Anonymous at Sat, 6 Jul 2024 08:51:31 UTC No. 16270087

>>16270077
What have you tried?

Anonymous at Sat, 6 Jul 2024 08:57:36 UTC No. 16270088

>>16270087
Counting how many zeroes there are from Rolle's theorem. If we assume none of the [math]k_i[/math] are zero then after one differentiation there will be [math]n+1+n[/math] zeroes. But I don't know how to apply this generally for [math]k_i[/math].

Anonymous at Sat, 6 Jul 2024 08:59:42 UTC No. 16270089

>>16270088
Can you prove this for the simplest nontrivial case of two points, k_0 = 0, k_1 = 1? I think working it out will be enlightening

Anonymous at Sat, 6 Jul 2024 09:01:22 UTC No. 16270090

>>16270089
There exists [math]c \in (a_0, a_1)[/math] such that [math]f''(c)=0[/math]. There are two zeroes after one differentiation, and one more after another.

Anonymous at Sat, 6 Jul 2024 09:01:53 UTC No. 16270091

>>16270077
It's trivial if N = 1. (Rolle's theorem)

If N>1 then consider the derivative of f. It vanishes with multiplicities ki at a0, a1, ... ,an and once each between those points so counted with multiplicities a total of N-1 times which is enough to do induction.

Anonymous at Sat, 6 Jul 2024 09:03:48 UTC No. 16270093

>>16270090
Do you see the general pattern yet? It's not a problem when some of k_i's are zero, because you get new points that are zero.
If you're still unsure, try three points with following multiplicities: (0,2,1)

Anonymous at Sat, 6 Jul 2024 09:06:32 UTC No. 16270095

>>16270091
But once you've differentiated it [math]max{k_i}[/math] times, there will be no more new zeroes. It's not like differentiating it [math]k_1[/math] times gets rid of the zeroes at [math]a_1[/math] but leaves the rest.

Anonymous at Sat, 6 Jul 2024 09:09:35 UTC No. 16270098

>>16270093
Perhaps how I'm thinking about it is inefficient, because I imagine a number line with the [math]a_i[/math] marked with dots, and then I draw another line above that, representing a differentiation. I mark the new line with dots at all the zeroes which includes some of the [math]a_i[/math] and all the points in between. And so on.

Anonymous at Sat, 6 Jul 2024 09:13:08 UTC No. 16270102

>>16270095
That's not true. Think of problem a. All the k_i's are zero.

Anonymous at Sat, 6 Jul 2024 09:26:38 UTC No. 16270109

>>16270102
I mean that the number of zeroes will decrease by 1 with each differentiation.

Anonymous at Sat, 6 Jul 2024 11:45:38 UTC No. 16270198

>>16270091
>>16270093
Ok, what about let [math]g[/math] be a function such that [math]f(x)=(x-a_0)^{k_0+1} \dots (x-a_n)^{k_n+1} g(x)[/math] and [math]g(a_i) \neq 0[/math] for [math]i=0, \dots , n[/math]. Then [math]f[/math] has [math]n + 1 + k_0 + \dots + k_n[/math] roots, counting with multiplicity. The derivative of [math]f[/math] is a sum of products like [math](x-a_i)^{k_i} \prod_{j \neq i}^{n} (x-a_j)^{k_j+1} g(x)[/math] and [math](x-a_0)^{k_0+1} \dots (x-a_n)^{k_n+1} g'(x)[/math]. Now, [math]f'[/math] has [math]n + 1 + (k_0 -1) + ... + (k_n -1)[/math] roots, that is [math]k_0 + ... + k_n[/math] roots counted with multiplicity from the [math]a_i[/math]. In addition, it has [math]n[/math] roots between each of the [math]a_i[/math] by Rolle, so [math]f'[/math] has [math]n + k_0 + \dots + k_n[/math] roots counted with multiplicity. That is, differentiating a function [math]f[/math] with [math]N+1[/math] roots gives a function [math]f'[/math] with [math]N[/math] roots. So now [math]f'[/math] is a function with [math]N[/math] roots (counted with multiplicity), and so [math]f''[/math] has [math]N-1[/math] roots. Differentiating [math]N[/math] times we get [math]f^{(N)}[/math] has one root.

Is this valid?

Anonymous at Sat, 6 Jul 2024 11:55:04 UTC No. 16270203

>>16270077
all you idiots with your formalities ain't got shit on pictorial proofs

Anonymous at Sat, 6 Jul 2024 12:39:39 UTC No. 16270238

Bump

Anonymous at Sat, 6 Jul 2024 16:20:21 UTC No. 16270428

Bump

Anonymous at Mon, 8 Jul 2024 02:43:35 UTC No. 16272374

>>16270077
this is super easy. here is the answer from gemini

Absolutely! This is a classic problem related to Rolle's Theorem and its extension. Here's the proof:

**Proof**

1. **Base Case (n = 1):** If a function *f* is continuous on [a0, a1] and differentiable on (a0, a1), and it vanishes at the endpoints (f(a0) = f(a1) = 0), then by Rolle's Theorem, there exists a point c ∈ (a0, a1) where f'(c) = 0.

2. **Induction Step:** Assume the statement is true for some n = k. That is, if a function *f* is continuous on [a0, aₖ] and k times differentiable on (a0, aₖ), and it vanishes at k+1 distinct points, then there exists a point c ∈ (a0, aₖ) such that f(ᵏ)(c) = 0.

3. **Prove for n = k + 1:** Consider a function *f* that is continuous on [a0, aₖ+1] and (k+1) times differentiable on (a0, aₖ+1), vanishing at the k+2 distinct points a0, a1, ..., aₖ+1.

* Apply Rolle's Theorem to the intervals [a0, a1], [a1, a2], ..., [aₖ, aₖ+1]. This gives us k points c1, c2, ..., cₖ, where cᵢ ∈ (aᵢ-1, aᵢ) and f'(cᵢ) = 0 for each i.

* Now, consider the function f'. It is continuous on [c1, cₖ] and k times differentiable on (c1, cₖ), and it vanishes at the k+1 points c1, c2, ..., cₖ.

* By the induction hypothesis, there exists a point c ∈ (c1, cₖ) ⊂ (a0, aₖ+1) where f(ᵏ)(c) = 0.

* Since f(ᵏ)(c) = 0, this implies f(ᵏ+1)(c) = 0.

**Conclusion**

By the principle of mathematical induction, the statement holds true for all n ∈ ℕ. Therefore, if a function *f* is continuous on [a0, aₙ] and n times differentiable on (a0, aₙ), and it vanishes at n+1 distinct points, then there exists a point c ∈ (a0, aₙ) such that f(n)(c) = 0.