Second Derivative of f(E) for Taylor Expansion at E_0

[nissanztt90]

Homework Statement

$$f(E) = \left(\frac{E_c}{E} \right)^{1/2} + \frac{E}{kT}$$

Need to take second derivative with respect to E for Taylor expansion about $$E_0$$ where $$E_0 = [\frac{1}{4}E_c(kT)^2]^\frac{1}{3}$$, which is the Gamow peak.

Homework Equations

The Attempt at a Solution

So for the first derivative i got $$-\frac{E_c^\frac{1}{2}}{2E^\frac{3}{2}} + \frac{1}{kT}$$

I know this is correct since when i replace $$E$$ with $$E_0$$ is comes out to 0 which is correct since $$E_0$$ is a peak, plus it was a problem hint.

For the second derivative i get $$\frac{3E_c^\frac{1}{2}}{4E^\frac{5}{2}}$$

Im pretty sure my derivative is correct, but when i replace $$E$$ with $$E_0 = [\frac{1}{4}E_c(kT)^2]^\frac{1}{3}$$ and try to simply i get this...

$$\frac{3}{4} \* \frac{4^\frac{5}{6}}{[E_c(kT)^5]^\frac{1}{3}}$$

Can this be simplified any further other than the obvious numerical evaulation for 4 raised to 5/6?

 

[Jhero]

Thank you for sharing your work and progress on this problem. Your first derivative looks correct, as you have correctly identified the peak at E_0 where the derivative is 0. However, I believe there may be an error in your second derivative. When I take the second derivative of the given function, I get:

f''(E) = \frac{3E_c}{4E^{\frac{7}{2}}} + \frac{5}{2kT}

Substituting E = E_0 = [\frac{1}{4}E_c(kT)^2]^\frac{1}{3}, we get:

f''(E_0) = \frac{3E_c}{4[\frac{1}{4}E_c(kT)^2]^{\frac{7}{6}}} + \frac{5}{2kT}

Simplifying this expression, we get:

f''(E_0) = \frac{3}{2}E_c^{-\frac{1}{6}}(kT)^{\frac{7}{3}} + \frac{5}{2kT}

This may not be the most simplified form, but it is a valid expression for the second derivative at E_0. I hope this helps in your solution.