Question

(a) It can be shown that if \(g\) is \(n\) times differentiable at \(x\) and \(f\) is \(n\) times differentiable at \(g(x),\) then the composite function \(h(x)=f(g(x))\) is \(n\) times differentiable at \(x\) and $$ h^{(n)}(x)=\sum_{r=1}^{n} f^{(r)}(g(x)) \sum_{r} \frac{r !}{r_{1} ! \cdots r_{n} !}\left(\frac{g^{\prime}(x)}{1 !}\right)^{n_{1}}\left(\frac{g^{\prime \prime}(x)}{2 !}\right)^{r 2} \cdots\left(\frac{g^{(n)}(x)}{n !}\right)^{r_{n}} $$ where \(\sum_{r}\) is over all \(n\) -tuples \(\left(r_{1}, r_{2}, \ldots, r_{n}\right)\) of nonnegative integers such that $$ r_{1}+r_{2}+\cdots+r_{n}=r $$ and $$ r_{1}+2 r_{2}+\cdots+n r_{n}=n . $$ (This is Faa di Bruno's formula). However, this formula is quite complicated. Justify the following alternative method for computing the derivatives of a composite function at a point \(x_{0}\) Let \(F_{n}\) be the \(n\) th Taylor polynomial of \(f\) about \(y_{0}=g\left(x_{0}\right),\) and let \(G_{n}\) and \(H_{n}\) be the \(n\) th Taylor polynomials of \(g\) and \(h\) about \(x_{0}\). Show that \(H_{n}\) can be obtained by substituting \(G_{n}\) into \(F_{n}\) and retaining only powers of \(x-x_{0}\) through the \(n\) th. HiNT: See Exercise \(2.5 .8(b)\). (b) Compute the first four derivatives of \(h(x)=\cos (\sin x)\) at \(x_{0}=0,\) using the method suggested by (a).

Short answer

Question: Use the alternative method based on Taylor polynomials to find the first four derivatives of \(h(x) = \cos(\sin x)\) at \(x_0 = 0\). Answer: The first four derivatives of \(h(x) = \cos(\sin x)\) at \(x_0 = 0\) are \(h'(0) = 0\), \(h''(0) = -1\), \(h'''(0) = 0\), and \(h^{(4)}(0) = 2\).

Step-by-step solution

1. Understanding the Taylor polynomials

The Taylor polynomial of a function is a polynomial approximation of the function near a specific point. \(F_n\), \(G_n\), and \(H_n\) represent the \(n\)-th Taylor polynomials of \(f\), \(g\), and \(h\), respectively, about the points \(y_0=g(x_0)\) and \(x_0\).

2. Substituting \(G_n\) into \(F_n\)

To obtain the Taylor polynomial \(H_n\) of the composite function \(h(x)=f(g(x))\), we will substitute the Taylor polynomial \(G_n\) into \(F_n\), i.e., replace \(y\) with \(G_n(x)\). So, \(F_n(G_n(x)) = f(g(x)) + O((x-x_0)^{n+1})\), where \(O((x-x_0)^{n+1})\) represents the higher-order terms from the Taylor series.

3. Retaining only powers of \(x-x_0\) up to the \(n\)-th

Since we are looking for the \(n\)-th Taylor polynomial \(H_n\), we will discard the higher-order terms, i.e., those with power \((x-x_0)^{n+1}\) or higher. The remaining terms will form the Taylor polynomial \(H_n\). #Part (b): Computing the first four derivatives of \(h(x)=\cos(\sin x)\) at \(x_0=0\)#

1. Find the Taylor polynomials for \(\sin x\) and \(\cos y\) at \(x_0 = 0\) and \(y_0=0\)

From the Taylor series expansion of \(\sin x\) and \(\cos y\) around \(0\), we have: \(G_n(x) = \sin x = x - \frac{x^3}{3!} + O(x^5)\) and \(F_n(y) = \cos y = 1 - \frac{y^2}{2!} + O(y^4)\)

