# Taylor Expansion on d/d

• Jun 15th 2008, 09:23 PM
was1984
Taylor 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, 09:56 PM
Mathstud28
Quote:

Originally Posted by was1984
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.

$\sqrt{\psi_{s0}+\phi_{t}e^{\frac{\psi_{s0}-2\cdot\phi_f-V_{SB}}{\phi_t}}}$??
• Jun 15th 2008, 09:57 PM
was1984
Yes, thank you, that is correct.
• Jun 15th 2008, 10:00 PM
Mathstud28
Quote:

Originally Posted by was1984
Yes, thank you, that is correct.

What is the variable in which we are differentiating in respect to? I will assume it is $\psi$

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

$f(0)+f'(0)x+\frac{f''(0)x^2}{2}$
• Jun 15th 2008, 10:03 PM
was1984
The variable we are differentiating with respect to is Xi, which is the second term under the square root.
• Jun 15th 2008, 10:11 PM
was1984
I'll try to elaborate. We are setting $\xi = \phi_{t}e^{\frac{\psi_{s0}-2\cdot\phi_f-V_{SB}}{\phi_t}}$

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

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