What is the 26th Derivative of the Given Function at 0?

Frillth
Messages
77
Reaction score
0
Homework Statement

Given that f(x) = sin(x) for x =/= 0 and f(x) = 1^x for x=0, find the 26th derivative of f at 0. Hint: can you find a power series for f(x)?

The attempt at a solution

I have no idea how to solve this problem. Since 1^x is always 1, the first derivative at 0 is 0, so ALL derivatives must be 0, right? I'm confused as to how a power series even comes into play in this problem.
 
Physics news on Phys.org
That's a very strange defintion. Note that f(0) is just a number, so all they had to say was f(x)=1 for x=0, the 1^x bit is superfluous. But moreover, the function is not continuous at x=0, so doesn't have any derivatives, let alone 26. Which leads me to ask, are you sure you copied the question correctly?
 
I just noticed that somebody erased a line in my book! It should have been sin(x)/x for x=/=0 and 1 for x=0. That makes a lot more sense.
 
I advise you to use the hint given!
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top