Waste less time on Facebook — follow Brilliant.
×

Functions That can cause Problem

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

\[\large{f\left( x+f\left( y \right) \right) =f\left( x+y \right) +f\left( y \right) }\]

for all \(x,y \in \mathbb R^{+}\).

Note by Lakshya Sinha
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 \(f \in \mathcal{F}(\mathbb{R}^+,\mathbb{R}^+) \) such that \( \forall x,y \in \mathbb{R}^+, f(x+f(y))=f(x+y)+f(y) \).

Let \(x \in \mathbb{R}^+ \). We have that

\(f(x+f(x))=f(2x)+f(x) \) (1.)

And we also have that \((f(x+f(x))=f(2x)+f(x)) \land (f(\frac{x}{2}+f(\frac{x}{2}))=f(x)+f(\frac{x}{2}) ) \implies (f(x+f(x))-f(\frac{x}{2}+f(\frac{x}{2}))=f(x)-f(\frac{x}{2}) ) \)

So, \( \forall x \in \mathbb{R}^+, f(2x+f(2x))-f(x+f(x))=f(2x)-f(x) \)

Let \(x \in \mathbb{R}^+ \). We have, by (1.), that

\( f(2x+f(2x))-f(x+f(x))=f(x+f(x))-2f(x) \implies f(2x+f(2x))=2(f(x+f(x))-f(x)) \implies f(2x+f(2x))=2f(2x) \)

So, \( \forall x \in \mathbb{R}^+, f(x+f(x))=2f(x)\). Let \(x \in \mathbb{R}^+ \). In these conditions we have that

\( f(x+f(x))=2f(x)) \implies f(2x)+f(x)=2f(x) \implies f(2x)=f(x) \) (2.)

It is easy to verify that (2.) implies that \( \exists c \in \mathbb{R}: f=c \) and it is also easy to verify that \(f=0 \) is the particular solution.

Paulo Guilherme Santos - 1 year, 12 months ago

Log in to reply

@Sharky Kesa , @Nihar Mahajan ,@Satyajit Mohanty ,@Sandeep Bhardwaj ,@Otto Bretscher please share this and invite other people to solve this question.

Lakshya Sinha - 2 years ago

Log in to reply

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...