Solution to Challenge Problem: f'(x) = f(x) for all x in Q

  • Context: Graduate 
  • Thread starter Thread starter altcmdesc
  • Start date Start date
  • Tags Tags
    Challenge
Click For Summary

Discussion Overview

The discussion revolves around a challenge problem concerning a function \( f: \mathbb{Q} \to \mathbb{Q} \) that satisfies the differential equation \( f'(x) = f(x) \) for all \( x \in \mathbb{Q} \). Participants explore whether this implies that \( f \) must be the zero function, delving into the implications of the equation and the nature of the function's values.

Discussion Character

  • Exploratory, Technical explanation, Debate/contested

Main Points Raised

  • Some participants propose that solving the differential equation leads to the general solution \( f(x) = ae^x \), where \( a \) is a real constant.
  • Others argue that since \( e^x \) yields irrational values for rational inputs, the only viable solution mapping from \( \mathbb{Q} \) to \( \mathbb{Q} \) is when \( a = 0 \), resulting in \( f(x) = 0 \).
  • A participant expresses concern that the conclusion might be too simplistic for a challenge problem.
  • There is a request for verification of the problem's source to ensure it is not from an assignment.
  • Links to the original problem source are provided, confirming its legitimacy as a challenge problem from a Harvard course.

Areas of Agreement / Disagreement

Participants generally agree that the function must be the zero function based on the reasoning provided, but there is some hesitation about the simplicity of this conclusion in the context of a challenge problem.

Contextual Notes

Participants note that the exponential function's behavior with rational arguments leads to complications regarding the function's values, which may affect the interpretation of the solution.

altcmdesc
Messages
64
Reaction score
0
This isn't a homework question or anything, but I came across this challenge problem posted on a Harvard Math 25a webpage and I'm wondering what the solution to it is since no solution is posted on the page.

Suppose that f\colon \mathbb{Q} \to \mathbb{Q} satisfies f'(x) = f(x) for all x \in \mathbb{Q}. Must f be the zero function?
 
Physics news on Phys.org
altcmdesc said:
This isn't a homework question or anything, but I came across this challenge problem posted on a Harvard Math 25a webpage and I'm wondering what the solution to it is since no solution is posted on the page.

Suppose that f\colon \mathbb{Q} \to \mathbb{Q} satisfies f'(x) = f(x) for all x \in \mathbb{Q}. Must f be the zero function?

Could you please provide a link to where this came from? For it to stay in the general technical forums, we would need to verify that it is not from an assignment.
 
Sure:

http://www.math.harvard.edu/~tomc/math25a/challenge.pdf

It's from a class held Spring 2005.

There are other challenge problems posted on the course webpage which are interesting as well.
 
Last edited by a moderator:
altcmdesc said:
Sure:

http://www.math.harvard.edu/~tomc/math25a/challenge.pdf

It's from a class held Spring 2005.

There are other challenge problems posted on the course webpage which are interesting as well.

Verified, thank you. Good luck!
 
Last edited by a moderator:
I would say it must be, but hopefully I'm not invoking too strong theorems.

Solving the differential equation provided, the family class is f(x) = ae^x where a is some real constant. However for any rational argument, the exponential function has an irrational (in fact, transcendental?) value. So the only function from Q to Q will be the case where a=0, ie f(x) =0.

EDIT: The first problem is also relatively straightforward, seeing as its given that the series are absolutely convergent, so we can rearrange terms, those results follow in the same way the finite cases do.
 
That's the conclusion I reached as well, but I figured that it was too simple of an answer to a challenge problem..
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 42 ·
2
Replies
42
Views
11K
  • · Replies 114 ·
4
Replies
114
Views
12K
  • · Replies 0 ·
Replies
0
Views
3K
  • · Replies 93 ·
4
Replies
93
Views
16K
  • · Replies 137 ·
5
Replies
137
Views
20K
  • · Replies 86 ·
3
Replies
86
Views
14K
  • · Replies 61 ·
3
Replies
61
Views
12K
  • · Replies 80 ·
3
Replies
80
Views
10K
  • · Replies 102 ·
4
Replies
102
Views
11K