Linear-Algebra-Done-Right · Unit 17 · Video 1 · Interactive Practice

The Closest Point Is the Projection: Fitting Sine on [−π, π]

IKey Formulas

Formula	Name	When it applies
$\\|v - P_U v\\| \le \\|v - u\\|$ for all $u \in U$ , with equality iff $u = P_U v$	Closest point is the projection	$U$ finite-dimensional, $v \in V$
$\\|v - u\\|^2 = \\|v - P_U v\\|^2 + \\|P_U v - u\\|^2$	Pythagoras behind the proof	$v - P_U v \in U^\perp$ , $P_U v - u \in U$
$P_U v = \langle v, e_1 \rangle e_1 + \cdots + \langle v, e_m \rangle e_m$	Computing the closest point	$e_1, \dots, e_m$ orthonormal basis of $U$
$\langle f, g \rangle = \displaystyle\int_{-\pi}^{\pi} f g$	Inner product on $C[-\pi, \pi]$	Makes $\\|\sin - u\\|^2$ the integrated squared error

Key Insight: Projecting $\sin$ onto $\mathcal{P}_5(\mathbf{R})$ gives $u(x) = 0.987862x - 0.155271x^3 + 0.00564312x^5$ , the degree-5 polynomial with the smallest $\int_{-\pi}^{\pi} |\sin x - u(x)|^2\,dx$ — the theorem guarantees the integral, not the error at each point.

IIThe Closest Point Is the Foot of the Perpendicular

For every $u \in U$ , the legs $v - P_U v$ and $P_U v - u$ are orthogonal, so $\|v - u\|$ is a hypotenuse.

IIIProjecting $\sin$ onto $\mathcal{P}_5(\mathbf{R})$ One Term at a Time

Each orthonormal term $\langle \sin, e_k \rangle e_k$ lowers the squared error by exactly $\langle \sin, e_k \rangle^2$ .

Terms k 0

$e_1 = 0.398942$ is even, so $\langle \sin, e_1 \rangle = 0$ .
$e_2 = 0.219948x$ is odd, and $\langle \sin, e_2 \rangle = 1.38198$ .
$e_3 = 0.135577x^2 - 0.446031$ is even, so $\langle \sin, e_3 \rangle = 0$ .
$e_4 = 0.0851039x^3 - 0.503965x$ is odd, and $\langle \sin, e_4 \rangle = -1.09734$ .
$e_5 = 0.0537539x^4 - 0.454740x^2 + 0.448810$ is even, so $\langle \sin, e_5 \rangle = 0$ .
$e_6 = 0.0340493x^5 - 0.373392x^3 + 0.789692x$ is odd, and $\langle \sin, e_6 \rangle = 0.165734$ .

IVProjection Versus Taylor

The projection minimizes the integrated squared error, yet the Taylor polynomial still wins pointwise near $x = 0$ .

Degree Probe x 3.00

VQuiz Questions

Problem 1 · Distance to a Plane

Given: In $\mathbf{R}^3$ with the dot product, $U = \operatorname{span}(e_1, e_2)$ where $e_1 = \tfrac{1}{\sqrt{2}}(1, 1, 0)$ and $e_2 = (0, 0, 1)$ are orthonormal, and $v = (3, 1, 2)$ . Find $\min_{u \in U} \|v - u\|$ .

✅ Correct!

P_U v = (2, 2, 2)

, so the minimum is

\|v - P_U v\| = \|(1, -1, 0)\| = \sqrt{2}

❌ Not quite.

2

is the squared distance

\|v - P_U v\|^2

; take its square root.

❌ Not quite.

\sqrt{14} = \|v\| = \|v - 0\|

is the distance to one particular point of

U

, the origin, not the closest one.

❌ Not quite.

