Generate the Taylor series expansion for any function around a specified point. Taylor series are used to approximate complex functions in engineering, physics, and economics. Input your function, select the expansion point, and specify the degree for an accurate series representation, aiding in simplifying complex calculations and approximations.
The intuition behind the Taylor series function is a powerful mathematical tool, used to represent functions with an infinite sum of representing polynomial terms. The Taylor series, named after Brook Taylor, breaks a function into a sum of terms using its derivatives at one point. The Taylor series for e^x, sin(x) and cos(x): Proof & Solutions Here we are going to explore the Taylor series of e^x, sin(x), cos(x), and other examples along with proofs and solutions.
The Taylor series of a function \( f(x) \) about a given point \( x = a \) is given by:
\( f(x) = f(a) + f'(a)(x - a) + \frac{f''(a)}{2!}(x - a)^2 + \frac{f'''(a)}{3!}(x - a)^3 + \dots \)
In summation form:
\( f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!} (x - a)^n \)
Where:
If the series is a Maclaurin series, where \( a = 0 \), then it is a special case of the Taylor series.
The function \( e^x \) has derivatives \( f^{(n)}(x) = e^x \). At \( x = 0 \):
\( f^{(n)}(0) = 1 \)
The Taylor series of \( e^x \) centered at \( x = 0 \) (Maclaurin series) is:
\( e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \dots = \sum_{n=0}^{\infty} \frac{x^n}{n!} \)
Use the first four terms to approximate \( e^x \) at \( x = 1 \):
If we try to obtain \( e^1 \), it gives us \( e^1 \approx 1 + 1 + \frac{1^2}{2!} + \frac{1^3}{3!} = 1 + 1 + 0.5 + 0.1667 = 2.6667 \)
The function \( \sin(x) \) has derivatives of alternating form \( \sin(x), \cos(x), -\sin(x), -\cos(x) \). At \( x = 0 \):
The trigonometric function \( \sin(0) = 0 \), \( \cos(0) = 1 \).
The Maclaurin series is:
\( \sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \dots = \sum_{n=0}^{\infty} (-1)^n \frac{x^{2n+1}}{(2n+1)!} \)
Use the approximate value of \( \sin(\frac{\pi}{6}) \):
\( \sin(\frac{\pi}{6}) \approx \frac{\pi}{6} - \frac{(\frac{\pi}{6})^3}{6} = 0.5236 - 0.0237 = 0.4999 \) (Approx. \( 0.5 \))
The function \( \cos(x) \) has derivatives \( \cos(x), -\sin(x), -\cos(x), \sin(x) \). At \( x = 0 \):
\( \cos(0) = 1, \quad \sin(0) = 0 \)
The Maclaurin series is:
\( \cos(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \dots = \sum_{n=0}^{\infty} (-1)^n \frac{x^{2n}}{(2n)!} \)
Let \( f(x) = e^{x^2} \), and every derivative can be computed iteratively from it. At \( x = 0 \):
\( e^{x^2} = 1 + x^2 + \frac{x^4}{2!} + \frac{x^6}{3!} + \dots = \sum_{n=0}^{\infty} \frac{x^{2n}}{n!} \)
The Taylor series turns a function into an infinite polynomial using its derivatives at one point.
\( f(x) = f(a) + f'(a)(x - a) + \frac{f''(a)}{2!}(x - a)^2 + \dots \)
It is done using higher-order derivatives and applying the limiting case where the remainder term tends toward 0 as the number of terms tend toward infinity.
\( e^x = 1 + x + \frac{x^2}{2!} \) At \( x = 1 \): \( e^1 \approx 1 + 1 + \frac{1}{2} = 2.5 \)
\( \sin(x) \approx x - \frac{x^3}{3!} \) At \( x = \frac{\pi}{4} \): \( \sin(\frac{\pi}{4}) \approx \frac{\pi}{4} - \frac{(\frac{\pi}{4})^3}{6} \)
The Taylor series of \( e^x \) at \( x = 0 \) is:
\( e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!} = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \dots \)
\( \sin(x) = \sum_{n=0}^{\infty} (-1)^n \frac{x^{2n+1}}{(2n+1)!} = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \dots
\( \cos(x) = \sum_{n=0}^{\infty} (-1)^n \frac{x^{2n}}{(2n)!} = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \dots
\( e^{x^2} = \sum_{n=0}^{\infty} \frac{x^{2n}}{n!} = 1 + x^2 + \frac{x^4}{2!} + \dots
If you understand Taylor series, you are equipped with one of the most powerful tools in mathematics to approximate functions, solve equations, and study models in maths.
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