Tuesday, 8 January 2019

multivariable calculus - How to prove differentiability of g(x)=xTAx?



How to prove differentiability of g(x)=xTAx?




What I've started with is the definition of differentiability:



Let GRn be open. g:GRn is differentiable at xG if exists a linear transformation L:RnRm s.t.



limh0g(x+h)g(x)Lh||h||=0



Now since xTAx is not a linear form, then I don't understand how can do anything with the above definition.


Answer



Here, g(x+h)=(x+h)TA(x+h)=xTAx+xTAh+hTAx+hTAh.




So that g(x+h)g(x)||h||=xTAh+hTAx+hTAh||h||

.



Let L(h)=xTAh+hTAx. You can check that this is linear (in h).



Then, g(x+h)g(x)L(h)||h||=hTAh||h||



Finally, note that limh0hTAh||h||=0. Hence, the linear map L(h)=xTAh+hTAx is the derivative of the map.



The way of picking L was simple: I just took out all the linear terms of h from the numerator using L. What remained were higher order terms, and these dominate ||h|| as the quotient goes to zero. Hence, L is very logically picked.




There are many ways of doing this problem. One is via what Denis suggested above. Another is via chain rule. But this is the elementary method.


No comments:

Post a Comment

real analysis - How to find limhrightarrow0fracsin(ha)h

How to find limh0sin(ha)h without lhopital rule? I know when I use lhopital I easy get $$ \lim_{h\rightarrow 0}...