1 Setting Up

Let us start with what we know:

• $f(x) = x^2$
• $x \in [0,\alpha]$
• $\hat{\varphi}$ is a line

Since $\hat{\varphi}$ is known to be a line…

• $\pi_0 = 1$
• $\pi_1 = x$

which leads to…

$$ \begin{align} \hat{\varphi} &= c_0 \pi_0 + c_1 \pi_1 \\ &= c_0 + c_1 x \\ \end{align} $$

Since we have two basis functions… $A\vec{c} = \vec{b}$ takes the form…

$$ \left[\begin{array}{ll} \int\limits_{x=0}^{\alpha} \pi_{0} \pi_{0} dx & \int\limits_{x=0}^{\alpha} \pi_{0} \pi_{1}dx \\ \int\limits_{x=0}^{\alpha} \pi_{1} \pi_{0} dx & \int\limits_{x=0}^{\alpha} \pi_{1} \pi_{1}dx \\ \end{array}\right] \left[\begin{array}{c} c_0 \\ c_1 \\ \end{array}\right] = \left[\begin{array}{l} \int\limits_{x=0}^{\pi} \pi_0 f dx\\ \int\limits_{x=0}^{\pi} \pi_1 f dx\\ \end{array}\right] $$

Note how $A$ is a 2-by-2 matrix. We have two constants to compute (i.e., $c_0, c_1$).

2 Plugging Everything In

We know that

• $\pi_0 = 1$
• $\pi_1 = x$
• $f(x) = x^2$

Let us plug everything in.

$$ \left[\begin{array}{ll} \int\limits_{x=0}^{\alpha} dx & \int\limits_{x=0}^{\alpha} x dx \\ \int\limits_{x=0}^{\alpha} x dx & \int\limits_{x=0}^{\alpha} x^2 dx \\ \end{array}\right] \left[\begin{array}{c} c_0 \\ c_1 \\ \end{array}\right] = \left[\begin{array}{l} \int\limits_{x=0}^{\pi} x^2 dx\\ \int\limits_{x=0}^{\pi} x^3 f dx\\ \end{array}\right] $$

2.1 Power Rule

Since all integrals take the form

$$ \int\limits_{x=0}^{\alpha} x^n dx \\ $$

We can specialize the definition of the power rule to this problem…

$$ \begin{align} \int\limits_{x=0}^{\alpha} x^n dx &= \frac{1}{n+1} x^{n+1} \Big|_{x=0}^{\alpha} \\ &= \frac{1}{n+1} \left(\alpha^{n+1} - (0)^{n+1}\right) \\ &= \frac{\alpha^{n+1}}{n+1} \\ \end{align} $$

2.2 Using the Extension

Now… we evaluate everything…

$$ \left[\begin{array}{ll} \alpha & \frac{1}{2} \alpha^2 \\ \frac{1}{2}\alpha^2 & \frac{1}{3} \alpha^3 \\ \end{array}\right] \left[\begin{array}{c} c_0 \\ c_1 \\ \end{array}\right] = \left[\begin{array}{l} \frac{1}{3}\alpha^3 \\ \frac{1}{4}\alpha^4 \\ \end{array}\right] $$

3 Linear Algebra (Solving)

The next step is to apply some elementary row operations. Let us start with:

• $\alpha^{-1}\vec{r_0}$
• $2\alpha^{-2}\vec{r_1}$

$$ \left[\begin{array}{ll} 1 & \frac{1}{2} \alpha \\ 1 & \frac{2}{3} \alpha \\ \end{array}\right] \left[\begin{array}{c} c_0 \\ c_1 \\ \end{array}\right] = \left[\begin{array}{l} \frac{1}{3}\alpha^2 \\ \frac{2}{4}\alpha^2 \\ \end{array}\right] $$

Let us rewrite the fractions.

$$ \left[\begin{array}{ll} 1 & \frac{3}{6} \alpha \\ 1 & \frac{4}{6} \alpha \\ \end{array}\right] \left[\begin{array}{c} c_0 \\ c_1 \\ \end{array}\right] = \left[\begin{array}{l} \frac{2}{6}\alpha^2 \\ \frac{3}{6}\alpha^2 \\ \end{array}\right] $$

