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

In summary, the conversation discusses deriving a formula for a physics problem by expanding the function f(E) as a Taylor series. The first three terms of the series are given, and the hint that A1=0 because f(E) peaks at Eo is discussed. The process of finding the first and second derivatives of F with respect to Eo is explained, and it is determined that A1=0 and A2=0. The conversation also covers finding Eo by setting f'(E) = 0 and solving for E, and using this value to plug into the derivatives of F to find A1 and A2.
  • #1
leonne
191
0

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
 
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  • #2
You say "Taylor series", do you not know what a Taylor series is?

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

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

Start by finding the first and second derivatives of F with respect to E.
 
  • #3
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
 
  • #4
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.
 
  • #5
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?
 
  • #6
leonne said:
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 .
 
  • #7
yeah thanks
 

What is a Taylor series?

A Taylor series is a mathematical representation of a function as an infinite sum of terms. It is used to approximate a function near a specific point by calculating the value of the function and its derivatives at that point.

Why is the Taylor series important?

The Taylor series is important because it allows us to approximate complex functions with simpler polynomial functions. This can make it easier to solve problems in physics, engineering, and other fields.

How do you find the coefficients of a Taylor series?

The coefficients of a Taylor series can be found by taking the derivatives of the function at the point of approximation and plugging them into the formula for the Taylor series. The coefficients represent the rate of change of the function at that point.

What is the purpose of truncating a Taylor series?

Truncating a Taylor series means stopping the infinite sum of terms at a certain point. This is often done in practice because it is not practical to calculate an infinite number of terms. Truncating a Taylor series can still provide a good approximation of a function, but the accuracy decreases as you go further from the point of approximation.

What is the difference between a Taylor series and a Maclaurin series?

A Taylor series is a more general form of a Maclaurin series. A Maclaurin series is a special case of a Taylor series where the point of approximation is 0. This means that all the derivatives of the function at 0 are used to calculate the coefficients, making it a simplified form of the Taylor series.

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