MA1132 (Advanced Calculus) Tutorial sheet 1
[January 22-23, 2013]
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1. Write the Maclaurin series of f (x) = e3x . Write the series with the
an enumeration of (the first few) terms of the series.
P
notation, as well as
Solution. We start by computing the derivatives of f .
f 0 (x) = 3e3x , f 00 (x) = 32 e3x , f 000 (x) = 33 e3x , f (4) (x) = 34 e3x ...,
hence, f (n) (x) = 3n e3x . We obtain that the Maclaurin series for the given function is
∞
X
f (n) (0)
n=0
n!
n
x =
∞
X
3n
n=0
n!
xn = 1 + 3x +
32 2 33 3
x + x + ...
2
6
2. Write the fourth order Taylor polynomial for f (x) = ln(2 − x) at x = 1.
Solution. The formula for the fourth order Taylor polynomial about x = 1 is given by:
Pn (x) = f (1) + f 0 (1)(x − 1) +
We have that
f 0 (x) =
f 00 (1)
f 000 (1)
f (4) (1)
(x − 1)2 +
(x − 1)3 +
(x − 1)4 .
2
3!
4!
−1
1
=
,
2−x
x−2
f 000 (x) =
2
,
(x − 2)3
f 00 (x) =
f (4) (x) =
−1
(x − 2)2
−6
,
(x − 2)4
so f 0 (1) = −1, f 00 (1) = −1, f 000 (1) = −2, f (4) (1) = −6.
Hence
1
1
1
Pn (x) = ln 1 − (x − 1) − (x − 1)2 − (x − 1)3 − (x − 1)4
2
3
4
or
Pn (x) = −(x − 1) −
(x − 1)2 (x − 1)3 (x − 1)4
−
−
.
2
3
4
Please turn over
3. Find the n-th order Taylor polynomial about x = 1 for
f (x) = a0 + a1 (x − 1) + a2 (x − 1)2 + ... + an (x − 1)n .
Solution. We compute the derivatives of f as follows:
f 0 (x) = a1 + 2a2 (x − 1) + ... + nan (x − 1)n−1
f 00 (x) = 2a2 + 3 · 2a3 (x − 1) + ... + n · (n − 1)an (x − 1)n−2 ...
f (n) (x) = n!an .
After evaluating the derivatives at x = 1 we obtain that the Taylor polynomial of order n
about x = 1 is given by
Pn (x) =
n
X
f (k) (x)
k=0
k!
(x − 1)k = a0 + a1 (x − 1) + a2 (x − 1)2 + ... + an (x − 1)n .
Remark: Another way (much simpler!) to obtain the solution is to notice that the function
is already a polynomial of order n centered at x = 1 (i.e. the powers are of (x − 1)). Since
the Taylor polynomial the order n about a given point is unique, we obtain that the function
coincides with its nth order approximation.
4. Find the local quadratic approximation of tan x at x0 = π3 . Use this result to write an
approximation for tan 58◦ . You do not need to compute the final number, just write the
expression for the approximation.
1
= 1 + tan2 x, hence the second
Solution. The first derivative of tan x is cosec2 x =
cos2 x
derivative is 2 tan x(1 + tan2 x) (we could simplify it further, but it is not necessary).
The quadratic approximation will be given by
P2 (x) = tan
Since tan π3 =
√
π
π
π
1
π
π
π
+ (1 + tan2 )(x − ) + 2 tan (1 + tan2 )(x − )2 .
3
3
3
2
3
3
3
3 we obtain
P2 (x) =
We use the fact that 58◦ =
58π
180
√
=
3 + 4(x −
29π
90
√
π
π
) + 4 3(x − )2 .
3
3
radians, hence
√
√ 29π π 2
29π
29π π
) = 3 + 4(
− ) + 4 3(
− ) = 1.60086627.
90
90
3
90
3
The final equality is the number given by the calculator, so it is not really an equality,
but rather an approximation (since we deal with irrational numbers). With a help of a
calculator we get that tan 58◦ = 1.60033452904 (again approximation up to the first 8
decimals).
P2 (
Please turn over
2
One more example:
5. Find the Maclaurin series for g(x) = sin 2x. Write the general term of the series, as well
as enough terms in the series to show a pattern.
Solution. We begin by computing the derivatives of g.
g 0 (x) = 2 cos 2x, g 00 (x) = −22 sin 2x, g 000 (x) = −23 cos 2x, g (4) (x) = 24 sin 2x,
g (5) (x) = 25 cos 2x, g (6) (x) = −26 sin 2x...
From these formulas we deduce the general formulas for the derivatives of g (note that we
have 4 possible situations):
g 4k (x) = 24k sin 2x, g 4k+1 (x) = 24k+1 cos 2x, g 4k+2 (x) = −24k+2 sin 2x,
g 4k+3 (x) = −24k+3 cos 2x, for every k nonnegative integer.
The Maclaurin series is given by
∞
X
g (n) (0)
n=0
n!
n
x =
∞
X
k=0
(−1)k
22k+1 2k+1
x
.
(2k + 1)!
Petronela Radu
3