Master Taylor Series with Expert Tips | F(E) = E/(KT) + (Ec/E)^1/2

[leonne]

Homework Statement

Expand the function f(E) as a Taylor series.

Homework Equations

f(E)=E/(KT)+(Ec/E)^1/2

The Attempt at a Solution

E=Eo
So it says that
F(E)~Ao+A1(E-Eo)+A2(E-Eo)²...
I need to find out what Ao A1 and A2 are, but not sure how to do that. It says as a hint that A1=0 becasue f(E) peaks at Eo

Any help/hints on what to do? This is from aphysics problem need to derive a formula. I checked like pauls calc but no real help
thanks

 

[HallsofIvy]

You say "Taylor series", do you not know what a Taylor series is?

The Taylor series for f(x), about $x_0$ is
$$\displaytype \sum_{n=0}^\infty \frac{f^{(n)}(x_0)}{n!}(x- x_0)^n$$
where "$f^{(n)}(x_0)$" is the nth derivative of f evaluated at $x_0$.

In particular, the first three terms are
$$f(x_0)+ f'(x_0)(x- x_0)^2+ \frac{f''(x_0)}{2}(x- x_0)^2$$

Start by finding the first and second derivatives of F with respect to E.

 

[leonne]

o ok i see now was getting confussed what the problem was asking so it would be
F(E)=(E/kt)+(Ec/E)^1/2 +0(E-Eo)+0(E-Eo)²

is this right? says as a hint that a1=0 so than a2=0 also

 

[SammyS]

If f(E) peaks at E₀, then f '(E₀) = 0.

That explains why A₁ = 0.

Find: f '(E), f ''(E), f '''(E), etc. & evaluate them at E=E₀.

To find E₀, set f '(E) = 0 & solve for E. That's E₀.

 

[leonne]

cool thxs, so i got
f'=1/kt-((Ec)^1/2)/2(E)^3/2
f"=3(Ec)^1/2/4(E)^5/2
also in my notes it says that Eo=(1/4 Ec(kT)²)^1/3
so i just plug in Eo for E into the f' and f" and that will give me a1 and a2 right?

 

[SammyS]

so i just plug in Eo for E into the f' and f" and that will give me a1 and a2 right?

Right.

Of course, that should give you: a₁ = 0, b/c f '(E₀) = 0 .

 

[leonne]

yeah thanks