Problems in Calculus II

Chapter 3. Logarithms and Exponential Function.

    Problem 2. Given ln 2 = 0.6931, ln 3 = 1.0986. Find ln 4, ln 6, ln 96, ln 144, ln Ö2, ln (6^1/5), without using calculators. You can use them to check for correct answers.
    Ans.: ln 4 = ln (2²) = 2 ln 2 = 1.9362.
ln 6 = ln (2 . 3) = ln 2 + ln 3 = 1.7917
ln 96 = ln (2⁵. 3) = 5 ln 2 + ln 3 = . . .
ln 144 = ln (16×9) = 4 ln 2 + 2 ln 3 = . . .
ln Ö2 = (½) ln 2 = 0.3466.
ln (6^1/5) = (1/5) (ln 2 + ln 3) = 0.3583...
    In fact: ln 2 = 0.69314 71805 59945 30941 7....

    Problem 3. Prove that D_x ln x = 1/x.
    Ans.: Using fundamental theorem of calculus, by Newton,
D_x ln x = D_x ò₁^x (1/t) dt = 1/x. (QED, quite easily done!)

    Problem 4. Property of a constant function. If D_xf(x) = 0 for all values of x with a £ x £ b, prove that f(x) is a constant function.
    Ans.: [Calm Down! You do not have to memorize this proof yet. You do so in Advanced Calculus.]
     For any p < q, we define "average slope of the interval [p,q]", or slope(p,q), as [f(q)-f(p)]/(q-p).
Let e be some small number > 0.
    Now since f '(a) = 0, there is some d, such that the following condition, called K(d), is true for d:
| slope(a,x) | £ e for all x such that a < x <d.
    Let d₀ be some lub (least upper bound) of such d,
i.e., K(d) is true if d < d₀, but not true if d > d₀.
    Then the curve y = f(x) must cross
the line L₁: y = e(x-a) or the line L₂: y = -e(x-a) at x = a+d₀.
    For if |f(a+d₀)| > ed₀, then due to continuity of f(x) around a+d₀,
K(d) already failed for some d slightly < d₀.
    On the other hand, if |f(a+d₀)| < ed₀, then due to continuity of f(x) around a+d₀,
K(d) is true even if d is a little bit > d₀.
    Therefore f(a+d₀) has to be exactly ±ed₀.
    Then if d > d₀, K(d) will be false, no matter how close to d₀ is d.
This means f(a+d) will be outside the lines L₁ or L₂.
    But this violates the fact that f '(d₀) = 0.
     So d₀ simply cannot exist. That means the curve
y = f(x) lies between the lines L₁ and L₂ for all x with a < x < b.
     But e is just any arbitrary positive number.
Hence the graph of y = f(x) is a straight line parallel to the x-axis.
    The proof is complete.
    Remark: This kind of proof is water tight logically. It is based on the concepts of limits that Archimedes to Newton had developed, but it was Newton who defined the concept of derivatives. However, it was until the 1800's that Weierstrass came up with these epsilon-delta definitions and such long-winded logical proofs. Of course most books use the Mean Value Theorem or Rolle's Theorem to prove this, but if you chase through the proofs of those theorems you will find the same epsilon-delta argument somewhere. A third way to prove it, not so messy, is to use point-set topology, which is an undergrad math subject in many colleges.

    Problem 5. Prove that ln (1+x) = x -x²/2 +x³/3 -x⁴/4 +- . . . , if |x| < 1.
    Ans.: (This is not a water-tight proof, but good enough to convince us now.)
Consider f(x) = ln (1+x) - [x -x²/2 +x³/3 -x⁴/4 +- ...]
Then D_x f(x) = 1/(1+x) - lim_n®¥ [1 - x + x² - x³ + - . . . - (-1)ⁿxⁿ/n]
= 1/(1+x) - lim _n\->¥ [1+(-1)ⁿxⁿ⁺¹]/(1+x)
= 1/(1+x) - 1/(1+x) = 0, because lim xⁿ = 0.
Hence by Theorem on property of constant functions, f(x) is a constant function. Furthermore, as x®0, f(x) ® ln 1 - 0 = 0. Hence f(x) = 0 for all x.
Hence ln (1+x) = x - x²/2 + x³/3 - + . . . is the logarithmic series, which converge if -1 < x £ 1.
We also have discovered the practically useless formula that
ln 2 = 0.6931... = 1 - 1/2 + 1/3 -+ . . . = 1 - [ 1/(2 . 3) + 1/(4 . 5) + . . . ]
    Remark. This series is used inside your pocket calculators. In the old days, they can hand-calculate ln (3/2), with x = 0.5, then ln (4/3) then ln 2 = ln (3/2) + ln (4/3), then ln 4, then ln (5/4), then ln 5 = ln 4 + ln (5/4), then ln 10 = ln 2 + ln 5.
It is important to memorise: ln 10 = 2.30258509...