2\sqrt{3} = \|P_U v\|

is the length of the projection itself; the distance is the length of the perpendicular piece

v - P_U v

Show solution

By the theorem the closest point is $P_U v$ , computed with the orthonormal basis:

\langle v, e_1 \rangle = \tfrac{3 + 1}{\sqrt{2}} = 2\sqrt{2}, \qquad \langle v, e_2 \rangle = 2,

P_U v = 2\sqrt{2}\cdot\tfrac{1}{\sqrt{2}}(1, 1, 0) + 2(0, 0, 1) = (2, 2, 2).

Then $v - P_U v = (1, -1, 0)$ , which is orthogonal to both $e_1$ and $e_2$ , and

\min_{u \in U} \|v - u\| = \|v - P_U v\| = \sqrt{1 + 1 + 0} = \sqrt{2}.

Problem 2 · The Formula Needs an Orthonormal Basis

Given: In $\mathbf{R}^3$ with the dot product, $U = \operatorname{span}\big((1, 1, 0), (1, 0, 0)\big)$ and $v = (1, 2, 3)$ . Find the point of $U$ closest to $v$ .

✅ Correct!

U

is the

xy

-plane, and Gram–Schmidt gives the orthonormal basis

(1, 0, 0), (0, 1, 0)

, so

P_U v = (1, 2, 0)

❌ Close, but…

\langle v, w_1 \rangle w_1 + \langle v, w_2 \rangle w_2

with

w_1 = (1, 1, 0)

w_2 = (1, 0, 0)

uses a basis that is neither orthogonal nor normalized; run Gram–Schmidt first.

❌ Not quite.

\big(\tfrac{3}{2}, \tfrac{3}{2}, 0\big)

is the projection onto the line spanned by

(1, 1, 0)

alone, a smaller subspace than

U

❌ Not quite.

(0, 0, 3) = v - P_U v

is the piece of

v

U^\perp

; the closest point is the piece in

U

Show solution

Run Gram–Schmidt on $w_1 = (1, 1, 0)$ , $w_2 = (1, 0, 0)$ :

e_1 = \tfrac{1}{\sqrt{2}}(1, 1, 0), \qquad w_2 - \langle w_2, e_1 \rangle e_1 = (1, 0, 0) - \tfrac{1}{2}(1, 1, 0) = \big(\tfrac{1}{2}, -\tfrac{1}{2}, 0\big), \qquad e_2 = \tfrac{1}{\sqrt{2}}(1, -1, 0).

Now apply the formula:

\langle v, e_1 \rangle = \tfrac{3}{\sqrt{2}}, \quad \langle v, e_2 \rangle = -\tfrac{1}{\sqrt{2}}, \quad P_U v = \tfrac{3}{2}(1, 1, 0) - \tfrac{1}{2}(1, -1, 0) = (1, 2, 0).

Check: $v - P_U v = (0, 0, 3)$ is orthogonal to both $(1, 1, 0)$ and $(1, 0, 0)$ , so it lies in $U^\perp$ , and the closest point is $(1, 2, 0)$ at distance $3$ .

Problem 3 · Best Line for Sine

Given: $V = C[-\pi, \pi]$ with $\langle f, g \rangle = \int_{-\pi}^{\pi} f g$ , $U = \mathcal{P}_1(\mathbf{R})$ , and $v = \sin$ . You may use $\int_{-\pi}^{\pi} x \sin x \, dx = 2\pi$ , $\int_{-\pi}^{\pi} x^2 \, dx = \tfrac{2\pi^3}{3}$ and $\int_{-\pi}^{\pi} \sin^2 x \, dx = \pi$ . Find the closest point $u = P_U(\sin)$ and the minimum value of $\int_{-\pi}^{\pi} |\sin x - u(x)|^2 \, dx$ .

What is $P_U(\sin)$ ?

What is the minimum of $\int_{-\pi}^{\pi} |\sin x - u(x)|^2 \, dx$ ?

✅ Correct!

P_U(\sin) = \tfrac{3}{\pi^2}x \approx 0.304x

, and the squared error is

\pi - \tfrac{6}{\pi} \approx 1.2317

❌ Not quite.

x

is the degree-1 Taylor polynomial, which matches

\sin

0

; it is not the closest point of

U

in the integral norm.

