Themes > Science > Mathematics > Algebra > Foci of a conic section > Topics and Problems > Taylor and Maclaurin

..Taylor and Maclaurin polynomials
..Maclaurin polynomial
..ex and its Maclaurin polynomial
..sin(x) and its Maclaurin polynomial
..cos(x) and its Maclaurin polynomial
..ln(1+x) and its Maclaurin polynomial
..(1+x)q and its Maclaurin polynomial
..Taylor polynomial
..ex and the Taylor polynomial
..sin(x) and its Taylor polynomial
..cos(x) and its Taylor polynomial
..Taylor formula with derivative form of remainder
..Maclaurin formula with derivative form of remainder
..Expansion of ex
..Expansion of sin(x)
..Expansion of cos(x)

Taylor and Maclaurin polynomials

The purpose is to approach a function with a polynomial in the environment of x=a.

We denote the consecutive derivatives of a function g(x) as 

 
g'(x); g"(x); g(3)(x); g(4)(x); ...; g(n)(x); ...

Maclaurin polynomial

Theorem:
If the n-th derivative of f(x) exists in a environment of x=0, there is exactly one polynomial V(x), of degree n or lower, such that 
 
V(0)=f(0); V'(0)=f'(0); V"(0)=f"(0); ... ; V(n)(0)=f(n)(0)

Proof 

 
Take
  V(x)  = a + b x + c x2 + ... + l xn
Then
  V'(x) =     b   + 2c x  + ... + n l xn-1

  V"(x) =           2 c   + 3.2.d + ... + n(n-1) l xn-2

  ...


  V(n) =  n! l

and then

  V(0)  = a

  V'(0) = b

  V"(0) = 2! c

  ...


  V(n)(0) =  n! l
Now we require 
 
    V(0)=f(0); V'(0)=f'(0); V"(0)=f"(0); ... ; V(n)(0)=f(n)(0)

<=>
    a = f(0);

    b = f'(0);

  2!c = f"(0);

  ...

  n! l = f(n)(0)
So all coefficients of V(x) are known and we can conclude :

The only polynomial of degree n or lower, such that 

 
V(0)=f(0); V'(0)=f'(0); V"(0)=f"(0); ... ; V(n)(0)=f(n)(0)
is 
 
f(0) + x.f'(0)/1! + x2.f"(0)/2! + x3.f(3)(0)/3! +...+ xn.f(n)(0)/n!
This is called the Maclaurin polynomial of order n for the function f(x).

Notation: Tnf(x)

The Maclaurin polynomial of order n for the function f(x) is 
 
Tn f(x) = f(0) + x.f'(0)/1! + x2.f"(0)/2! + x3.f(3)(0)/3! +...+ xn.f(n)(0)/n!

ex and its Maclaurin polynomial

 
We know that if f(x) = ex then f(n)(x) = ex and f(n)(0) = 1

Thus

Tn ex = 1 + x/1! + x2/2! + x3/3! + ... + xn/n!

sin(x) and its Maclaurin polynomial

 
f(x)    = sin(x)                     => f(0)  = 0
f'(x)   = cos(x)  = sin(x + pi/2)    => f'(0) = 1
f"(x)   = -sin(x) = sin(x + 2.pi/2)  => f"(0) = 0
f"'(x)  = -cos(x) = sin(x + 3.pi/2)  => f"'(0)= -1
...
f(n)(x) = sin(x + n.pi/2)         => f(n)(0) = sin(n.pi/2)

Thus

Tn sin(x) = x - x3/3! + x5/5! - x7/7! + ... + xn.sin(n.pi/2)/n!
If x is very close to 0, we see that x - x3/6 is a good approach for sin(x).

cos(x) and its Maclaurin polynomial

In the same way as above we find for cos(x) 
 
Tn cos(x) = 1 - x2/2! + x4/4! - x6/6! + ... + xn.cos(n.pi/2)/n!

ln(1+x) and its Maclaurin polynomial

 
f(x) = ln(1+x)             => f(0)   = 0
f'(x) = (1+x)-1         => f'(0)  = 1
f"(x) = -1(1+x)-2       => f"(0)  = -(1!)
f"'(x)= 2(1+x)-3        => f"'(0) = 2!
f(4)= -2.3(1+x)-4    => f(4) = -(3!)
...

