Expert: Sherry Wallin - 10/31/2011

Question
QUESTION: if f(x)=x1.x2.x3 and f(y)=y1.y2.y3 and y1=x1+dx1, y2=x2+dx2, y3=x3+dx3

then, f(x)-f(y) is not equal to summation of [(y1.x2.x3)-(fx)], [(x1.y2.x3)-f(x)], [(x1.x2.y3)-f(x)]
why so?
kindly explain. is this a mathematical crude?

ANSWER: f(x)-f(y)
=x1x2x3-(x1x2x3+x1x3dx2+x2x3dx1+x3dx1dx2+x1x2dx3+x1dx2dx3+x2dx1dx3+dx1dx2dx3)
= -(x1x3dx2+x2x3dx1+x3dx1dx2+x1x2dx3+x1dx2dx3+x2dx1dx3+dx1dx2dx3).
Are you using '.' to mean multiplication? Why not '*', it is a little more universal. Also is dxn the derivative of x??

Math Prof

---------- FOLLOW-UP ----------

QUESTION: QUESTION: if f(x)=x1*x2*x3 and f(y)=y1*y2*y3 and y1=x1+dx1, y2=x2+dx2, y3=x3+dx3, where dxn is the delta increament/decreament on xn

then, f(x)-f(y) is not equal to summation of [(y1.x2.x3)-(fx)], [(x1.y2.x3)-f(x)], [(x1.x2.y3)-f(x)]
why so?
kindly explain. is this a mathematical crude?

Answer
You can take what I did earlier then and in place of the dxn but xn'. I'm not sure what you are wanting me to do. I have shown you what it is and it isn't what you have asked.  Is this a question being asked of you or is it an answer you think it is and want to know why it isn't??

Math Prof