Numerical Analysis Notes

Math 342 - Spring 2026

Week 1 Notes

Mon, Jan 12

We talked about how computers store floating point numbers. Most modern programming languages store floating point numbers using the IEEE 754 standard.

In the IEEE 754 standard, a 64-bit floating point number has the form

$$\mathbf{value} = (-1)^\mathbf{sign} \, \times \, 2^{(\mathbf{exponent} - 1023)} \, \times \, (1 + \mathbf{fraction})$$

where

• $\mathbf{sign}$ is a single bit: 0 for positive, 1 for negative.
• $\mathbf{fraction}$ is a 52-bit fraction (in binary) with $0 \le \mathbf{fraction} < 1$.
• $\mathbf{exponent}$ is the 11-bit binary integer which ranges from 0 to 2047. Only 1 to 2046 are used for regular floating point numbers, $\mathbf{exponent}=0$ is reserved for zero and subnormal numbers, and $\mathbf{exponent}=2047$ is reserved for infinity and NaN (“not a number”).

We talked about how to convert between binary numbers and decimal numbers. This system works very well, but it leads to weird outcomes like 0.1 + 0.1 + 0.1 = 0.30000000000000004.

When you convert $\tfrac{1}{10} = 0.1$ to binary, you get a infinitely repeating binary decimal: $$\tfrac{1}{10} = (0.000110011001100\ldots)_2.$$ So any finite binary representation of $\tfrac{1}{10}$ will have rounding errors.

We did the following examples in class:

1. Convert (10110)₂ to decimal. (https://youtu.be/a2FpnU9Mm3E)

2. Convert 35 to binary. (https://youtu.be/gGiEu7QTi68)

3. What is the 64-bit string that represents the number 35 in the IEEE standard?

4. What are the largest and smallest 64-bit floating point numbers that can be stored?

5. In Python, compute 2.0**1024 and 2**1024. Why do you get different results?

6. In Python, compare 2.0**1024 with 2.0**(-1024) and 2.0**(-1070). What do you notice?

Wed, Jan 14

Today we talked about significant digits. Here is a quick video on how these work.

Next we defined absolute and relative error:

The base-10 logarithm of the relative error is approximately the number of significant digits, so you can think of significant digits as a measure of relative error.

Intuitively, addition & subtraction “play nice” with absolute error while multiplication and division “play nice” with relative error. This can lead to problems:

1. Catastrophic cancellation. When you subtract two numbers of roughly the same size, the relative error can get much worse. For example, both 53.76 and 53.74 have 4 significant digits, but $$53.76 - 53.74 = 0.02$$ only has 1 significant digit.

2. Useless precision. If you add two numbers with very different magnitudes, then having a very low relative error in the smaller one will not be useful.

We did these examples in class.

1. $\pi = 3.141592...$. What is the absolute and relative error if your round $\pi$ to $3.14$?

You can sometimes re-write algorithms on a computer to avoid issues with floating point numbers such as overflow/underflow and catastrophic cancellation.

2. Consider the function $f(x) = \dfrac{1 - \cos x}{\sin x}$. Use Python to compute $f(10^{-7})$.

3. The exact answer to previous question is $0.00000005 = 5 \times 10^{-8}$ (accurate to 22 decimal places). Use this to find the relative error in your previous calculation.

4. A better way to compute $f(x)$ is to use a trick to avoid the catastrophic cancellation: $$f(x) = \dfrac{1-\cos x}{\sin x} = \dfrac{1 - \cos x}{\sin x} \cdot \left( \frac{1+ \cos x}{1+\cos x} \right) = \dfrac{\sin x}{1 + \cos x}.$$ Use this new formula to compute $f(10^{-7})$. What is the relative error now?

Stirling’s formula is a famous approximation for the factorial function. $$n! \approx \sqrt{2 \pi n} \frac{n^n}{e^n}.$$ We approximated Stirling’s formula using the following code.

import math
n = 100
print(float(math.factorial(n)))
f = lambda n: math.sqrt(2 * math.pi * n) * n ** n / math.exp(n)
print(f(n))

Our formula worked well until $n=143$, then we got an overflow error. The problem was that $n^n$ got too big to convert to a floating point number. But you can prevent the overflow error by adjusting the formula slightly to.

n = 143
f = lambda n: math.sqrt(2 * math.pi * n) * (n / math.e) ** n
print(f(n))

Fri, Jan 26

Today we reviewed Taylor series. We recalled the following important Maclaurin series (which are Taylor series with center $c = 0$):

• $e^x = \sum_{n=0}^\infty \dfrac{x^n}{n!}$
• $\sin(x) = \sum_{n=0}^\infty \dfrac{(-1)^n \, x^{2n+1}}{(2n+1)!}$
• $\cos(x) = \sum_{n=0}^\infty \dfrac{(-1)^n \, x^{2n}}{(2n)!}$
• $\dfrac{1}{1-x} = \sum_{n = 0}^\infty x^n$
• $\ln(1+x) = \sum_{n = 0}^\infty \dfrac{(-1)^n \, x^{n+1}}{n+1}$

We graphed Maclaurin polynomials for $\cos x$ and $\dfrac{1}{1-x}$ on Desmos to see how they converge with different radii of convergence.

We also use the Maclaurin series for $\dfrac{\sin x}{x}$ to approximate the integral

$$\displaystyle \int_{-\pi}^\pi \frac{\sin x}{x} \, dx.$$

Then we did the following workshop in class.

• Workshop: Taylor series

Week 2 Notes

Wed, Jan 21

Today we reviewed some theorems that we will need throughout the course. The first is probably the most important theorem in numerical analysis since it lets us estimate error when using Taylor series approximations.

A special case of Taylor’s theorem is when $n = 0$. Then you get the Mean Value Theorem (MVT):

We did this example:

1. Use Taylor’s theorem to estimate the error in using the 0th and 2nd degree Maclaurin polynomials to estimate $\cos(0.03)$ and $\cos(0.6)$.

Then we started this workshop

• Workshop: Error bounds

Fri, Jan 23

Last time we started this workshop about using Taylor’s remainder formula and the triangle inequality to find upper bounds for functions. Today we revisited that workshop, but first we talked about the following.

This error formula gives a way to estimate the worst case (absolute) error when you use a Taylor polynomial approximation.

1. The 3rd degree Taylor polynomial for $\cos x$ centered at $c = \pi$ is $P_3(x) = -1 + \tfrac{1}{2}(x-\pi)^2$ (the coefficient on the 3rd degree term is zero). What is the worst case absolute error using this polynomial to estimate $\cos 3$?

After that we talked about the triangle inequality.

We talked about how you can use the triangle inequality to find upper bounds for functions. We also talked about tight upper bounds versus upper bounds that are not tight. We did this example.

2. Use the triangle inequality to find an upper bound for $|x^2 + 3x \cos x|$.

Week 3 Notes

Week 4 Notes

Week 5 Notes

Week 6 Notes

Week 7 Notes

Week 8 Notes

Week 9 Notes

Week 10 Notes

Week 11 Notes

Week 12 Notes

Week 13 Notes

Week 14 Notes