Quadratic Approximation of Potential Function using Taylor Expansion Method

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Homework Help Overview

The discussion revolves around finding the quadratic approximation of a potential function using the Taylor expansion method. The potential function is defined as U(x) = U0((a/x)+(x/a)), with specific values provided for U0 and a.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to understand the Taylor expansion method for approximating the potential function, expressing uncertainty about the process due to lack of prior exposure. Some participants suggest that the method is straightforward and encourage the poster to apply the formula using derivatives at a specific point.

Discussion Status

Participants are actively engaging with the original poster's attempts, providing guidance on how to approach the problem. There is a recognition of the poster's initial efforts, and some validation of their calculations has been offered, although no consensus on the final form has been reached.

Contextual Notes

The original poster mentions that this question is part of a larger engineering homework assignment, indicating potential constraints in their understanding of the topic. There is also a reference to the need for clarity on the Taylor expansion method, which has not been previously covered in their coursework.

Crazy Gnome
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Homework Statement



What is the quadratic approximation to the potential function?

Homework Equations



U(x) = U0((a/x)+(x/a))
U0= 20
a=4

The Attempt at a Solution



This is just the last part of a question on my engineering homework, I never learned Taylor expansions before even though I have taken all the class prerequisites. So if you could just walk me through it that would be much appreciated.
 
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Read this http://en.wikipedia.org/wiki/Taylor_series and try it out yourself. It's just a cookbook formula. You have to add various derivatives of U(x) at x=4 times powers of (x-4) up to quadratic. It's not that mysterious. We'll be glad to look at your efforts.
 
Dick said:
Read this http://en.wikipedia.org/wiki/Taylor_series and try it out yourself. It's just a cookbook formula. You have to add various derivatives of U(x) at x=4 times powers of (x-4) up to quadratic. It's not that mysterious. We'll be glad to look at your efforts.

So for my first try I got U=(1.25)x2-10x+60. I am pretty sure I remember something about only going out to the second order place to make it "quadratic" so I got.

40 for the first place, 0 zero for the first order, and 1.25x2-10x+20 for the second order.

Does this look right?
 
Yes, that's right. For most purposes you probably want to leave that in the form 40+5*(x-4)^2/4, since you are thinking of x's near 4.
 

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