Quantitative Reasoning

1015SCG

Lecture 8


Enrico Fermi

Enrico Fermi (1901-1954) was an Italian physicist known for his major contributions to nuclear physics, quantum theory, and statistical mechanics.

  • Nobel Prize in Physics (1938).
  • Key figure in the development of the first nuclear reactor.
  • Worked on both theoretical and experimental physics.

The Fermi Paradox

The Fermi Paradox asks why, given the vast size of the universe and the likelihood of extraterrestrial life, we have not yet observed any clear evidence of it.

  • “Where is everybody?” — attributed to Fermi.
  • Highlights the tension between probability and observation.
  • Leads to scientific and philosophical discussions.

The Fermi Paradox

We know that we live in a big old galaxy, in a big old universe

Consider for a moment the Milky Way Galaxy

The Fermi Paradox

- The Milky Way Galaxy has about 400 billion suns 🌞

- Most of those suns have planetary systems around them.

  So, there are trillions of planets.






The galaxy it's been around for about 10 billions years. Even though there have been billions of years on billions of worlds for civilizations to arise,
we see no evidence of any them in the galaxy at all. 😬



The Fermi Paradox

Where is everybody?










Source: Hubble Space Telescope - Ultra-Deep Field galaxies to Legacy field zoom out (May 2, 2019) Wikipedia

Fermi Problems

Fermi problems are estimation questions that use rough assumptions and simple calculations to obtain approximate answers.

  • Emphasise order-of-magnitude reasoning.
  • Encourage critical thinking and modelling.
  • Common in physics and mathematics education.

An example is Enrico Fermi's estimate of the strength of the atomic bomb that detonated at the Trinity test, based on the distance traveled by pieces of paper he dropped from his hand during the blast. Fermi's estimate of 10 kilotons of TNT was well within an order of magnitude of the now-accepted value of 21 kilotons.


Not good for Fermi problems ❌

  • How much Panadol can I take in a day?
  • How long will the air in my scuba tank last?
  • How much weight can this bridge support?



Good for Fermi problems ✅

  • Is it more expensive to go to Melbourne by plane or by car?
  • How many people are on Facebook at the moment?
  • How much water do people flush down the toilet each year?



Fermi problems: Strategy

  1. Work out what the question is asking you.
  2. Work out your steps, and how they go together:
    • What do you know?
    • What are you assuming?
    • Use dimensional analysis
  3. Estimate the unknowns:
    • Geometric mean of underestimate and overestimate.
  4. Combine to obtain an order of magnitude estimate
    • Use rounding (usually to 1 s.f.) and scientific notation.
  5. Interpret answer through comparison with known quantities.
  6. Refine your steps.

Arithmetic mean

$AM = \bar{x} =\dfrac{1}{n}\ds \sum_{i=1}^{n}x_i$

Average of $n$ numbers $x_i$

$\bar{x} =\dfrac{0.81+0.9+0.94+0.85+0.88}{5}$

$\quad =0.876 \text{ m}$
Experiment $x_i$ (m)
1 0.81
2 0.9
3 0.94
4 0.85
5 0.88

Geometric mean

$GM = \ds \left(\prod_{i=1}^{n} x_i\right)^{1/n}$

Multiplicative average of $n$ positive numbers $x_i$

$GM = \left(0.81 \times 0.9 \times 0.94 \times 0.85 \times 0.88\right)^{1/5}$

$\;\;\;\quad \approx 0.875 \text{ m}$

Experiment $x_i$ (m)
1 0.81
2 0.9
3 0.94
4 0.85
5 0.88

Geometric mean

How to estimate unknown values?

  • Find underestimate.
  • Find overestimate.
  • Calculate their geometric mean.
🤔


📝 Examples

🚽 Toilet tank

What is the volume of a toilet cistern (the tank where water is stored)?

