Consider a function f(x,y)which can be expanded in a two-variable power series, (2.3) or (2.7). Let x-a=h=x,y-b=k=y; then localid="1664363195022" x=a+x,y=b+y, so that f(x,y)becomes f(a+x,b+y). The change zin z=f(x+y)when x changes from a to a+xand y changes from b to b+yis then

z=f(a+x,b+y)-f(a,b).

Use the series (2.7) to obtain (3.11) and to see explicitly what 1and2are and that they approach zero as xandy0andy0.

Short Answer

Expert verified

Hence, we conclude 1,2tend to zero as bothxandytends to zero respectively.

Step by step solution

01

Form two variables power series

We have a function withx=a+xandy=b+ysubjected to change as:

f(x,y)=f(a+x,b+y)=zf(x,y)=f(a+x,b+y)=z

And

z=f(a+x,b+y)-f(a,b)(1)

Now, the two variable power series is given by:

f(x,y)=n=0hx+kynf(a,b)(2)

02

Differentiate the given function

From equations (1) and (2), we get:

z=n=11n!xx+yynf(a,b)-f(a,b)=xfx+yfy+n=11n!xx+yynf(a,b)=xfx+x1+yfy+x2x,y0asx,y0=dz+x1+y2

03

Define limit

From above evaluation, we obtained:

z=dz+x1+y2

From this, we conclude:

1,20,x,y0

Hence, we conclude 1,2tend to zero as bothxandy tends to zero respectively.

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