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      Calculus II

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© The scientific sentence. 2010

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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