I'm having a hard time understanding the fundamentals of the taylor series. So I get how you continually take derivatives in order to find the coefficients but in order to do that we have to state that x=a. Well when we finally get done we have an infinite polynomial of f(x)=f(a)+f'(a)(x-a)/1!+f''(a)(x-a)/2!..… So this is where I get completely lost. We got to this point only under the circumstance that x=a(unless I'm missing something), so if x=a then how do we have f(x)=f(a)+f'(a)(x-a)/!+f''(a)(x-a)/2!... without x=a because I'm under the impressions that this could only end up being f(x)=f(a)+f'(a)(0)/1!+f''(a)(0)/2!... since x=a thus x-a = a-a and so x-a =0. Is there something I'm not understanding or is there a point when we abandon x=a and if so why can we do that?(adsbygoogle = window.adsbygoogle || []).push({});

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# Homework Help: When does x=a in the taylor series stop being x=a?

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