Taylor Series Calculator

The Taylor series calculator calculates all coefficients of a Taylor series expansion for a function centred at point n. Also, you can set point n as zero (0) to get the Maclaurin series representation.

What is the Taylor Series?

In mathematics, the Taylor series is defined as the representation of a given function. It is an infinite series that represents the value of the derivative of a function at a certain point.

There is a special case for the Taylor and Maclaurin series calculator. This series helps to reduce the number of mathematical proof and is used for power flow analysis.

Taylor Series Expansion Formula:

The formula used by taylor series formula calculator for calculating a series for a function is given as:

$$ F (x) = ∑^ ∞_{n=0} f^k (a) / k! (x – a) ^k $$

Where f^(n)(a) is the nth order derivative of function f(x) as evaluated at x = a, n is the order, and a is where the series is centered. The series will be most precise near the centering point.

A Taylor expansion may be infinite, but we can select to make our series or function as little or long terms as we want. We can set the maximum n value to make it an n order series.

Example:

Calculate Taylor expansion of (x^2+4)^{1/2} up to n = 4?

Solution:

Given function f(m)= (x^2+4)^{1/2} and order point n = 1 to 4

Maclaurin equation for the function is:

$$ f(y)=∑k=0^∞ f (k) (a)/ k! (x – a)^k $$

$$ f(y)≈ P (x) = ∑_k=0^4 f^(k) (a) / k! (x – a)^k = ∑k=0^∞ f (k) (a)/ k! (x – a)^k $$

So, find taylor series calculator evaluates the derivatives and calculate them at the given point, and substitute the obtained values into the series formula.

$$ F^0(y) = f (y) = \sqrt {x^2 + 4 } $$

Evaluate function:

$$f (1) = \sqrt {5} $$

Take the first derivative \( f^1(y) = [f^0(y)]’ \)

$$ [\sqrt {x^2 + 4 }]’ = \frac {x} { \sqrt {x^2 + 4 }} $$

$$(f (1))’ = \frac { \sqrt { 5}} {5}$$

Find the second Derivative:

$$f^2 (y) = [f^1 (y)]’ = \frac {x} { \sqrt {x^2 + 4 }} = 4 / (x^2 + 4) ^{3/2} $$

Calculate the second derivative at given point:

F (1)’’ = 4\sqrt{5} / 25

Now, take the third derivative:

$$ f^3(y) = [f^2(y)]’ = (4/ (x^2 + 4) ^{3/2}) = – 12x / (x^2 + 4) ^{5/2} $$

Calculate the third derivative of \( (f (0))”’ = – 12 \sqrt {5} / 125 \)

Fourth derivative:

$$f^4 (y) = [f^3 (y)]’ = [- 12x / (x^2 + 4) ^{5/2}]’ = 48x^2 – 48 / \sqrt {x^2 + 4} (x^6 + 12x^4 + 48x^2 + 64) $$

Then, find the fourth derivative of function (f(0))””  = 0

$$ f(y) ≈ \sqrt {5}/0! (x – 1)^0 + \frac { \sqrt{5} / 5} {1!} (x – 1)^1 + … + 0 / 4! (x – 1)^4 $$

After simplification, we get the final results:

$$ f(y) ≈ P(x) = \sqrt {5} + \sqrt {5} (x-1) / 5 + 2 \sqrt {5} (x-1)^2 / 25 – 2 \sqrt {5} (x – 1)^3 / 125  $$

How Our Calculator Works?

Input:

• Firstly, substitute a function with respect to a specific variable.
• Now, enter a particular point to evaluate the Taylor series of functions around this point.
• Then, add the order n for approximation.
• By using this taylor series error calculator, find the series and determine the error at the given point. (optional)
• Click the calculate button for further solution.

Output:

• The sum of Taylor series calculator with steps shows the series after simplification.
• It computes the series of entered functions around the given order number n.
• The third degree taylor polynomial calculator takes the derivative for getting the polynomials and puts the results into the taylor series formula.
• It displays the results after the simplification of polynomials.

Reference:

From the source of Wikipedia: Analytic functions, Approximation error, and convergence, Generalization, List of Maclaurin series of some common functions, Exponential function, Natural logarithm, Geometric series, Binomial series.