❌ Close, but…

\langle \sin, x \rangle x

forgets to divide by

\|x\|^2 = \tfrac{2\pi^3}{3}

; the formula needs a unit vector.

❌ Not quite. The projection onto

\operatorname{span}(x)

\tfrac{\langle \sin, x \rangle}{\|x\|^2}\, x

; compute that ratio carefully.

❌ Not quite. By Pythagoras,

\|\sin - P_U \sin\|^2 = \|\sin\|^2 - \langle \sin, e_2 \rangle^2

, where

e_2 = x / \|x\|

Show solution

Orthonormal basis of $U$ . On the symmetric interval $\langle 1, x \rangle = 0$ , so Gram–Schmidt only normalizes: $e_1 = \tfrac{1}{\sqrt{2\pi}}$ and $e_2 = x \big/ \sqrt{2\pi^3/3}$ .

Coefficients. $\langle \sin, e_1 \rangle = 0$ because $\sin$ is odd. Then

P_U(\sin) = \langle \sin, e_2 \rangle e_2 = \frac{\langle \sin, x \rangle}{\|x\|^2}\, x = \frac{2\pi}{2\pi^3/3}\, x = \frac{3}{\pi^2}\, x \approx 0.304x.

Minimum squared error. Since $\sin = P_U(\sin) + (\sin - P_U \sin)$ with orthogonal pieces,

\|\sin - P_U \sin\|^2 = \|\sin\|^2 - \langle \sin, e_2 \rangle^2 = \pi - \frac{(2\pi)^2}{2\pi^3/3} = \pi - \frac{6}{\pi} \approx 1.2317.

Problem 4 · What the Theorem Guarantees

Given: $U = \mathcal{P}_5(\mathbf{R}) \subseteq C[-\pi, \pi]$ with the integral inner product, $u = P_U(\sin)$ , and the Taylor polynomial $p(x) = x - \tfrac{x^3}{3!} + \tfrac{x^5}{5!}$ , which lies in $U$ and is not equal to $u$ . Which statement follows from the theorem?

✅ Correct! The inequality

\|v - P_U v\| \le \|v - p\|

is strict because equality would force

p = P_U v

; numerically the two integrals are about

0.000116

and

0.119

❌ Not quite. The theorem compares norms, and this norm is an integral. At

x = 0.5

the Taylor error is about

0.0000015

while

|\sin 0.5 - u(0.5)| \approx 0.0047

❌ Not quite. Equality in the theorem holds only when the competitor is

P_U v

itself, and

p \ne u

❌ Not quite. The hypothesis is that

U

is finite-dimensional, which

\mathcal{P}_5(\mathbf{R})

is (

\dim U = 6

);

V

may be infinite-dimensional.

Show solution

The theorem applies because $U = \mathcal{P}_5(\mathbf{R})$ is finite-dimensional, and $p \in U$ . It gives $\|\sin - u\| \le \|\sin - p\|$ , with equality iff $p = u$ . Since $p \ne u$ , the inequality is strict, and squaring the norms gives

\int_{-\pi}^{\pi} |\sin x - u(x)|^2\,dx \lt \int_{-\pi}^{\pi} |\sin x - p(x)|^2\,dx.

Numerically the left side is about $0.000116$ and the right side about $0.119$ . The theorem says nothing about single points: for $|x|$ below about $1.21$ the Taylor polynomial is the closer of the two, while at $x = 3$ the error of $u$ is about $0.0014$ against $0.38$ for $p$ , a fact we checked, not one the theorem promises.

Solved: 0 / 4

The Closest Point Is the Projection: Fitting Sine on [−π, π]

IKey Formulas

IIThe Closest Point Is the Foot of the Perpendicular

IIIProjecting sin⁡\sinsin onto P5(R)\mathcal{P}_5(\mathbf{R})P5​(R) One Term at a Time

IVProjection Versus Taylor

VQuiz Questions

IIIProjecting $\sin$ onto $\mathcal{P}_5(\mathbf{R})$ One Term at a Time