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Enter the values to find the Taylor series representation of a function.

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Use this Taylor series calculator to represent your function as a Taylor series step by step. It allows you to expand the function by specifying:

- The center point (a) around which you want to center the Taylor series. By default, this is typically indicated to be x = 0 (when the point is zero, then it finds the Maclaurin series for the given function)
- The desired degree (n) of the Taylor series polynomial, this will help to determine the number of terms that are considered for the approximation
- Error bounds or convergence analysis that depends on the degree of the polynomial

**Limitation: **This calculator is suitable for representing the Taylor and Maclaurin series. It cannot handle advanced features like analyzing convergence or exploring alternative series representations.

The Taylor series is an infinitely long sum of terms derived from the function's derivatives at a specified point. It is widely used in calculus to approximate the values of complex functions, especially near the chosen point. This Taylor series is particularly useful for representing complex functions with simpler polynomials.

The general formula for Taylor series expansion is:

**\(\ f(x)=\sum_{n=0}^{\infty}\frac{f^{(n)}(a)}{n!}(x-a)^n=f(a)+f^{\prime}(a)(x-a)+\frac{f^{\prime\prime}(a)}{2!}(x-a)^2+\frac{f^{\prime\prime\prime}(a)}{3!}(x-a)^3+\ldots+\frac{f^{(n)}(a)}{n!}(x-a)^n+\ldots\)**

Where

- “n” is the total number of terms included in the Taylor Series
- “a” is the center point of the function
- 𝑓(𝑎) represents the value of the function at the point x = 𝑎
- \(\ 𝑓^{′}(𝑎)\) is the first derivative
- \(\ 𝑓^{′′} (𝑎)\) represents the second derivative
- \(\ 𝑓′′′(𝑎)\) shows the third derivative

The Taylor series is infinite, but you can set the degree of the polynomial (n) to specify. It can also be done with our Taylor series calculator which allows you to specify the “n” value for the approximation(adding a higher degree leads to a more accurate approximation of the function).

To calculate the Taylor series expansion for the function, look at the example using a formula:

The function is “\(\sqrt{x^{2} + 4}\) up to n = 2, where Point =1”, find its Taylor series.

**Solution:**

\(\ f (x) = \sum\limits_{k=0}^{\infty} \frac{f^{(k)}(a)}{k!}(x - a)^{k}\)

\(\ f (x) ≈ P(x) = \sum\limits_{k=0}^{\infty} \frac{f^{(k)}(a)}{k!}(x - a)^{k} = \sum\limits_{k=0}^{2} \frac{f^{(k)}(a)}{k!}(x - a)^{k}\)

\(\ f^{(0)}(x) = f(x)= \sqrt{x^{2} + 4}\) \(\ f(1) = \sqrt{5}\)

Calculate 1st Derivative:

\(\ f^{(1)}(x) = \left(f^{(0)}(x)\right)^{'}= \left(\sqrt{x^{2} + 4}\right)^{'} = \frac{x}{\sqrt{x^{2} + 4}}\)

1st Dervitaive at The Given Point:

\(\left(f(1)\right)^{'} = \frac{\sqrt{5}}{5}\)

2nd Derivative:

\(\ f^{(2)}(x) = \left(f^{(1)}(x)\right)^{'}= \left(\frac{x}{\sqrt{x^{2} + 4}}\right)^{'} = \frac{4}{\left(x^{2} + 4\right)^{\frac{3}{2}}}\)

2nd Dervitaive at The Given Point:

\(\left(f(1)\right)^{''} = \frac{4 \sqrt{5}}{25}\)

Use these Values to Get The Polynomial:

\(\ f(x) ≈ \frac{\sqrt{5}}{0!}(x- (1))^{0} + \frac{\frac{\sqrt{5}}{5}}{1!}(x- (1))^{1} + \frac{\frac{4 \sqrt{5}}{25}}{2!}(x- (1))^{2}\)

Here you can also perform polynomial Taylor expansion by specifying the values in the Taylor polynomial calculator.

After Simplification:

\(\ f(x)≈P(x)=\sqrt{5}+\frac{\sqrt{5} \left(x - 1\right)}{5}+\frac{2 \sqrt{5} \left(x - 1\right)^{2}}{25}\)

Taylor Remainder is the error that occurred approximating a function f(x) using its nth-degree series polynomial around a specified point. The difference between the approximation and actual value of the function f(x) is the remainder. It is denoted by the function Rn(x).

\(\ f(x)=f(a)+f^{\prime}(a)(x-a)+\frac{f^{\prime\prime}(a)}{2!}(x-a)^2+...+ \frac{f^{\prime\prime\prime}(a)}{n!}(x-a)^n+R_n(x)\)

Where

\(\ R_n(x)=\sum _{n=0}^{\infty }\left(\frac{f^{(n+1)}\left(c\right)}{(n+1)!}\left(x-a\right)^{(n+1)}\right)\)

The error that is associated with approximating a function can be easily determined by using the Taylor series error calculator.

People use the Taylor series because of a few reasons, including:

**Approximating Functions:**The partial sums of the Taylor series let you approximate functions. These partial sums are (finite) polynomials, and they can be easily computed. With its help, you can create a simple polynomial function that resembles the complex function around a specific point.**Analyzing Function Behavior:**The terms of the series provide you with information on how the function behaves near the center point (Each term of the Taylor polynomial is obtained by taking the function's derivatives at a single point)**Series Representations:**It helps to represent crucial functions such as sine, cosine, and exponential functions. Therefore it is widely used in science and engineering**Solving Differential Equations:**In some cases, the Taylor series can also be used to get the approximate solutions of differential equations. It is particularly beneficial when finding the exact solution looks difficult

It represents the number as a simple polynomial that has a similar value to the original function around a specified x value.

\(\ f(x)=f(a)+f^{\prime}(a)(x-a)+\frac{f^{\prime\prime}(a)}{2!}(x-a)^2+\frac{f^{\prime\prime\prime}(a)}{3!}(x-a)^3+\ldots+\frac{f^{(n)}(a)}{n!}(x-a)^n+\ldots\)

This polynomial is often much easier to work as compared to the original function, especially when you are dealing with complex functions.

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