Multiple Derivatives

[ƒ(x)]

1. Find the first ten values of f (k)(0) and determine whether or not there is an equation for the fkth(0) term.



f(x) = 1/(x2+1)
x = 1, 2, 3,....

Find f(k)(0) when k = 1, 2, 3,... for the first 10 values of k.


I got that
f'(x) = 0
f''(x) = -2
f'''(x) = 0
f4(x) = 24
f5(0) = 0

And...that is as far as I got.

[Wretchosoft]

Are you allowed to dip into complex numbers?

[Dick]

Are you allowed to dip into complex numbers?

Why? Write out the geometric series expansion of 1/(1-(-x^2)). Match that up with the taylor series expansion of 1/(1+x^2) with the x^k*f^(k)(x)/k! things in it. That will give you a formula for f^(k).

[ƒ(x)]

Why? Write out the geometric series expansion of 1/(1-(-x^2)). Match that up with the taylor series expansion of 1/(1+x^2) with the x^k*f^(k)(x)/k! things in it. That will give you a formula for f^(k).

Could you please elaborate on this a little bit more? I'm not familiar with the Taylor series.

[HallsofIvy]

Then look at what you have so far:
f'(0)= 0
f"(0)= -2= -2!
f"'(0)= 0
f""(0)= 24= 4!
f""'(0)= 0
It should be easy to see what f(n)(0) is for n odd!

I might guess that f"""(0)= -720= (-1)37!. Can you check that?

[Dick]

The geometric series expansion is 1-x^2+x^4-x^6+... The Taylor series expansion is f(0)+f'(0)x+f''(0)*x^2/2!+f'''(0)x^3/3!+f''''(0)x^4/4!+... Match up equal powers of x and read off the derivatives of f.