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Physics 2984: Basic Tools for Physics
Spring 2009, First Midterm Part 1 Sample Solutions
- (20 points) Find the sums of the following series.
(a)
∑^ ∞
n=
2 n^
(b)
∑^ ∞
n=
2 nn!
∑^ ∞
n=
(1/2)n n! = e^1 /^2 =
e.
(c)
∑^ ∞
n=
n(n + 1)
∑^ N n=
n(n + 1)
∑^ N
n=
n
n + 1
[(
N − 1
N
N
N + 1
)]
N + 1
Therefore,
∑^ ∞ n=
n(n + 1) = lim N →∞
∑^ N
n=
n(n + 1) = lim N →∞
N + 1
(d)
∑^ ∞
n=
n(n + 1)(n + 2)
∑^ N
n=
n(n + 1)(n + 2)
∑^ N
n=
n
n + 1
n + 2
N − 2
N − 1
N
N − 1
N
N + 1
N
N + 1
N + 2
2 −^
N + 1 +^
N + 2.
Therefore,
∑^ ∞ n=
n(n + 1)(n + 2) = lim N →∞
∑^ N
n=
n(n + 1)(n + 2) = lim N →∞
N + 1
N + 2
- (30 points) The Taylor expansion of a function f (x) around x = 0 is given by
f (x) =
∑^ ∞
n=
anxn^ =
∑^ ∞
n=
f (n)(0) xn n! , f (n)(x) = dn dxn^ f (x).
Find the first five coefficients a 0 through a 4 for the following functions:
(a) f (x) = sin x x
. Hint: derive the Taylor expansion of sin x first.
The Taylor expansion of sin x is
sin x = x − x
3 3!
5 5!
Therefore, sin x x
x^2 3!
x^4 5!
(b) f (x) = ln
1 + x 1 − x
. Hint: ln
1 + x 1 − x
= ln(1 + x) − ln(1 − x).
The Taylor expansions of ln(1 + x) and ln(1 − x) are
ln(1 + x) = x − x^2 2
x^3 3
x^4 4
ln(1 − x) = −x − x^2 2
x^3 3
x^4 4
Therefore,
ln
1 + x 1 − x
= 2x +^2 x
3 3
(c) f (x) = tan−^1 (x). Hint: tan−^1 (x) is an odd function of x.
f (x) = tan−^1 (x) → f (0) = 0 , f ′(x) =
1 + x^2 → f ′(0) = 1 , f ′′(x) = − 2 x (1 + x^2 )^2 → f ′′(0) = 0 ,
f ′′′(x) = −2 + 6x^2 (1 + x^2 )^3 → f ′′′(0) = − 2 ,
f ′′′′(x) = 24 x − 24 x^3 (1 + x^2 )^4 → f ′′′′(0) = 0.
Therefore,
tan−^1 (x) = 0 + (1) x + (0) x^2 2!
x^3 3!
x^4 4!
