Hi! It's the anon that had trouble with an induction proof. The links you directed me to were really informative, but I still can't figure it out. The proposition is: prove that if f_1(x)+f_2(x)+...+f_n(x) are differentiable functions, then: d/dx(f_1(x)+f_2(x)+...f_n(x))=f'_1(x)+f'_2(x)+...+f'_n(x) for any natural number n.
Okay, so first you want to prove your “base case” - i.e. prove the statement for n=2. (Normally you start with n=1, but for things like sums, it makes more sense to start with n=2, as using n=1 would lead to a trivial statement and wouldn’t help in the next part.) So basically you just have to show that d/dx(f_1(x)+f_2(x))=f’_1(x)+f’_2(x). (Now, I don’t know what you’ve already covered in your class, but this is probably something you’re allowed to just accept as true.)
Next is the “induction step.” This means that we just assume the statement is true for n=k (i.e. d/dx(f_1(x)+…+f_k(x))=f’_1(x)+…+f’_k(x)) for some finite natural number k greater than or equal to 2 (our base case). We don’t have to show this at all, we can just assume it’s true and use it to show our statement still holds for n=k+1:
- d/dx(f_1(x)+…+f_k(x)+f_k+1(x))=d/dx(g(x)+f_k+1(x)), where g(x)=f_1(x)+…+f_k(x) (since a sum of functions is itself a function)
- from our base case, we know that the derivative of the sum of two functions is the sum of their derivatives; in our case: d/dx(g(x)+f_k+1(x))=g’(x)+f’_k+1(x)
- using substitution, g’(x)+f’_k+1(x)=(d/dx(f_1(x)+…+f_k+1(x)))+f’_k+1(x)
- now here’s where our induction hypothesis comes into play; we can rewrite (d/dx(f_1(x)+…+f_k+1(x)))+f’_k+1(x) as (f’_1(x)+…+f’_k(x))+f’_k+1(x)=f’_1(x)+…+f’_k+1(x)
- therefore d/dx(f_1(x)+…+f_k+1(x))=f’_1(x)+…+f’_k+1(x)
So now we’re actually done, as this is enough to prove that the statement is true for all finite values of n. Usually you just stop after proving the base case and induction hypothesis (save for perhaps a final sentences or two wrapping up the entire proof), but for the sake of further explanation…
We’ve showed that this statement holds for n=2 in our base case. So what about n=3? Well, in the induction step, we showed that the statement holds for n=k+1 so long as its true for n=k. Well, let k=2 (since we know this to be true) and this implies that it’s true for n=2+1=3. Now, if we let k=3 (which we now know to be true), then we know it’s also true for n=3+1=4. Letting k=4 implies its true for n=4+1=5, and so on for any value of n.
Hopefully that helped!