So here is what i have for a solution to the heat eqn.

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Homework Help Overview

The discussion revolves around solving the heat equation, specifically focusing on determining the coefficients \( A_n \) in the series representation of the solution. Participants are exploring the implications of orthogonality of cosine functions in this context.

Discussion Character

  • Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • One participant discusses using the orthogonality of cosine functions to derive the coefficients \( A_n \) by integrating both sides of the equation after multiplying by a cosine term. They express uncertainty about handling the summation and integration when \( m \neq n \).

Discussion Status

Participants are actively engaging with the mathematical reasoning involved in the problem. One participant has provided clarification on the integration process, emphasizing the importance of fixing \( m \) while summing over \( n \). There is a collaborative atmosphere with participants offering insights and references to standard procedures.

Contextual Notes

One participant notes that they are working on this problem for personal interest rather than as part of a class assignment, which may influence the depth of exploration in the discussion.

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Homework Statement


i am solving the heat equation and so far i know what i have is correct. basically, i am down to this [tex]\sum_{n=0}^{\infty}A_n\cos(\frac{n\pi x}{L})=273+96(2L-4x)[/tex] where all i need is to solve for [itex]A_n[/itex]

The Attempt at a Solution


i was thinking about taking advantage of the orthogonality of the cosine function and multiplying both sides by [itex]\cos(\frac{m\pi x}{L})[/itex] and then integrate over the interval [itex][0,L][/itex]. my question is, if [itex]m\neq n[/itex] then i can move this cosine into the sum, integrate term wise, yet the left side equals zero ([itex]m\neq n[/itex]). Thus, [itex]m = n[/itex], and then if i multiply both sides by [itex]\cos(\frac{n\pi x}{L})[/itex] i cannot put this cosine term inside the sum, and thus i have lost the idea of how to solve for [itex]A_n[/itex]. any help/advice is awesome!

for what it's worth, this is not a class i am in, I'm just doing the problem for fun. thanks for your help!
 
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joshmccraney said:

Homework Statement


i am solving the heat equation and so far i know what i have is correct. basically, i am down to this [tex]\sum_{n=0}^{\infty}A_n\cos(\frac{n\pi x}{L})=273+96(2L-4x)[/tex] where all i need is to solve for [itex]A_n[/itex]

The Attempt at a Solution


i was thinking about taking advantage of the orthogonality of the cosine function and multiplying both sides by [itex]\cos(\frac{m\pi x}{L})[/itex] and then integrate over the interval [itex][0,L][/itex].

This is the standard procedure.

my question is, if [itex]m\neq n[/itex] then i can move this cosine into the sum, integrate term wise, yet the left side equals zero ([itex]m\neq n[/itex]).

No. The left hand side is
[tex] \int_0^L \cos(m\pi x/L) \sum_{n=0}^\infty A_n \cos(n \pi x /L)\,dx<br /> = \sum_{n= 0}^\infty \left(A_n \int_0^L \cos(m \pi x/L) \cos(n\pi x/L)\,dx\right) \\<br /> = A_m \int_0^L \cos^2(m \pi x/L)\,dx[/tex]
Remember that [itex]n[/itex] varies in the summation, but [itex]m[/itex] is fixed. At some point [itex]n[/itex] must take the value [itex]m[/itex], and this is the only term in the sum which doesn't vanish on integration.
 
Last edited:
thanks!
 
i have another post in the pde/ode theory part on a similar topic. if youre not too busy, perhaps you could take a look?
 

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