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Taylor Series of a function is an infinite sum of terms, expressed in terms of the function's derivatives at any one point, with each following term having a larger exponent like x, x2, x3, and so on. As a result, the Taylor series formula helps to describe the Taylor series mathematically. Taylor Series is the expansion of a function into an infinite sum of terms. In 1715, Scottish mathematician James Gregory discovered it, and English mathematician Brook Taylor formally introduced it. This series is used in a variety of fields, including biology, astronomy, advanced arithmetic, algebra, and computer engineering.
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Key Terms: Taylor Series, Maclaurin Series, Functions, Derivatives, Exponent, Variable, Complex Function, Differentiable Function, Taylor Series Expansion
What Is Taylor Series Formula?
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Using the derivatives of a function, the Taylor series formula may be used to extend a function around a variable value. It can be written as,
\(f(x)=f(a)+f^{\prime}(a)(x-a)+\left[\frac{f^{\prime \prime}(a)}{2 !}(x-a)^{2}\right]+\left[\frac{f^{\prime \prime \prime}(a)}{3 !}(x-a)^{3}\right]+\ldots . .+\left[\frac{f^{(n)}(a)}{n !}(x-a)^{n}\right]\)
OR
\(f(x)=\sum_{n=0}^{\infty} \frac{f^{(n)} a}{n !} \times(x-a)^{n}\)
Here,
f(x) = Infinitely differentiable real or complex-valued function at a real or complex number. The power series is denoted by "a."
n = The number of terms in the series
Taylor Series Formula Proof
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Taylor's Series Theorem Statement
Assume that f(x) is a real or composite function that is a differentiable function of a real or composite neighbourhood number. The Taylor series then goes on to explain the following power series:
\(f(x)=f(a) \frac{f^{\prime}(a)}{1 !}(x-a)+\frac{f^{p}(a)}{2 !}(x-a)^{2}+\frac{f^{(i)}(a)}{3 !}(x-a)^{3}+\ldots\)
The Taylor series can be expressed in sigma notation as
\(\sum_{n=0}^{\infty} \frac{f^{n}(a)}{n !}(x-a)^{n} \$\)
Where,
F(n)(a) = nth derivative of f
n! = factorial of n
Read More: Taylor Series Formula: Meaning, Theorem, Steps, Application
Taylor Series Formula Proof
The series of power of any function can be written as:
f(x) = ∑∞n = 0 anxn = a0 + a1x + a2x2 + a3x3 + …
When the value of x is 0
f(x) = a0
Differentiate the given function,
f′(x) = a1 + 2a2x + 3a2x2 + 4a4x3 + …
Put x = 0, then,
f′(0) = a1
Differentiate the given function again,
f′′(x) = 2a2 + 6a3x + 12a4x2 + …
Put x = 0, then,
f′′(0) = 2a2
Hence,
f′′(0) / 2! = a2
Forming an relation for fn(0),
fn(0) / n! = an
replace all the values from the above equations,
\(f(x)=f(0)+f^{\prime}(0) x+\frac{f^{\prime \prime}(0)}{2 !} x^{2}+\frac{f^{\prime \prime \prime}(0)}{3 !} x^{3}+\ldots\)
generalise this equation,
f(x) = b + b1(x−a) + b2(x−a)2 + b3(x−a)3 + …
Put, x = a, we obtain
bn = fn(a) / n!
Put this value of bn in the equation,
\(f(x)=f(a)+f^{\prime}(a)(x-a)+\left[\frac{f^{\prime \prime}(a)}{2 !}(x-a)^{2}\right]+\left[\frac{f^{\prime \prime \prime}(a)}{3 !}(x-a)^{3}\right]+\ldots . .+\left[\frac{f^{(n)}(a)}{n !}(x-a)^{n}\right]\)
This proves the Taylor Series.
Read More: Relations and Functions
Taylor Series and Maclaurin Series
Taylor Series Expansion
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The Taylor Series expansion assumes that every smooth function may be rewritten into an unlimited number of polynomial terms.
