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# Solution of Term Exam 4

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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 .
Prove the following:
1. is everywhere differentiable and .
2. for all . The only lemma you may assume is that if a function satisfies for all then is a constant function.

1. The fact that means that . Hence, using we get

This proves both that is differentiable at and that .
2. Consider . Differentiating we get

Hence is a constant function. But , hence this constant must be . So and thus .

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

1. .
2. (assume that and that and ).
3. .
4. . (This, of course, is ).

1. .
2. Let denote the anti-derivative we are interested in; i,e,, . Integrating by parts twice we get

so

or

(with the neglected).
3. Taking hence and using a formula from class, , we get

Problem 3.

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

1. 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.
2. 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.

1. 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).
2. 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 ,

1. 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

Proof: See your class notes from March 10, 2005.
2. By the remainder formula with and with we have that for any , there is a for which

But implies and so

Now take big enough so that (this is possible because ) and get

as required.

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

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Dror Bar-Natan 2005-03-23