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Chapter 4: Q1P (page 196)

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 ∈1and∈2are and that they approach zero as ∆xand∆y→0and∆y→0.

Short Answer

Expert verified

Hence, we conclude ∈1,∈2tend to zero as both∆xand∆ytends 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=0∞h∂∂x+k∂∂ynf(a,b)(2)

02

Differentiate the given function

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

∆z=∑n=1∞1n!∆x∂∂x+∆y∂∂ynf(a,b)-f(a,b)=∆x∂f∂x+∆y∂f∂y+∑n=1∞1n!∆x∂∂x+∆y∂∂ynf(a,b)=∆x∂f∂x+∆x∈1+∆y∂f∂y+∆x∈2x,y→0as∆x,∆y→0=dz+∆x∈1+∆y∈2

03

Define limit

From above evaluation, we obtained:

∆z=dz+∆x∈1+∆y∈2

From this, we conclude:

∈1,∈2→0,∶Ä∆x,∆y→0

Hence, we conclude ∈1,∈2tend to zero as both∆xand∆y tends to zero respectively.

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