Calculus II: algebraic operations on power series
1. Definitions
As polynomials, we can add, subtract, multiply or divide
power series. If the series we operate have the same radius
of convergence R, this R is remains the same and less than R
for the division.
Let two power series that converge for |x - c| < R
f(x) = Σn=0∞
an (x - c)n
and
g(x) = Σn=0∞
bn (x - c)n
a) f(x) ± g(x) =
Σn=0∞
an (x - c)n ± Σn=0∞
bn (x - c)n
b) f(x) . g(x) =
Σn=0∞
an (x - c)n . Σn=0∞
bn (x - c)n
c) f(x) / g(x) =
Σn=0∞
an (x - c)n / Σn=0∞
bn (x - c)n
with g(x) ≠ 0.
2. Examples: Hyperbolic functions
The hyperbolic functions are:
Hyperbolic sine:
sinh(x) = (ex - e-x)/2
Hyperbolic cosine:
cosh(x) (ex + e-x)/2
Hyperbolic tangent:
tanh(x) = sinh(x) /cosh(x)
Hyperbolic cotangent:
coth(x) = cosh(x) / sinh(x)
According to:
ex = 1 + x/1! + x2/2! +
x3/3! + ...
we can write the expansion in power series of x for
sinh(x):
sinh(x) = (ex - e-x)/2 =
(1/2)[(1 + x/1! + x2/2! +
x3/3! + ... ) -
(1 - x/1! + x2/2! -
x3/3! + ...)]
= x + x3/3! + x5/5! +
x7/7! + ...
3. Examples: Approximation of functions
Using power series expansion, we can approximate some function f(x)
over a certain interval of radius of convergence R, centred in a
certain point x = c .
If f(x) = Σn=0∞
an (x - c)n
Around the point c, we can write:
f(x) = an for the first approximation,
f(x) = an + a1 (x - c) for the second approximation,
f(x) = an + a1 (x - c) + a2 (x - c)2
for the third approximation,
and so on ...
The more we move away from the point c , the more we use the
terms of the series in order to the approximation remains
valid.
Examples:
1.
Around 0:
cos (x) = 1 - x2/2! + x4/4! - ...
cos(0) = 1 - 0 = 1
(one terme is sufficient)
2.
cos(0.5) = 1 - 1/8 + 1/16x24 = 0.878(one terme is
not sufficient. Even three are not enough)
cos(0.5) = 0.999
cos 3
Here we use Taylor expansion aroud x = π
3.
cos (π - 3) = - cos 3 .
A calculator gives - 0.998
The expansion of cos (π - 3) in Taylor series is:
cos (π - 3) = 1 - (π - 3)2/2! + (π - 3)4/4! - ...
= 1 - 0.0100 + ... = 0.989
so
cos 3 = - 0.989, which is a good approximation at 0.8 %.
4. Errors on approximation of functions
Let f(x) = Σn=0∞
an (x - c)n =
a0 + an1 (x - c) + a2 (x - c)2
+ ... + an (x - c)n + ...
with an = f(n)(c)/n!
The maximal error in approximating the function f(x) by
a part of a power series (at the order n - 1) is:
E = |(x - c)n . f(n)(t)/n!|
The variable t is located between x and c.
Example:
The Maclaurin (c = 0) series expansion of sin (x) is:
sin (x) = x - x3/3! + x5/5! - x7/7! + ...
A calculator gives: sin(0.5 radians) = 0.4794
If we want to calculate sin(0.5), taking just three terms
(order n = 3) of the series, we can write:
sin (0.5 ) = 0.5 - (0.5)3/3! + (0.5)5/5!.
= 0.47946
The maximal error done is:
E = |(0.5 - 0)3 . sin(3)(t)/3!| =
|(1/8) . sin(3)(t)/6| =
|- cos(t)/48| = (1/48) |cos (t)|
t is between 0 and 0.5
We have cos t < 1 , so (1/48) |cos (t)| < (1/48) = 0.021 = 2%.
5. Other usefulness of power series
5.1. Power series can also be used to
calculate limites
Example:
lim sin(x)/x = 0/0 : undetermined.
x → 0
The Maclaurin series expansion of sin (x) is:
sin (x) = x - x3/3! + x5/5! - x7/7! + ...
Therefore :
sin(x)/x = 1 - x2/3! + x4/5! - x6/7! + ...
Then:
lim sin(x)/x = 1
x → 0
5.2. Power series can also be used to
integrate expressions
Example:
∫ cos(x2) dx
x: 0 to 1
The Maclaurin series expansion of cos (x) is:
cos (x) = 1 - x2/2! + x4/4! - ...
So :
cos (x2) = 1 - x4/2! + x8/4! - ...
Then:
∫ cos(x2) dx =
x: 0 to 1
x - x5/5.2! + x9/9.4! =
x: 0 to 1
[1 - 1/5.2! + 1/9.4!] - [0] ≈ 0.95
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