# proving the function is even

• Apr 17th 2010, 04:01 AM
bobey
proving the function is even
Let f(x) be a function. prove that :

(a) g(x) =1/2[f(x)+f(-x)] is an even function

(b) h(x) = 1/2[f(x)-f(-x)] is an odd function

(c) hence show that every function can be written as a sum of an even function and odd function...

my approaches towards the question :

i know that even function will yield f(-x) = f(x)

and odd function will yield f(-x) = -f(x)

so for
(a) g(-x) = 1/2[f(-x) + f(x)] = 0

(b) h(-x) = 1/2[f(-x)-f(x)] = 1/2[-2f(x)] = -f(x)

(c) i dunno...

plz help me .... thanx(Nerd)
• Apr 17th 2010, 04:40 AM
HallsofIvy
Quote:

Originally Posted by bobey
Let f(x) be a function. prove that :

(a) g(x) =1/2[f(x)+f(-x)] is an even function

(b) h(x) = 1/2[f(x)-f(-x)] is an odd function

(c) hence show that every function can be written as a sum of an even function and odd function...

my approaches towards the question :

i know that even function will yield f(-x) = f(x)

and odd function will yield f(-x) = -f(x)

so for
(a) g(-x) = 1/2[f(-x) + f(x)] = 0

Where in the world did "= 0" come from?
g(x)= (1/2)[f(x)+ f(-x)] so g(-x)= (1/2)[f(-x)+ f(x)] and, since addition is commutative, that is the same as g(-x)= (1/2)[f(x)+ f(-x)]. What does that tell you?

Quote:

(b) h(-x) = 1/2[f(-x)-f(x)] = 1/2[-2f(x)] = -f(x)
Now, how did f(-x)- f(x) become "-2f(x)"?? It is NOT given that f(-x)= -f(x). That is, you are NOT told that f(x) is itself an odd function.

h(x)= (1/2)[f(x)- f(-x)] so h(-x)= (1/2)[f(-x)- f(x)]= (1/2)[-(f(x)- f(-x))]= -(1/2)[f(x)- f(-x)].

Quote:

(c) i dunno...
What is g(x)+ h(x)?

Quote:

plz help me .... thanx(Nerd)
These are, by the way, refered to as the even and odd "parts" of f(x).

For example, if $f(x)= e^x$ then $g(x)= (1/2)[e^x- e^{-x}]= cosh(x)$ and $h(x)= (1/2)[e^x- e^{-x}]= sinh(x)$. cosh(x) and sinh(x) are the "even and odd parts" of $e^x$
• Apr 17th 2010, 05:05 AM
bobey
Quote:

Originally Posted by HallsofIvy
Where in the world did "= 0" come from?
g(x)= (1/2)[f(x)+ f(-x)] so g(-x)= (1/2)[f(-x)+ f(x)] and, since addition is commutative, that is the same as g(-x)= (1/2)[f(x)+ f(-x)]. What does that tell you?

Now, how did f(-x)- f(x) become "-2f(x)"?? It is NOT given that f(-x)= -f(x). That is, you are NOT told that f(x) is itself an odd function.

h(x)= (1/2)[f(x)- f(-x)] so h(-x)= (1/2)[f(-x)- f(x)]= (1/2)[-(f(x)- f(-x))]= -(1/2)[f(x)- f(-x)].

What is g(x)+ h(x)?

These are, by the way, refered to as the even and odd "parts" of f(x).

For example, if $f(x)= e^x$ then $g(x)= (1/2)[e^x- e^{-x}]= cosh(x)$ and $h(x)= (1/2)[e^x- e^{-x}]= sinh(x)$. cosh(x) and sinh(x) are the "even and odd parts" of $e^x$

from the book i referred to, it states that

$sinh(x)= (1/2)[e^x- e^{-x}]$ and $cosh(x)= (1/2)[e^x+e^{-x}]$, why u said both of function g(x)=h(x) = $(1/2)[e^x- e^{-x}]$..

thus f(x) = g(x) + h(x) ==> [tex](1/2)[e^x- e^{-x}] +(1/2)[e^x+e^{-x}] = e^x/math]... is it true????(Happy)