SigmaField
Mathematical Foundations for Quant Finance
0 of 256 lessons
      1. 1.4.1 Exponents and their rulesConcept · 25 min
      2. 1.4.2 Exponential functions and the number eConcept · 25 min
      3. 1.4.3 Logarithms and their rulesConcept · 25 min
      4. 1.4.4 Solving exponential and logarithmic equationsConcept · 25 min
      5. 1.4.5 Practice set: exponents and logarithmsPractice set · 25 min
      6. 1.4.6 Case study: compounding and continuous ratesCase study · 35 min
      7. 1.4.7 Case study: simple returns and log returnsCase study · 30 min

Level 1, Precalculus 1.4 Exponents and logarithms Lesson 2 of 7

ConceptLesson 1.4.2 in Level 1, Precalculus. About 25 minutes.

Exponential functions and the number e

From powers to functions

In lesson 1.4.1 the exponent was a fixed number. Now let the exponent be the variable:

f(x)=bx(b>0,  b≠1)f(x) = b^x \qquad (b > 0,\; b \neq 1)

This is an exponential function with base bb. In x2x^2 the variable is the base and the exponent is fixed. In 2x2^x the base is fixed and the variable is the exponent.

The base must be positive. With b=−2b = -2, the value at x=12x = \frac{1}{2} would be −2\sqrt{-2}, which has no real value (lesson 1.4.1). A base of 1 is left out because 1x=11^x = 1 for every xx: a flat line.

With a positive base, any real number works as the exponent. So the domain of bxb^x is all real numbers.

Try it

2x2^x starts ahead, falls behind at x=3x = 3 (8 against 9), draws level at 4, and leads from then on.

By x=10x = 10, x2x^2 has reached 100 while 2x2^x has reached 1024. An exponential eventually overtakes any power of xx.

⌘/Ctrl + Enter runs the cell

Adding against multiplying

A straight line adds the same amount at each step. An exponential multiplies by the same factor at each step. Here are both, starting at 100:

xx01234
100+10x100 + 10x100110120130140
100⋅1.1x100 \cdot 1.1^x100110121133.1146.41

The line rises by 10 every time. The exponential rises by 10, then 11, then 12.1, then 13.31, because each rise is 10% of a bigger number. What stays fixed is the ratio: each value is 1.1 times the one before.

The product rule from lesson 1.4.1 says why: bx+1=b⋅bxb^{x+1} = b \cdot b^x. One more step in xx is one more factor of bb.

Rule: Properties of exponential functions

b0=1bx>0bx+1=b⋅bx\begin{gathered} b^0 = 1 \qquad b^x > 0 \\ b^{x+1} = b \cdot b^x \end{gathered}

Every exponential passes through (0,1)(0, 1) and is always positive. Each step of 1 in xx multiplies the value by bb.

Try it

The cell prints, for each step of 100⋅1.1x100 \cdot 1.1^x, the rise and the ratio to the step before.

The rises grow; the ratio stays at 1.1.

⌘/Ctrl + Enter runs the cell

Growth by a constant percentage

If a quantity grows by a fraction gg at every step, each step multiplies it by 1+g1 + g, as in lesson 0.3.1. After xx steps it has been multiplied by (1+g)x(1 + g)^x.

Rule: Percentage growth

P(x)=P0 (1+g)xP(x) = P_0\,(1 + g)^x

P0P_0 is the starting amount, gg is the growth per step as a decimal (5% is 0.05), and xx is the number of steps. PP has the units of P0P_0. A negative gg gives decay.

Try it

Start at 100 and grow by 5% a step. After 10 steps the value is about 162.89, not 150.

Adding 5 each time would give 150. The extra comes from growth on top of earlier growth.

⌘/Ctrl + Enter runs the cell

A fall works the same way. Losing 20% per step keeps 80%, so the factor is 1−0.2=0.81 - 0.2 = 0.8.

Check yourself
What does P(x)=50⋅0.8xP(x) = 50 \cdot 0.8^x describe?

The shape of bxb^x: growth and decay

Three exponentials and the flat line 1x1^x. Every curve passes through (0,1)(0, 1) and through (1,b)(1, b). The decay curve 0.5x0.5^x is the mirror image of 2x2^x in the yy-axis.

A base above 1 rises from left to right, and a larger base rises faster. A base between 0 and 1 falls: that is decay.

No curve touches the xx-axis. Going left, 2x2^x halves at every step: 2−1=0.52^{-1} = 0.5, then 2−2=0.252^{-2} = 0.25, closer and closer to 0 but never 0. So the range of bxb^x is all positive numbers (lesson 1.1.1). The xx-axis is a horizontal asymptote: a line the graph gets ever closer to without reaching.

Decay is growth run backwards

Since 0.5=2−10.5 = 2^{-1}, the power-of-a-power rule gives 0.5x=2−x0.5^x = 2^{-x}. Replacing xx with −x-x reflects a graph in the yy-axis (lesson 1.1.2). So decay is growth read from right to left.

Rule: Negative exponents give decay

b−x=(1b)xb^{-x} = \left(\frac{1}{b}\right)^x

A negative exponent on a base above 1 is the same as a positive exponent on a base below 1. So e−xe^{-x}, which you will meet often, is a decay curve.