Thus

Tn ln(1+x) = x/1 - x2/2 + x3/3 - x4/4 + ... + (-1)n-1.xn/n

(1+x)q and its Maclaurin polynomial

 
f(x) = (1+x)q       => f(0) = 1
f'(x) = q(1+x)q-1  => f'(0) = q
f"(x) = q(q-1)(1+x)q-2 => f"(0) = q(q-1)
f"'(x)= q(q-1)(q-2)(1+x)q-3  => f"'(0) = q(q-1)(q-2)
...

Thus

Tn (1+x)q = 1 + q x + q(q-1)x2/2! + q(q-1)(q-2) x3/3! + ...

analogous

Tn (1+x)q = 1 - q x + q(q-1)x2/2! - q(q-1)(q-2) x3/3! + ...

For q = -1 we have

Tn (1+x)-1 = 1 + x + x2 + x3 + ... + xn

for q = 1/2 we have

Tn sqrt(1+x) = 1 + x/2 - x2/8 + ...

So, if x is very close to 0, we see that 1 + x/2 - x2/8 is a good approach
for sqrt(1+x).

analogous

Tn sqrt(1-x) = 1 - x/2 - x2/8 + ...

If x is very close to 0, we see that 1 + x/2 - x2/8 is a good approach
for sqrt(1-x).

Taylor polynomial

Theorem:

If the n-th derivative of f(x) exists in an environment of x=a, there is exactly one polynomial V(x), of degree n or lower, such that 
 
V(a)=f(a); V'(a)=f'(a); V"(a)=f"(a); ... ; V(n)(a)=f(n)(a)

Proof: 

 
Take
  V(x)  = b + c (x - a) + d (x - a)2 + ... + l (x - a)n
In the same way as for the Maclaurin polynomial, you can show that 
 
  V(a) = f(a)  <=>  b = f(a)
  V'(a) = f'(a) <=> c = f'(a)/1!
  V'(a) = f"(a) <=> d = f"(a)/2!
  ...
  V(n)(a) = f(n)(a)  <=>  l = f(n)(a)/n!
So all coefficients of V(x) are known and we can conclude :

The only polynomial of degree n or lower, such that 

 
V(a)=f(a); V'(a)=f'(a); V"(a)=f"(a); ... ; V(n)(a)=f(n)(a)

is

f(a) + (x-a).f'(a)/1! + (x-a)2.f"(a)/2! + (x-a)3.f(3)(a)/3! +...+ (x-a)n.f(n)(a)/n!
This is called the Taylor polynomial of order n for the function f(x) in an environment of x=a.

Notation: Tn,af(x)

The Taylor polynomial of order n for the function f(x) in an environment of x=a is 
 
Tn,a f(x) = f(a) + (x-a).f'(a)/1! + (x-a)2.f"(a)/2!
           + (x-a)3.f(3)(a)/3! +...+ (x-a)n.f(n)(a)/n!

Let x = a + h and the previous formula becomes 
 
Tn,a f(a + h) = f(a) + h.f'(a)/1! + h2.f"(a)/2!
           + h3.f(3)(a)/3! +...+ hn.f(n)(a)/n!

ex and the Taylor polynomial

In the same way as for the Maclaurin polynomial you'll find: 
 
Tn,a ex = ea( 1 + (x-a)/1! + (x-a)2/2! + (x-a)3/3! + ... + (x-a)n/n!)

sin(x) and its Taylor polynomial

In the same way as for the Maclaurin polynomial you'll find: 
 

Tn,a sin(x)

       sin(a + i.pi/2)
= sum ----------------- .(x-a)i   for i=0...n
   i        i!

cos(x) and its Taylor polynomial

In the same way as for the Maclaurin polynomial you'll find: 
 

Tn,a cos(x)

       cos(a + i.pi/2)
= sum ----------------- .(x-a)i   for i=0...n
   i        i!