Make a universal nth-degree polynomial. Here's what it is:
f(x) = a0 + a (x - c) + a2 (x - c)2 + a3 (x - c)3 +...
Where a0, a1... are the coefficients of each polynomial term, and c is a constant indicating where on the x-axis to begin the approximation.
In closed form, this series is as follows:
\(f(x)=\displaystyle\sum_{j=0}^{\infty} a_{j}(x-c)^{j}\)
Find the first few derivatives of the given polynomial.
f' (x) = a1 + 2a2 (x - c) + 3a3 (x - c)2 + 4a4 (x - c)3 +...
f" (x) = 2a2 + 6a3 (x - c) + 12a4 (x - c)2+ 20a5 (x - c)3 …
f"' (x) = 6a3 + 24a4 (x - c) + 60a5 (x - c)2 + 120a6 (x - c)3…
Above, we see a pattern that we'll employ later. Now that we have n derivatives of f, we will assess them for a number that will cause the majority of their terms to disappear.
This is the most important phase.
f(c) = a0 = a00!
f'(c) = a1 = a11!
f"(c) = 2 × a2 = a22!
f"'(c) = 3 × 2 × a3 = a33!
f""(c) = 4 × 3 × 2 × a4 = a44!
n! is divided on both sides. This has the following consequences:
\(\begin{aligned} &a_{0}=\frac{f(c)}{0 !} \\ &a_{1}=\frac{f^{\prime}(c)}{1 !} \\ &a_{2}=\frac{f^{\prime \prime}(c)}{2 !} \\ &a_{3}=\frac{f^{\prime \prime \prime}(c)}{3 !} \\ &a_{4}=\frac{f^{\prime \prime \prime \prime}(c)}{4 !} \\ \end{aligned}\)
f (x) = a0 + (a1 * (x - c)) + (a2 * (x - c)2) + (a3 * (x - c)3)+…
\(f(x)=\frac{f(c)}{0 !}+\frac{f^{\prime}(c)}{1 !}(x-c)+\frac{f^{\prime \prime}(c)}{2 !}(x-c)^{2}+\frac{f^{\prime \prime \prime}(c)}{3 !}(x-c)^{3}+\ldots\)
The polynomial expansion for the function f is given by the preceding equation. First, compute the n derivatives of f, calculate them at c, divide each derivative by n! and total the terms. The resulting function will be a close approximation to the primary purpose. Additional terms would only aid in obtaining a more precise estimate.
Read More: Differential Equation
Taylor Series of Sin x
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f(x) = sin x
f'(x) = cos x
f"(x) = - sin x
The Taylor series for Sin x at x = 0 is:
\(\begin{aligned} \sin x &=\sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n !} x^{n} \\ &=\frac{1}{1 !} x-\frac{1}{3 !} x^{3}+\frac{1}{5 !} x^{5}-\frac{1}{7 !} x^{7}+\cdots \end{aligned}\)
Taylor Series in Several Variables
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The Taylor series can also be expressed as a function with several variables. The Taylor series with several variables has the following general form:
\(T\left(x_{1}, x_{2}, x_{3}, \ldots x_{m}\right)=f\left(a_{1}, a_{2}, a_{3}, \ldots a_{m}\right)+\sum_{j=1}^{m} \frac{\partial f\left(a_{1}, a_{2}, a_{3}, \ldots a_{m}\right)}{\partial x_{j}}\left(x_{j}-a_{j}\right)+\frac{1}{2 ! \sum_{j=1}^{m}}\)
Read More: Analytic Function
Maclaurin Series
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When the Taylor Series is placed at 0, it is referred to as the Maclaurin series. In the Taylor series, if a = 0, we obtain;
\(f(x)=f(0)+f^{\prime}(0) x+\frac{f^{\prime \prime}(0)}{2 !} x^{2}+\frac{f^{ \prime \prime \prime}(0)}{3 !} x^{3}+\ldots\)
This is referred to as the Maclaurin series.
Applications of Taylor Series
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- If the functional values and derivatives are known at a single point, the Taylor series is used to calculate the value of the complete function at each point.