Underestimate: 1L

Underestimate: 50L

$GM = \sqrt{1\text{ L} \times 50 \text{ L}}$ $\approx 7 \text{ L}$

Underestimate: 25L

$GM = \sqrt{1\text{ L} \times 25 \text{ L}}$ $\approx 5 \text{ L}$



😴 People sleeping

How many people in the world are sleeping at the moment?

Assume people sleep for 6hrs a day.

Fraction of the time sleeping: \(\dfrac{6\text{ h}}{24\text{ h}}\) \(=\dfrac{1}{4}\)

Amount sleeping = Number of people $\times$ fraction sleeping $\qquad$

$\qquad\qquad$\(\approx\) 8 billion people $\times \dfrac{1}{4} $ \(\approx\) 2 billion people



Recall: Scientific notation

A number is in normalized scientific notation if it is in the form \[ \large \pm r \times 10^n \]

  • $r$ is a real number (called the significand) with $1 \leq r \lt 10$.
  • $n$ is an integer (called the exponent).




Recall: Scientific notation

It is easy to compare numbers! 😃

Example:

How many orders of magnitude there is between 2 mL and 80 L?

\(\dfrac{80 \text{ L}}{2 \text{ mL}}\) \(=\dfrac{80 \text{ L}}{2 \text{ mL}} \times \dfrac{1000 \text{ mL}}{1 \text{ L}}\) \(=\dfrac{80\,000 }{2 }\) \(=40\, 000\) \(=4\times 10^4\)


There are about 4 orders of magnitude between 2 mL and 80 L




Recall: Scientific notation

It is easy to find the geometric mean

\(\sqrt{10^a \times 10^b}\) \(=\sqrt{10^{a+b}}\) \(=10^{\frac{a+b}{2}}\)

Example:

Find the geometric mean of $40000$ and $0.00006$.

\(\sqrt{4\times 10^4 \times 6\times 10^{-5}}\) \(=\sqrt{24\times 10^{-1}}\) \(=\sqrt{2.4}\) \(\approx 1.5\)

Find the geometric mean of $10,000,000$ and $0.004$.

\(\sqrt{10^7\times 4\times ^{-3}}\) \(=\sqrt{4\times 10^4 }\) \(=\sqrt{4}\sqrt{ 10^{4}}\) \(=2\times 10^{4/2}\) \(=2\times 10^2\) \(=200\)


🧻 Toilet paper

How much toilet paper do we use in Australia each year?

We are looking for $\dfrac{\text{rolls}}{\text{year}}$

How many times a person visits the toilet?

Underestimate: $\;1 \dfrac{\text{visit}}{\text{day}}$
Overestimate: $\;25 \dfrac{\text{visit}}{\text{day}}$

$N_t = \sqrt{25}$ $=5$


🧻 Toilet paper

How much toilet paper do we use in Australia each year?

We are looking for $\dfrac{\text{rolls}}{\text{year}}$

What is the amount used each visit?

Underestimate: $\;1 \dfrac{\text{sheet}}{\text{visit}}$
Overestimate: $\;20 \dfrac{\text{sheet}}{\text{visit}}$

$N_p = \sqrt{20}$ $\approx 4.47$ $\approx 4$


🧻 Toilet paper

How much toilet paper do we use in Australia each year?

Thus we have $\;N_t =5 \dfrac{\text{visit}}{\text{day}}\;$ and $\;N_p = 4 \dfrac{\text{sheet}}{\text{visit}}$

How many sheets of toilet paper does a person use in a year?

$N_t \times N_p \times \dfrac{365.25\text{ days}}{\text{year}}$

$5 \dfrac{\text{visit}}{\text{day}} \times 4 \dfrac{\text{sheet}}{\text{visit}} \times \dfrac{365.25\text{ days}}{\text{year}}$ $=7305\dfrac{\text{sheets}}{\text{year}}$


🧻 Toilet paper

How much toilet paper do we use in Australia each year?

