Taylor and Maclaurin

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); ...
 
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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)

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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!
 
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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).
 
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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!

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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!
 
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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'.

 

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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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