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Magazine Chapter 9: Problem 22
Prove: The Taylor series for \(\sin x\) about any value \(x=x_{0}\) converges to \(\sin x\) for all \(x\)
Short Answer
Expert verified
The Taylor series for \(\sin x\) converges to \(\sin x\) for all \(x\).
Step by step solution
01
Understanding the Taylor Series for Sin
The Taylor series for a function \(f(x)\) at a point \(x_0\) can be expressed as: \[f(x) = f(x_0) + \frac{f'(x_0)}{1!}(x-x_0) + \frac{f''(x_0)}{2!}(x-x_0)^2 + \frac{f'''(x_0)}{3!}(x-x_0)^3 + \cdots \] For \(\sin x\), we need to apply this formula starting with the first few derivatives evaluated at \(x_0\).
02
Calculating Derivatives of Sin
The derivatives of \(\sin x\) are cyclic:- \(\frac{d}{dx}\sin x = \cos x\)- \(\frac{d^2}{dx^2}\sin x = -\sin x\)- \(\frac{d^3}{dx^3}\sin x = -\cos x\)- \(\frac{d^4}{dx^4}\sin x = \sin x\)Thus, they repeat every four terms. We will substitute these into the Taylor series formula.
03
Substitute Derivatives into Taylor Series
Substitute the cyclic derivatives into the Taylor series expansion:\[\sin x \approx \sin x_0 + \cos x_0(x-x_0) - \frac{\sin x_0}{2!}(x-x_0)^2 - \frac{\cos x_0}{3!}(x-x_0)^3 + \cdots\] This series continues indefinitely, representing \(\sin x\) in terms of \(x_0\) and \(x-x_0\).
04
Investigating Convergence Using Lagrange's Remainder
The Lagrange error bound for Taylor series is given by:\[R_n(x) = \frac{f^{(n+1)}(c)}{(n+1)!}(x-x_0)^{n+1}\]Where \(c\) is some value between \(x_0\) and \(x\). For \(\sin x\), the absolute value of the \((n+1)\)-th derivative is at most 1, thus\[|R_n(x)| \leq \frac{|x-x_0|^{n+1}}{(n+1)!}\]As \(n \to \infty\), this remainder approaches zero, confirming convergence.
05
Conclude the Convergence
Since the remainder \(R_n(x)\) approaches zero as \(n\) becomes large, the series converges to \(\sin x\) irrespective of the choice of \(x_0\). This proves that the Taylor series for \(\sin x\) converges to \(\sin x\) for all \(x\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Sin Function
The sine function, denoted as \( \sin x \), is a fundamental trigonometric function that describes the ratio of the length of the opposite side to the hypotenuse in a right-angled triangle. It is an essential component in mathematics, particularly in calculus and trigonometry, due to its unique periodic and wave-like nature.
The sine function is periodic with a period of \( 2\pi \), meaning every \( 2\pi \) units, the wave pattern repeats. This property makes it a key function for modeling periodic phenomena such as sound waves, light waves, and alternating current in electrical circuits.
Using the Taylor series, we can represent \( \sin x \) as an infinite sum of its derivatives evaluated at a specific point, providing a polynomial approximation of the function that becomes more precise with more terms included.
The sine function is periodic with a period of \( 2\pi \), meaning every \( 2\pi \) units, the wave pattern repeats. This property makes it a key function for modeling periodic phenomena such as sound waves, light waves, and alternating current in electrical circuits.
Using the Taylor series, we can represent \( \sin x \) as an infinite sum of its derivatives evaluated at a specific point, providing a polynomial approximation of the function that becomes more precise with more terms included.
Convergence
Convergence refers to the behavior of a sequence or series of numbers or functions getting closely and progressively approximated towards a specific value as more terms are added. This concept is crucial in the context of Taylor series.
When we evaluate the Taylor series for a function like \( \sin x \), we express \( \sin x \) as an infinite series of polynomial terms. The idea is that as we add more terms from this series, the approximation gets closer to the actual value of \( \sin x \).
For the sine function, the convergence of its Taylor series is guaranteed over all real numbers. This means that no matter what value of \( x \) we choose, the series will eventually approximate \( \sin x \) accurately as more terms are added. The convergence of the sine function's series is particularly robust due to the alternating pattern of its derivatives.
When we evaluate the Taylor series for a function like \( \sin x \), we express \( \sin x \) as an infinite series of polynomial terms. The idea is that as we add more terms from this series, the approximation gets closer to the actual value of \( \sin x \).
For the sine function, the convergence of its Taylor series is guaranteed over all real numbers. This means that no matter what value of \( x \) we choose, the series will eventually approximate \( \sin x \) accurately as more terms are added. The convergence of the sine function's series is particularly robust due to the alternating pattern of its derivatives.
Derivatives
In the context of the Taylor series, derivatives are essential in determining the coefficients of the series expansion. The sine function has derivatives that are cyclic every four terms, which simplifies the task of creating its Taylor series.
The first derivative of \( \sin x \) is \( \cos x \), the second derivative is \( -\sin x \), the third derivative is \( -\cos x \), and the fourth derivative returns to \( \sin x \).
This cyclic nature enables us to predict and use derivatives beyond the fourth degree without recalculation. When substituted into the Taylor series formula, the derivatives help build the polynomial expansion that makes up the Taylor series of the sine function.
The first derivative of \( \sin x \) is \( \cos x \), the second derivative is \( -\sin x \), the third derivative is \( -\cos x \), and the fourth derivative returns to \( \sin x \).
This cyclic nature enables us to predict and use derivatives beyond the fourth degree without recalculation. When substituted into the Taylor series formula, the derivatives help build the polynomial expansion that makes up the Taylor series of the sine function.
Lagrange Remainder
The Lagrange remainder provides an estimate of the error involved in truncating a Taylor series at a finite number of terms. This is crucial in proving the convergence of a Taylor series.
For the sine function \( \sin x \), the Lagrange remainder \( R_n(x) \) represents the difference between the actual function value and the Taylor polynomial approximation. It is expressed as \( R_n(x) = \frac{f^{(n+1)}(c)}{(n+1)!}(x-x_0)^{n+1} \), where \( c \) is some value between \( x_0 \) and \( x \).
The beauty of the sine function is that the absolute value of any derivative is bounded by 1. This leads to an error term that becomes smaller and smaller as \( n \) increases. Ultimately, as \( n \to \infty \), the Lagrange remainder tends toward zero, affirming that the Taylor series converges to \( \sin x \) for all real numbers.
For the sine function \( \sin x \), the Lagrange remainder \( R_n(x) \) represents the difference between the actual function value and the Taylor polynomial approximation. It is expressed as \( R_n(x) = \frac{f^{(n+1)}(c)}{(n+1)!}(x-x_0)^{n+1} \), where \( c \) is some value between \( x_0 \) and \( x \).
The beauty of the sine function is that the absolute value of any derivative is bounded by 1. This leads to an error term that becomes smaller and smaller as \( n \) increases. Ultimately, as \( n \to \infty \), the Lagrange remainder tends toward zero, affirming that the Taylor series converges to \( \sin x \) for all real numbers.
