Practice Midterm Examination 3 – Math 142B
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Practice Midterm Examination 3 – Math 142B
Question 1. Let an ∈ R for n ≥ 1, and suppose that 对 an converges. Decide which of the following is true or false. If true, give a proof, if false give an example of a sequence (an )n≥1 for which the statement is false.
(a) 对(− 1)n an converges. (c) 对 an(3) converges. |
Solution (a). This is false. If an = ( − 1)n /n then 对 an converges (by the alternating series theorem), but 对(− 1)n an is the harmonic series, which we
proved diverges.
Solution (b). This is false. If an = ( − 1)n /^n, then 对 an converges (by the alternating series theorem), but 对 an is the harmonic series, which we proved diverges.
Solution (c)* This is false but an example is not so easy to find. Try to see where the following proof fails: by Cauchy critetion, for all e > 0, there exists N such that |对n>N an | < ^e, by Cauchy, and |an | < ^e for n > N since limn→∞ an = 0. Since an(2) = |an |2 > 0,
I an(3) I ≤n(s)pN |an |2 I an I.
Both quantities are at most ^e, and so the above expression is less than e. By Cauchy, this means 对 an(3) converges. Can you spot the error?
Question 2. For n ≥ 1, let fn (北) = sin(北n)北n and let
∞
f(北) =工 fn (北).
n=1
(a) Prove that f is continuous on ( − 1, 1).
(b) Is f(北) differentiable on ( − 1, 1)? If so prove it, otherwise disprove it.
(c) Is f(北) continuous at 北 = 1? If so prove it, otherwise disprove it.
Solution (a). For this it is enough to prove that for any 0 < s < 1, there
exists Mn such that 之 Mn converges and |fn (北)| ≤ Mn for all 北 ∈ [ −s,s].
This is true since |fn (北)| ≤ |sin 北n ||北 |n ≤ sn and so with Mn = sn we
have 之 Mn converges (it is a geometric series and 0 < s < 1). Therefore
by the Weierstrass M-test, 之 fn (北) converges uniformly to f on [ −s,s] for
0 < s < 1 . Since fn is a polynomial it is continuous on [ −s,s], so by the
theorem in the notes, f is continuous on [ −s,s] . To see that f is continuous
at any 北 ∈ ( − 1, 1), we note f is continuous on [ − |北 |, |北 |] and hence f is
Solution (b). We have f(北) = ncos(北n)北n + nsin(北n)北n − 1 for n ≥ 1. Our guess should be that
∞
f\ (北) =工 ncos(北n)北n + nsin(北n)北n − 1 .
n=1
If gn (北) = ncos(北n)北n +nsin(北n)北n − 1 and 北 ∈ [ −s,s] where 0 < s < 1, then
|gn (北)| ≤ ns +nsnn − 1 ≤ 2nsn − 1 . Since 之(2n)sn − 1 is a convergent power series
for 0 < s < 1, the Weierstrass M-test with Mn = (2n)sn − 1 shows 之 gn (北)
converges and in fact is continuous on [ −s,s]. See the proof of Theorem 2
(ii) in the book to see how to prove 之 gn (北) = f\ (北).
Solution (c). We have f(1) = 之 sin(n), which is a sum that does not converge, so f(1) is not even defined (therefore f is clearly not continuous at
北 = 1.)
Question 3. Let f(x) = (ex − 1)/x for x ∈ R/{0} and f(0) = 1. Determine
a sequence ak such that the polynomials fn (x) = 对k(n)=0 ak xk (1 − x)n −k for
n ≥ 1 converge uniformly to f(x) on [0, 1]. Does fn (x) converge uniformly to f(x) on the whole of R?
Solution. The function f is continuous on [0, 1], the only issue being when x = 0: in that case we have continuity since
lim(ex − 1)/x = lim ex = 1
fn (x) = f(k/n) ( )k(n) xk (1 − x)n −k .
Since f(k/n) = (ek/n − 1)n/k, we have
ak = (ek/n − 1)n/k · ( )k(n) .
By the Weierstrass approximation theorem, fn → f uniformly on [0, 1]. The convergence is not uniform on all of R: let us determine that
sup{|fn (x) − f(x)| : x ∈ R}
does not converge to zero. One should pick x = xn very large so that in fact the difference diverges. We leave this to the reader! Therefore fn → f pointwise but not uniformly.
2022-08-18