February 21, 2011
18.01 Problem Set 2 solutions
Part II: 15 points
1a) For n a positive integer, calculate the derivative of the function f n
( x ) = n [ x
1 /n − 1] ( x > 0) .
This function is nx
1 /n derivative
− n . The second term is a constant, with derivative zero. The first has n ·
1 n x
1 n
− 1
= x
− 1+
1 n , so that’s the derivative of f n
.
b) Calculate lim n →∞
As n → ∞ , 1 /n → f
0 n
( x ) . (This is meant to be very easy.)
0, so the exponent in on the exponent a , this means that f
0 n is approaching 1. Since x a depends continuously lim n →∞ x
− 1+1 /n
= x
− 1
( x > 0) .
c) As n gets bigger and bigger, the derivative of f n derivative of ln . You might suspect that f n is getting closer and closer to the is getting closer and closer to ln . To check whether this is reasonable, use a calculator to compute f
10
(2) , f
100
(2) , and ln(2) .
You’re just asked to push buttons: f
10
(2) ≈ .
716635 ., f
100
(2) ≈ .
695555 , ln(2) ≈ .
693147 .
You might take this as experimental evidence that f n
(2) − ln(2) ≈ .
24 /n, which does indeed go to zero.
d) Suppose n = 2 k is a power of 2 . Explain how to calculate calculate f n by repeated extraction of square roots (together with subtraction and multiplication by 2 ). Could this be a reasonable way for computers to calculate natural logarithms?
Since the fourth power of y is the square of the square of y , the fourth root of x is the square root of the square root. (This sounds a bit like a mystical incantation, but it’s just a restatement of the fact that n th roots are defined as undoing n th powers. You should think about it until it makes sense.)
For the same reason, the eighth root of x is the square root of the square root of the square root. For instance:
√
65536 = 256 ,
√
256 = 16 ,
√
16 = 4; and therefor 65536 1 / 8 = 4.
If you repeat this argument k times, you find x
1 / 2 k
= r q
· · ·
√ x .
| {z } k times

That is, you can calculate the root in f
2 k
( x ) = 2 k ( x 1 / 2 k times.
− 1) by taking successive square roots k
So is this a reasonable way for computers to calculate natural logarithms? The preceding part suggested that the error in the approximation f
2 k
( x ) to ln( x ) is on the order of 2
− k study approximations more carefully, you’ll learn how to estimate the error as ln
2
( x ) / 2
. (After we k +1
.) To get
9 decimal places of accuracy for ln(2), you therefore want 2 k to be around 10
− 9
; that is, k around
30. So the computer should extract square roots about thirty times, to an accuracy of nearly twenty decimal places. (The reason you need extra accuracy is that the square roots end up very close to one; then we subtract one and multiply by 2
30
, losing about nine decimal places of accuracy in the process.) It isn’t completely ridiculous, but there are much faster methods available, that we’ll talk about at the end of this course.
2. This problem is about the circle x 2 + y 2 = 1 .
a) Use implicit differentiation to calculate the slope of the tangent line to the circle at ( x, y ) .
Differentiating the defining equation gives dy
2 x + 2 y dx
= 0 , dy dx
= − x y
.
So the slope of the tangent line is − x y
.
b) Calculate the slope of the line through the origin and the point ( x, y ) .
The slope is (change in y ) divided by (change in x ), or y − 0 x − 0
= y x
.
(The two zeros are the coordinates (0 , 0) of the origin.) c) Two lines L
1 and L
2 are perpendicular to each other if their slopes m
1 and m
2 are negative reciprocals; that is, m
1 m
2
= − 1 . Can you find a geometric reason for the lines in (a) and (b) to be perpendicular to each other?
The two lines in question are the tangent to the circle at ( x, y ) and the radius at that point.
These lines meet at the point ( x, y ). You can flip the circle over this radius without changing it, so flipping the tangent line over the radius doesn’t change the tangent line. It follows that the angles made by the tangent line on the two sides of the radius must be equal (since one is carried to the other by the flip over the radius). Since these two angles add up to 180
◦
, each must be 90
◦
.
There are probably many other ways to deduce this fact.
3a) The derivative of the function 2 u
2 − 1 with respect to u is 4 u . Use this fact and the chain rule to calculate the derivative of 2 cos 2 ( θ ) − 1 with respect to θ .
The chain rule says d (2 u 2 − 1) dθ
= 4 u du
, dθ d (2 cos 2 ( θ ) − 1)
= 4 cos( θ )( − sin( θ ) = − 4 sin( θ ) cos( θ ) .
dθ b) Calculate the derivative of 1 − 2 sin
2
( θ ) with respect to θ . What does this have to do with (a)?
Exactly the same argument gives the derivative
− 4 sin( θ ) cos( θ ) ,

which is the same as the derivative in (a).
One way to see this is that the functions we are differentiating are both equal to cos(2 θ ) (by the doubling formula for cosine); same function, so same derivative. Or you can subtract the two functions, getting
(2 cos
2
( θ ) − 1) − (1 − 2 sin
2
( θ )) = 2(cos
2
( θ ) + sin
2
( θ )) − 2 = 2 − 2 = 0 .
c) A trigonometric polynomial is a sum of terms of the form a cos m
( θ ) sin n
( θ ) . Explain why the derivative of any trigonometric polynomial is another trigonometric polynomial.
According to the product rule and the chain rule, d dθ
(cos m
( θ ) sin n
( θ )) = − m cos m − 1
( θ ) sin n +1
( θ ) + n cos m +1
( θ ) sin n − 1
( θ ) , which is a trigonometric polynomial. So the derivative of any trigonometric polynomial is a sum of multiples of trigonometric polynomials, which is a trigonometric polynomial.
d) It’s not hard to see that any polynomial is the derivative of another polynomial.
Is this true for trigonometric polynomials?
The answer turns out to be no , although we don’t quite have the tools to prove this. The trigonometric polynomial 1 (take n = m = 0 in the definition) is the derivative only of θ + a , with a a constant. (Of course you know that 1 is the derivative of θ + a with respect to θ ; what we haven’t yet proved is that it can’t be the derivative of anything else. So the question is whether
θ + a is a trigonometric polynomial, for some value of a . It doesn’t look like one, but that isn’t a reason; cos 2 ( θ ) + sin
2
( θ ) doesn’t look like 1, but it’s equal to it.
What happens is that any trigonometric polynomial is bounded : since sin and cos are never larger than 1 or smaller than − 1, b cos m
( θ ) sin n
( θ ) is never larger than | b | or smaller than −| b | . If you add several of these terms, you can add the bounds. But the function θ + a is always unbounded : it tends to ∞ as θ → ∞ , and to −∞ as θ → −∞ . So θ + a cannot be a trigonometric polynomial, so 1 is not the derivative of a trigonometric polynomial.
A non-zero constant is the only example of a trigonometric polynomial that isn’t the derivative of another trigonometric polynomial (although you can make this example look different using
1 = cos
2
( θ ) + sin
2
( θ )). One precise statement is for any trigonometric polynomial T ( θ ) there is a unique constant a and another trigonometric polynomial S ( θ ) so that S
0
= T − a .
We’ll learn how to calculate S later in the course.