The last section discussed bx and logb y. The base b was arbitrary—it could be 2 or 6 or 9.3 or any positive number except 1. But in practice, only a few bases are used. I have never met a logarithm to base 6 or 9.3. Realistically there are two leading candidates for b, and 10 is one of them. This section is about the other one, which is an extremely remarkable number. This number is not seen in arithmetic or algebra or geometry, where it looks totally clumsy and out of place. In calculus it comes into its own.
The number is e. That symbol was chosen by Euler (initially in a fit of selfishness, but he was a wonderful mathematician). It is the base of the natural logarithm. It also controls the exponential ex, which is much more important than ln x. Euler also chose π to stand for perimeter—anyway, our first goal is to find e.
Remember that the derivatives of bx and logb y include a constant c that depends on b. Equations (10) and (11) in the previous section were
| d | bx = cbx and | d | logb y = | 1 | . (1) |
| dx | dy | cy |
At x = 0, the graph of bx starts from b0 = 1. The slope is c. At y = 1, the graph of logb y starts from logb 1 = 0. The logarithm has slope 1/c. With the right choice of the base b those slopes will equal 1 (because c will equal 1).
For y = 2x the slope c is near .7. We already tried Δx = .1 and found Δy ≈ .07. The base has to be larger than 2, for a starting slope of c = 1.
We begin with a direct computation of the slope of logb y at y = 1:
| 1 | = slope at 1 = limh→0 | 1 | [logb (1 + h) − logb 1] = limh→0 logb[(1 + h)1/h]. (2) |
| c | h |
Always logb 1 = 0. The fraction in the middle is logb (1 + h) times the number 1/h. This number can go up into the exponent, and it did.
The quantity (1 + h)1/h is unusual, to put it mildly. As h → 0, the number 1 + h is approaching 1. At the same time, 1/h is approaching infinity. In the limit we have 1∞. But that expression is meaningless (like 0/0). Everything depends on the balance between "nearly 1" and "nearly ∞." This balance produces the extraordinary number e:
DEFINITION The number e is equal to limh→0 (1 + h)1/h. Equivalently e = limn→∞ (1 + 1/n)n.
Before computing e, look again at the slope 1/c. At the end of equation (2) is the logarithm of e:
1/c = logb e. (3)
When the base is b = e, the slope is logee = 1. That base e has c = 1 as desired:
The derivative of ex is 1⋅ex and the derivative of loge y is 1/(1⋅y). (4)
This is why the base e is all-important in calculus. It makes c = 1.
To compute the actual number e from (1 + h)1/h, choose h = 1, 1/10, 1/100, …. Then the exponents 1/h are n = 1, 10, 100, …. (All limits and derivatives will become official in Section 6.4.) The table shows (1 + h)1/h approaching e as h → 0 and n → ∞:
| n | h = 1/n | 1 + h = 1 + 1/n | (1 + h)1/h = (1 + 1/n)n |
| 1 | 1.0 | 2.0 | 2.0 |
| 2 | 0.5 | 1.5 | 2.25 |
| 10 | 0.1 | 1.1 | 2.593742 |
| 100 | 0.01 | 1.01 | 2.704814 |
| 1000 | 0.001 | 1.001 | 2.716924 |
| 10000 | 0.0001 | 1.0001 | 2.718146 |
The last column is converging to e (not quickly). There is an infinite series that converges much faster. We know 125,000 digits of e (and a billion digits of π). There are no definite patterns, although you might think so from the first sixteen digits:
e = 2.718281828459045… (and 1/e ≈ .37).
The powers of e produce y = ex. At x = 2.3 and 5, we are close to y = 10 and 150.
The logarithm is the inverse function. The logarithms of 150 and 10, to the base e, are close to x = 5 and x = 2.3. There is a special name for this logarithm—the natural logarithm. There is also a special notation "ln" to show that the base is e:
ln y means the same as loge y. The natural logarithm is the exponent in ex = y.
