# Graphs, approximation, and numerical sense

Physics I · Calculus of a Single Variable · https://tryals.app/learn/physics-i/graphs-approximation-and-numerical-sense

## Taylor Approximations

A **Taylor polynomial** approximates a smooth function near an anchor point $a$:

$$P_n(x) = \sum_{k=0}^{n}\frac{f^{(k)}(a)}{k!}(x-a)^k$$

A degree 1 polynomial is just the tangent line; higher degrees capture local curvature.

| Function | Maclaurin Expansion ($a=0$) |
| --- | --- |
| **$e^x$** | $1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \dots$ |
| **$\sin x$** | $x - \frac{x^3}{3!} + \frac{x^5}{5!} - \dots$ |
| **$\cos x$** | $1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \dots$ |

**Common pitfall:** Linear approximation is a **local** promise. $\sqrt{16.4} \approx 4.05$ works, but using that same tangent line for $\sqrt{25}$ fails. Always check your drift from $a$.

## Euler and Graph Reading

**Euler's formula** links exponentials to trigonometry in the **complex plane** $\mathbb{C}$:

$$e^{i\theta} = \cos\theta + i\sin\theta$$

A complex number $z = a + bi$ in **polar form** is $z = r\,e^{i\theta}$, where $r=|z|$ and $\theta=\arg(z)$.

| Feature | Graphical Indicator |
| --- | --- |
| **Concavity** | $f''>0$ bends up; $f''<0$ bends down |
| **Inflection point** | Where concavity changes sign |
| **Asymptotes** | Horizontal as $x\to\pm\infty$; vertical where denominator $\to 0$ |

## Practice questions

7 of this lesson's 13 practice questions, with answers. The full set is in the app.

### 1. Newton’s method races to a root when it works, but pick the starting point badly and it fails. Which situation **breaks** it?

A. A function that is always increasing
B. A root that is not at a whole number
C. Starting where $f'(x_0) \approx 0$
D. Starting too close to the root, the steps become too small

**Answer:** C. Starting where $f'(x_0) \approx 0$

**Why:** A flat tangent barely tilts toward zero, so its root-crossing lies absurdly far away: division by $f' \approx 0$ catapults the iteration. Away from flat spots, though, the method roughly **doubles the correct digits every step**: flawed and brilliant at once.

Page: https://tryals.app/practice/physics-i/graphs-approximation-and-numerical-sense/newtons-method-races-to-a-root-when-it-works-but-pick-the-starting

### 2. The linearization $L(x) = f(a) + f'(a)(x-a)$ is exact for every function on a small enough interval.

**Answer:** False

**Why:** False. It is exact only where the error term $\tfrac12 f''(c)(x-a)^2$ vanishes, that is, for linear functions. For everything else it is an approximation whose error grows with the square of the distance, at a rate set by the curvature. Small intervals make it *good*, never exact.

Page: https://tryals.app/practice/physics-i/graphs-approximation-and-numerical-sense/the-linearization-l-x-f-a-f-a-x-a-is-exact-for-every

### 3. Use linearisation to estimate sin(0.1). (Use sin(0) = 0 and cos(0) = 1.) **Set the slider to your answer.**

**Answer:** 0.1 (within ±0.01)

**Why:** $L(x) = \sin(0) + \cos(0)\cdot(x - 0) = x$. So $\sin(0.1) \approx 0.1$ (actual value $\approx 0.0998$).

Page: https://tryals.app/practice/physics-i/graphs-approximation-and-numerical-sense/use-linearisation-to-estimate-sin-0-1-use-sin-0-0-and-cos-0

### 4. A linear approximation is excellent near the expansion point and poor far from it. Why does the approximation degrade faster for some functions than others?

A. The distance from the origin rather than from the expansion point
B. The magnitude of the function value at the expansion point
C. The first derivative: a steeper tangent gives a worse approximation
D. The curvature, which sets the leading error term

**Answer:** D. The curvature, which sets the leading error term

**Why:** The error is governed by $f''$ and the square of the distance. This is why the small-angle approximation for a pendulum holds well to about 15° and then fails quickly.

Page: https://tryals.app/practice/physics-i/graphs-approximation-and-numerical-sense/a-linear-approximation-is-excellent-near-the-expansion-point-and-poor

### 5. Which statements about Newton's method are true?

A. It can fail if the initial guess is far from a root
B. Each step uses the formula $x_{n}$₊₁ = $x_{n}$ − f($x_{n}$)/f'($x_{n}$)
C. It always converges to a root in finitely many steps
D. It requires f'($x_{n}$) ≠ 0 at each step

**Answer:** A. It can fail if the initial guess is far from a root; B. Each step uses the formula $x_{n}$₊₁ = $x_{n}$ − f($x_{n}$)/f'($x_{n}$); D. It requires f'($x_{n}$) ≠ 0 at each step

**Why:** Newton's method is not guaranteed to converge; a poor initial guess or a zero derivative causes failure.

Page: https://tryals.app/practice/physics-i/graphs-approximation-and-numerical-sense/which-statements-about-newtons-method-are-true

### 6. Order the steps in one iteration of Newton's method for root-finding.

**Answer:**

1. Start with an initial guess $x_{0}$ close to the root
2. Evaluate f($x_{0}$) and f'($x_{0}$)
3. Compute the next iterate: $x_{1} = x_{0} - f(x_{0})/f'(x_{0})$
4. Repeat until |$x_{n}$₊₁ − $x_{n}$| is smaller than the desired tolerance

**Why:** Newton's method has quadratic convergence near simple roots.

Page: https://tryals.app/practice/physics-i/graphs-approximation-and-numerical-sense/order-the-steps-in-one-iteration-of-newtons-method-for-root-finding

### 7. Use the **linear approximation** at $x=16$ to estimate $\sqrt{16.4}$, then set the slider. ($f(x)=\sqrt{x}$, $f'(x) = \frac{1}{2\sqrt{x}}$.)

**Answer:** 4.05 (within ±0.015)

**Why:** $\sqrt{16.4} \approx 4 + 0.05 = 4.05$ (true value 4.0497…). Near the anchor, the tangent line **is** the function to within a whisker, the idea behind every calculator’s first guess and every "small correction" in physics.

Page: https://tryals.app/practice/physics-i/graphs-approximation-and-numerical-sense/use-the-linear-approximation-at-x-16-to-estimate-16-4-then-set