$5 \dfrac{\text{visit}}{\text{day}} \times 4 \dfrac{\text{sheet}}{\text{visit}} \times \dfrac{365.25\text{ days}}{\text{year}}$ $=7305\dfrac{\text{sheets}}{\text{year}}$

🧻 Assume: $\;200 \dfrac{\text{sheets}}{\text{roll}}$

Then we have $\;\dfrac{7305\dfrac{\text{sheets}}{\text{year}}}{200 \dfrac{\text{sheets}}{\text{roll}}}$ $\approx 37 \dfrac{\text{rolls}}{\text{year}}$ per person

Population of Australia $\approx 28$ million $=2.8\times 10^7$


🧻 Toilet paper

How much toilet paper do we use in Australia each year?

Population of Australia $\approx 28$ million $=2.8\times 10^7$

👉$\;\; 37 \dfrac{\text{rolls}}{\text{year}}$ per person

Therefore $\;37 \dfrac{\text{rolls}}{\text{year}}\times 2.8\times 10^7 $ $\approx 10^9 \dfrac{\text{rolls}}{\text{year}}$




💨 Air we breathe

How much air would you breathe in a year?


Try it yourself! 📝 😃







Archimedes' Sand Reckoner

In The Sand Reckoner, Archimedes asked:

“How many grains of sand would be needed to fill the universe?”

  • Written around 260 BCE.
  • Required estimating the size of the cosmos.
  • Required creating a new system for expressing very large numbers.

A 3rd-Century BCE Fermi Problem

Archimedes' strategy closely mirrors modern Fermi reasoning:

  • Estimate the size of a grain of sand.
  • Estimate the size of the Earth.
  • Estimate the size of the known universe.
  • Use geometric reasoning to scale up.

The goal was not precision — but order of magnitude.



The Result

Archimedes concluded that the universe could be filled with fewer than

\(10^{63}\) grains of sand.

To express such numbers, he developed a system extending beyond the Greek “myriad” (10,000).

Like modern Fermi problems, this was an exercise in structured estimation, scaling, and mathematical imagination.

🤯

The Result

Even though $10^{63}$ is unimaginably large, it is still 17 orders of magnitude smaller than the number of atoms in the observable universe, estimated at roughly $10^{80}$.

Archimedes was already thinking on a cosmic scale — 2200 years before modern astronomy.




Taylor series

Any function $f(x)$ can be rewritten as an infinite sum \[f(x)= a_0 + a_1 x + a_2x^2 + a_3x^3+\cdots a_nx^n + \cdots\]

Example:

\(e^x \) \(= 1 +x + \dfrac{x^2}{2!}+ \dfrac{x^3}{3!} + \cdots + \dfrac{x^n}{n!}+ \cdots\)

where $n! = 1 \times 2 \times 3 \times \cdots \times (n-1) \times n.$




Taylor series: Useful approximations, if $x$ is small

Truncating the Taylor series at second order gives useful approximations:

Function 2nd-order Taylor approx.
$e^x$ $1 + x + \dfrac{x^2}{2}$
$(1+x)^n$ $1 + nx + \dfrac{n(n-1)}{2}x^2$
$\ln(1+x)$ $x - \dfrac{x^2}{2}$
$\sin x$ $x$
$\cos x$ $1 - \dfrac{x^2}{2}$
$\tan x$ $x$

These approximations are valid when $|x| \ll 1$.

Taylor series

Change the function in the input box. Drag slider to increase the number of terms in the Taylor series.


📝 Practice


Using Taylor approximations:

  • Estimating $e^{0.05}$.
  • Estimate $\cos(0.1)$.




Estimating $e^{0.05}$ using Taylor approximations

\( \ds e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots \)

Approximation order Expression Estimate for $x=0.05$
First order $1 + x$ $1 + 0.05$$\,=1.05$
Second order $1 + x + \dfrac{x^2}{2}$ $1 + 0.05 + \dfrac{(0.05)^2}{2}$$\,=1.05125$
Third order $1 + x + \dfrac{x^2}{2} + \dfrac{x^3}{6}$ $1.05125 + \dfrac{(0.05)^3}{6} \approx 1.05127$
Calculator $e^{0.05}$ $\approx 1.051271096376024\ldots$

Higher-order approximations give better accuracy for small $x$.