Now… it is time for $\vec{r_1} - \vec{r_0}$.

$$ \left[\begin{array}{ll} 1 & \frac{3}{6} \alpha \\ 0 & \frac{1}{6} \alpha \\ \end{array}\right] \left[\begin{array}{c} c_0 \\ c_1 \\ \end{array}\right] = \left[\begin{array}{l} \frac{2}{6}\alpha^2 \\ \frac{1}{6}\alpha^2 \\ \end{array}\right] $$

Let us scale row 1 using $\frac{6}{\alpha}\vec{r_1}$

$$ \left[\begin{array}{ll} 1 & \frac{3}{6} \alpha \\ 0 & 1 \\ \end{array}\right] \left[\begin{array}{c} c_0 \\ c_1 \\ \end{array}\right] = \left[\begin{array}{l} \frac{2}{6}\alpha^2 \\ \alpha \\ \end{array}\right] $$

4 Switching to Algebra

Since we have an upper-triangular matrix with ones on the diagonal… let us switch to basic Algebra.

$$ \begin{array}{rrr} (1) c_0 &+& \frac{3}{6}\alpha c_1 &=& \frac{2}{6}\alpha^2 \\ (0) c_0 &+& c_1 &=& \alpha \\ \end{array} $$

The second equation simplifies to

$$ c_1 = \alpha $$

Now… for the first equation…

$$ \begin{array} c_0 + \frac{3}{6}\alpha c_1 &=& \frac{2}{6}\alpha^2 \\ c_0 + \frac{3}{6}\alpha (\alpha) &=& \frac{2}{6}\alpha^2 \\ c_0 + \frac{3}{6}\alpha^2 &=& \frac{2}{6}\alpha^2 \\ c_0 &=& \frac{2}{6}\alpha^2 - \frac{3}{6}\alpha^2\\ c_0 &=& -\frac{1}{6}\alpha^2\\ \end{array} $$

5 The Result

Now that we have $c_0$ and $c_1$ the approximation function can be written as…

$$ \hat{\varphi} = -\frac{1}{6}\alpha^2\ + \alpha x $$

6 What is the Error

Now… for the important question: How “good” is our approximation?

6.1 Error Metric

Least Squares Approximation Error can be evaluated using

$$ \begin{eqnarray} E^2(\alpha,x) &=& ||f(x) - \hat{\varphi}(\alpha,x)|| \\ &=& \int\limits_{\mathbb{R}} \left(f - \hat{\varphi} \right)^2 d\lambda(x) \\ &=& \int\limits_{\mathbb{R}} f^2 d\lambda(x) -2\int\limits_{\mathbb{R}} \hat{\varphi}f d\lambda(x) +\int\limits_{\mathbb{R}} \hat{\varphi}^2 d\lambda(x) \\ &=& \int\limits_{\mathbb{R}} \hat{\varphi}^2 d\lambda(x) -2\int\limits_{\mathbb{R}} \hat{\varphi}f d\lambda(x) + \int\limits_{\mathbb{R}} f^2 d\lambda(x) \\ \end{eqnarray} $$

Since we have a continuous approximation… with the domain $0 \leq x \leq \alpha$…

$$ \begin{eqnarray} E^2(\alpha,x) &=& \int\limits_{\mathbb{R}} \hat{\varphi}^2 d\lambda(x) &-2\int\limits_{\mathbb{R}} \hat{\varphi}f d\lambda(x) &+ \int\limits_{\mathbb{R}} f^2 d\lambda(x) \\ &=& \int\limits_{x=0}^{\alpha} \hat{\varphi}^2 dx &-2\int\limits_{x=0}^{\alpha} \hat{\varphi}f dx &+ \int\limits_{x=0}^{\alpha} f^2 dx \\ \end{eqnarray} $$

6.2 Plugging Everything In

Now… we plug in $f$ and $\hat{\varphi}$ (after preparing ourselves for some Calculus, Algebra, and Arithmetic).

