We denote the consecutive derivatives of a function g(x) as
g'(x); g"(x); g^{(3)}(x); g^{(4)}(x); ...; g^{(n)}(x); ...
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 x^{2} + ... + l x^{n} Then V'(x) = b + 2c x + ... + n l x^{n-1} V"(x) = 2 c + 3.2.d + ... + n(n-1) l x^{n-2} ... V^{(n)} = n! l and then V(0) = a V'(0) = b V"(0) = 2! c ... V^{(n)}(0) = n! lNow 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! + x^{2}.f"(0)/2! + x^{3}.f^{(3)}(0)/3! +...+ x^{n}.f^{(n)}(0)/n!This is called the Maclaurin polynomial of order n for the function f(x).
Notation: T_{n}f(x)
The Maclaurin polynomial of order n for the function f(x) is
T_{n} f(x) = f(0) + x.f'(0)/1! + x^{2}.f"(0)/2! + x^{3}.f^{(3)}(0)/3! +...+ x^{n}.f^{(n)}(0)/n! |
We know that if f(x) = e^{x} then f^{(n)}(x) = e^{x} and f^{(n)}(0) = 1 Thus T_{n} e^{x} = 1 + x/1! + x^{2}/2! + x^{3}/3! + ... + x^{n}/n!
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 T_{n} sin(x) = x - x^{3}/3! + x^{5}/5! - x^{7}/7! + ... + x^{n}.sin(n.pi/2)/n!If x is very close to 0, we see that x - x^{3}/6 is a good approach for sin(x).
T_{n} cos(x) = 1 - x^{2}/2! + x^{4}/4! - x^{6}/6! + ... + x^{n}.cos(n.pi/2)/n!
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 T_{n} ln(1+x) = x/1 - x^{2}/2 + x^{3}/3 - x^{4}/4 + ... + (-1)^{n-1}.x^{n}/n
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 T_{n} (1+x)^{q} = 1 + q x + q(q-1)x^{2}/2! + q(q-1)(q-2) x^{3}/3! + ... analogous T_{n} (1+x)^{q} = 1 - q x + q(q-1)x^{2}/2! - q(q-1)(q-2) x^{3}/3! + ... For q = -1 we have T_{n} (1+x)^{-1} = 1 + x + x^{2} + x^{3} + ... + x^{n} for q = 1/2 we have T_{n} sqrt(1+x) = 1 + x/2 - x^{2}/8 + ... So, if x is very close to 0, we see that 1 + x/2 - x^{2}/8 is a good approach for sqrt(1+x). analogous T_{n} sqrt(1-x) = 1 - x/2 - x^{2}/8 + ... If x is very close to 0, we see that 1 + x/2 - x^{2}/8 is a good approach for sqrt(1-x).
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) |
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: T_{n,a}f(x)
The Taylor polynomial of order n for the function f(x) in an
environment of x=a is
T_{n,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! |
T_{n,a} f(a + h) = f(a) + h.f'(a)/1! + h^{2}.f"(a)/2! + h^{3}.f^{(3)}(a)/3! +...+ h^{n}.f^{(n)}(a)/n! |
T_{n,a} e^{x} = e^{a}( 1 + (x-a)/1! + (x-a)^{2}/2! + (x-a)^{3}/3! + ... + (x-a)^{n}/n!)
T_{n,a} sin(x) sin(a + i.pi/2) = sum ----------------- .(x-a)^{i} for i=0...n i i!
T_{n,a} cos(x) cos(a + i.pi/2) = sum ----------------- .(x-a)^{i} for i=0...n i i!
If f^{(n+1)}(x) exists in an environment of a containing a+h, then
f(a+h) = f(a) + h.f'(a)/1! + h^{2}.f"(a)/2! + h^{3}.f^{(3)}(a)/3! +...+ h^{n}.f^{(n)}(a)/n! + h^{n+1}.f^{(n+1)}(c)/(n+1)!with c between a and a+h. |
f(a+h) = f(a) + h.f'(a)/1! + h^{2}.f"(a)/2! + h^{3}.f^{(3)}(a)/3! +...+ h^{n}.f^{(n)}(a)/n! + h^{n+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! + x^{2}.f"(a)/2! + x^{3}.f^{(3)}(a)/3! +...+ x^{n}.f^{(n)}(a)/n! + x^{n+1}.r/(n+1)! )and calculate the n+1 derivatives.
g'(x) = f'(a+x) -( f'(a) + x.f"(a)/1! + ... + x^{n-1}.f^{(n)}(a)/(n-1)! + x^{n}.r/n! ) g"(x) = f"(a+x) -( f"(a) + ... + x^{n-2}.f^{(n)}(a)/(n-2)! + x^{n-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)-rSince 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 c_{1} between 0 and h such that g'(c_{1}) = 0. g'(0)=g'(c_{1}) => there is a c_{2} between 0 and h such that g"(c_{2}) = 0. ... g^{(n)}(0)=g^{(n)}(c_{n}) =>there is a c_{n+1} between 0 and h such that g^{(n+1)}(c_{n+1}) = 0.And from this last conclusion, we can write
0 = f^{(n+1)}(a+c_{n+1})-r <=> f^{(n+1)}(c) = r for a value c between a and a+h.The term h^{n+1}.f^{(n+1)}(c)/(n+1)! is called 'the derivative form of the remainder'.
If f^{(n+1)}(x) exists in an environment of 0 containing x, then
f(x) = f(0) + x.f'(0)/1! + x^{2}.f"(0)/2! + x^{3}.f^{(3)}(0)/3! +...+ x^{n}.f^{(n)}(0)/n! + x^{n+1}.f^{(n+1)}(c)/(n+1)!with c between 0 and x. |
e^{x} = 1 + x/1! + x^{2}/2! + ... + x^{n}/n! + e^{c}.x^{n+1}/(n+1)! If n --> infinity then x^{n+1}/(n+1)! has limit 0 for all x.Therefore we can write
e^{x} = 1 + x/1! + x^{2}/2! + ... + x^{n}/n! + ... for all x |
sin(x) = x/1! - x^{3}/3! + x^{5}/5! - ... + sin(n.pi/2).x^{n}/n! + sin(c + (n+1).pi/2).x^{n+1}/(n+1)! If n --> infinity then x^{n+1}/(n+1)! has limit 0 for all x.Therefore we can write
sin(x) = x/1! - x^{3}/3! + x^{5}/5! -x^{7}/7! + ... |
cos(x) = 1 - x^{2}/2! + x^{4}/4! -x^{6}/6! + ... |