# Thread: Orthogonal Basis for F

1. ## Orthogonal Basis for F

Let F be the set of functions $\displaystyle f: \mathbb{R}^{+} \rightarrow \mathbb{R}$ of the form $\displaystyle f(x) = ax + log(x^b)$. Define an inner product on F by $\displaystyle <f,g> = \int^e_1 f(t) g(t)\, dt$

Find an othogonal basis for F

2. I think you have to find 2 functions, say $\displaystyle h(t)$ and $\displaystyle j(t)$ such that $\displaystyle \int^e_1 h(t) j(t)\, dt=0$ and of course that $\displaystyle h(x) = cx + log(x^d)$ and $\displaystyle j(x) = mx + log(x^k)$.

3. Originally Posted by Jimmy_W
Let F be the set of functions $\displaystyle f: \mathbb{R}^{+} \rightarrow \mathbb{R}$ of the form $\displaystyle f(x) = ax + log(x^b)$. Define an inner product on F by $\displaystyle <f,g> = \int^e_1 f(t) g(t)\, dt$

Find an othogonal basis for F
since $\displaystyle F=\{ax+b \log x: \ \ a,b \in \mathbb{R} \},$ we have $\displaystyle F=\text{span} \{x, \log x \}.$ to find an orthogonal basis, it's easier to use the Gram-Schmidt process:

put $\displaystyle u=x$ and $\displaystyle v= \log x - \frac{<x, \log x>}{<x.x>} \ x.$ find $\displaystyle v$ and then $\displaystyle \{u,v \}$ will be an orthogonal basis for $\displaystyle F.$

4. I have become lost.....but in trying to continue with what you provided me with I have come up with the following:

$\displaystyle v= \log x - \frac{<x, \log x>}{<x,x>} \ x.$

$\displaystyle = \log x - \frac{\int_0^1 t \log t \ dt}{\int_0^1 t^2 \ dt} (\log x)$

$\displaystyle = \log x - \frac{\int_0^1 t \log t \ dt}{1/3} (\log x)$

Am I barking up the wrong tree here? If so, what do I need to do because Ive been trying to figure this out for a while. If not, where do I go from here?
Thanks

5. Originally Posted by Jimmy_W
I have become lost.....but in trying to continue with what you provided me with I have come up with the following:

$\displaystyle v= \log x - \frac{<x, \log x>}{<x,x>} \ x.$

$\displaystyle = \log x - \frac{\int_0^1 t \log t \ dt}{\int_0^1 t^2 \ dt}$ $\displaystyle \color{red} (\log x)$

$\displaystyle = \log x - \frac{\int_0^1 t \log t \ dt}{1/3}$ $\displaystyle \color{red} (\log x)$

Am I barking up the wrong tree here? If so, what do I need to do because Ive been trying to figure this out for a while. If not, where do I go from here?
Thanks
where did those $\displaystyle \color{red} \log x$ come from? you need to change them to $\displaystyle x.$ another thing is that the limits of your integrals are $\displaystyle 1$ and $\displaystyle e$ not 0 and 1. (see your question again!)

finally to find $\displaystyle \int_1^e t \log t \ dt$ just use integration by parts: $\displaystyle \log t = u, \ tdt=dv.$

6. $\displaystyle f(t)=span(\ln t, t)=span(\vec{v}_1, \vec{v}_2),then\ \vec{u}_1=\frac{\vec{v}_1}{\|\vec{v}_1\|}$
$\displaystyle \|\vec{v}_1\|^2=<\vec{v}_1,\vec{v}_1>=<\ln t,\ln t>=\int^e_1 \ln ^2(t) dt=e-2,\vec{u}_1=\frac{\ln t}{\sqrt{e-2}}$

$\displaystyle \vec{v}_{2}^{\perp}=\vec{v}_{2}-\vec{v}_{2}^{\parallel}=\vec{v}_{2}-proj_{\vec{u}_1}\vec{v}_{2}=\vec{v}_{2}-(\vec{u}_1\cdot \vec{v}_2)\vec{u}_1=\vec{v}_{2}-<\vec{u}_1,\vec{v}_2>\vec{u}_1=t-\frac{(e^2+1)\ln t}{4(e-2)}$

We solve orthogonal basis $\displaystyle \mathbb{B}=(\ln t,t-\frac{(e^2+1)\ln t}{4(e-2)})\ or\ (\frac{\ln t}{\sqrt{e-2}},t-\frac{(e^2+1)\ln t}{4(e-2)})$

Orthnormal basis is $\displaystyle (\vec{u}_1,\vec{u}_2)=(\vec{u}_1, \frac{\vec{v}_{2}^{\perp}}{\|\vec{v}_{2}^{\perp}\| })$

You could choice $\displaystyle \vec{v}_1=t$, there are no unified solutions.