According to Taylor’s theorem, a polynomial of degree n, known as the nth-order Taylor polynomial, approximates an n-times differentiable function around a given point.
Mathematically, it states that if f(x) is a function that is differentiated n + 1 times on an open interval ‘I’ containing ‘a,’ then for each positive integer ‘n’ and for each x Є I,
${f\left( x\right) =f\left( a\right) +f’\left( a\right) \left( x-a\right) +\dfrac{f”\left( a\right) }{2!}\left( x-a\right) ^{2}+\ldots +\dfrac{f^{n}\left( a\right) }{n!}\left( x-a\right) ^{n} +R_{n}\left( x\right)}$
⇒ ${f\left( x\right) =p_{n}\left( x\right) +R_{n}\left( x\right)}$,
Here,
pn(x) is the Taylor polynomial of f(x) at ‘a,’ and
${R_{n}\left( x\right) =\dfrac{f^{n+1}\left( c\right) }{\left( n+1\right) !}\left( x-a\right) ^{n+1}}$ for some unknown real number c Є (a, x) is known as Taylor’s remainder theorem and the Taylor polynomial form is known as Taylor’s theorem with Lagrange form of the remainder.
Rn(x) is often estimated without knowing the value of ‘c’ using the Remainder Estimation Theorem, which states that:
If there exists a positive real number M by which M is an upper bound for the absolute value of the (n+1)th derivative of f over the interval I, that is, |fn + 1(x)| ≤ M for x Є I, then ${\left| R_{n}\left( x\right) \right| \leq \dfrac{M}{\left( n+1\right) !}\left| x-a\right| ^{n+1}}$ for all x Є I.
Taylor’s theorem is used to determine a function’s value at one point in terms of its value and derivatives at a nearby point, which is useful for approximating complex functions into simpler ones.
Proof
To prove Taylor’s theorem, let us first fix a point x Є I, and ‘t’ be another point in ‘I.’
Now, g(t) is a function such that:
${g\left( t\right) =f\left( x\right) -f\left( t\right) -f’\left( t\right) \left( x-t\right) -\dfrac{f^{11}\left( t\right) }{2!}\left( x-t\right) ^{2}-\ldots -\dfrac{f^{n}\left( t\right) }{n!}\left( x-t\right) ^{n} -R_{n}\left( x\right) \dfrac{\left( x-t\right) ^{n+1}}{\left( x-a\right) ^{n+1}}}$
Since ‘g’ is a polynomial function in ‘I,’ it is a differentiable function.
Also, ${g\left( a\right) =f\left( x\right) -f\left( a\right) -f’\left( a\right) \left( x-a\right) -\dfrac{f”\left( a\right) }{2!}\left( x-a\right) ^{2}+\ldots +\dfrac{f^{n}\left( a\right) }{n!}\left( x-a\right) ^{n} -R_{n}\left( x\right)}$
⇒ g(a) = f(x) – pn(x) – Rn(x)
⇒ g(a) = 0
g(x) = f(x) – f(x) – 0 – … – 0
⇒ g(x) = 0
Here, we observe that g(t) = 0 at t = a and t = x, which means g satisfies Rolle’s theorem.
Thus, a real number c Є (a, x) exists such that g’(c) = 0.
