We are looking for a theorem like this: Let n and k be positive integers. Let f

Question

We are looking for a theorem like this: Let n and k be positive integers.

Let f be a function defined on an open interval I. Assume that f is n-times differentiable on I.

• IF the equation f ^(n)(x) = 0 has k solutions on I,

• THEN the equation f(x) = 0 has at most ??? solutions on I.

There are various quantities we can write instead of ??? that will make the theorem true. For example, if the theorem were true when we write ??? = nk2 (we are not saying it is), then it would also be true when we write ??? = nk2 + 1. We want you to find the smallest expression we can write instead of ??? that will still make the theorem true.

Once you have that quantity, you need to do two things. First prove that the theorem is true. Second, show with an example, that your expression is the smallest one that makes the theorem true.

Explanation / Answer

Statement for minimum value of ??? is below:

If f(n)(x) = 0 has k solutions, then f(x) = 0 has at most (n+k-1) solutions.

We can consider a polynomial function as an example. For example, f(x) = x4 is 5 times differentiable. And f5(x) = 0 has 0 solutions. Therefore, as per our claim, f(x) = 0 should have at most (n+k-1) zeroes.

That is, x4 = 0 has (5+0-1) = 4 solutions. Which is a true statement, because x4 = 0 has 4 solutions. (x=0 with multiplicity 4).

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