The notation In y (or In x—it is the function that matters, not the variable) is standard in calculus courses. After calculus, the base is generally assumed to be e. In most of science and engineering, the natural logarithm is the automatic choice. The symbol "exp (x)" means ex, and the truth is that the symbol "log x" generally means ln x. Base e is understood even without the letters ln. But in any case of doubt—on a calculator key for example—the symbol "ln x" emphasizes that the base is e.
Come back to derivatives and slopes. The derivative of bx is cbx, and the derivative of log, y is 1/cy. If b = e then c = 1. For all bases, equation (3) is 1/c = logb e. This gives c—the slope of bx at x = 0:
6E The number c is 1/logbe = logeb. Thus c equals ln b. (5)
c = ln b is the mysterious constant that was not available earlier. The slope of 2x is ln 2 times 2x. The slope of ex is ln e times ex (but ln e = 1). We have the derivatives on which this chapter depends:
6F The derivatives of ex and ln y are ex and 1/y. For other bases
(d/dx)bx = (ln b)bx and (d/dy)logb y = 1 / (ln b)y. (6)
To make clear that those derivatives come from the functions (and not at all from the dummy variables), we rewrite them using t and x:
(d/dx)ex = ex and (d/dx)ln x = 1/x. (7)
Remark on slopes at x = 0: It would be satisfying to see directly that the slope of 2x is below 1, and the slope of 4x is above 1. Quick proof: e is between 2 and 4. But the idea is to see the slopes graphically. This is a small puzzle, which is fun to solve but can be skipped.
2x rises from 1 at x = 0 to 2 at x = 1. On that interval its average slope is 1. Its slope at the beginning is smaller than average, so it must be less than 1—as desired. On the other hand 4x rises from ½ at x = -½ to 1 at x = 0. Again the average slope is ½ / ½ = 1. Since x = 0 comes at the end of this new interval, the slope of 4x at that point exceeds 1. Somewhere between 2x and 4x is ex, which starts out with slope 1.
This is the graphical approach to e. There is also the infinite series, and a fifth definition through integrals which is written here for the record:
The connections between 1, 2, and 3 have been made. The slopes are 1 when e is the limit of (1 + 1/n)n. Multiplying this out wlll lead to 4, the infinite series in Section 6.6. The official definition of ln x comes from ∫ dx/x, and then 5 says that ln e = 1. This approach to e (Section 6.4) seems less intuitive than the others.
Figure 6.6b shows the graph of e-x. It is the mirror image of ex across the vertical axis. Their product is exe-x = 1. Where ex grows exponentially, e-x decays exponentially—or it grows as x approaches -∞. Their growth and decay are faster than any power of x. Exponential growth is more rapid than polynomial growth, so that ex/xn goes to infinity (Problem 59). It is the fact that ex has slope ex which keeps the function climbing so fast.
The other curve is y = e-x²/2. This is the famous "bell-shaped curve" of probability theory. After dividing by √2π,it gives the normal distribution, which applies to so many averages and so many experiments. The Gallup Poll will be an example in Section 8.4. The curve is symmetric around its mean value x = 0, since changing x to -x has no effect on x².
About two thirds of the area under this curve is between x = -1 and x = 1. If you pick points at random below the graph, 2/3 of all samples are expected in that interval. The points x = -2 and x = 2 are "two standard deviations" from the center, enclosing 95% of the area. There is only a 5% chance of landing beyond. The decay is even faster than an ordinary exponential, because -½x² has replaced x.
The slope of ex is ex. This opens up a whole world of functions that calculus can deal with. The chain rule gives the slope of e3x and esin x and every eu(x):
6G The derivative of eu(x) is eu(x) times du/dx. (8)
Special case u = cx: The derivative of ecx is cecx. (9)
EXAMPLE 1 The derivative of e3x is 3e3x (here c = 3). The derivative of esin x is esin xcos x (here u = sin x). The derivative of ƒ(u(x)) is dƒ/du times du/dx. Here ƒ = eu so dƒ/du = eu. The chain rule demands that second factor du/dx.
EXAMPLE 2 e(ln 2)x is the same as 2x. Its derivative is In 2 times 2x. The chain rule rediscovers our constant c = ln 2. In the slope of bx it rediscovers the factor c = ln b.
