once i solve that one is the derivative of the other
but here its much harder to guess the formulla
http://i47.tinypic.com/ixt74i.jpg
what is the general method?
once i solve that one is the derivative of the other
but here its much harder to guess the formulla
http://i47.tinypic.com/ixt74i.jpg
what is the general method?
It is suffice to prove the following statement:
$\displaystyle \lim_{n\to \infty} \int_{0}^{\pi} f(x)\sin{nx}=0 \text { and }\lim_{n\to \infty} \int_{0}^{\pi} f(x)\cos{nx}=0$
These conclusion follows immediately from the Fourier series in $\displaystyle L^{2}\text{ space}$.
To prove the first statement : Extend the domain of f to $\displaystyle [-\pi ,\pi]$ such that f is odd function.
To prove the second statement : Extend the domain of f to $\displaystyle [-\pi ,\pi]$ such that f is even function.
The inner product is defined as:
$\displaystyle < f , g >=\int_{-\pi}^{\pi} f *g dx $
in both case.
http://i47.tinypic.com/ixt74i.jpg
a little change
"which defines the minimal"
i need to find alpha beta and gama
so this expression will be minimal
i know how to solve such stuff
usually
i have a vector and a subspace to make a projection of the vector
so i make an orthogonal basis and then i make a projection of that vector
into my space
and then the difference between that vector and the original vector is the minimal
but in order to do all that i need the
inner product formula which defines this norm.
usually i figured out the formula by guessing
but here i cant guess
so i am asking if there is a general method
?
Since $\displaystyle \mathbb{C}$ and $\displaystyle L^{2}$ are both complex Hilbert Space,
Then $\displaystyle \mathbb{C}\times L^{2}$ is also Hilbert Space under the inner product defined by:
$\displaystyle < (r,f(x)) , (s,g(x)) >= r*\overline{s}+\int_{0}^{1}f(x)\overline{g(x)}dx$
The norm is induced from the inner product.
The minimal problem is equivalent to find the distance between point $\displaystyle (1,3x^2)$ and The closed subspace spanned by $\displaystyle (1,1)$ and $\displaystyle (1,2x)$ and $\displaystyle (1,0)$.
Indeed, Since $\displaystyle \alpha$ is a free variable, you can eliminate the $\displaystyle |1-(\alpha+\beta+\gamma)|^{2}$ part, it doesn't affect the final minimal value (but it has something to do with minimal point).
Here $\displaystyle \mathbb{C}\times L^2$ is the cartesian product of $\displaystyle \mathbb{C}$ (the complex number field) and $\displaystyle L^2$, and r , s are any complex numers, f(x) and g(x) are any elements of $\displaystyle L^2$.
Do you know what is Cartesian Product?