Problem 32
Question
Use the fact that the Taylor series of \(g(x)=\sin \left(x^{2}\right)\) is $$x^{2}-\frac{x^{6}}{3 !}+\frac{x^{10}}{5 !}-\frac{x^{14}}{7 !}+\cdots$$ to find \(g^{\prime \prime}(0), g^{\prime \prime \prime}(0),\) and \(g^{(10)}(0) .\) (There is an easy way and a hard way to do this!)
Step-by-Step Solution
Verified Answer
\( g''(0) = 2 \), \( g'''(0) = 0 \), \( g^{(10)}(0) = 30240 \).
1Step 1: Identify the general form of a Taylor series
The Taylor series for a function \( g(x) \) centered at zero is given by \( g(x) = \sum_{n=0}^{\infty} \frac{g^{(n)}(0)}{n!}x^n \). This involves the derivatives of \( g(x) \) evaluated at 0.
2Step 2: Match terms with the Taylor series
Compare the given series \( x^2 - \frac{x^6}{3!} + \frac{x^{10}}{5!} - \frac{x^{14}}{7!} + \cdots \) with the Taylor series to identify the coefficients \( \frac{g^{(n)}(0)}{n!} \).
3Step 3: Identify terms relevant to the needed derivatives
Since \( g^{(n)}(0) \) represents the coefficient of \( \frac{x^n}{n!} \), set each coefficient equal to the respective derivative terms from the series:- For \( g''(0) \), identify the coefficient of \( x^2 \).- For \( g'''(0) \), identify the coefficient of \( x^3 \).- For \( g^{(10)}(0) \), identify the coefficient of \( x^{10} \).
4Step 4: Derive the values for specific derivatives
- For \( x^2 \): \( \frac{g''(0)}{2!} = 1 \), so \( g''(0) = 2! \cdot 1 = 2 \).- Since \( x^3 \) does not appear, \( \frac{g'''(0)}{3!} = 0 \), so \( g'''(0) = 0 \).- For \( x^{10} \): \( \frac{g^{(10)}(0)}{10!} = \frac{1}{5!} \), so \( g^{(10)}(0) = 10! \cdot \frac{1}{5!} \).
5Step 5: Calculate the derivative value for \( g^{(10)}(0) \)
Use the factorial relationship to simplify \( g^{(10)}(0) \):- \( g^{(10)}(0) = 10! \cdot \frac{1}{5!} = \frac{10!}{5!} \).- Calculate it as \( 10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 = 30240 \).
Key Concepts
Higher DerivativesTrigonometric FunctionsFactorials
Higher Derivatives
In mathematics, a derivative measures how a function changes as its input changes. The higher derivatives are simply repeated derivatives of that function. For example, the first derivative refers to the function's slope, while the second derivative indicates how the slope itself changes over an interval.
A function's Taylor series can help identify these derivatives at a specific point, often zero. In the context of the Taylor series, you can determine the coefficients of the series to uncover specific higher derivatives.
A function's Taylor series can help identify these derivatives at a specific point, often zero. In the context of the Taylor series, you can determine the coefficients of the series to uncover specific higher derivatives.
- For a term like \( x^n \), the coefficient in front of the term helps find \( g^{(n)}(0) \) using the formula \( \frac{g^{(n)}(0)}{n!} \).
- Determining derivatives can be less intimidating by focusing on each term of the Taylor series.
Trigonometric Functions
Trigonometric functions form the cornerstone of many mathematical concepts, such as waveform analysis and calculus. Functions like \( \sin x \), \( \cos x \), and \( \tan x \) describe relationships in triangles but are also essential in periodic function descriptions.
When dealing with Taylor series, applying these functions can aid in expressing complex equations in simpler polynomial formats.
When dealing with Taylor series, applying these functions can aid in expressing complex equations in simpler polynomial formats.
- For instance, \( \sin(x^2) \) can be expanded into a Taylor series that tells us a lot about the function near the point zero.
- These expansions help approximate the function's behavior through infinite sums, where each term simplifies calculations and derivative extractions.
Factorials
Factorials, denoted with an exclamation point (\( n! \)), are the product of all positive integers up to a given number \( n \). Factorials frequently arise within the context of permutations, combinations, and series like the Taylor series.
They play an integral role in balancing the series expansion by determining the weight of each term. For example, in the Taylor series expression \( \sum \frac{g^{(n)}(0)}{n!}x^n \), the factorial \( n! \) helps normalize each term's contribution.
They play an integral role in balancing the series expansion by determining the weight of each term. For example, in the Taylor series expression \( \sum \frac{g^{(n)}(0)}{n!}x^n \), the factorial \( n! \) helps normalize each term's contribution.
- Factorials grow rapidly as \( n \) increases, drastically reducing the magnitude of terms in a Taylor series for any function \( g(x) \) evaluated at zero.
- Involving factor terms like \( \frac{1}{5!} \) in calculations balances long expansions by ensuring each derivative's contribution is scaled appropriately.
Other exercises in this chapter
Problem 32
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