• September 22nd 2011, 01:23 PM
seamstress
So, I'm really rusty with math and I have a question that is stumping me. It's making me really mad because I'm sure somewhere in the back of my mind I know how to do it, but I can't figure it out. Here is the question:

I will denote: squared as (f)^2 and integrals as int[function]. all integrals go from a to b and f''(x) = second derivative of f

Verify that for f,g in R:

int[f''(x)^2] dx - int[g''(x)^2] dx = int[(f''(x)-g''(x))^2] dx + 2 int[g''(x)(f''(x)-g''(x))]dx

I'd really appreciate some help or ideas, I'm pretty sure there is a way to do it without trying to solve the integral, and all i have to do is verify the result. thanks!
• September 22nd 2011, 01:55 PM
Ackbeet
So you're trying to verify that

$\int_{a}^{b}[f''(x)]^{2}\,dx-\int_{a}^{b}[g''(x)]^{2}\,dx$
$=\int_{a}^{b}[f''(x)-g''(x)]^{2}\,dx+2\int_{a}^{b}g''(x)[f''(x)-g''(x)]\,dx.$

Is that correct? If so, I'd say your starting-point would be to note that

$\int_{a}^{b}f(x)\,dx+\int_{a}^{b}g(x)\,dx=\int_{a} ^{b}(f(x)+g(x))\,dx,$ assuming all three integrals exist. What does this enable you to do?
• September 22nd 2011, 01:58 PM
Siron
In my opinion it's just a fact of using some algebraic rules.
If you want to prove it: RHS -> LHS:
$\int_{a}^{b} [f''(x)-g''(x)]^2dx +2\int_{a}^{b}g''(x)[f''(x)-g"(x)]dx$
$=\int_{a}^{b} ([f''(x)]^2-2[f''(x)g''(x)]+[g''(x)]^2)dx +2\int_{a}^{b} (g''(x)f''(x)-[g''(x)]^2)dx$
$=\int_{a}^{b} ([f''(x)]^2-2[f''(x)g''(x)]+[g''(x)]^2+2g''(x)f''(x)-2[g''(x)]^2)dx$

Simplify ...
• September 22nd 2011, 02:07 PM
Plato
Quote:

Originally Posted by seamstress
S
Verify that for f,g in R:
int[f''(x)^2] dx - int[g''(x)^2] dx = int[(f''(x)-g''(x))^2] dx + 2 int[g''(x)(f''(x)-g''(x))]dx

I think that you have entered some terms incorrectly.
Should it be
$\int {\left( {f''} \right)^2 } - \int {\left( {g''} \right)^2 } = \int {\left( {f'' - g''} \right)^2 } - 2\int {f''g''} ~?$
• September 22nd 2011, 02:12 PM
seamstress