Practice Midterm Examination 1 – Math 142B
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Practice Midterm Examination 1 – Math 142B
Question 1. Let f(x) = . (a) Prove that the series converges for |x| < 1. (b) Determine whether the series converges for x = 1 and for x = −1. (c) Does the series converge uniformly for x ∈ (−1, 1)? |
Solution (a). This is not a power series, but we can use the Weierstrass M-test. Note that for n ≥ 2, |xn /(n+xn )| ≤ |x| /(nn − 1) for |x| < 1, and so if Mn = |x| /(nn − 1) for n ≥ 2 and M1 = |x|/|1+x| then之 Mn converges (it is at most a geometric series) and Weierstrass shows 之 |xn /(n+xn )| converges for |x| < 1.
Solution (b). For x = 1, the series is 之 1/(n + 1) which is the harmonic series, and it diverges. For x = −1, it is 之(−1)n /(n+1) which converges by the alternating series theorem.
Solution (c). No, the series does not converge uniformly on (−1, 1). If fn (x) is the nth partial sum then
sup{|f(x)−fn (x)| : |x| < 1} ≥ sup{工 xk /(k+xk ) : |x| < 1} ≥工 1/(k+1).
k>n k>n
This is the end of a divergent (harmonic) series, which therefore does not go to zero, by Cauchy’s criterion. Or observe directly
2n 2n
k=n+1 k=n+1
and therefore sup{|f(x) − fn (x)| : |x| < 1} does not converge to zero.
Question 2. Let f(x) = e −1/|x| for x 0 and f(x) = 0 for x = 0.
(a) Prove that f\ (0) = 0.
(b) Prove that f(n)(0) = 0 for n ≥ 2.
(c) Determine a number r > 0 such that the Taylor series for f(x) about x = 1 converges to e −1/|x| for 1 − r < x < 1 + r .
Solution (a). By definition of the derivative:
f\(0) = lim f(h) − f(0) = lim he−|h| .
Recall ex ≥ x2 /2 from Taylor series for ex when x ≥ 0. Therefore h/e|h| ≤ h/2|h|2 and this converges to zero as h → ∞ . Therefore f\(0) = 0.
Solution (b). We claim f(n)(h) = pn(1/h)e−1/|h| for h 0, where pn is a monic polynomial of degree 2n. For n = 0 this is true since f(h) = f(0)(h) = e−1/|h| so we can take p0 (x) = 1. Suppose by induction that it is true for some n ≥ 0. Then for h 0,
f(n+1)(h) = − p( )e−1/|h| + pn( )e−1/|h| = pn+1( )e−1/|h|
where pn+1(x) = x2 (pn(x) − p(x)). This proves the claim since the degree of pn+1 is 2n + 2 from the x2pn(x) term. In particular, f(n)(h) ≤ h−2n−1e−1/|h| for small enough h. Equivalently, f(n)(1/h) ≤ h2n+1e −h for large enough h, Finally, assuming we have proved f(n)(0) = 0 (part (a) does the case n = 1):
f(n)(h) − f(n)(0)
h→0 h h→∞
Since hf(n)(1/h) ≤ h2n e−|h| for large enough h, and e|h| ≥ h2n+1 for large enough h, we find that
the limit above is zero. Therefore f(n)(0) = 0 for all n ≥ 0. 1
Solution (c). This the tricky one. We omit it.
Question 3.
(a) Prove that (1k(/)2) = (−1)k−1/k22k−1 () for k ≥ 1. (b) Prove that (k(2k)) ≤ 4k for k ≥ 1 by mathematical induction.
(c) Determine carefully for which x the Taylor series for (1 − x)1/2
about 0 converges to (1 − x)1/2 .
Solution (a). By definition
( 1k(/)2) = = .
Multiply numerator and denominator by (−2)(−4)(−6) ... (2 − 2k). Then the numerator becomes (−1)2k (2k − 2)! and the denominator is (−1)k−1 22k−1(k − 1)!k!. So we get for k ≥ 1:
( 1k(/)2) = = ).
Solution (b). For k = 1, (k(2k)) = (1(2)) = 2 ≤ 4. Suppose the statement is true for some k ≥ 1. Then
) = (k(2k)) ≤ · 4k
by induction. The fraction is at most 4 so the whole thing is at most 4k+1 as required. Solution (c). The Taylor series is computed to be
工(∞) ( 1k(/)2)(−1)kxk = 1 − 工(∞) )xk
using (a). Since () ≤ 4k−1 by (b), we have convergence for |x| < 1 and divergence for |x| > 1. Alternatively we could compute the remainder term and conclude the same.
2022-08-18