Taylor series of f(x,y)=x*e^(cosxy−1) around the point (1,0), my attempt so far: Obviously just by checking first and second partial derivatives it's only getting more complicated.

profesorluissp

profesorluissp

Answered question

2022-09-08

Taylor series of f ( x , y ) = x e cos x y 1 around the point ( 1 , 0 ), my attempt so far:
Obviously just by checking first and second partial derivatives it's only getting more complicated. I'm not sure how it is possible to determine the general form of D k f ( 1 , 0 ) ( ( x 1 , y ) , ( x 1 , y ) . . . ( x 1 , y ) ) (where there are k pairs of ( x 1 , y ). And without it I'm not sure how to determine the infinite series needed in this problem.

Answer & Explanation

Monserrat Ellison

Monserrat Ellison

Beginner2022-09-09Added 22 answers

Consider the auxiliary function
g ( z ) := e cos z 1
of one variable z. This function satisfies the IVP
(1) g ( z ) = sin z g ( z ) , g ( 0 ) = 1   .
The known series for sin allows to obtain a finitary recursive formula for the coefficients a n of the Taylor series of g, so that you in principle have
g ( z ) = n = 1 a n z n
with known rational coefficients a n . Mathematica says that
g ( z ) = 1 z 2 2 + z 4 6 31 z 6 720 + 379 z 8 40 320 1639 z 10 907 200 +   ? z 12   .
Given this you can write
f ( x , y ) = x n = 0 a n ( x y ) n   ,
which is the "entire thing" you wanted.

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