What is the Maclaurin Series of Tanh(x)?

[thomas49th]

Let's find the Maclaurin Series of tanhx up to powers of x^5

Yeah! Good idea!

I know

Right, f(x) = tanh
$$f'(x) = sech^{2}(x)$$
$$f''(x) = -2sech^{2}(x)tanh(x)$$
$$f''(x) = 4sech^{2}(x)tanh^{2}(x) - 2sech^{4}(x)$$

giving f(0) = 0, f'(0) = 1, f''(0) = 0 f'''(-2)

but according to my textbook apparently f'(0) = -1, how can this be. Especially as f'''(0) = -2, not just 2.

Thanks
Thomas

 

[tiny-tim]

Hi Thomas!

Looks like they're doing cothx instead of tanhx

 

[Count Iblis]

Yes, something is wrong here...

Another thing is that the Taylor formula is not the most efficient way to derive such series expansions. You can speed things up by dividing the series expansions of sinh(x) and cosh(x). To do that, you can use long division, or similarly, you can equate the fraction to an undetermined series a1 x + a3 x^3 + a5 x^5 + ... (as we know that tanh(x) is an aodd function) and then multiply both sides by the series for cosh(x) and then solve for a1, a3, etc by equating the coefficients of equal powers of x on both sides.

A third way is to use a more advanced algorithm for division, like e.g. Newton-Raphson division. Such algorithms will double the number of correct coefficients at each step, so you can obtain the first million terms in just 20 steps.

 

[Mute]

Hi Thomas!

Looks like they're doing cothx instead of tanhx

Not if the series is about x = 0!

OP: Are you sure the book claims f'(0) = -1? The first term in the series for tanh(x) is indeed x.

 

[tiny-tim]

Not if the series is about x = 0!

oops!^∞