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find the coefficient of $xy^3$ in taylor's polynomial of the function $f(x,y)=\frac{1}{3-x+y}$ near $(0,0)$.

My Solution
I know the series $\frac{1}{1-x}=1+x+x^2+x^3+... $

I just did $f(x,y)=\frac{1}{3}*\frac{1}{1-\frac{x+y}{3}}=\frac{1}{3}(1+\frac{x+y}{3}+(\frac{x+y}{3})^2+(\frac{x+y}{3})^3+(\frac{x+y}{3})^4+...)$
and basically, $(x+y)^4=x^4+4x^3y+6x^2y^2+4xy^3+y^4$ , focus on $xy^3$
I can see that the coefficient of it is $\frac{4}{3^4}$ multiplied by $\frac{1}{3}$ outside the parenthesis, so my final answer would be $\frac{4}{3^5}$.

Extra Questions:
Incase what I did above is all legitimate, if I'm asked to find a taylor expansion of the function near the point $(-1,2)$, is there any trick to do it, maybe transfer it to near $(0,0)$ of a known function series? or the only way would be to take derivatives and substitute?
What about points of the form $(-k,k)$, $x+y$ there goes to zero, can I do what I did right here, if it's a legit way?


Would appreciate any feedback, thanks in advance!

Pwaol
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1 Answers1

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Your first question: yes, you can do that, it is a neat solution. You may need to watch out for convergence-interval in practical use: $|x+y|<3$.

For the general case, you questioned, it is answered in Derivation of multivariable Taylor series

Another good source would be: https://mathinsight.org/taylors_theorem_multivariable_introduction

Peter Lang
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  • Appreciate the help, I didn't quite get the convergence-interval sentence, could you elaborate a little more on it? (Note that I haven't dealt with convergence of series with two variable functions, I am just familiar with radius of convergence of power series of one variable). – Pwaol May 28 '21 at 08:03
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    @Pwaol sorry for late response. It is because the individual terms of your expansion are: (x+y)/3, these terms have to be between -1 and 1, otherwise as the exponent goes to infinity, the higher order terms would not be vanishing so the sum of the series would diverge as well. That is: |(x+y)/3| < 1 – Peter Lang May 28 '21 at 14:33