Exercise 4

BGU Physics Dept. Introduction to Mathematical Methods in Physics
Exercise 4
Taylor expansion
1. Remainder estimation The approximation
Estimate the error when |x| < 0.01.
√
1 + x ≈ 1 + x/2 is used when x is small.
2. Estimating π using taylor expansion of arctan x.
We saw in class that in order to calculate π within an error of magnitude less than 10−3 , one
needs to calculate at least 500 terms of the series arctan(1). In this problem we will see a
more efficient method.
Define two angles
1
α = arctan ,
2
β = arctan
1
3
(?)
calculate tan(α + β), find α + β and express π/4 in terms of α + β. By expanding to a Taylor
series the expressions (?), how many terms in the Taylor expansion of arctan(x) do you need
to assure an error of magnitude less than 10−3 ?
Solution
Using the hint we see that tan(α + β) = 1, hence α + β = π/4 and finally
arctan
1
π
1
+ arctan =
2
3
4
In general, we should expand both arctan functions in a taylor series, however, since arctan 1/2
converges slower than arctan 1/3, it is enough to calculate the error in arctan 1/2.
∞
1 X
arctan =
(−1)n
2
n=0
2n+1
1
1
2
2n + 1
Since this is an alternating sign sequence, the error is bound by the first discarded term so
2n+3
1
1
< 10−3
2n + 3 2
by trial and error we get that n = 3 terms is sufficient for an error of 10−3 .
3. Obtain the taylor series for 1/(1 + x)2 from the series for −1/(1 + x).
Solution
−
1
1+x
1
(1 + x)2
∞
= −
X
1
=
(−1)n+1 xn
1 − (−x)
n=1
X
∞
∞
X
d
1
−
=
(−1)n+1 nxn−1 =
(−1)n (n + 1)xn
dx
1 − (−x)
=
n=1
1
n=0
4. (a) Find the first four nonzero terms of the Maclaurin series for
sinh−1 x =
Z
0
x
√
dt
x3 3x5 5x7
≈ x−
+
−
6
40
112
1 + t2
(b) Use the first three terms of the series in (a) to estimate sinh−1 (0.25). Give an upper
bound for the magnitude of the estimation error.
sinh−1 (0.25) ≈ 0.247469 ± 2.7 × 10− 6
5. Approximation of non elementary integrals Estimate the following integrals’ values with
an error of magnitude less than 10−8
Z 0.1
Z 0.1
x4 x6
x3 x5 x7
−x2
(1 − x2 +
e dx ≈
(a)
− )dx ≈ x −
+
−
= 0.0996687
2
3
3
10 21
0
0
Z 0.1 p
Z 0.1
x4
x5
(1 + )dx = x +
(b)
1 + x4 dx ≈
≈ 0.100001
2
10
0
0
2