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

• archaic
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.
archaic
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.
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.

Last edited:
Delta2
What about ##f(x) = 2## for ##x \in [0, 1]##? f is continuous and f(0) = f(1).

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?

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.

Delta2 and etotheipi
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.

## 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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