© | Dror Bar-Natan: Classes: 2004-05: Math 157 - Analysis I: | (102) |
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**Problem 1. ** Agents of
CSIS have secretly developed a
function that has the following properties:

- for all .
- is differentiable at 0 and .

- is everywhere differentiable and .
- for all . The only lemma you may assume is that if a function satisfies for all then is a constant function.

**Solution. ** (Graded by Brian Pigott)

- The fact that means that
. Hence,
using
we get
- Consider
. Differentiating we get

**Problem 2. ** Compute the following integrals: (a few lines of
justification are expected in each case, not just the end result.)

- .
- (assume that and that and ).
- .
- . (This, of course, is ).

**Solution. ** (Graded by Shay Fuchs)

- .
- Let denote the anti-derivative we are interested in; i,e,,
. Integrating by parts twice we get

so - Taking hence and using a formula from class,
, we get

**Problem 3. **

- State (without proof) the formula for the surface area of an object defined by spinning the graph of a function (for ) around the axis.
- Compute the surface area of a sphere of radius 1.

**Solution. ** (Graded by Brian Pigott)

- That surface area, excluding the ``caps'', is . Including the caps it is the same plus the area of the caps, . For the purpose of this exam, both solutions are acceptable.
- Here is the function whose graph is a semi-circle, so , and and . By direct computation, so the surface area of a sphere of radius 1 is .

**Problem 4. **

- State and prove the remainder formula for Taylor polynomials (it is sufficient to discuss just one form for the remainder, no need to mention all the available forms).
- It is well known (and need not be reproven here) that the th
Taylor polynomial
of around 0 is given by
. It is also well known (and need not be
reproven here) that factorials grow faster than exponentials, so for any
fixed we have
. Show that for large enough ,

**Solution. ** (Graded by Derek Krepski)

- Statement: If is differentiable times on the interval
between and and if denotes the th Taylor polynomial of
at , then there is some between and for which
- By the remainder formula with and with we have that
for any , there is a
for which

**The results. ** 66 students took the exam; the average
grade was 58.2 and the standard deviation was about 24.

The generation of this document was assisted by
L^{A}TEX2`HTML`.

Dror Bar-Natan 2005-03-23