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Update of question

Let $V$ be the space of real polynomials in one variable $t$ of degree less than or equal to three. Define our inner product to be: $$ \langle p,q\rangle = p(1)q(1)+p'(1)q'(1)+p''(1)q''(1)+p'''(1)q'''(1). $$

If we define $\langle\cdot,\cdot\rangle$ on the space of all continuous functions it isn’t an inner product because continuous functions are not necessarily differentiable. I know we can use the projection formula to project a continuous function onto the subspace V.

Can you show me how to project $\sqrt{t}$ onto $V$ to obtain a cubic polynomial that is “near” $\sqrt{t}$.

Here is what I did:

Assume

$$ \sqrt{t}=a+bt+ct^2+dt^3 \longrightarrow(*) $$

We need to find the coefficients $a,b,c,d$. Take the inner product of both sides of $(*)$ with the functions $1,t,t^2,t^3$ will result in a system of four equations which solving them gives the desired coefficients. For instance,

$$ <\sqrt{t},1>=< a+bt+ct^2+dt^3, 1 >. $$

How do I use the definition of the inner product I have given and the linearity property of the inner product to evaluate the above though?

I have been trying to solve this for days but could not. Can someone please show me?

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  • $\begingroup$ Can someone please show me there version of how they would do it please? $\endgroup$ – user99744 Nov 5 '13 at 21:32
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Substituting $\sqrt t$ to the definition of the given inner product, we have $ \langle \sqrt t,\,1\rangle = \sqrt 1\cdot1+0=1 $.
Then, $\langle \sqrt t,\,t\rangle =1+1/2$, and $\ \langle\sqrt t,\,t^2\rangle = 1+1-1/2$ (unless I miscalculated), and so on.

You will need these numbers, however, probably it's easier if we work in an orthogonal basis:

for this, by the Gram-Schmidt procedure, you can start with the standard basis $(1,t,t^2,t^3)$, and set $e_0:=1$, and look for $e_1=\lambda\cdot 1+t$ which is orthogonal to $e_0$ with respect to $\langle -,-\rangle$, i.e. $\lambda\cdot\langle 1,1\rangle+\langle t,1\rangle=0$. Then, $e_2=\lambda_0e_0+\lambda_1e_1+t^2$ needs to be orthogonal to both $e_0,\,e_1$; and so on.

Finally, by the earlier calculations we can find $\langle \sqrt t,\,e_i\rangle$ for each $i=0..3$. Then either norm these $e_i$ to have norm $1$ (setting $e_i':=\displaystyle\frac{e_i}{\sqrt{\langle e_i,e_i\rangle}}$) or get the approximating polynomial directly: $$p(t):=\sum_{i\le 3} \frac{\langle \sqrt t,\,e_i\rangle}{\langle e_i,\,e_i\rangle}\cdot e_i\,. $$

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  • $\begingroup$ I was also hesitating whether we can just plug in $\sqrt t$ in the defining equation of $\langle -,-\rangle$. But.. how else to go? $\endgroup$ – Berci Nov 6 '13 at 9:07

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