The Taylor Series is a way of writing a function as a sum of many terms, almost like building it piece by piece with polynomials. Each new term in the series has a higher power (exponent) than the one before it. This method is very useful in mathematics and science, as it allows many complicated functions to be replaced with simpler polynomial expressions.
For example, suppose you want to find values of functions like e^x, ln(x), or sin(x), but you don’t have a calculator. In that case, the Taylor series helps you estimate their values using only basic arithmetic and derivatives.
Table of Contents
The Taylor series definition tells us that:
If a function f(x) is infinitely differentiable at a point a, then it can be written as an infinite sum of its derivatives at that point.
Mathematically, it is written as:
f(x) = f(a) + f'(a)(x - a) + f''(a)/2! · (x - a)² + f'''(a)/3! · (x - a)³ + ...
This is called the Taylor series formula, and it’s central to understanding function approximation.
The Taylor Series Theorem allows us to create the Taylor series. It gives us a formula that links a function to its polynomial approximation around a point a.
Assume that f(x) is a real or composite function, which is differentiable in the neighborhood of a number a. Then the Taylor series expansion of f(x) about the point a is given by:
f(x) = f(a) + f’(a)/1! (x – a) + f’’(a)/2! (x – a)² + f’’’(a)/3! (x – a)³ + …
In sigma notation:
f(x) = Σ (from n=0 to ∞) [ fⁿ(a) / n! ] (x – a)ⁿ
Where:
fⁿ(a) = nth derivative of f at a
n! = factorial of n
Proof of Taylor’s Theorem
We know that any function can be expressed as a power series. So, assume:
f(x) = Σ (from n=0 to ∞) aₙxⁿ = a₀ + a₁x + a₂x² + a₃x³ + …
Step 1: Substituting x = 0
f(0) = a₀
Step 2: First derivative
f’(x) = a₁ + 2a₂x + 3a₃x² + 4a₄x³ + …
At x = 0 → f’(0) = a₁
Step 3: Second derivative
f’’(x) = 2a₂ + 6a₃x + 12a₄x² + …
At x = 0 → f’’(0) = 2a₂ → a₂ = f’’(0)/2!
Step 4: Third derivative
f’’’(x) = 6a₃ + 24a₄x + …
At x = 0 → f’’’(0) = 6a₃ → a₃ = f’’’(0)/3!
Step 5: General case
aₙ = fⁿ(0)/n!
So, the expansion becomes:
f(x) = f(0) + f’(0)/1! x + f’’(0)/2! x² + f’’’(0)/3! x³ + …
This is the Maclaurin Series (Taylor series at 0).
Now, let expansion be around a (not 0). Assume:
f(x) = b₀ + b₁(x–a) + b₂(x–a)² + b₃(x–a)³ + …
At x = a → f(a) = b₀
f’(a) = b₁
f’’(a) = 2!b₂
f’’’(a) = 3!b₃
So, bₙ = fⁿ(a)/n!
Final Result:
f(x) = f(a) + f’(a)/1!(x–a) + f’’(a)/2!(x–a)² + f’’’(a)/3!(x–a)³ + …
Hence, Taylor’s series is proved.
To write the Taylor series formula in a shorter way, we use sigma notation (∑). This helps when we write long series:
f(x) = ∑ (from n = 0 to ∞) [f⁽ⁿ⁾(a) / n!] · (x - a)ⁿ
This simply means we’re summing all the terms from n = 0 to infinity.
Example:
For exe^xex, the Taylor series is:
e^x = ∑ (from n = 0 to ∞) xⁿ / n!
= 1 + x + x²/2! + x³/3! + x⁴/4! + ...
This sigma notation is often used in calculus, programming, and applied mathematics.
Let’s go step-by-step to understand the taylor series proof.
Start with a function f(x) that is infinitely differentiable at a point a.
Use its derivatives to form a polynomial:
1st derivative gives the linear term: f'(a)(x - a)
2nd derivative gives the quadratic term: f''(a)/2! (x - a)²
Continue this process for higher derivatives.
The function can then be written as:
f(x) = f(a) + f'(a)(x - a) + f''(a)/2! (x - a)² + ... + f⁽ⁿ⁾(a)/n! (x - a)ⁿ + Rₙ(x)
The remainder term Rₙ(x) can be made small by increasing the number of terms.
If Rₙ(x) approaches zero as n becomes very large, the function becomes equal to its Taylor series.0
So, this Taylor series proof shows us why the method works for smooth functions.
The general form of the Taylor series at a point aaa is:
f(x) = ∑ (from n = 0 to ∞) [f⁽ⁿ⁾(a) / n!] · (x - a)ⁿ
This formula shows us how to write any function as a power series, as long as all the derivatives exist and the series converges.