$$ \begin{eqnarray} E^2(\alpha,x) &=& \int\limits_{x=0}^{\alpha} \hat{\varphi}^2 dx &-2\int\limits_{x=0}^{\alpha} \hat{\varphi}f dx &+ \int\limits_{x=0}^{\alpha} f^2 dx \\ &=& \int\limits_{x=0}^{\alpha} \left(-\frac{1}{6}\alpha^2 + \alpha x\right)^2 dx &-2\int\limits_{x=0}^{\alpha} \left(-\frac{1}{6}\alpha^2 + \alpha x\right) x^2 dx &+ \int\limits_{x=0}^{\alpha} (x^2)^2 dx \\ \end{eqnarray} $$

6.3 The Math…

Now… let us apply some math…

$$ \begin{eqnarray} E^2(\alpha,x) &=& \int\limits_{x=0}^{\alpha} \left(-\frac{1}{6}\alpha^2 + \alpha x\right)^2 dx &-2\int\limits_{x=0}^{\alpha} \left(-\frac{1}{6}\alpha^2 + \alpha x\right) x^2 dx &+ \int\limits_{x=0}^{\alpha} x^4 dx \\ &=& \int\limits_{x=0}^{\alpha} \left(-\frac{1}{6}\alpha^2 + \alpha x\right)^2 dx &-2\int\limits_{x=0}^{\alpha} \left(-\frac{1}{6}\alpha^2 + \alpha x\right) x^2 dx &+ \frac{1}{5}\alpha^5 \\ &=& \int\limits_{x=0}^{\alpha} \left(-\frac{1}{6}\alpha^2\ + \alpha x\right)^2 dx &-2\int\limits_{x=0}^{\alpha} \left(-\frac{1}{6}\alpha^2 x^2 + \alpha x^3\right) dx &+ \frac{1}{5}\alpha^5 \\ &=& \int\limits_{x=0}^{\alpha} \left(-\frac{1}{6}\alpha^2\ + \alpha x\right)^2 dx &-2\left( \int\limits_{x=0}^{\alpha} \left(-\frac{1}{6}\alpha^2 x^2\right) dx + \int\limits_{x=0}^{\alpha} \left(\alpha x^3\right) dx \right) &+ \frac{1}{5}\alpha^5 \\ &=& \int\limits_{x=0}^{\alpha} \left(-\frac{1}{6}\alpha^2\ + \alpha x\right)^2 dx &-2\left( -\frac{1}{6}\alpha^2\int\limits_{x=0}^{\alpha} \left(x^2\right) dx + \alpha \int\limits_{x=0}^{\alpha} \left(x^3\right) dx \right) &+ \frac{1}{5}\alpha^5 \\ &=& \int\limits_{x=0}^{\alpha} \left(-\frac{1}{6}\alpha^2\ + \alpha x\right)^2 dx &-2\left( -\frac{1}{6}\alpha^2 \left(\frac{1}{3}\alpha^3\right) + \alpha\left(\frac{1}{4}\alpha^4\right) \right) &+ \frac{1}{5}\alpha^5 \\ &=& \int\limits_{x=0}^{\alpha} \left(-\frac{1}{6}\alpha^2\ + \alpha x\right)^2 dx &-2\left( -\frac{1}{18}\alpha^5 + \frac{1}{4}\alpha^5 \right) &+ \frac{1}{5}\alpha^5 \\ &=& \int\limits_{x=0}^{\alpha} \left(-\frac{1}{6}\alpha^2\ + \alpha x\right)^2 dx &-2\left( -\frac{2}{36}\alpha^5 + \frac{9}{36}\alpha^5 \right) &+ \frac{1}{5}\alpha^5 \\ &=& \int\limits_{x=0}^{\alpha} \left(-\frac{1}{6}\alpha^2\ + \alpha x\right)^2 dx &-2\left( -\frac{2}{36}\alpha^5 + \frac{9}{36}\alpha^5 \right) &+ \frac{1}{5}\alpha^5 \\ &=& \int\limits_{x=0}^{\alpha} \left(-\frac{1}{6}\alpha^2\ + \alpha x\right)^2 dx &-2\left( \frac{7}{36}\alpha^5 \right) &+ \frac{1}{5}\alpha^5 \\ &=& \int\limits_{x=0}^{\alpha} \left(-\frac{1}{6}\alpha^2\ + \alpha x\right)^2 dx &-\left( \frac{7}{18}\alpha^5 \right) &+ \frac{1}{5}\alpha^5 \\ &=& \int\limits_{x=0}^{\alpha} \left(-\frac{1}{6}\alpha^2\ + \alpha x\right)^2 dx &-\frac{7}{18}\alpha^5 &+ \frac{1}{5}\alpha^5 \\ &=& \int\limits_{x=0}^{\alpha} \left(-\frac{1}{6}\alpha^2\ + \alpha x\right)^2 dx &-\frac{35}{90}\alpha^5 &+ \frac{18}{90}\alpha^5 \\ &=& \int\limits_{x=0}^{\alpha} \left(-\frac{1}{6}\alpha^2\ + \alpha x\right)^2 dx &-\frac{17}{90}\alpha^5 \\ \end{eqnarray} $$

