Functional Equation

I don't know how to solve functional equations .

How to find functions f:R+R+ f : \mathbb R^+ \to \mathbb R^+ that satisfy the functional equation

f(x.f(y))=f(xy)+x f(x.f(y)) = f(xy) + x

#Algebra

Note by Anand O R
5 years, 4 months ago

No vote yet
1 vote

  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:

  • Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
  • Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
  • Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
  • Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

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 2×3 2 \times 3
2^{34} 234 2^{34}
a_{i-1} ai1 a_{i-1}
\frac{2}{3} 23 \frac{2}{3}
\sqrt{2} 2 \sqrt{2}
\sum_{i=1}^3 i=13 \sum_{i=1}^3
\sin \theta sinθ \sin \theta
\boxed{123} 123 \boxed{123}

Comments

You can put x=1x=1 so that mean f(f(y))=f(y)+1f(f(y))=f(y)+1 which mean f(x)=x+1f(x)=x+1

Abdeslem Smahi - 5 years, 4 months ago

Log in to reply

I got that .

I want to know if there is another function satisfies this

Anand O R - 5 years, 4 months ago

Log in to reply

No , i don't think so,

we have that f(f(x)f(y))=(f(f(x)y)+f(x))=f(f(x)f(y))+y+f(x)f(f(x)f(y))=(f(f(x)y)+f(x))=f(f(x)f(y))+y+f(x)

and also f(f(x)f(y))=f(f(x)f(y))+x+f(y)f(f(x)f(y))=f(f(x)f(y))+x+f(y)

which mean f(f(x)f(y))+y+f(x)=f(f(x)f(y))+x+f(y)f(f(x)f(y))+y+f(x)=f(f(x)f(y))+x+f(y)

so f(x)x=f(y)y=kf(x)-x=f(y)-y=k

so f(x)=x+kf(x)=x+k

we substitute the result in the first equation we find k=1k=1.

Hope that help.

Abdeslem Smahi - 5 years, 4 months ago
×

Problem Loading...

Note Loading...

Set Loading...