- #1
MathewsMD
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For the question attached in the file, how exactly does one go about finding a solution? Problem 28 says that if n does not equal m, then ## \int_{-1} ^{1} {P_n}{P_m} = 0 ##
With that statement, I've tried treating this as a Taylor series (centred at 0, arbitrarily) and then trying to find a general term that would allow me to isolate for ## a_k ## but that was unsuccessful.
Since ## P_k (x) ## and ## f(x) ## have similar orders, can I assume ## \int_{-1} ^{1} {P_k (x)}{f(x)} = 0 ##? This doesn't really help, unless I want the trivial solution for ## a_k = 0##
I also tried substituting the ##a_k## term into the sum to show the two sides are equivalent, but end up with ## f(x) = \sum [{P_k} \int_{-1} ^{1}f(x){P_k}(x)]## but I can't quite do anything with this either.
I just don't quite see how the above statement from Problem 28 is useful in this problem, and am uncertain on how to proceed. Any hints or explanations would be greatly appreciated!
With that statement, I've tried treating this as a Taylor series (centred at 0, arbitrarily) and then trying to find a general term that would allow me to isolate for ## a_k ## but that was unsuccessful.
Since ## P_k (x) ## and ## f(x) ## have similar orders, can I assume ## \int_{-1} ^{1} {P_k (x)}{f(x)} = 0 ##? This doesn't really help, unless I want the trivial solution for ## a_k = 0##
I also tried substituting the ##a_k## term into the sum to show the two sides are equivalent, but end up with ## f(x) = \sum [{P_k} \int_{-1} ^{1}f(x){P_k}(x)]## but I can't quite do anything with this either.
I just don't quite see how the above statement from Problem 28 is useful in this problem, and am uncertain on how to proceed. Any hints or explanations would be greatly appreciated!
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