📝 Practice: Estimate $\cos(0.1)$ using Taylor approximations

\( \ds \cos x = \; \large ??? \)

Approximation order Expression Estimate for $x=0.1$
First order
Second order
Third order
Calculator

Odd-power terms vanish for $\cos x$, so first- and third-order approximations coincide.


Estimating $\cos(0.1)$ using Taylor approximations

\( \ds \cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots \)

Approximation order Expression Estimate for $x=0.1$
First order $1$ $1$
Second order $1 - \dfrac{x^2}{2}$ $1 - \dfrac{(0.1)^2}{2} = 0.995$
Third order $1 - \dfrac{x^2}{2}$ $0.995$
Calculator Exact value $\cos(0.1) \approx 0.995004165\ldots$

Odd-power terms vanish for $\cos x$, so first- and third-order approximations coincide.

📝 Practice: Population

The population of the bacteria in the sample is given by \[ \large P(t) = 500 e^{0.001t} \] where $t$ is given in weeks.

Consider a second order approximation of $e^t$.

  • What will be population after 2 weeks?
  • When will the population reach 505?
  • Compare results to the exact values.


📝 Practice: Population — Solution

Using a second-order Taylor approximation for \( e^t\): \( \ds \;e^x \approx 1 + x + \frac{x^2}{2}. \)

  • Population after 2 weeks:
    \[ P(2) \approx 501.001 \quad (\text{Exact: } 501.001001) \]
  • Time when the population reaches 505:
    \[ t \approx 9.95 \text{ weeks} \quad (\text{Exact: } 9.95033 \text{ weeks}) \]
  • Comparison:
    The Taylor approximation gives values virtually identical to the exact model.


📝 Practice: Population — Step 1: Taylor Approximation

We approximate the exponential function using a second-order Taylor expansion around 0:
$$e^x \approx 1 + x + \frac{x^2}{2}$$

For our population model: $\,P(t) = 500 e^{0.001t}\,$ we use $$e^{0.001t} \approx 1 + 0.001t + \frac{(0.001t)^2}{2}$$


📝 Practice: Population — Step 2: Population after 2 Weeks

Using the approximation:

$P(2) \approx 500 \cdot \left(1 + 0.001 \cdot 2 + \dfrac{(0.001 \cdot 2)^2}{2}\right)$

Step-by-step calculation:

$0.001 \cdot 2 = 0.002$

$\dfrac{(0.002)^2}{2} = 0.000002$

$1 + 0.002 + 0.000002 = 1.002002$

$P(2) \approx 500 \cdot 1.002002 \approx 501.001$


📝 Practice: Population — Step 3: Time to Reach 505

Solve $P(t) \approx 505$ using the second-order approximation:

$500 \cdot \left(1 + 0.001t + \dfrac{(0.001t)^2}{2}\right) = 505$

Divide both sides by 500:

$1 + 0.001t + \dfrac{(0.001t)^2}{2} \approx 1.01$

Let $x = 0.001t,$ then solve:

$\ds 1 + x + \frac{x^2}{2} = 1.01 $ $\Ra \ds \frac{x^2}{2} + x - 0.01 = 0$

Solve quadratic: $x \approx 0.00995 \rightarrow t \approx 9.95$ weeks


📝 Practice: Population — Step 4: Comparison

Comparing the second-order Taylor approximation with exact values:

  • Population after 2 weeks: $501.001$ (approx) vs $501.001001$ (exact)
  • Time to reach 505: $9.95$ weeks (approx) vs $9.95033$ weeks (exact)

The approximation gives results virtually identical to the exact model for such small exponents.

That's all for today!

See you in Week 9!

👾