- Many mathematical proofs are made easier by using the Taylor series representation.
- To estimate the overall series, the sum of partial series might be employed.
- Multivariate Taylor series is used in several optimization methods.
- This series analyses the power flow of electrical power networks.
Read More: Differentiation and Integration Formula
Things to Remember
- A Taylor Series is a function extension with an infinite sum of terms.
- The Taylor Series formula is a powerful tool for estimating difficult-to-calculate functions. It is also capable of calculating endless sums and integrals.
- When the Taylor Series is centred at 0, it is known as the Maclaurin series.
- If the functional values and derivatives are known at a single point, the Taylor series is used to calculate the value of the complete function at each point.
- It is a kind of power series supplied only for functions that are infinitely differentiable on an open interval.
- Any functions that are infinitely differentiable have a Taylor series at a given point 0. The Cauchy Riemann equations apply to holonomic (analytic) functions.
Sample Questions
Ques. Expand the function, f(x) = 2x - 2x2 at a = -3 using the Taylor series. (3 Marks)
Ans. f(x) = 2x - 2x2
At a = -3
Pn(x) = f(a) + (f′(a) * (x − a)) + ((f′′(a) / 2!) * (x − a)2) + ((f′′′(a) / 3!) * (x − a)3) + ((f(4)(a) / 4!) * (x − a)4) + ... + ((f(n)(a) / n!) * (x − a)n)
Derivatives of the function are -
f(x) = 2x − 2x2
f′(x) = 2 − 4x
f′′(x) = −4x
f′′′(x) = −4
Since a = −3 and n = 3, the required expansion is:
f(x) = f(−3) + (f′(−3) * (x − (−3))) + ((f′′(−3) / 2!) * (x − (−3))2) + ((f′′′(−3) / 3!) * (x − (−3))3)
f(x) = f(−3) + (f′(−3) * (x + 3)) + ((f′′(−3) / 2!) * (x + 3)2) + ((f′′′(−3) / 3!) * (x + 3)3)
At x = - 3
f(−3) = (2 * (−3)) – (2 * (−3)2) = -24
f′(−3) = 2 – (4 * (-3)) = -10
f′′(−3) = − 4 * (−3) = 12
f′′′(−3) = − 4
Expansion of the function is:
P3(x) = -24 – (10 * (x + 3)) + ((12 / 2) * (x + 3)2) – (4 * (x + 3)3)
Ques. For f(x) = cos x, at x = 0, find its Taylor series expansion. (3 Marks)
Ans. f(x) = Cos x
Using Taylor series formula,
f(x) = f(a) + (f′(a) * (x − a)) + ((f′′(a) / 2!) * (x − a)2) + ((f′′′(a) / 3!) * (x − a)3) + ((f(4)(a) / 4!) * (x − a)4) + ... + ((f(n)(a) / n!) * (x − a)n)
We evaluate the function and its derivatives:
f(x) = cos(x)
f'(x) = −sin(x)
f''(x) = −cos(x)
f'''(x) = sin(x)
Thus,
cos(x) = cos(a) – ((sin(a) / 1!) * (x – a)) – ((cos(a) / 2!) * (x - a)2) + ((sin(a) / 3!) * (x - a)3) + ...
Put a = 0.
cos(x) = 1 – ((0 / 1!) * (x – 0)) – (1 / 2!) * (x - 0)2) + ((0 / 3!) * (x - 0)3) + ((1 / 4!) * (x - 0)4) + ...
cos(x) = 1 – (x2 / 2!) + (x4 / 4!) − ...
Ques. Expand f(x) = x3 - 10x2 + 6 at x = 3. (3 Marks)
Ans. f(x) = x3 − 10x2 + 6 = f(3) = -57
f’(x) = 3x2 − 20x = f’(3) = 33
f’’(x) = 6x – 20 = f’’(3) = -2
f’’’(x) = 6 = f’’’(3) = 6
f’’’’(x) = 0
x3 − 10x2 + 6 is the series required.