Try it

The table shows the mirror: each value of 2x2^x appears for 0.5x0.5^x at the opposite xx.

The plot draws both. Change the bases in the code and run it again; try 1.2 and 3.

⌘/Ctrl + Enter runs the cell

Where ee comes from

Suppose something grows by 100% in one step: it doubles. Now split that growth into nn smaller pieces of 1n\tfrac{1}{n} each, applied one after another. The total growth factor is

(1+1n)n\left(1 + \frac{1}{n}\right)^n

With two pieces, each adds 50%: 1.5×1.5=2.251.5 \times 1.5 = 2.25, more than 2. The second piece grows the growth the first piece added.

Try it

With more pieces the factor rises further, but by less each time. Going from 1 piece to 2 adds 0.25; going from 365 pieces to a million adds about 0.0037.

It does not rise forever. It settles near 2.71828.

⌘/Ctrl + Enter runs the cell

Splitting 100% growth into nn pieces. The factor climbs quickly at first, then flattens just under the dashed line at ee, which it never passes.

The number it settles on is called ee, or Euler's number.

Rule: The number e

e=lim⁡n→∞(1+1n)ne≈2.718281828\begin{gathered} e = \lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^n \\ e \approx 2.718281828 \end{gathered}

Read lim⁡n→∞\lim_{n \to \infty} as "the value this gets closer and closer to as nn grows without end". Lesson 2.1.3 makes that precise; for now the table and the figure are the evidence.

The natural exponential exe^x

The same splitting works for any growth xx, not just 100%. Split it into nn pieces of x/nx/n and the factor (1+x/n)n(1 + x/n)^n settles on exe^x.

It works for a negative xx too. With x=−0.5x = -0.5 each piece shrinks the value a little, and the factor settles on e−0.5≈0.606531e^{-0.5} \approx 0.606531: decay.

Try it

math.exp(x) computes exe^x. NumPy has np.exp for whole arrays.

The cell splits growth of 0.5 and then of −0.5-0.5 into more and more pieces. Each list ends at math.exp(x): about 1.648721 and 0.606531.

⌘/Ctrl + Enter runs the cell

exe^x is an ordinary exponential with base ee, so every rule from lesson 1.4.1 applies.

Rule: Rules for e

eaeb=ea+beaeb=ea−b(ea)b=eabe−x=1ex\begin{gathered} e^a e^b = e^{a+b} \qquad \frac{e^a}{e^b} = e^{a-b} \\ (e^a)^b = e^{ab} \qquad e^{-x} = \frac{1}{e^x} \end{gathered}

Same rules, same reasons. exe^x grows, and e−xe^{-x} decays.

Common mistakes

  1. Mixing up x2x^2 and 2x2^x. In 2x2^x the variable is the exponent. At x=10x = 10 they are 100 and 1024.
  2. Writing the percentage as a whole number. Growth of 5% multiplies by 1.05, not by 5.
  3. Using the loss as the factor. A fall of 20% multiplies by 0.8, not by 0.2.
  4. Adding percentages across steps. Ten steps of 5% multiply by 1.0510≈1.62891.05^{10} \approx 1.6289, a rise of 62.89%, not 50%.
  5. Expecting decay to reach 0. 0.5x0.5^x gets as small as you like but stays positive.
  6. Reading e−xe^{-x} as a negative number. e−2≈0.1353e^{-2} \approx 0.1353: a negative exponent means one over.

Exponential functions and e

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How to type answers
  • Powers with ^: x^7, 2^(1/3), a^-2.
  • Multiply with * or by writing side by side: 3x^2, 2(x+1).
  • ln(x) is the natural log, log(x, 2) is log base 2, e is Euler's number, sqrt(x) is a square root.
  • sin(x), cos(x) and tan(x) take radians; write pi for π and sin(x)^2 for sin²x.
  • Your first answer to each question is saved: it counts towards your score and brings the question back for review.
  • Exact answers like ln(20)/ln(3) are best. Decimals need about 4 significant figures.

Summary card

  • An exponential function bxb^x (b>0b > 0, b≠1b \neq 1) has the variable in the exponent. Its domain is all real numbers; its range is the positive numbers.
  • A line adds a fixed amount per step; an exponential multiplies by a fixed factor: bx+1=b⋅bxb^{x+1} = b \cdot b^x.
  • b>1b > 1 grows and 0<b<10 < b < 1 decays. Every graph passes through (0,1)(0, 1) and never reaches the xx-axis. b−x=(1/b)xb^{-x} = (1/b)^x.
  • Growth by a fraction gg per step: P(x)=P0(1+g)xP(x) = P_0 (1 + g)^x. A fall of 20% is a factor of 0.8.
  • e=lim⁡n→∞(1+1/n)n≈2.71828e = \lim_{n\to\infty}(1 + 1/n)^n \approx 2.71828, and (1+x/n)n(1 + x/n)^n settles on exe^x for any xx, negative included.
  • exe^x obeys every exponent rule. In Python, math.exp(x) or np.exp(x).

Free further reading

Optional. This lesson covers what you need; these add more examples and practice. All are free to read.