Download Quiz 6 Solution: Convergence of the Series ∞ n=1 x2n 2n and more Exercises Calculus in PDF only on Docsity!
Answer Key for Quiz 6 (section B)
If an = x^2 n 2 n , then an+1 = x2(n+1) 2(n + 1)
x^2 n+ 2 n + 2
Note that an and an+1 are always positive, because of the even powers of x, so we won’t need absolute values when we use the ratio test. We have to calculate
nlim→∞
∣∣^ an+ an
∣∣ = lim n→∞
x^2 n+ 2 n+ x^2 n 2 n = (^) nlim→∞ x^2 n+ x^2 n
2 n 2 n + 2 = (^) nlim→∞ x^2 n n + 1 = x^2 ,
where to do the remaining limit we can either write
nlim→∞
n n + 1 = (^) nlim→∞
1 + (^1) n
or
nlim→∞
n n + 1 =^ nlim→∞
(n + 1) − 1 n + 1 =^ nlim→∞ 1 −^
n + 1 = 1^ −^ 0 = 1 or use L’Hopital’s rule, since it has the form ∞∞ :
nlim→∞
n n + 1
LH = lim n→∞
So the series converges if x^2 < 1, it diverges if x^2 > 1, and we don’t know yet what it does if x^2 = 1. If x^2 = 1 then x = 1 or x = −1, so these are the endpoints that it converges in between. If we plug x = 1 into the series we get ∑∞ n=
12 n 2 n
∑^ ∞
n=
2 n
∑^ ∞
n=
n
since the p-series with p = 1 diverges. If we plug x = −1 into the series we get the same thing, since (−1)^2 n^ = 1 because the power of −1 is always even. So the interval of convergence is − 1 < x < 1. One way to get the function that the series converges to is to calculate the derivative:
∑^ ∞ n=
2 n x^2 n−^1 2 n
∑^ ∞
n=
x^2 n−^1 = x + x^3 + x^5 + x^7 +...
This is geometric with ratio x^2 since each term is x^2 times the preceding term. Since we know x^2 < 1, this series converges to x/(1 − x^2 ). Therefore the original series must be the integral of this; that is,
∑^ ∞ n=
x^2 n 2 n
x dx 1 − x^2
We can do this integral by letting u = 1 − x^2 , so that du = − 2 x dx and therefore − 12 du = x dx and we have
∑^ ∞ n=
x^2 n 2 n
∫ (^) x dx 1 − x^2
= − 1 2
du u
ln |u| + C
= −
ln | 1 − x^2 | + C = −
ln(1 − x^2 ) + C;
we don’t need the absolute values since we know x^2 < 1. We also don’t need C, since if we set x = 0 above we get 0 = 0 + C and therefore C = 0. So we finally have
(A)
∑^ ∞
n=
x^2 n 2 n
ln(1 − x^2 ) if − 1 < x < 1.
This also holds at the endpoints, in the sense that both sides are infinite. But we actually knew (A) already, because we knew
(B)
∑^ ∞
n=
xn n = − ln(1 − x) if − 1 ≤ x < 1,
and we only have to replace x by x^2 in (B) and divide through by 2 to get (A).
- The function ex^ + e−x 2 has a name; it is called the^ hyperbolic cosine^ of^ x^ and denoted by cosh^ x. Its derivative is 1 2
ex^ + e−x(−1)
ex^ − e−x 2 and this function also has a name; it is called the hyperbolic sine of x and denoted by sinh x. Its derivative is 1 2
ex^ − e−x(−1)
= e
x (^) + e−x 2 = cosh x.
In other words, the derivatives of these functions repeat in a cycle of 2; so they are better than sin x and cos x in this sense in that the trig functions repeat in a cycle of 4. If f (x) = cosh x, then
f (x) = cosh x = f ′′(x) = f ′′′′(x) = f (6)(x) = f (8)(x) = f (10)(x) = f (12)(x) =...
and f ′(x) = sinh x = f ′′′(x) = f (5)(x) = f (7)(x) = f (9)(x) = f (11)(x) = f (13)(x) =...
Since
cosh 0 = e^0 + e−^0 2 =
2 = 1^ and^ sinh 0 =^
e^0 − e−^0 2 =
we have
ex^ + e−x 2 = 1 + 0x^ +^
x^2 2! + 0 +^
x^4 4! + 0 +^
x^6 6! + 0 +^
x^8 8! + 0 +^ · · ·^ =
∑^ ∞
n=
x^2 n (2n)! = cosh^ x.
We also have ex^ − e−x 2 =^ x^ +^
x^3 3! +^
x^5 5! +^
x^7 7! +^ · · ·^ =
∑^ ∞
n=
x^2 n+ (2n + 1)! = sinh^ x.
