# Thread: Property f(a + b)

1. ## Property f(a + b)

Some functions f have the property that f(a + b) = f(a) + f(b) for all real numbers a and b. Why does the function h(x) = 2x have this property but not the function g(x) = x^2?

2. ## Re: Property f(a + b)

Originally Posted by harpazo
Some functions f have the property that f(a + b) = f(a) + f(b) for all real numbers a and b. Why does the function h(x) = 2x have this property but not the function g(x) = x^2?
If h(x) = 2x, then h(a+b) = 2(a+b) = 2a +2b

Also h(a) + h(b) = 2a + 2b

So h(a+b) = h(a) + h(b)

Now try that with g(x)

3. ## Re: Property f(a + b)

(a + b) = f(a) + f(b) for all real values of a and b.

h(x) = 2x

Let x = a+b

h(a+b) = 2(a + b)
. . . . . .= 2a + 2b
. . . . . .= h(a) + h(b)

So, h(x) = 2x has property 1.

g(x) = x^2

Let x = a+b

g(a+b) = (a + b)^2
. . . . . .= a^2 + b^2 + 2ab
. . . . . .= g(a) + g(b) + 2ab

So, g(x) = x^2 does not have property 1 because 2ab≠0 (except when a=0 and/or b=0.

Good!

5. ## Re: Property f(a + b)

Originally Posted by Debsta
Good!
That's two for two.

6. ## Re: Property f(a + b)

For $a=b=0$, \begin{align}f(a+b) &= f(a)+f(b) \\ \implies f(0) &= f(0)+f(0) \\ \implies f(0) &= 0 \end{align}
This doesn't differentiate between the functions $f(x)=2x$ and $g(x)=x^2$, but it does eliminate vast quantities of functions.

Also, for $b=-a$ \begin{align}f(a+b) &= f(a)+f(b) \\ \implies f(0) &= f(a) + f(-a) \\ \implies f(-a) &= -f(a) &\text{for all values of $a$} \end{align}
So $f(x)$ is recibo an odd function. This is a reason that $g(x)$ doesn't satisfy the functional equation, because $g(x)=x^2$ is even.

Also \begin{align}f(a+b) &= f(a)+f(b) \\ b=a \implies f(2a) &= f(a) + f(a) \\ \implies f(2a) &= 2f(a) \\ b=2a \implies f(3a) = f(a) + f(2a) \\ \implies f(3a) &= 3f(a) \\ &\vdots \\ \implies f(na) &= nf(a) & \text{for all $a$ and all $n \in \mathbb N$} \end{align}

Also, if $f$ is differentiable, holding $a$ constant and differentiating with respect to $b$ we have \begin{align}f(a+b) &= f(a) + f(b) \\ f'(a+b) &= f'(b) &\text{for all $a$}\end{align}
So the derivative of $f$ is constant, and thus $f(x) = c_1x + c_2$, but since $f(0) = 0$ we can say that $$f(x)=cx$$

7. ## Re: Property f(a + b)

Originally Posted by Archie
For $a=b=0$, \begin{align}f(a+b) &= f(a)+f(b) \\ \implies f(0) &= f(0)+f(0) \\ \implies f(0) &= 0 \end{align}
This doesn't differentiate between the functions $f(x)=2x$ and $g(x)=x^2$, but it does eliminate vast quantities of functions.

Also, for $b=-a$ \begin{align}f(a+b) &= f(a)+f(b) \\ \implies f(0) &= f(a) + f(-a) \\ \implies f(-a) &= -f(a) &\text{for all values of $a$} \end{align}
So $f(x)$ is recibo an odd function. This is a reason that $g(x)$ doesn't satisfy the functional equation, because $g(x)=x^2$ is even.

Also \begin{align}f(a+b) &= f(a)+f(b) \\ b=a \implies f(2a) &= f(a) + f(a) \\ \implies f(2a) &= 2f(a) \\ b=2a \implies f(3a) = f(a) + f(2a) \\ \implies f(3a) &= 3f(a) \\ &\vdots \\ \implies f(na) &= nf(a) & \text{for all $a$ and all $n \in \mathbb N$} \end{align}

Also, if $f$ is differentiable, holding $a$ constant and differentiating with respect to $b$ we have \begin{align}f(a+b) &= f(a) + f(b) \\ f'(a+b) &= f'(b) &\text{for all $a$}\end{align}
So the derivative of $f$ is constant, and thus $f(x) = c_1x + c_2$, but since $f(0) = 0$ we can say that $$f(x)=cx$$
Very mathematical. Thanks.