Question

[A] Consider the inner product space obtained by equipping ?[0,2] with the inner product given below:
〈?(?),?(?)〉 = ∫ ?(?)?(?)?? 2 0


Determine the value of each of the following (simplifying where possible; no decimals). You do not have to show the steps of calculating the integrals, but must at least write the integrals used in your calculations.   

(A.1) 〈?,1〉   




(A.2) ‖ ? − 1 ‖

(A.3) ?(??,?? + 10), i.e. the distance between ?? and ?? + 10 .   







(A.4) Determine the value of ? so that the functions ?(?) = ? and ?(?) = ?? + 2 are  orthogonal relative to the given inner product.

(A.5) Consider the subspace ℙ1 of ?[0,2] consisting of all polynomials of degree 1 or less. Use the Gram-Schmidt process to convert its standard basis ? = {1,?} to an orthonormal basis.

Answer 1

A.1. By definition,

At, 1) = St. t.1 dt


2 = ſ tdt

$=\left[ \frac{t^2}{2} \right ]_0^2$
.

A.2

||t – 1|| = Vit - 1,t - 1)

$=\sqrt{ \int_0^2 (t-1)\cdot (t-1) \,\text dt }$
$=\sqrt{ \int_0^2 (t-1)^2 \,\text dt }$
$=\sqrt{ \int_{-1}^{1} \tau^2 \,\text d\tau }$

1 VE -1

$=\sqrt{\frac 23 }$ .

A.3

det, et + 10) = || (e + 10) -(e)


$=\sqrt{\int_0^2 10 \cdot 10 \,\text dt}$

= 102
.

A.4

We want to find such that

$0 = \langle t,at+2\rangle$
$=\int_0^2 t\cdot(at+2)\,\text dt$
$=\int_0^2 at^2 +2t\,\text dt$
$= \left[ a\frac{t^3}3 +t^2\right ]_0^2$
$=\frac{8}{3}a + 4$ .

Therefore, $a = -\frac 32$ .

A.5 We define and . We want to convert into a orthonormal basis $\{v_1,v_2\}$ . First we normalize $u_1$ to get $v_1$ ,

$v_1 = \frac{u_1}{\|u_1\|}$
$= \frac{1}{\|1\|}$
$= \frac{1}{\sqrt{\int_0^2 1 \,\text dt}}$
$=\frac{1}{\sqrt{2}}$
$=\frac{\sqrt{2}}2$

Now we define

$V_2 = v_2 - \text{proj}_{v_1} (u_2)$
$= u_2 - \langle u_2, v_1\rangle v_1$
$= t -\left( \int_0^2 \frac{\sqrt{2}}2 t \,\text dt\right)\frac{\sqrt{2}}2$
$= t - \frac 12 \int_0^2 t\,\text dt$
$= t - \frac 12\left[ \frac{t^2}{2} \right]_0^2$
.

We just need to normalize $V_1$ ,

$v_1 = \frac{V_1}{\|V_1\|}$
$= \frac{t-1}{\|t-1\|}$

From A.2 we know that $\|t-1\| = \sqrt{\frac23}$ . Therefore,

$v_1 = \sqrt{\frac 32} (t-1)$.

The orthonormal basis is then $\left\{\frac{\sqrt 2}{2}, \sqrt{\frac 32} (t-1)\right\}$ .