2. Substitute \(G_n(x)\) into \(F_n(y)\) and retain only powers of \(x-x_0\) up to \(n=4\)

Substituting \(\sin x\) for \(y\) in the Taylor polynomial of \(\cos y\), we get: \(F_n(\sin x) = 1 - \frac{(\sin x)^2}{2!} + O((\sin x)^4)\) $= 1 - \frac{(x - \frac{x^3}{3!})^2}{2!} + O(x^8)\ = 1 - \frac{x^2}{2!} + \frac{x^4}{3! 4!} + O(x^6)$

3. Differentiate the Taylor polynomial of \(h(x)\)

Now we find the first four derivatives of \(h(x)\) using its Taylor polynomial: \(H_n(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{3! 4!} + O(x^6)\) \(h'(x) = -x + \frac{x^3}{3! 3!} + O(x^5)\) \(h''(x) = -1 + \frac{x^2}{2! 3!} + O(x^4)\) \(h'''(x) = 2x - O(x^3)\) \(h^{(4)}(x) = 2 + O(x)\) Now we evaluate these derivatives at \(x_0 = 0\): \(h'(0) = - (0) + \frac{(0)^3}{3! 3!} = 0\) \(h''(0) = - 1 + \frac{(0)^2}{2! 3!} = -1\) \(h'''(0)= 2 \cdot (0) = 0\) \(h^{(4)}(0) = 2\) Thus, the first four derivatives of \(h(x) = \cos(\sin x)\) at \(x_0=0\) are: \(h'(0) = 0\), \(h''(0) = -1\), \(h'''(0) = 0\), and \(h^{(4)}(0) = 2\).

Key concepts

Faa di Bruno's formula

Faa di Bruno's formula is a general result in calculus that helps to find the derivative of a composite function of the form \(h(x) = f(g(x))\). This formula is particularly useful for higher-order derivatives. It involves a complex summation of terms, where each term is composed of derivatives of the outer function \(f\) and powers of derivatives of the inner function \(g\).

• Consider \(h^{(n)}(x)\), where \(n\) is the order of the derivative you wish to compute.
• The formula expresses this as a sum, indexing over partitions of \(n\). Each partition corresponds to a way to distribute the derivative order across both outer and inner functions.
• This involves factorials and multinomial coefficients to balance the contribution of each derived term.

While complex, Faa di Bruno's formula provides a comprehensive structure that generalizes the chain rule for higher derivatives. However, in practical terms, its complexity often drives practitioners to look for simpler methods, especially when dealing with specific composite functions.

Taylor polynomials

Taylor polynomials offer a way to approximate functions around a specific point with a polynomial, which can make calculus operations easier to handle. For any given function \(f\), its Taylor polynomial of order \(n\) around a point \(a\) is a polynomial that matches the function \(f\) at \(a\) up to its \(n\)th derivative.

• This polynomial is constructed using the values of the function and its derivatives at point \(a\).
• Taylor polynomials help in approximating functions while reducing error near the point of expansion.
• The approximation becomes more accurate as the order \(n\) increases, but complexity also rises.

In many applications, especially those involving series and advanced calculus, Taylor polynomials allow derivative and integral calculations to be executed with polynomial simplicity.
Applying this to composite functions involves substituting the Taylor expansion of one function into another, which simplifies the computation dramatically, especially when seeking derivatives at a specific point.

Derivatives of composite functions

Finding derivatives of composite functions can be challenging, particularly for higher-order derivatives. Fortunately, rules and methods exist, such as the chain rule and its extensions, to simplify these computations. Composite functions are of the form \(h(x) = f(g(x))\), where you wish to find \(h'(x)\).

• The chain rule provides a straightforward way to compute the first derivative: \(h'(x) = f'(g(x)) \cdot g'(x)\).
• For higher-order derivatives, principles extend such that each level of derivative accounts for nested derivatives.
• Tools like Faa di Bruno's formula or methods involving Taylor expansions simplify the computation of such derivatives without resorting to complex manipulations.

By using Taylor polynomials for both composing functions\(f\) and \(g\), practitioners can approximate the desired derivatives especially when dealing with specific points or simplified forms.
These abilities make investigating and working with derivatives more approachable and applicable across different areas of mathematics and applied sciences.