\(=\displaystyle\sum_{n=0}^{\infty} \frac{f^{(n)}(3)}{n !}(x-3)^{n}\)
\(=f(3)+f^{\prime}(3)(x-3)+\frac{f^{\prime \prime}(3)}{2 !}(x-3)^{2}+\frac{f^{\prime \prime}(3)}{3 !}(x-3)^3\)
= − 57 – (33 * (x−3)) – (x−3)2 + (x−3)3
Ques. Find the Taylor series for f(x) = sin(x), a = π. (3 Marks)
Ans. f(x) = sin x and a = π.
sin(x) = x – (x3/3!) + (x5/5!) – (x7/7!) + …… for all x
f(x) = f(a) + (f′(a) * (x − a)) + ((f′′(a) / 2!) * (x − a)2) + ((f′′′(a) / 3!) * (x − a)3) ... + ((f(n)(a) / n!) * (x − a)n)
f(a) = sin π = 0, first term of series is zero
differentiate the function,
⇒ f’(x) = π
= cos(π) = -1
⇒ f’’(x) = π
= -sin(π) = 0
⇒ f’’’(x) = π
= -cos(π) = 1
⇒ f’’’’(x) = π
= -sin(π) = 0
even powers of f(x) will be 0
Substitute the values
sin(x) = -(x - π) + 1/6((x - π)3) - 1/120((x - π)5) + 1/5040((x - π)7) + ...
Ques. Find t3(x) for the function f(x) = 1 / x , a = 2. (3 Marks)
Ans. Series for the given function is:
∑∞k=0 ((f(k)(a) / k!) * (x−a)k) = f(a) + (f′(a) * (x − a)) + ((f′′(a) / 2!) * (x − a)2) ... + ((f(k)(a) / k!) * (x − a)k)
t3(x) = ((f’’(a) / 2!) * (x - a)2)
f(x) = 1/x
= x-1
f’(x) = - x-2
f’’(x) = 2x-3
f’’(x) = 2
= 2 * (2)–3
= 2 / 8
= 1 / 4
t3(x) = ((f’’(2) / 2!) * (x - a)2)
= (1/4) * (1/2!) * (x - 2)2
= (1/8) (x - 2)2
= (x - 2)2 / 8
Ques. Expand t3(x) for the function f(x) = 7 tan-1(x), a = 1. (5 Marks)
Ans. f(x) = tan-1(x)
f(x) = f(a) + (f′(a) * (x − a)) + ((f′′(a) / 2!) * (x − a)2) + ((f′′′(a) / 3!) * (x − a)3) ...
a = 1
at x = 1 is-:
f(x) = f(1) + (f′(1) * (x − 1)) + ((f′′(1) / 2!) * (x − 1)2) + ((f′′′(1) / 3!) * (x − 1)3) ...
f(x) = tan-1(x)
f(1) = tan-1(1)
f(1) = 90°
1st derivative
f’(x) = 1 / (1 + x2)
⇒ f’(1) = 1/(1 + 12) = 1 / (1 + 1)
⇒ f’(1) = 1 / 2
2nd derivative
⇒ f’’(x) = - (2 * 2x) / (1 + x2)2
⇒ f’’(x) = - 4x / (1 + x2)2
⇒ f’’(1)= - 4 / (1 + 12)2
= -(4) / 22
⇒ f’’(1) = - 4 / 4
= -1
3rd derivative
⇒ f’’’(x) = [((1 + x2)2 * (-4)) – ((-4x) * (2 (1 + x2)2x))] / (1 + x2)4
⇒ f’’’ (x) = (6x2 - 2) / (1 + x2)3
⇒ f’’’(1) = (6(1)2 - 2) / (1 + 12)3
⇒ f’’’(1) = (6 - 2) / (2)3
⇒ f’’’(1) = 4 / (2)3
⇒ f’’’(1) = 4 / 8 = 1 / 2
Expansion up to 3 terms:
⇒f(x) = 90° + ((x - a) / 1!) * f’(a)) + ((x - a)2 / 2!) * f’’(a)) + ((x - a)3 / 3!) * f’’’(a)) +....