This general form is very helpful for solving problems, making numerical methods, and modeling functions.
Let’s now see an example. What is the Taylor series of sin(x) around 0?
Let’s compute the derivatives of sin(x):
f(x) = sin(x)
f'(x) = cos(x)
f''(x) = -sin(x)
f'''(x) = -cos(x)
... and so on.
Evaluating them at 0:
f(0) = 0
f'(0) = 1
f''(0) = 0
f'''(0) = -1
...
Putting into the formula:
sin(x) = x - x³/3! + x⁵/5! - x⁷/7! + ...
This is one of the most famous Taylor series examples, and is widely used in trigonometry and physics.
When a function has two or more variables, we can still use the Taylor series to approximate it.
Let’s say f(x, y) is a function of x and y.
Then the Taylor series around point (a, b) is:
f(x, y) = f(a, b) + fₓ(a, b)(x - a) + fᵧ(a, b)(y - b)
1/2! [fₓₓ(a, b)(x - a)² + 2fₓᵧ(a, b)(x - a)(y - b) + fᵧᵧ(a, b)(y - b)²] + ...
Here,
fₓ means partial derivative with respect to x
fₓₓ, fᵧᵧ, fₓᵧ are second-order derivatives
This is very useful in machine learning, economics, and engineering for modeling functions with many variables.
The Maclaurin series is a special type of Taylor series where the center point is 0 (i.e., a = 0). This makes calculations easier.
f(x) = f(0) + f'(0)x + f''(0)/2! x² + f'''(0)/3! x³ + ...
This is used for many common functions such as e^x, sin(x), cos(x), ln(1 + x), etc.
Example: Maclaurin Series of 1/(1 - x)
Let’s find the Maclaurin series of f(x) = 1 / (1 - x)
We use the formula:
1 / (1 - x) = ∑ (from n = 0 to ∞) xⁿ
= 1 + x + x² + x³ + x⁴ + ...
(valid for |x| < 1)
This is a simple but very powerful Taylor series example, often used in calculus and computer programming.
The Taylor series is more than just theory - it is used everywhere:
Calculators use it to compute sin, cos, log, exp, and more.
Physics uses it to approximate energy, force, and motion.
Engineering applies the Taylor series in simulations and design.
Machine learning uses it for gradient-based optimization.
Finance and economics apply it in interest rate models, stock market predictions, and cost estimations.
Problem 1: Find the Taylor Series for e^x at a = 0
Solution:
All derivatives of e^x are e^x, and e^0 = 1
So:
e^x = 1 + x + x²/2! + x³/3! + x⁴/4! + ...
Problem 2: Find the Maclaurin Series of cos(x)
Solution:
cos(x) = 1 - x²/2! + x⁴/4! - x⁶/6! + ...
= ∑ (from n = 0 to ∞) [(-1)ⁿ x²ⁿ / (2n)!]
Problem 3: Find the Taylor Series of ln(1 + x)
Solution:
ln(1 + x) = x - x²/2 + x³/3 - x⁴/4 + ...
= ∑ (from n = 1 to ∞) [(-1)ⁿ⁺¹ xⁿ / n]
The Taylor Series is a useful method for approximating complex functions with polynomials based on their derivatives. By understanding the Taylor series definition, formula, and theorem, we can express functions like sinx\sin xsinx, e^x, and ln(1+x) as infinite series. This approach makes calculations easier, especially when finding exact values is difficult. Whether using the general Taylor series or the specific Maclaurin series, this concept is important in mathematics, physics, engineering, and technology. In summary, the Taylor Series allows us to understand and work with functions more easily and accurately.
Answer: Yes, a Maclaurin series is a special case of a Taylor series centered at x = 0. In general:
Taylor Series: Expanded around any point a.
Maclaurin Series: Expanded around a = 0.
Answer: The Taylor series theorem states that a function f(x) can be represented as an infinite sum of derivatives at a single point, provided all derivatives exist and the series converges to the function.
Answer: The Taylor series of a function f(x) centered at a is:
f(x) = f(a) + f'(a)(x−a)/1! + f''(a)(x−a)²/2! + f'''(a)(x−a)³/3! + …
Answer: The Maclaurin series is the Taylor series at a = 0:
f(x) = f(0) + f'(0)x/1! + f''(0)x²/2! + f'''(0)x³/3! + …
Answer: The Maclaurin series is a power series that represents a function as an infinite sum of its derivatives evaluated at 0. It simplifies the Taylor series by assuming the expansion is around x = 0.
Answer: To find the n-th derivative of a function:
Differentiate the function n times using standard differentiation rules.
Evaluate at the point of expansion if using Taylor or Maclaurin series.
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