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Homework Assignment 20

Assigned Tuesday March 8; not to be submitted.

this document in PDF: HW.pdf

Required reading. All of Spivak's Chapter 20.

In class review problem(s) (to be solved in class on Tuesday March 15). Chapter 20 problem 16:

  1. Prove that if $ f''(a)$ exists, then

    $\displaystyle f''(a)=\lim_{h\to 0}\frac{f(a+h)+f(a-h)-2f(a)}{h^2}. $

    The limit on the right is called the Schwarz second derivative of $ f$ at $ a$. Hint: Use the Taylor polynomial $ P_{2,a}(x)$ with $ x=a+h$ and with $ x=a-h$.
  2. Let $ f(x)=x^2$ for $ x\geq 0$ and $ -x^2$ for $ x\leq 0$. Show that

    $\displaystyle \lim_{h\to 0}\frac{f(0+h)+f(0-h)-2f(0)}{h^2} $

    exists, even though $ f''(0)$ does not.
  3. Prove that if $ f$ has a local maximum at $ a$, and the Schwartz second derivative of $ f$ at $ a$ exists, then it is $ \leq 0$.
  4. Prove that if $ f'''(a)$ exists, then

    $\displaystyle \frac{f'''(a)}{3} = \lim_{h\to 0}\frac{f(a+h)-f(a-h)-2hf'(a)}{h^3}. $

Recommended for extra practice. From Spivak's Chapter 20: Problems 3, 4, 5, 6, 9, 18 and 20.

Just for fun. According to your trustworthy professor, $ \displaystyle P_{2n+1,0,\sin}(x)=\sum_{k=0}^n(-1)^k\frac{x^{2k+1}}{(2k+1)!}$ should approach $ \sin x$ when $ n$ goes to infinity. Here are the first few values of $ P_{2n+1,0,\sin}(157)$:

$ n$ $ P_{2n+1,0,\sin}(157)$
0 157.0
1 -644825.1666
2 794263446.1416
3 -465722259874.7894
4 159244913619814.5429
5 -35629004757275297.7787
6 5619143855101017161.3172
7 -658116552443218272478.0047
8 59490490719826164706638.3418
9 -4275606060900548165855463.4918
10 250142953226934230105633222.4574
100 $ \sim 5.653\cdot 10^{63}$

In widths of hydrogen atoms that last value is way more than the diameter of the observable universe. Yet surely you remember that $ \vert\sin 157\vert\leq 1$; in fact, my computer tells me that $ \sin 157$ is approximately -0.0795485. In the light of that and in the light of the above table, do you still trust your professor?

The Small Print. For $ n=(200, 205, 210, 215, 220)$ we get $ P_{2n+1,0,\sin}(157)=(1.8512\cdot 10^8, -13102.9, 0.648331, -0.0795805, -0.0795485)$.

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Dror Bar-Natan 2005-03-09