Why can't I define f(x)=x+x+x? Would you please explain that?

I don't know I'm wrong. In fact I'm sure I'm right because despite asking you repeatedly for an example of N where the equations don't hold you still haven't answered it.

You and Dr Math are both making the same mistake. f(x)=x+x+x...(N times); N=x; does

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***>>> NOT <<< ***(highlighted for clarity)

mean f(x)=x^2.

x+x+x...(N times) = x^2 ONLY when N=x and at no other time. Suppose N=4 and f(x)=x+x+x+x. Then f(x)=16 and g(x)=x^2=16. But if we increase x to 5 while keeping N fixed at 4 then x+x+x+x=20 and x^2=25. So f(x) does not equal g(x) when x!=4.

Here are two examples of N where the equations DO hold.

Reminder: define f(x)=f1(x)+f2(x)+f3(x)+...+fN(x) where fp(x)=x for p=1..N. Also we set x=N.

(Example1) N=3 so f(x)=x+x+x and if x=N then f'(x)=1+1+1=3, g(x)=x^2=9 and g'(x)=2x=6. So f'(x) != g'(x) because 3!=6.

(Example2) N=4 so f(x)=x+x+x+x and if x=N then f'(x)=1+1+1+1=4, g(x)=x^2=16 and g'(x)=2x=8. So again f'(x)!=g'(x) because 4!=8.

So Saswat, mayjune and everyone else, I have given you TWO examples where f(x)=g(x) and f'(x)!=g(x). All I'm asking for is ONE SINGLE VALUE OF N where the above does not work and I will concede that I am wrong.