Download Quiz 5 Solution: Convergence of a Series and Taylor Series Expansion and more Exercises Calculus in PDF only on Docsity!
Answer Key for Quiz 5 (section A)
- Start by using the ratio test:
an = (x − 5)^2 n−^1 2 n − 1 , so an+1 = (x − 5)2(n+1)−^1 2(n + 1) − 1
(x − 5)^2 n+ 2 n + 1
and therefore
nlim→∞
∣∣^ an+ an
∣∣ = lim n→∞
∣∣^ (x^ −^ 5) 2 n+ 2 n + 1
2 n − 1 (x − 5)^2 n−^1
= (x − 5)^2 nlim→∞ 2 n − 1 2 n + 1 = (x − 5)^2 nlim→∞ 1 − (^21) n 1 + (^21) n = (x − 5)^2
= (x − 5)^2.
So the series converges if (x − 5)^2 < 1, it diverges if (x − 5)^2 > 1, and we don’t know yet what it does if (x − 5)^2 = 1. If (x − 5)^2 = 1 then either x − 5 = 1 or x − 5 = −1; in other words, either x = 6 or x = 4. We know so far that the series converges if 4 < x < 6 and that the only other places where it might converge are x = 4 and x = 6. Plugging x = 6 into the series we get
∑^ ∞ n=
(6 − 5)^2 n−^1 2 n − 1
∑^ ∞
n=
2 n − 1
∑^ ∞
n=
2 n
since it is half of a diverging p-series. So we have divergence when x = 6. When x = 4 the series becomes
∑^ ∞ n=
(4 − 5)^2 n−^1 2 n − 1
∑^ ∞
n=
(−1)^2 n−^1 2 n − 1
∑^ ∞
n=
2 n − 1
since an odd power of −1 equals −1. This is the negative of the series we got for x = 6, so it diverges too.
Therefore the series
n=
(x−5)^2 n−^1 2 n− 1 converges only for 4^ < x <^ 6.
We can find what function the series converges to by looking at its derivative, which is
∑^ ∞ n=
(2n − 1)(x − 5)^2 n−^2 2 n − 1 =
∑^ ∞
n=
(x − 5)^2 n−^2 = 1 + (x − 5)^2 + (x − 5)^4 + (x − 5)^6 +...
This is geometric with ratio (x − 5)^2 and first term 1, so it converges to
1 1 − (x − 5)^2
[1 − (x − 5)][1 + (x − 5)]
(6 − x)(x − 4) for − 1 < x − 5 < 1, which means 4 < x < 6.
Therefore
∑^ ∞ n=
(x − 5)^2 n−^1 2 n − 1 =
dx (6 − x)(x − 4)
=^1 2
(6 − x) + (x − 4) (6 − x)(x − 4) dx
=^1 2
dx x − 4
dx x − 6 =
(ln |x − 4 | − ln |x − 6 |) + C
for some constant C. If we plug in x = 5 then the series becomes zero and so does everything on the right except C, so C = 0. Since we also know that 4 < x < 6, we can say |x − 4 | = x − 4 and |x − 6 | = 6 − x, so we finally have ∑∞ n=
(x − 5)^2 n−^1 2 n − 1
=^1
ln
x − 4 6 − x
if 4 < x < 6.
- We have to plug into the formula
f (a) + f ′(a)(x − a) + f ′′(a) (x^ −^ a)
2 2!
3 3!
with a = 1 and f (x) =
x + 2. We can plug the 1 in right away: we need to work out
(T) f (1) + f ′(1)(x − 1) + f ′′(1) (x^ −^ 1)
2 2!
3 3! In particular we need the first three derivatives of f (x) = (x + 2)−^1 , which are
f ′(x) = (−1)(x + 2)−^2 = −
(x + 2)^2 f ′′(x) = −(−2)(x + 2)−^3 = 2(x + 2)−^3 = 2 (x + 2)^3 f ′′′(x) = 2(−3)(x + 2)−^4 = −
(x + 2)^4 Plugging in x = 1 we get
f (1) =
3 ,^ f^
9 ,^ f^
27 ,^ f^
Putting these numbers in formula (T) we get the answer
1 3
(x − 1) +
(x − 1)^2 2!
(x − 1)^3 3!
(x − 1) 9
(x − 1)^2 27
(x − 1)^3 81
We can get the full Taylor series by realizing that
f (n)(x) = (−1)n^ n! (x + 2)n+^ and therefore f (n)(1) = (−1)n^ n! (1 + 2)n+^
)n n! 3
So the general formula ∑^ ∞ n=
f (n)(1) (x − 1)n n! becomes
∑^ ∞
n=
n! 3
(x − 1)n n!
)n
∑^ ∞
n=
1 − x 3
)n .
Another way to get this is to recall that
(G)
1 − r =
∑^ ∞
n=
rn^ if − 1 < r < 1.
If we rewrite 1 x + 2
x − 1 + 3
=^1
1 + x− 31
=^1
1 − 1 − 3 x
and use (G) with r = 1 − 3 xthen we get
1 x + 2
∑^ ∞
n=
1 − x 3
)n
as before, and the series converges if
− 1 < 1 − x 3 < 1 which implies − 2 < x < 4.
