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Midterm Answers Fall 2011 .pdf
Midterm Answers Fall 2011 .pdf

Midterm Answers Fall 2011 .pdf

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McGill University

Mathematics & Statistics (Sci)

MATH 222

Christa Scholtz

Fall

Description

MATH 222 MIDTERM SOLUTIONS
1. SOLUTIONS
Problem 1 Consider the sequencenfa g deﬁned recursively by
p
a1= 11; a n+1= 2an+ 3 for n ▯ 1:
p p
[So 1 = 25 = 5, 2 = 13, etc.] Which of the following is true?
(a) lin!1 an= 0:
(b) lin!1 an= 11:
(c) lim a = 3:
n!1 n
(d) lin!1 an= 2:
(e) The sequence does not converge to any real number.
Solution. The correct solution is (c). Notice that
n!1 an+1 = n!1 an
Let X = limn!1 an, then
p
X = lim 2an+ 3
p!1
= 2X + 3:
We ﬁnd that
2
X ▯ 2X ▯ 3 = 0:
Therefore X = 3 or X = ▯1. The sequence is strictly positive, so we choose X = 3.▯
Problem 2 The value of
2x 2 3
3e ▯ 3 ▯ 6x ▯ 6x ▯ 4x
x!0 x4 is
(a) 1.
(b) 2.
(c) 3.
(d) None of the above, because the limit is some other real number.
(e) None of the above, because the limit doesn’t exist.
Solution. The correct solution is (b). We can use L’Hopital’s rule, or we can substitute the
2x
Maclaurin series for 3e . We know that
X1 n
x x
e = :
n=0n!
1 2 MATH 222 MIDTERM SOLUTIONS
Therefore
▯ ▯
X1 (2x)n (2x)2 (2x) 3 (2x)4 (2x) 5
3e2x = 3 = 3 1 + (2x) + + + + + :::
n! 2! 3! 4! 5!
n=0
2 3 4 96x 5
= 3 + 6x + 6x + 4x + 2x + + :::
5!
This becomes
2x 2 3
3e ▯ 3 ▯ 6x ▯ 6x ▯ 4x = 2 + 96x + :::;
x 4 5!
and now it is possible to calculate the limit. We conclude that
2x 2 3
lim 3e ▯ 3 ▯ 6x ▯ 6x ▯ 4x = 2 + 0;
x!0 x4
and therefore the correct answer is (b). ▯
Problem 3 The interval of convergence of the power series
1
X 5 + 2 n n
(x ▯ 2) is
n=1 n
(a) (▯:2;:2)
(b) [▯:2;:2)
(c) (▯1;1)
(d) (1:8;2:2)
(e) [1:8:2:2)
Solution. The correct answer is (e). Use the ratio test. Let
5 + 2 n
an= (x ▯ 2) :
n
Then ▯ ▯ ▯ ▯▯ ▯ ▯
▯a ▯ ▯ 5 n+1+ 2 n+1 n (x ▯ 2)n+1 ▯
lim ▯ n+1▯ = lim ▯ ▯
n!1 ▯ a n ▯ n!1 ▯ 5 + 2 n n + 1 (x ▯ 2) n ▯
▯ n+1 n+1 ▯ ▯ ▯
= lim ▯5 + 2 ▯▯ lim ▯ n ▯ ▯ jx ▯ 2j
n!1 ▯ 5 + 2 n ▯ n!1 ▯n + 1▯
▯ ▯ ▯ ▯
▯1 + limn!1 2 n+1▯
= 5 ▯ ▯ ▯n ▯ ▯ 1 ▯ jx ▯ 2j
▯ 1 + limn!1 5 ▯
▯ ▯
▯1 + 0▯
= 5 ▯ ▯▯ 1 ▯ jx ▯ 2j
1 + 0
= 5jx ▯ 2j
The ratio test implies that we have absolute convergence if 5jx▯2j < 1. Thus we must have
1 1
▯ < x ▯ 2 < ;
5 5 4 MATH 222 MIDTERM SOLUTIONS
Problem 5 The sum of the series

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