    Problem 6. Define common logarithms, and explain why log_ab = ln b / ln a. Explain why common logarithms are useful.
    Ans.: log₁₀x is defined as ln x / ln 10 = (ln x) / 2.30258 50929 94043 68401 7....
= k ln x where k = 0.43429 44819 03251 82765 ... (Memorize it)
    In general log_ax is defined as ln x / ln a. Then log_a (x y) = log_a x + log_a y and log_a (xⁿ) = n log_a x
    There is an advantage of log₁₀x, namely, log₁₀ (10ⁿx) = n + log₁₀x. Thus: log₁₀2 = 0.69314718 × 0.43429448 = 0.301029997. Hence log₁₀2000 = 3.301029997. Historically, they employed hundreds of apprentices like human calculators to calculate log₁₀ of 1.00001, 1.00002, etc. until 9.99999, and printed them as a 200 page book. If you want to find log₁₀ 0.0002345 you rewrite it as 10_-4 × 2.345, so its log₁₀ is -4 + log₁₀2.345 etc.
    Back in the 1950's I would carry a book called "Tables of logarithms and trig functions" with me all the time, in grades 11, 12, and in college. In chemistry,
if I wanted to compute x = 0.3456 * 345.7 / (76.54 * 543.21) I can first compute
log₁₀x = log₁₀0.3456 + log₁₀345.7 -log₁₀76.54 -log₁₀345.7,
then search the whole table for x whose logarithm is this value. This is the fastest way to compute these things to 4 or 7 significant figures, before the invention of the pocket calculator. Slide rule is another way, but they are accurate to no more than 3 figures.

    Problem 7. Define exponential function, and prove its basic properties.
    Ans.: We use the rectangular hyperbola y = 1/t in the ty-plane. Given any real number x, we define exp x to be the value t' on the t-axis such that x = ln t'.
Thus exp x satisfies ln (exp x) = x . . . . (7.1)
    On the other hand, for any value t > 1, if x = ln t, then by definition of exp x,
t = exp x. Hence t = exp (ln t) . . . . (7.2)
    For values of x < 0, we can also define exp x to be the value t with 0 < t < 1 such that ln t = x < 0. Thus 0 < exp x < ¥, and exp x is a monotonically increasing function of x. It is called an inverse function of ln x. Their graphs on the same xy-plane will be mirror images of each other across the line y = x.
    Now define e to be the value on t-axis such that ln e = 1. Hence for any rational value x, x ln e = x, or ln (e^x) = x. But ln exp x = x.
Hence exp x = e^x . . . . (7.3) for all x.
    It follows that e^x e^y = e^x+y, e^x / e^y = e^x-y, e^{ln x} = x and ln (e^x) = x.

    Problem 8. Prove that e = 1 + 1 + 1/2! + 1/3! + . . .
    Ans.: Consider ln (1+ 1/n)ⁿ. It = n ln(1 + 1/n)
= n (1/n - 1/2n² + 1/3n³ -+ . . .)
= 1 - 1/2n + 1/3n² - 1/4n³ +- . . . = x_n, say.
Now 1/2n + 1/3n² + 1/4n³ + ... < (½)(1/n + 1/n² + 1/n³ + . . . )
which = (½) (1/n) / (1 - 1/n) = (½) 1/(n-1).
It follows that x_n is between 1 ± (½) 1/(n-1).
Therefore lim _n®¥ ln [(1 + 1/n)ⁿ] is between 1 ± lim (½) 1/(n-1), which is 1.
We have thus proved that lim (1 + 1/n)ⁿ = e.
Using Binomial theorem, e =
lim_n®¥ [ 1 + n (1/n) + (n(n-1)/2!)/n² + (n(n-1)(n-2)/3!)n³ + . . . ]
= 1 + 1 + 1/2! + 1/3! + . . . = å_{i = 0 to ¥}(1/i!).
By hand we can compute as follows:

 1st term = 1 =           1.0000000

 2nd term = 1 =           1.0000000

 3rd term = 2nd term/2    0.5000000

 4th term = 3rd term/3    0.1666667

 5th term = 4th term/4    0.0416667

 6th ,,  = 5th ,, /5      0.0083333

 7th ,,  = 6th ,, /6      0.0013889

 8th ,,  = 7th ,, /7      0.0001984

 9th ,,  = 8th ,, /8      0.0000248

 10th ,, = 9th ,, /9      0.0000028

 11th ,, = 10th ,,/10     0.0000003

 12th term and others are negligible.

 Now add them up:         2.7182819 which is accurate to 7 places.

    Problem 9. Prove that D_x e^x = e^x and D_x a^x = a^x ln a.
Also e^x = 1 + x + x²/2! + x³/3! + . . .
    Ans.: From ln e^x = x, we get D_x(ln (e^x)) = 1. Using chain rule, that gives (1/e^x) D_x(e^x) = 1. Hence D_x e^x = e^x.
    Warning D_x e^x is not x e^x-1. Why? because
D_x xⁿ = n x^n-1 is true when x is the variable and the superscript n is constant.
But D_x a^x = ln a a^x is different, because a is constant, and x is now the superscript.
    Now a^x = (e^{ln a})^x =
Using chain rule, D_x a^x then = e ^{x ln a} ln a = a^x ln a.
    Now consider f(x) = 1 + x + x²/2! + x³/3! + . . . Obviously D_xf(x) = f(x). . . . (9.2)
Now consider D_x ln(f(x)). Using chain rule, it is (1/f(x))×D_xf(x) = 1 due to (9.2).
Hence ln(f(x)) = x + C. Hence f(x) = e^x+C = e^x e^C = C₂e^x where C₂ = e^C.
Now f(0) = 1, hence C₂ e⁰ = 1, or C₂ = 1. Hence f(x) = e^x.

    Problem 10. Find D_x x^x.
    Ans.: It is not x^x-1 because the formula D_xxⁿ = n x^n-1 is valid only when n is a constant.
    It is not x^x-1ln x, because the formula D_xa^x = a^x ln a is true only when a is a constant.
My trick is: D_xx^x = D_xe^{x ln x} = by chain rule, e^{x ln x} [x (1/x) + 1 ln x] = x^x (1 + ln x).

    Exercise 9.1. Find D_x x^xx. Find D_x (x^x)^x. Are they the same? They should not be same.

    Exercise 9.2. Fine D_x (sin x)^{sin x} and D_xx^{sin x}.

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