⇒f(x) = 90° + ((x - 1) / 1!) * (1 / 2)) + ((x - a)2 / 2!) * (-1)) + ((x - a)3 / 3!) +....
⇒ f(x) = 90° + ((x - 1) / 2) - ((x - 1)2 / 2) + ((x - 1)3 / 12 +….
Ques. Expand f(x) = ex at x=0 (3 Marks)
Ans.
f’(x) = ex
f’’(x) = ex
f’’’(x) = ex
At x = 0, we get
f’(0) = e0 =1
f’’(0) = e0=1
f’’’(0) = e0 = 1
The Maclaurin series at x = 0 is
\(f(x)=f(0)+f^{\prime}(0) x+\frac{f^{\prime \prime}(0)}{2 !} x^{2}+\frac{f \prime \prime(0)}{3 !} x^{3}+\ldots\)
ex = 1+ x(1) + (x2 / 2!)(1) + (x3 / 3!)(1) + …
Ques. Find the Maclaurin series for f(x) = cos(x2) and use it to determine f8(0). (5 Marks)
Ans. We know that,
\(f(x)=f(0)+x f^{\prime}(0)+\left(x^{2} / 2 !\right) f^{\prime \prime}(0)+\left(x^{3} / 3 !\right) f^{\prime \prime \prime}(0)+\ldots \ldots \ldots\left(x^{n} / n !\right) f^{n}(0)\)
f(x) = cos x,
f(0) = cos 0 = 1
f'(x) = - sin x
f'(0) = 0
f''(x) = - cosx
f''(0) = - cos 0
= -1
f'''(x) = sinx = 0
f''''(x) = cosx
f''''(0) = cos 0
= 1
f(x) = 1 + x(0) + ((x2/2!) * (-1)) + ((x3/3!) * (0)) + ((x4/ 4!) * (1)) + …
cos x = 1 + 0 - (x2/2) + 0 + (x4/ 24) - (x6/720) +... =
= ∑∞n=0(−1)n(x2n / (2n)!) ---(1)
Replace x by x2
cos(x2) = 1 - ((x2)2 / 2) + ((x2)4 / 24 - ((x2)6 / 720) +…
cos(x2) = 1 - (x4 / 2) + (x8 / 24) - (x12 / 720) +... =
= ∑∞n=0(−1)n(x4n / (2n)!) ---(2)
f(x) = cos x2
f(8)(0) is the 8th derivative of f(x) = cos x2 evaluated at x = 0.
(x8 / 8!) f8(0)= x8/ 24
Comparing the coefficients of x8 from equations (1) and (2), we obtain
f8(0) / 8! = 1 / 24
f8(0) = 8! / 24
f8(0) = (8 × 7 × 6 × 5 × 4 × 3 × 2 ×1) / 24
f8(0) = 8 × 7 × 6 × 5 = 1680
Ques. Expand f(x) = 1/x, c = 1. (3 Marks)
Ans. The expansion of the function will be
F(x) = ∑∞n=0 ((f(n)(c) / n!) * (x − c)n) = f(c) + (f′(c) * (x − c)) + ((f′′(c) / 2!) * (x − c)2) ... + ((f(n)(c) / n!) * (x − c)n)
f(x) = x-1, f(1) = 1-1 = 1 / 1
f’(x) = -x-2, f’(1) = - 1/12
f”(x) = 2! x-3, f”(1)2! = 1-3 = 1/13
f”’(x) = -3! x-4, f’’’(1) / 3! = 1-4 = 1/14
\(f^{n}(x)=(-1)^{n} n ! x^{-(n+1)}, f^{n}(1) / n !=(-1)^{n} / 1^{(n+1)}\)
Expanded series is
f(1) + (f’(1) * (x – 1)) + (f”(1) * (x - 1)2 / 2!) +...+ fn(1)(x - 1)n / n! +…
= 1 - (x - 1) + (x -1)2 13 + … + (-1)n(x - 1)n (1)(n + 1) + …
= 1 - (x - 1) + (x - 1)2 + …+ (x - 1)n
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