Generally ecx is preferred to the original bx. The derivative just brings down the constant c. It is better to agree on e as the base, and put all complications (like c = ln b) up in the exponent. The second derivative of ecx is c2ecx.
EXAMPLE 3 The derivative of e-x²/2 is -xe-x²/2 (here u = -x²/2 so du/dx = -x).
EXAMPLE 4 The second derivative off= e - x2/2, by the chain rule and product rule, is ƒ'' = (-1)⋅e-x²/2 + (-x)²e-x²/2 = (x² − 1)e-x²/2. (10)
Notice how the exponential survives. With every derivative it is multiplied by more factors, but it is still there to dominate growth or decay. The points of inflection, where the bell-shaped curve has ƒ'' = 0 in equation (10), are x = 1 and x = -1.
EXAMPLE 5 (u = n ln x). Since en ln x is xn in disguise, its slope must be nxn-1:
slope = en ln x (d/dx)(n ln x) = xn(n/x) = nxn-1. (11)
This slope is correct for all n, integer or not. Chapter 2 produced 3x² and 4x³ from the binomial theorem. Now nxn-1 comes from ln and exp and the chain rule.
EXAMPLE 6 An extreme case is xx = (eln x)x. Here u = x ln x and we need du/dx:
(d/dx)(xx) = ex ln x(ln x + x⋅(1/x)) = xx(ln x + 1).
The integral of ex is ex. The integral of ecx is not ecx. The derivative multiplies by c so the integral divides by c. The integral of ecx is ecx/c (plus a constant).
| EXAMPLES | ∫ | e2xdx = | 1 | e2x + C | ∫ | bxdx = | bx | + C |
| 2 | ln b |
| ∫ | e3(x+1)dx = | 1 | e3(x+1) + C | ∫ | e-x²/2dx → failure |
| 3 |
6H The indefinite integral ∫ eu(du/dx)dx equals eu(x) + C.
Here are three examples with du/dx and one without it:
| ∫ | esin xcos x dx = esin x + C | ∫ | xex²/2dx = ex²/2 + C |
| ∫ | e√xdx | = e√x + C | ∫ | exdx | = | −1 | + C |
| √x | (1 + ex)² | 1 + ex |
The first is a pure eudu. So is the second. The third has u = √x and du/dx = 1/2√x, so only the factor 2 had to be fixed. The fourth example does not belong with the others. It is the integral of du/u², not the integral of eudu. I don't know any way to tell you which substitution is best—except that the complicated part is 1 + ex and it is natural to substitute u. If it works, good.
Without an extra ex for du/dx, the integral ∫ dx/(1 + ex)² looks bad. But u = 1 + ex is still worth trying. It has du = exdx = (u−1)dx:
| ∫ | dx | = | ∫ | du | = | ∫ | du | 1 | − | 1 | − | 1 | . (12) | |||
| (1 + ex)² | (u − 1)u² | u − 1 | u | u² |
That last step is "partial fractions." The integral splits into simpler pieces (explained in Section 7.4) and we integrate each piece. Here are three other integrals:
| ∫ | e1/xdx | ∫ | ex(4 + ex)dx | ∫ | e-x(4 + ex)dx |
The first can change to -∫ eudu/u²,which is not much better. (It is just as impossible.) The second is actually ∫ udu, but I prefer a split: ∫ 4ex and ∫ e2x are safer to do separately. The third is ∫ (4e-x + 1)dx, which also separates. The exercises offer practice in reaching eudu/dx —ready to be integrated.
Warning about dejinite integrals When the lower limit is x = 0, there is a natural tendency to expect ƒ(0) = 0 —in which case the lower limit contributes nothing. For a power ƒ = x³ that is true. For an exponential ƒ = e3x it is definitely not true, because ƒ(0) = 1:
| ∫ | 1 | e3xdx = | 1 | e3x | 1 | = | 1 | (e³ − 1) | ∫ | 1 | xex²dx = | 1 | ex² | 1 | = | 1 | (e − 1). | ||
| 0 | 3 | 0 | 3 | 0 | 2 | 0 | 2 |
