Proving f(x)=0 in [-1,1] | Analytical Function Homework Statement

In summary, the student is looking for help with a homework question that they are having difficulty solving. They have been told by their professor that he will not be providing solutions and they are starting to get frustrated.
  • #1
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Homework Statement


I had this question on a test. No one I talked to had a clue how to solve it, and our professor will not be publishing solutions. Any insight as to how to approach this would be great because I am totally stumped.

Let [tex] f \in C^\infty[-1,1] [/tex] such that for all [tex] j \in N |f^{(j)}|<M [/tex]. given that [tex] f(1/k)=0 \forall k \in N [/tex] prove that [tex] f=0 [/tex] in [tex] [-1,1] [/tex]



The Attempt at a Solution



The only idea I've had is that I need to use the taylor series. But I have no idea what to do with it.
Thanks
Tal
 
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  • #2
hmm. can only offer comments really as I am not much of an analyst. but if its zero at every rational number in the interval then by the density of the rationals, it would need to be zero everywhere else since the derivatives are bounded which prevents things such as discontinuities etc.
 
  • #3
I thought of that, but it's npot the case. we know nothing about, say, F(5/7)
 
  • #4
Can you make use of the theorem that says analytic functions have at least one singular point or are constant?
 
  • #5
Never heard that one, so I guess not.
 
  • #6
Hints:

1) What is f(0) ?

2) What are the derivatives of f at zero?
 
  • #7
By the way, you title this "analytic functions" but refer \(\displaystyle f\in C^\infty [-1, 1]\). That is NOT the set of "analytic functions on [-1, 1]. The fact that a function has all derivatives continuous does NOT imply that it is analytic. For example, the function
[tex]f(x)= \begin{array}{c}0 if x= 0 \\ e^{-\frac{1}{x}} if x\ne 0[/tex]
is in [tex]C^\infty[-1, 1][/tex] but is not analytic on that set.
 
  • #8
@countbliss
I asume f(0) is 0 but I don't know how to prove it. I thought that since K can go to infity values of 1/k that are arbitrarily close to 0 are zero implies that f(0) is 0. But I'm not sure that really proves it and I don't have an idea as to how to prove it.

Halls ofIvy, duly noted. Thanks
 
  • #9
That's a general property of continuous functions. A function is continuous if and only if given any convergent sequence [tex]x_n \to x[/tex], you get [tex]f(x_n) \to f(x)[/tex]
 
  • #10
Gee, I wish I remebered thaat on the exam.
Ok, having established that f(0)=0 I'm clear on how to prove this for [0,1]. But what about [-1,0]? I'm assuming it has something to do with the bounded deriviatives, but that is just because we haven't used that information yet.
 
  • #11
You got the proof for x = 0 only so far. The next step is to use the Nth order Taylor formula with an appropriate form of the error term.
 
  • #12
I don't see it. I can prove f=0 for [0,1] using rolles theorem and induction. I don't see how to do this on [-1,0]. And I don't see where the error term can help. I'm stumped!
 

1. How can I prove that f(x)=0 in the interval [-1,1]?

To prove that f(x)=0 in the interval [-1,1], you can use the intermediate value theorem. This theorem states that if a continuous function f(x) is defined on an interval [a,b] and takes on values of both positive and negative sign at points a and b, then there must exist a point c in the interval (a,b) where f(c)=0. Therefore, by showing that f(x) is continuous and takes on values of both positive and negative sign at the endpoints of the interval [-1,1], you can prove that f(x)=0 in the interval.

2. Can I use the mean value theorem to prove that f(x)=0 in [-1,1]?

No, the mean value theorem cannot be used to directly prove that f(x)=0 in the interval [-1,1]. The mean value theorem only guarantees the existence of a point c in the interval (a,b) where f'(c) is equal to the average rate of change of f(x) on the interval [a,b]. It does not guarantee that f(c)=0 for any point c in the interval. Therefore, the mean value theorem alone cannot be used to prove that f(x)=0 in [-1,1].

3. Are there any other methods besides the intermediate value theorem to prove that f(x)=0 in [-1,1]?

Yes, there are other methods that can be used to prove that f(x)=0 in [-1,1]. One method is to use the properties of continuous functions, such as the fact that the sum, difference, product, and composition of continuous functions are also continuous. Another method is to use algebraic manipulations, such as factoring or substitution, to show that f(x)=0 in the interval. Additionally, if f(x) is a polynomial function, you can use the rational root theorem to find any rational roots in the interval and prove that f(x)=0 at those points.

4. Can I use a graph to prove that f(x)=0 in [-1,1]?

While a graph can provide visual evidence for why f(x)=0 in [-1,1], it is not a valid proof on its own. A graph can help you identify potential roots or intervals where f(x) changes sign, which can then be used to construct a formal proof using a mathematical theorem or method. However, a graph alone is not enough to prove that f(x)=0 in the interval [-1,1].

5. Do I need to use calculus to prove that f(x)=0 in [-1,1]?

No, calculus is not always necessary to prove that f(x)=0 in [-1,1]. Depending on the complexity of the function f(x), you may be able to use algebraic or numerical methods to prove that f(x)=0 in the interval. However, if f(x) is a polynomial function, calculus may be necessary to find the roots and prove that f(x)=0 at those points using the fundamental theorem of algebra.

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