of reversion of series

CB
The quoted Mathworld-Wolfram's page merits some comments...

a) the first 'series reversion algotrithm', dated 1961 and reported in the Abramowitz-Stegun's edition of 1972, is based on an exclusive algebraic procedure and permits to find the coefficients of the inverse funcion's expansion of degree not more than 7... but what does it happen if a greater precision is required?...

b) the second 'series reversion algotrithm' , dated 1953 and due to Morse and Feshbach, is very attractive but... simply unfeasible...

May be that is appropriate to inform Mathworld-Wolfram that one century ago efficient and relatively feasible ways to solve this problem were well known? ...

Kind regards

$\chi$ $\sigma$

5. I would write $\textrm{arcsinh}\,{x}$ as its logarithmic equivalent...

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

You should know that

$\ln{(1 + X)} = \sum_{n = 1}^{\infty}(-1)^{n + 1}\frac{X^n}{n}$ for $-1 < X \leq 1$.

Now let $X = x + \sqrt{x^2 + 1} - 1$ to get a series expression for $\ln{(x + \sqrt{x^2 + 1})}$. Then simplify.

6. Originally Posted by Prove It
I would write $\textrm{arcsinh}\,{x}$ as its logarithmic equivalent...

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

You should know that

$\ln{(1 + X)} = \sum_{n = 1}^{\infty}(-1)^{n + 1}\frac{X^n}{n}$ for $-1 < X \leq 1$.

Now let $X = x + \sqrt{x^2 + 1} - 1$ to get a series expression for $\ln{(x + \sqrt{x^2 + 1})}$. Then simplify.
Very nice and 'open-minded' idea!... but what to do with the terms $(\sqrt{1+x^{2}})^{k}$ that are in each term $(x + \sqrt{1+x^{2}} -1)^{n}$?...

Kind regards

$\chi$ $\sigma$