Proving that there exists a solution to ##f(x+\frac1n)-f(x)=0##

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In summary, the conversation discusses the use of the Intermediate Value Theorem (IVT) to prove that there are two opposite terms in a sum, which leads to the conclusion that the sum is null. This is based on the fact that the function ##h(x)=f(x+\frac1n)-f(x)## must have a non-negative and a non-positive term for the sum to be zero, which can be proven using the IVT.
  • #1
archaic
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Homework Statement
Given ##f:[0,1]\to\mathbb R## which is continuous, ##f(0)=f(1)##, and ##n\in\mathbb N^*## (doesn't include ##0##), show that ##f(x+\frac1n)-f(x)=0## has a solution.
Relevant Equations
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This was shown to me by an acquaintance who told me that if we consider ##h(x)=f(x+\frac1n)-f(x)##, then the fact that ##\sum_{k=0}^{n-1}f(\frac kn+\frac1n)-f(\frac kn)=0## somehow tells us that there are two opposite terms in that sum, and, as such, we can apply the IVT to ##h(x)##.

Clearly the sum is null because it is in the form ##f(1) + (f(a)-f(a)\text{ same argument a})+(f(b)-f(b)\text{ same argument b})+...-f(0)## and ##f(1)=f(0)##, not because there somehow exists two different in argument terms that cancel ##f(a)+f(b)=0##.
In either case, we require a ##h(a)## and a ##h(b)## such that ##h(a)h(b)<0##, not that but with ##f##.

But, from what I have understood, this is the correction given by their professor, so maybe I am missing something?

EDIT: Ignore the following nonsense. o:)
I thought of going about it by contradiction, but the use of a limit leaves me undecided as to whether this counts as a solution.
The negative of the proposition is ##\exists n\in\mathbb N^*\,|\,\forall x\in\mathbb R,\,f(x+\frac1n)-f(x)\neq0##, i.e ##f(x+\frac1n)<f(x)## or ##>##. In both cases, if we take ##n\to\infty##, we get an impossible inequality.

Any comments? Thank you!
 
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  • #2
What about ##f(x) = 2## for ##x \in [0, 1]##? f is continuous and f(0) = f(1).
 
  • #3
Mark44 said:
What about ##f(x) = 2## for ##x \in [0, 1]##? f is continuous and f(0) = f(1).

Isn't every ##x \in [0, 1- \frac{1}{n} ]## a solution for that example?
 
  • #4
You know that the sum must have a non-negative term and a non-positive term for it to be zero. This means that the function ##h(x)=f(x+1/n)-f(x)## is somewhere non-negative and somewhere non-positive. By IVT, this means that it is somewhere zero.
 
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  • #5
Infrared said:
You know that the sum must have a non-negative term and a non-positive term for it to be zero. This means that the function ##h(x)=f(x+1/n)-f(x)## is somewhere non-negative and somewhere non-positive. By IVT, this means that it is somewhere zero.
Yes.. I got that as I was going to sleep. It's so obvious, I don't know what happened.
 

Related to Proving that there exists a solution to ##f(x+\frac1n)-f(x)=0##

1. What does the equation ##f(x+\frac1n)-f(x)=0## mean?

The equation means that the function f has a solution when x is added to 1/n. In other words, there exists a value of x that satisfies the equation.

2. Why is it important to prove that there exists a solution to this equation?

Proving that there exists a solution to this equation is important because it shows that the function f has a point of stability, where the value of f does not change when x is added to 1/n. This can help in understanding the behavior of the function and making predictions.

3. How do you prove that there exists a solution to this equation?

There are several methods for proving that a solution exists to an equation. One way is to use the Intermediate Value Theorem, which states that if a continuous function has values of opposite signs at two points, then there must be a point in between where the function equals zero. Another method is to use the Mean Value Theorem, which states that if a function is continuous and differentiable on an interval, then there exists a point in that interval where the derivative of the function equals the average rate of change.

4. Can there be more than one solution to this equation?

Yes, there can be more than one solution to this equation. For example, if the function f(x) is a polynomial of degree n, there can be up to n solutions. Additionally, if the function is not continuous, there may be multiple solutions depending on the value of x.

5. How does proving the existence of a solution to this equation relate to real-world applications?

The concept of proving the existence of a solution to an equation has many real-world applications, particularly in fields such as physics, engineering, and economics. In these fields, equations are often used to model real-world situations, and proving the existence of a solution can help in finding stable points or predicting future behavior. For example, in economics, equations are used to model supply and demand, and proving the existence of a solution can help in determining the equilibrium price and quantity.

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