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:
Limitation:
This calculator is suitable for representing the Taylor 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
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:
\(\ f(x)≈P(x)=\sqrt{5}+\frac{\sqrt{5} \left(x - 1\right)}{5}+\frac{2 \sqrt{5} \left(x - 1\right)^{2}}{25}\)
People use the Taylor series because of a few reasons, including:
From the source of Wikipedia.org: Taylor Series.
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