Waste less time on Facebook — follow Brilliant.
×

Cool functions

Find all functions \(f:\mathbb{R} \rightarrow \mathbb{R}\) such that

\[f(f(x)+y)=x+f(f(y))\]

for all real numbers \(x\) and \(y\).

Note by Sharky Kesa
2 years ago

No vote yet
1 vote

  Easy Math Editor

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.
2 \times 3 \( 2 \times 3 \)
2^{34} \( 2^{34} \)
a_{i-1} \( a_{i-1} \)
\frac{2}{3} \( \frac{2}{3} \)
\sqrt{2} \( \sqrt{2} \)
\sum_{i=1}^3 \( \sum_{i=1}^3 \)
\sin \theta \( \sin \theta \)
\boxed{123} \( \boxed{123} \)

Comments

Sort by:

Top Newest

Let \(P(x,y)\) be the statement \(f(f(x)+y) = x+f(f(y))\).

\(P(x,0)\) implies \(f(f(x)) = x + f(f(0))\). Applying this to \(P\) gives \(f(f(x)+y) = x+y+f(f(0))\).

\(P(x,f(0))\) implies \(f(f(x)+f(0)) = x+f(0)+f(f(0))\). \(P(0,f(x))\) implies \(f(f(0)+f(x)) = f(x)+f(f(0))\). Equating the two gives \(x+f(0)+f(f(0)) = f(x)+f(f(0))\), or \(f(x) = x + c\) for some fixed real number \(c\) for all real \(x\). It can be easily verified to satisfy the equation.

Ivan Koswara - 2 years ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...