MATH 75 = CALCULUS I HOMEWORK #8, DUE MONDAY 18 NOV 2002 ----------------WARREN D. SMITH-------------------------- A) Reading. Basically we are done - ch 1-5, appendix on Newton method, ch 10.1-10.4 on Taylor series, ready reference and frontspiece and endspiece of book are the works. Rest of course is review and filling-in. --------------------------------------------------------------------- 2 1. Write the area (=pi r / 2) of a semicircle as a definite integral (the area under the curve of the semicircle). (Do not actually try to do the integral.) ANSWER: integral from -r to r of sqrt(r^2 - x^2) dx Also, since if the area is A(r) then arclength is A'(r) (consider putting a thin "coat of paint" dr thick on the surface of the semicircle making r grow to r+dr, the area increases by A'(r) dr and also by L dr, so L=A'(r) where L is the length and L=pi r, we have A(r) = integral from 0 to r of pi t dt. 2. Find the Taylor series (based at x=0) of arctan(x). That includes an infinite number of terms - you need to describe them all! Hint: It may help, as an tool, first to find a formula for the derivative (d/dx) of A x ----------- 2 B ( 1 + x ) for any nonnegative integers A,B. ANSWER: 3 5 7 9 arctan(x) = x - x / 3 + x / 5 - x / 7 + x / 9 - ... in which hopefuly you can see the pattern... this can be rewritten in several ways using sum notation, e.g. infinity k 2k arctan(x) = x SUM (-1) x / (2k+1). k=0 EASY WAY TO GET THIS: How did I get this? Well, the easiest way is to "cheat" (i.e. to avoid carrying out the whole Taylor series recipe, and get there via a shortcut) by remembering that 1 2 4 6 8 10 arctan'(x) = ------ = 1 - x + x - x + x - x + ... 2 1 + x (the left hand equality is in the book p.135, the right hand one is a "geometric series" and you can verify it by multiplying both sides by 2 (1+x ) whereupon the right hand side "telescopes": 2 4 6 8 10 1 - x + x - x + x - x + ... <--- 1 times series 2 4 6 8 10 2 + x - x + x - x + x + ... <--- x times series ----------------------------------------------------------------- = 1 + 0 + 0 + 0 + 0 + 0 +... <--- (1 + xx) times series = 1. So now we see that 3 5 7 9 arctan(x) = x - x / 3 + x / 5 - x / 7 + x / 9 - ... has to be the right answer since its derivative is the right answer for arctan'(x), and arctan(0)=0 works. EASIER STILL: The answer is found on page 448 of the book! HARD (BUT STRAIGHTFORWARD) WAY TO GET THIS: Start making a table of derivatives of F(x)=arctan(x): F(0)=arctan(0)=0 1 F'(x) = ------ , F'(0)=1 2 1 + x - 2 x F''(x) = ---------- , F''(0)=0 2 2 (1 + x ) 2 3 x - 1 F'''(x) = 2 --------- , F'''(0) = -2/8 = -2 2 3 (1 + x ) 3 (4 primes) x - x F (x) = -24 --------- , F''''(0)=0 2 4 (1 + x ) 4 2 (5 primes) 1 + 5 x - 10 x F (x) = 24 ---------------- , F'''''(0) = 24 2 5 (1 + x ) When doing the above, it helps to know that A A-1 A+1 d x A x + (A-2B) x -- ( --------- ) = --------------------- (handy tool) dx 2 B 2 B+1 (1 + x ) (1 + x ) which is a handy tool which you can get by using the product and power+chain rules, then simplifying the result. This tool makes it easy to keep differentiating 5 times in a row like I just did; without this tool it is hard. Well, can we see a pattern? Looks pretty hard to see a pattern! But you can at least see (by plugging F(0), F'(0), F''(0), etc into the master Taylor series formula) that 2 3 F(x) = F(0) + x F'(0) + F''(0) x / 2! + F'''(0) x / 3! + ... 