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I've try solve this question, but I haven't sucess...

The problem is the following:

A continuous functions $f:[a,b]\rightarrow \mathbb{R}$ assume positive and negative values in its domain, show that there exists $a_1,a_2,\ldots,a_k$, k numbers that are arithmetic progression and it is in the domain such that $$f(a_1)+f(a_2)+\cdots+f(a_k)=0$$

Someone can help me with this question?

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  • $\begingroup$ Please state how this relates to "Arithmethic Progression", as stated in your title. Are the $a_i$ supposed to be in AP? $\endgroup$ – Calvin Lin May 22 '13 at 18:17
  • $\begingroup$ You'd need the condition that the domain is connected. And I presume that you want the constant difference to be non-zero? $\endgroup$ – Calvin Lin May 22 '13 at 18:27
  • $\begingroup$ What is $k$ here? If I let $k=1$ or $k=2$ this is trivial. $\endgroup$ – Gyu Eun Lee May 22 '13 at 18:29
  • $\begingroup$ the domain is connected. Suppose $k>3$. $\endgroup$ – user78993 May 22 '13 at 18:31
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Fix $k$. Let the domain of the function include the interval $[a,b]$, where $f(a) \times f(b) < 0 $.

Consider the value of $ f(a) + f( a + \frac{b-a}{k-1}) + \ldots + f(b) $. If this is 0, we are done. Otherwise, WLOG we may assume that it has different sign as $f(a)$.

Now, consider the function

$$F(a, \epsilon) = f(a) + f(a + \epsilon) + f(a + 2 \epsilon) + \ldots + f(a + (k-1)\epsilon ), $$

where $\epsilon \in [0, \frac{b-a} {k-1}] $.

Apply the Intermediate-Value Theorem, and we are done.

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  • $\begingroup$ if the sum has the value of $f(a)$ I can consider the value $f(b)$? $\endgroup$ – user78993 May 22 '13 at 18:42
  • $\begingroup$ @user78993 Note that the sum has a different sign, so they can't have the same value. That's the WLOG part. If they have the same sign, then yes we consider $f(b)$, and modify the $F$ function slightly. $\endgroup$ – Calvin Lin May 22 '13 at 19:22

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