Proof Help: Directional Derivatives

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Prove the following is not true:

Let f : R^2->R^2 be a nonlinear function. For any vectors a,v in R^2;

f(a+v)-f(a)=[Df(a)]v


In terms of my attempt, I've been trying to find some combination of a and v that ensure this fails, but so far the best I've come up with is to start with:

[Df(a)]v=lim{h->0} (f(a+hv)-f(a))/h

And then trying to show that f(a+v)-f(a) is the same thing only as h->1; again, no luck. I really just need a hint or two as to which way to approach the proof.
 
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To show an equation is false you don't have to prove a big theorem. You just have to find an example where it doesn't work. Have you actually tried any examples? Take something simple with a squared term, an easy point and an easy unit vector. Almost anything you try should give a counterexample. Just show the equation you are given doesn't work using the formula for directional derivative.
 
Read the text; finding a counter example for one function f simply proves that it doesn't hold for that ONE function f; the problem asks to show this is true for ANY nonlinear function.
 
rman144 said:
Read the text; finding a counter example for one function f simply proves that it doesn't hold for that ONE function f; the problem asks to show this is true for ANY nonlinear function.

Read the text yourself. It asks you to prove a theorem is not true.
 
You can use that f is non-linear. What does that mean? Either
1) There exist x, y so that f(x+y) is NOT f(x)+f(y)

2) There exists x, c a constant so that f(cx) is NOT cf(x).

Cleverly pick your point a to be one at which f is 'not linear' so to speak (i.e. you can use a to demonstrate f is non-linear by one of these).
 
rman144 said:
Read the text; finding a counter example for one function f simply proves that it doesn't hold for that ONE function f; the problem asks to show this is true for ANY nonlinear function.
This is basic logic. You need to disprove a statement of the form "for all nonlinear functions f and vectors a,v: ...". This means finding one particular nonlinear function f and particular vectors a and v such that ... does not hold, is enough.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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