Now… let us take a moment to breathe. The first integral is fun (i.e., I made a bunch of silly mistakes the first time I solved it.)

$$ \begin{eqnarray} E^2(\alpha,x) &=& \int\limits_{x=0}^{\alpha} \left(-\frac{1}{6}\alpha^2\ + \alpha x\right)^2 dx &-\frac{17}{90}\alpha^5 \\ &=& \int\limits_{x=0}^{\alpha} \left(-\frac{1}{6}\alpha^2\ + \frac{6\alpha x}{6}\right)^2 dx &-\frac{17}{90}\alpha^5 \\ &=& \int\limits_{x=0}^{\alpha} \left(\frac{1}{6}^2\right)\left(-\alpha^2\ + 6\alpha x\right)^2 dx &-\frac{17}{90}\alpha^5 \\ &=& \int\limits_{x=0}^{\alpha} \left(\frac{1}{36}\right)\left(-\alpha^2\ + 6\alpha x\right)^2 dx &-\frac{17}{90}\alpha^5 \\ &=& \frac{1}{36}\left(\int\limits_{x=0}^{\alpha} \left(-\alpha^2\ + 6\alpha x\right)^2 dx\right) &-\frac{17}{90}\alpha^5 \\ &=& \frac{1}{36}\left(\int\limits_{x=0}^{\alpha} \left(\alpha^4 - 2\left(6\alpha^3x\right) + 36\alpha^2 x\right)^2 dx\right) &-\frac{17}{90}\alpha^5 \\ &=& \frac{1}{36}\left(\int\limits_{x=0}^{\alpha} \left(\alpha^4 - 2\left(6\alpha^3x\right) + 36\alpha^2 x^2\right) dx\right) &-\frac{17}{90}\alpha^5 \\ &=& \frac{1}{36}\left( \int\limits_{x=0}^{\alpha} \left(\alpha^4 - 12\alpha^3 x + 36\alpha^2 x^2\right) dx \right) &-\frac{17}{90}\alpha^5 \\ &=& \frac{1}{36}\left( \alpha^4 \int\limits_{x=0}^{\alpha} dx -12\alpha^3 \int\limits_{x=0}^{\alpha} x dx +36\alpha^2 \int\limits_{x=0}^{\alpha} x^2 dx \right) &-\frac{17}{90}\alpha^5 \\ &=& \frac{1}{36}\left( \alpha^5 -12\alpha^3 \frac{\alpha^2}{2} +36\alpha^2 \frac{\alpha^3}{3} \right) &-\frac{17}{90}\alpha^5 \\ &=& \frac{1}{36}\left( \alpha^5 -6\alpha^5 +12\alpha^5 \right) &-\frac{17}{90}\alpha^5 \\ &=& \frac{1}{36}\left( \alpha^5 6\alpha^5 \right) &-\frac{17}{90}\alpha^5 \\ &=& \frac{1}{36}\left( 7\alpha^5 \right) &-\frac{17}{90}\alpha^5 \\ &=& \frac{7}{36}\alpha^5 &-\frac{17}{90}\alpha^5 \\ &=& \frac{70}{360}\alpha^5 &-\frac{68}{360}\alpha^5 \\ &=& \frac{2}{360}\alpha^5 \\ &=& \frac{1}{180}\alpha^5 \\ \end{eqnarray} $$

7 The Result

$$ \begin{eqnarray} E^2(\alpha,x) &=& \frac{1}{180}\alpha^5 \\ \end{eqnarray} $$