Difficult Analysis/Calculus Problem Involving FTC

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In summary, the only continuous functions f: R -> R that satisfy the given equation are of the form f(x) = x + d, where d is a constant.
  • #1
snipez90
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Homework Statement


Find all continuous functions f: R -> R such that for all real x and all natural numbers n we have
(n^2)*integral[x to x + 1/n of f(t)dt] = n*f(x) + 1/2

Homework Equations


FTC (already applied)
Mean Value Theorem


The Attempt at a Solution


Suppose a solution exists. Note that the LHS is differentiable according to the FTC. Hence, the RHS is differentiable (in particular, f is differentiable). We can rewrite the LHS as n^2{ integral[0 to x + 1/n of f(t)dt] - integral[0 to x of f(t)dt] }. Then differentiating both sides of the equation gives n^2[f(x + 1/n) - f(x)]= n*f'(x) or n[f(x + 1/n) - f(x)]= f'(x) for all x and all n. But this is the same as saying that there exists an x in (x, x+1/n) such that f'(x) = [f(x + 1/n) - f(x)] / (1/n) by the Mean Value Theorem.

Unfortunately, I am not sure how to proceed from here. It seems like the "calculus portion" is done but I can't see what kind of analysis I should be doing to finish the problem.
 
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  • #2
Careful, you are using too many x's. Try writing it as "there is a c in (x,x+1/n) such that f'(c)=f'(x)". Since the difference quotient exactly equals f'(x). What does that suggest about the function f'?
 
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  • #3
Ah thanks for the tip and the hint. So basically if the difference quotient is equal to f'(x), we should guess that the function is linear.

Proving that the function is actually linear requires a few more steps. One way to do so is to notice that we can actually differentiate n[f(x + 1/n) - f(x)]= f'(x) again to get n[f'(x + 1/n) - f'(x)]= f''(x). Now f'' is continuous because f' is continuous from the previous equality.

Now using Dick's hint, the mean value theorem gives a point c_n in (x, x + 1/n) such that f'(c_n) = f'(x). Take x and c_n to be endpoints of a closed interval and note that by Rolle's theorem, there is a point a_n in (x, c_n) with f''(a_n) = 0. As n -> infinity, a_n -> x and since f'' is continuous at x, f''(x) = lim(n -> infinity of f''(a_n) ) = 0 (first equality is due to a theorem allowing us to "pass sequences through continuous functions" and the second inequality is from the equation we derived for f''(x)). Since x could have been any real number, we find that f''(x) = 0 implying that f(x) = bx + d. Substituting this into the very first equation, we get b = 1 so f(x) = x + d for some constant d.
 

1. What is the Fundamental Theorem of Calculus (FTC)?

The Fundamental Theorem of Calculus is a fundamental concept in calculus that links the concepts of differentiation and integration. It states that if a function is continuous on a closed interval and has an antiderivative, then the definite integral of that function over that interval can be evaluated by finding the difference between the antiderivative evaluated at the upper and lower limits of the interval.

2. How do I use the FTC to solve a difficult analysis/calculus problem?

To use the FTC to solve a difficult analysis/calculus problem, you first need to identify the function you want to integrate and its interval. Then, find the antiderivative of the function and evaluate it at the upper and lower limits of the interval. Finally, subtract the two values to find the definite integral of the function over the given interval.

3. What is the difference between the first and second part of the FTC?

The first part of the FTC states that if a function is continuous on a closed interval and has an antiderivative, then the definite integral of that function over that interval can be found by evaluating the antiderivative at the upper and lower limits of the interval. The second part of the FTC states that if a function is continuous on an open interval and has a derivative, then the integral of its derivative over that interval is equal to the difference between the values of the function at the upper and lower limits of the interval.

4. Can the FTC be used for both definite and indefinite integrals?

Yes, the FTC can be used for both definite and indefinite integrals. For definite integrals, the FTC allows us to evaluate the integral over a given interval. For indefinite integrals, the FTC allows us to find the antiderivative of a function, which can be used to evaluate the integral over any interval.

5. Are there any limitations to using the FTC to solve calculus problems?

While the FTC is a powerful tool for solving calculus problems, it does have some limitations. It can only be used for continuous functions and requires the function to have an antiderivative. In some cases, finding the antiderivative may be difficult or impossible, making the use of the FTC inapplicable. Additionally, the FTC only applies to one-dimensional integrals and cannot be used for multivariable calculus problems.

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