3 4 5 = 0 + x + 0 + x (-2)/6 + 0 x + 24 x / 120 +... 3 5 = x - x / 3 + x / 5 ... and NOW maybe we can see a pattern! Notice that in (handy tool), ODD exponents A on top get changed to EVEN ones (and vice versa) every time you differentiate, so that you have ALL-ODD and ALL-EVEN on top, alternately, which explains why we get (n primes) F (0) = 0 whenever n is EVEN! (Since only odd exponents on top!) Which explains why only ODD powers appear in the final series answer! ... 3. How could this series from exercise 2 be used to compute pi? Hint: think of some special value of x, such that pi could be easily found if you knew arctan(x). (Not allowed to use the extra credit answer below, must think of something else.) ANSWER: 4 arctan(1) = pi since a 45 degree angle (pi/4 radians) has tan=1. So: pi/4 = 1 - 1/3 + 1/5 - 1/7 + 1/9 - 1/11 + 1/13 - ... (Gregory's series) 4. What is the area under the curve y=1/x and above the x-axis, from x=1 to x=10? What about from x=1 to x=100? ANSWER: integral from 1 to 10 of 1/x dx. Because ln'(x) = 1/x, this is AREA = ln(10)-ln(1) = ln(10) [since ln(1)=0]. Similarly integral from 1 to 100 of 1/x dx = ln(100)-ln(1) = ln(100) = 2 ln(10) [since ln(AB)=ln(A)+ln(B) and here use A=B=10]. There is another reason why ln(100)=2ln(10). Divide the area into 2 pieces: from 1 to 10 [answer=ln(10)] and from 10 to 100. The area under the curve 1/x from x=10 to x=100 becomes the SAME THING as the area under the curve 1/x from x=1 to x=10 IF * we squish the x-axis by a factor of 10 * we expand the y-axis by a factor of 10 (net effect: area stays same) SO we have 2 pieces, each piece is area=ln(10), so total area is 2ln(10). OPTIONAL EXTRA CREDIT: Can you prove that Pi/4 = 4arctan(1/5) - arctan(1/239) ? Does this formula have advantages - in what way is it a "better" way to compute pi than the formula you found in exercise 3? ANSWER: Take the tan of both sides: tan(Pi/4)=1 = tan(4 arctan(1/5) - arctan(1/239)). ^ | the question is, is this true? We can use the tan(x+y) = [tan(x)+tan(y)]/[1-tan(x)tan(y)] ADDITION FORMULA now. We use it to see that 2 tan(2A) = 2tan(A) / (1-tan(A) ) so if A=arctan(1/5) then tan(A)=1/5 and so by the above formulas tan(2A)=5/12 and tan(4A)=tan(2A.2)=120/119. Now use the tan addition formula to see, with x=4arctan(1/5) and y=arctan(-1/239) and tan(x)=120/119 and tan(y)=-1/239, that our formula is equivalent to 120/119 - 1/239 28561/28441 1 = ---------------------- = ----------- 1 + (120/119) (1/239) 28561/28441 so it works - the claimed formula is true. The advantage of this is that you can use the two series 3 5 7 9 arctan(1/5) = (1/5) - (1/5) / 3 + (1/5) / 5 - (1/5) / 7 + (1/5) / 9 - ... 3 5 7 9 arctan(1/239) = (1/239) - (1/239) / 3 + (1/239) / 5 - (1/239) / 7 + (1/239) / 9 - ... which converge a lot FASTER. That is, their terms get small quickly, since, each term is at least 25 times smaller than the last term. That means you don't need to do a lot of work to get high accuracy. Meanwhile in the original series pi/4 = 1 - 1/3 + 1/5 - 1/7 + 1/9 - 1/11 + 1/13 - ... the terms get small SLOWLY. To get 6 decimal places accuracy you would need to take about a million terms this slow way, the fast way you only need about 10 terms. This is how pi was first calculated to several thousand decimal places.