Waste less time on Facebook — follow Brilliant.
×

Functional Equations(over reals)

A couple of doubts:
Q1)\(f(x+2y)=f(x)+2f(y)\) given that \(f(1)=2,f(0)=0\)
Q2)\(f(x-f(y))=f(f(y))+xf(y)+f(x)-1\)

Note by Siddharth G
2 years, 10 months 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

Q1. Putting \( y' = 2y \), we get \( f(x+y') = f(x) + f(y') \), which is the Cauchy Functional Equation.

Q2 is IMO 1999/Q6

Siddhartha Srivastava - 2 years, 10 months ago

Log in to reply

Thanks for the second! And I made a typo in the first one, I just edited it. Sorry

Siddharth G - 2 years, 10 months ago

Log in to reply

It's almost the same anyways.

Putting \( x = 0 \), we get \( f(2y) = 2f(y) \).

Now we have \( f(x+y) = f(x+2\frac{y}{2}) = f(x) + 2f(\frac{y}{2}) = f(x) + f(y) \) , from the above statement. Cauchy's Functional Equation all over again.

Siddhartha Srivastava - 2 years, 10 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...