By using the product rule, we get
${\dfrac{d}{dt}\left[ \dfrac{f^{n}\left( t\right) }{n!}\left( x-t\right) ^{n}\right] =\dfrac{-f^{n}\left( t\right) }{\left( n-1\right) !}\left( x-t\right) ^{n-1}+\dfrac{f^{n+1}\left( t\right) }{n!}\left( x-t\right) ^{n}}$
Now, calculating ‘g,’ we get
${g’\left( t\right) =-f’\left( t\right) +\left[ f’\left( t\right) -f”\left( t\right) \left( x-t\right) \right] +\left[ f”\left( t\right) \left( x-t\right) -\dfrac{f”’\left( t\right) }{2!}\left( x-t\right) ^{2}\right]+\ldots +\left[ \dfrac{f^{n}\left( t\right) }{\left( n-1\right) !}\left( x-t\right) ^{n-1}-\dfrac{f^{n+1}\left( t\right) }{n!}\left( x-t\right) ^{n}\right] +\left( n+1\right) R_{n}\left( x\right) \dfrac{\left( x-t\right) ^{n}}{\left( x-a\right) ^{n+1}}}$
⇒ ${g’\left( t\right) =-\dfrac{f^{n+1}\left( t\right) }{n!}\left( x-t\right) ^{n} +\left( n+1\right) R_{n}\left( x\right) \dfrac{\left( x-t\right) ^{n}}{\left( x-a\right) ^{n+1}}}$
According to Rolle’s theorem, we conclude that a real number c Є (a, x) exists such that g’(c) = 0
Since ${g’\left( c\right) =-\dfrac{f^{n+1}\left( c\right) }{n!}\left( x-c\right) ^{n} +\left( n+1\right) R_{n}\left( x\right) \dfrac{\left( x-c\right) ^{n}}{\left( x-a\right) ^{n+1}}}$
⇒ ${-\dfrac{f^{n+1}\left( c\right) }{n!}\left( x-c\right) ^{n} +\left( n+1\right) R_{n}\left( x\right) \dfrac{\left( x-c\right) ^{n}}{\left( x-a\right) ^{n+1}}=0}$
On adding ${-\dfrac{f^{n+1}\left( c\right) }{n!}\left( x-c\right) ^{n}}$ to both sides, we get
${\left( n+1\right) R_{n}\left( x\right) \dfrac{\left( x-c\right) ^{n}}{\left( x-a\right) ^{n+1}}=\dfrac{f^{n+1}\left( c\right) }{n!}\left( x-c\right) ^{n}}$
On dividing both sides by ${\left( n+1\right) \dfrac{\left( x-c\right) ^{n}}{\left( x-a\right) ^{n+1}}}$, we get
${R_{n}\left( x\right) =\dfrac{f^{n+1}\left( c\right) }{\left( n+1\right) !}\left( x-a\right) ^{n+1}}$, which follows that if there exists M such that |fn + 1(x)| ≤ M, for all x Є I, then
${\left| R_{n}\left( x\right) \right| \leq \dfrac{M}{\left( n+1\right) !}\left| x-a\right| ^{n+1}}$
Here, we also conclude that if this inequality holds for every ‘n’ and the other conditions of Taylor’s Theorem are also satisfied by f(x), then the series converges to f(x).
Solved Example
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Use Taylor’s theorem to estimate the maximum error when approximating f(x) = e2x, centered at a = 0 with n = 2 on the interval 0 ≤ x ≤ 0.2
Solution:
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As we know from Taylor’s Theorem:
${R_{n}\left( x\right) =\dfrac{f^{n+1}\left( c\right) }{\left( n+1\right) !}\left( x-a\right) ^{n+1}}$, here Rn(x) is the remainder of n-degree Taylor’s polynomial, and
fn + 1(c) is the (n + 1)th derivative of f(x) at some point c Є (a, x)
${f”’\left( e^{2x}\right) =\dfrac{d^{3}\left( e^{2x}\right) }{dx^{3}}=8e^{2x}}$
Now, the error is ${\left| R_{2}\left( x\right) \right| =\left| \dfrac{f”’\left( e^{2x}\right) }{3!}x^{3}\right|}$
= ${\left| \dfrac{8e^{2x}}{6}x^{3}\right|}$
At x = 0.2, we get ${\dfrac{8e^{0.4}}{6}\left( 0.2\right) ^{3}}$
Since e0.4 ≈ 1.492 (estimated to 3 decimal places)
(0.2)3 = 0.008
Thus, ${\left| R_{2}\left( 0.2\right) \right| =\left| \dfrac{8e^{2x}}{6}x^{3}\right|}$
${\approx \dfrac{8\times 1.492\times 0.008}{6}}$
≈ 0.159 (estimated to 3 decimal places)Thus, the estimated maximum error when approximating f(x) = e2x with a second-degree Taylor
polynomial centered at a=0 on the interval 0 ≤ x ≤ 0.2 is approximately 0.159
