Obtain Taylor Series at x0 = 0?

[PeteSampras]

Homework Statement

Is it possible obtain a Taylor serie at x₀=0?

Homework Equations

$$f(x)= (\frac{x^4}{x^5+1})^{1/2}$$[/B]

The Attempt at a Solution

I think that it is not possible , since f' is not differenciable at x=0, since f' have the factor

$$(\frac{x^4}{x^5+1})^{-1/2}$$

but, for example wolfram yield a solution f approx x²

http://www.wolframalpha.com/widget/...0&podSelect=&showAssumptions=1&showWarnings=1

 

[jedishrfu]

Isn't it when $x_0 = 1$ that is the problem where it's not differentiable?

 

[Mark44]

Homework Statement

Is it possible obtain a Taylor serie at x₀=0?

Homework Equations

$$f(x)= (\frac{x^4}{x^5+1})^{1/2}$$[/B]

The Attempt at a Solution

I think that it is not possible , since f' is not differenciable at x=0, since f' have the factor

f is continuous at 0, f' is continuous at 0, f'' is continuous at 0...
The function you're working with definitely has a Maclaurin series (i.e., a Taylor series in powers of x).

$$(\frac{x^4}{x^5+1})^{-1/2}$$

but, for example wolfram yield a solution f approx x²

http://www.wolframalpha.com/widget/widgetPopup.jsp?p=v&id=f9476968629e1163bd4a3ba839d60925&title=Taylor Series&theme=blue&i0=sqrt(x^4/(x^5+1))&i1=x&i2=0&i3=10&podSelect=&showAssumptions=1&showWarnings=1

 

[John Park]

I think that it is not possible , since f' is not differenciable at x=0, since f' have the factor

(x⁴/x⁵+1)−1/2

If you complete the calculation of f' by the chain rule, I think you'l find that factor isn't a problem.