I need to find the first 2 orders of the taylor series on the expression below.

sqrt(Psi_s0 + Phi_t*exp((Psi_s0-2*Phi_F - V_SB)/Phi_t))

where Phi_t*exp((Psi_s0-2*Phi_F - V_SB)/Phi_t) is defined as Xi, and the taylor series is around Xi = 0.

Printable View

- Jun 15th 2008, 08:23 PMwas1984Taylor Expansion on d/d(f(x))[sqrt(x+f(x))]
I need to find the first 2 orders of the taylor series on the expression below.

sqrt(Psi_s0 + Phi_t*exp((Psi_s0-2*Phi_F - V_SB)/Phi_t))

where Phi_t*exp((Psi_s0-2*Phi_F - V_SB)/Phi_t) is defined as Xi, and the taylor series is around Xi = 0. - Jun 15th 2008, 08:56 PMMathstud28
- Jun 15th 2008, 08:57 PMwas1984
Yes, thank you, that is correct.

- Jun 15th 2008, 09:00 PMMathstud28
What is the variable in which we are differentiating in respect to? I will assume it is $\displaystyle \psi$

Then let $\displaystyle f(\psi)$ be equal to the above expression, then the second order polynomial would be

$\displaystyle f(0)+f'(0)x+\frac{f''(0)x^2}{2}$ - Jun 15th 2008, 09:03 PMwas1984
The variable we are differentiating with respect to is Xi, which is the second term under the square root.

- Jun 15th 2008, 09:11 PMwas1984
I'll try to elaborate. We are setting $\displaystyle \xi = \phi_{t}e^{\frac{\psi_{s0}-2\cdot\phi_f-V_{SB}}{\phi_t}}$

Then we are solving the series around $\displaystyle \xi = 0$

So I actually want expand an equation of the form $\displaystyle \sqrt{\psi_{s0}+\xi(\psi_{s0})}$ for $\displaystyle \xi(\psi_{s0})$, and I only need the first two terms, fortunately. :)