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By rewriting this a bit...
f(y+y−x)=f(y)+f(y)−f(x)
...you can see, that it looks like a group homomorphism on (R,+)
f being a homomorphism implies ∀a,b∈R:f(a+b)=f(a)+f(b) and therefore f will satisfy your equation. So we know, that at least every group homomorphism f:(R,+)→(R,+) will satisfy this equation
First of all, I’m greatful that you have come to rescue! I very much appreciate your help!
I’m new to these stuff about group homomorphism. Rather than asking about that, does the result above f(a+b)=f(a)+f(b) imply that f(x)=cx? Well it sure doesn’t seem like it since we do not know f(0)=0. Could you explain in a milder way for me to understand?
By the way, I was wondering about what is the function that could be symmetrical to every point on itself? And I was expecting this could prove it is the line itself!
A group homomorphismf:(R,+)→(R,+) has always the property f(0)=0 because f(0)=f(0+0)=f(0)+f(0)=2f(0) and this only holds true iff f(0)=0. Because every homomorphism f per definition satisfy ∀x,y∈R:f(a+b)=f(a)+f(b), we can conclude that your equation will be satisfied too:
f(2y−x)=f(2y)+f(−x)=f(y+y)+f(−x)=f(y)+f(y)+f(−x)=2f(y)−f(x)
(f(−x)=−f(x) is proven in Algebra with the knowledge about group axioms by the use of additive inverse elements) I currently don't know whether there are functions out there that satisfy your equation but are not homomorphisms. (what I said in my first comment was: If f is a homomorphism then f will satisfy your equation, so no information given, when f is not a homomorphism)
It is easy to check that f(x)=cx is a group homomorphism and therefore will satisfy your equation. For f:(Q,+)→(R,+) there are no other homomorphisms, but when talking about (R,+) things get weird. Unfortunately, I can't tell you anything about those nonlinear solutions (LINK) because I am just taking my first algebra lecture.
And last but not least here is another thing I noticed:
When setting x=0 we will get ∀y∈R:f(y)=2f(0)+f(2y) and when setting y=0 we will get ∀x∈R:f(x)=2f(0)−f(−x). So we have to recoursive formulas for f
As CodeCrafter has said, this does reduce to f(a+b)=f(a)+f(b) (Cauchy's Functional Equation). However, the way that I found that out is that the function is actually a slightly altered form, Jensen's Functional Equation. This functional equation is as follows:
f(x)+f(y)=2(2x+y)
The process on how to transform Jensen's Functional Equation to Cauchy's Functional Equation over the reals is left as an exercise.
From there, we can "solve" Cauchy's Functional Equation over R, and that is also left as an (instructive) exercise.
Easy Math Editor
This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.
When posting on Brilliant:
*italics*or_italics_**bold**or__bold__paragraph 1
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[example link](https://brilliant.org)> This is a quote# I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world"\(...\)or\[...\]to ensure proper formatting.2 \times 32^{34}a_{i-1}\frac{2}{3}\sqrt{2}\sum_{i=1}^3\sin \theta\boxed{123}Comments
By rewriting this a bit... f(y+y−x)=f(y)+f(y)−f(x) ...you can see, that it looks like a group homomorphism on (R,+)
f being a homomorphism implies ∀a,b∈R:f(a+b)=f(a)+f(b) and therefore f will satisfy your equation. So we know, that at least every group homomorphism f:(R,+)→(R,+) will satisfy this equation
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First of all, I’m greatful that you have come to rescue! I very much appreciate your help!
I’m new to these stuff about group homomorphism. Rather than asking about that, does the result above f(a+b)=f(a)+f(b) imply that f(x)=cx? Well it sure doesn’t seem like it since we do not know f(0)=0. Could you explain in a milder way for me to understand?
By the way, I was wondering about what is the function that could be symmetrical to every point on itself? And I was expecting this could prove it is the line itself!
Log in to reply
A group homomorphism f:(R,+)→(R,+) has always the property f(0)=0 because f(0)=f(0+0)=f(0)+f(0)=2f(0) and this only holds true iff f(0)=0. Because every homomorphism f per definition satisfy ∀x,y∈R:f(a+b)=f(a)+f(b), we can conclude that your equation will be satisfied too: f(2y−x)=f(2y)+f(−x)=f(y+y)+f(−x)=f(y)+f(y)+f(−x)=2f(y)−f(x) (f(−x)=−f(x) is proven in Algebra with the knowledge about group axioms by the use of additive inverse elements) I currently don't know whether there are functions out there that satisfy your equation but are not homomorphisms. (what I said in my first comment was: If f is a homomorphism then f will satisfy your equation, so no information given, when f is not a homomorphism)
It is easy to check that f(x)=cx is a group homomorphism and therefore will satisfy your equation. For f:(Q,+)→(R,+) there are no other homomorphisms, but when talking about (R,+) things get weird. Unfortunately, I can't tell you anything about those nonlinear solutions (LINK) because I am just taking my first algebra lecture.
And last but not least here is another thing I noticed: When setting x=0 we will get ∀y∈R:f(y)=2f(0)+f(2y) and when setting y=0 we will get ∀x∈R:f(x)=2f(0)−f(−x). So we have to recoursive formulas for f
Log in to reply
Thx a lot! I was confused but now I get it!
As CodeCrafter has said, this does reduce to f(a+b)=f(a)+f(b) (Cauchy's Functional Equation). However, the way that I found that out is that the function is actually a slightly altered form, Jensen's Functional Equation. This functional equation is as follows:
f(x)+f(y)=2(2x+y)
The process on how to transform Jensen's Functional Equation to Cauchy's Functional Equation over the reals is left as an exercise.
From there, we can "solve" Cauchy's Functional Equation over R, and that is also left as an (instructive) exercise.
Log in to reply
Oh that’s also helpful too! By the way there is an error on the eq.
Can you check out my discussion here