Taylor formula with derivative form of remainder

Theorem:

If f(n+1)(x) exists in an environment of a containing a+h, then 
 
f(a+h) = f(a) + h.f'(a)/1! + h2.f"(a)/2!
 + h3.f(3)(a)/3! +...+ hn.f(n)(a)/n! + hn+1.f(n+1)(c)/(n+1)!
with c between a and a+h.

Proof:
There is always a suitable number r such that 

 
f(a+h) = f(a) + h.f'(a)/1! + h2.f"(a)/2!
 + h3.f(3)(a)/3! +...+ hn.f(n)(a)/n! + hn+1.r/(n+1)!
We'll prove that r = f(n+1)(c) with c between a and a+h.

Create the function 

 
g(x) = f(a+x)-( f(a) + x.f'(a)/1! + x2.f"(a)/2!
 + x3.f(3)(a)/3! +...+ xn.f(n)(a)/n! + xn+1.r/(n+1)! )
and calculate the n+1 derivatives. 
 
g'(x) = f'(a+x) -( f'(a) + x.f"(a)/1! + ...
       + xn-1.f(n)(a)/(n-1)! + xn.r/n! )

g"(x) = f"(a+x) -( f"(a) + ... + xn-2.f(n)(a)/(n-2)! + xn-1.r/(n-1)! )

...

g(n)(x) = f(n)(a+x) -( f(n)(a) + x r )

g(n+1)(x) = f(n+1)(a+x)-r
Since f, ... f(n) are continuous in the environment of a, we can use Rolle's theorem on g(x) and all the derivatives. 
 
g(0)=g(h) => there is a c1 between 0 and h such that g'(c1) = 0.

g'(0)=g'(c1) => there is a c2 between 0 and h such that g"(c2) = 0.

...

g(n)(0)=g(n)(cn) =>there is a cn+1 between 0 and h such
                                       that g(n+1)(cn+1) = 0.
And from this last conclusion, we can write 
 
        0 = f(n+1)(a+cn+1)-r

<=>    f(n+1)(c) = r  for a value c between a and a+h.
The term hn+1.f(n+1)(c)/(n+1)! is called 'the derivative form of the remainder'.

Maclaurin formula with derivative form of remainder

Take the Taylor formula with derivative form of remainder, choose a = 0 and write x instead of h.
If f(n+1)(x) exists in an environment of 0 containing x, then 
 
f(x) = f(0) + x.f'(0)/1! + x2.f"(0)/2!
 + x3.f(3)(0)/3! +...+ xn.f(n)(0)/n! + xn+1.f(n+1)(c)/(n+1)!
with c between 0 and x.

Expansion of ex

With the Maclaurin formula we can write 
 
ex = 1 + x/1! + x2/2! + ... + xn/n! + ec.xn+1/(n+1)!

If n --> infinity  then xn+1/(n+1)! has limit 0 for all x.
Therefore we can write
ex = 1 + x/1! + x2/2! + ... + xn/n! + ... for all x

Expansion of sin(x)

With the Maclaurin formula we can write 
 
sin(x) = x/1! - x3/3! + x5/5! - ... + sin(n.pi/2).xn/n! +
                              sin(c + (n+1).pi/2).xn+1/(n+1)!

If n --> infinity  then xn+1/(n+1)! has limit 0 for all x.
Therefore we can write
sin(x) = x/1! - x3/3! + x5/5! -x7/7! + ...

Expansion of cos(x)

In the same way as for sin(x) we can deduce the formula
cos(x) = 1 - x2/2! + x4/4! -x6/6! + ...


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