Chapter 4: Problem 32
Find the \(n\) th Taylor polynomial of \(f\) around \(a\), that is, $$ P_{n}(x)=f(a)+f^{\prime}(a)(x-a)+\cdots+\frac{f^{(n)}(a)}{n !}(x-a)^{n} \quad \text { for } x \in \mathbb{R} $$ when \(a=0\) and \(f(x)\) equals: (i) \(\frac{1}{1-x}\), (ii) \(\frac{1}{1+x}\), (iii) \(\frac{x}{1+x^{2}}\).
Short Answer
Expert verified
The \(n\) th Taylor polynomials for the given functions are:
(i) \(P_n(x) = 1 + x + \frac{2}{2!}x^2 + \frac{3}{3!}x^3 + \cdots + \frac{n}{n!}x^n\),
(ii) \(P_n(x) = 1 - x + \frac{(-1)^2 2}{2!}x^2 + \frac{(-1)^3 3}{3!}x^3 + \cdots + \frac{(-1)^n n}{n!}x^n\),
(iii) \(P_n(x) = \begin{cases}
0 & \text{for } n = 0, \\
x & \text{for } n = 1, \\
x & \text{for } n > 1.
\end{cases}\).
Step by step solution
01
Find the nth derivative
Using the geometric series formula, we can rewrite \(\displaystyle\frac{1}{1-x}\) as a power series:
$$
\frac{1}{1-x} = 1 + x + x^2 + \cdots + x^n + \cdots
$$
For each term in the series, the derivative is just the power of \(x\):
$$
f'(x) = 1 + 2x + 3x^2 + \cdots + nx^{n-1} + \cdots
$$
Notice that the general term of \(f'(x)\) has the form \(nx^{n-1}\), so we can say that the \(n\)th derivative \(f^{(n)}(x) = n!\).
02
Evaluate the nth derivative at a=0
For the given function \(f(x)\), the derivatives do not depend on the value of \(x\), so they remain the same at \(x=0\):
$$
f^{(n)}(0) = n!
$$
03
Find the nth Taylor polynomial
Using the given formula for \(P_n(x)\) and the results from Step 2, the \(n\)th Taylor polynomial of \(\frac{1}{1-x}\) is
$$
P_n(x) = 1 + x + \frac{2}{2!}x^2 + \frac{3}{3!}x^3 + \cdots + \frac{n}{n!}x^n
$$
(ii) \(\displaystyle\frac{1}{1+x}\)
04
Find the nth derivative
Using the geometric series formula, we can rewrite \(\displaystyle\frac{1}{1+x}\) as a power series:
$$
\frac{1}{1+x} = 1 - x + x^2 - \cdots + (-1)^n x^n + \cdots
$$
For each term in the series, the derivative is just the power of \(x\) multiplied by the corresponding sign:
$$
f'(x) = -1 + 2x - 3x^2 + \cdots + (-1)^n n x^{n-1} + \cdots
$$
Notice that the general term of \(f'(x)\) has the form (-1)^n nx^{n-1}\(, so we can say that the \)n\(th derivative \)f^{(n)}(x) = (-1)^n n!$.
05
Evaluate the nth derivative at a=0
For the given function \(f(x)\), the derivatives do not depend on the value of \(x\), so they remain the same at \(x=0\):
$$
f^{(n)}(0) = (-1)^n n!
$$
06
Find the nth Taylor polynomial
Using the given formula for \(P_n(x)\) and the results from Step 2, the \(n\)th Taylor polynomial of \(\frac{1}{1+x}\) is
$$
P_n(x) = 1 - x + \frac{(-1)^2 2}{2!}x^2 + \frac{(-1)^3 3}{3!}x^3 + \cdots + \frac{(-1)^n n}{n!}x^n
$$
(iii) \(\displaystyle\frac{x}{1+x^{2}}\)
07
Find the nth derivative
Since this function doesn't have a simple series representation like the previous ones, we will use the quotient rule to find the derivatives.
The first derivative of \(\displaystyle\frac{x}{1+x^{2}}\) is:
$$
f'(x) = \frac{1-x^2}{(1+x^2)^2}
$$
While computing higher-order derivatives is possible, it is complex and not required in this case, as we'll see in the next step.
08
Evaluate the nth derivative at a=0
We will evaluate the \(1^{st}\) derivative at \(a=0\) and check this for higher-order derivatives as well:
$$
f'(0) = \frac{1}{1} = 1
$$
Since higher-order derivatives have increasing powers of \(x\), they will always evaluate to zero at \(x=0\). So, for \(n>1\):
$$
f^{(n)}(0) = 0
$$
09
Find the nth Taylor polynomial
Using the given formula for \(P_n(x)\) and the results from Step 2, the \(n\)th Taylor polynomial of \(\frac{x}{1+x^{2}}\) is
$$
P_n(x) = \begin{cases}
0 & \text{for } n = 0, \\
x & \text{for } n = 1, \\
x & \text{for } n > 1.
\end{cases}
$$
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Geometric Series
A geometric series is a series with a constant ratio between successive terms. It takes the form:
This expression is very powerful because it allows the calculation of function values using a simple series formula. If the series converges, it can approximate functions more efficiently, avoiding complex calculations with derivatives.
- First term: \( a \)
- Common ratio: \( r \)
- Series: \( a, ar, ar^2, ar^3, \ldots \)
This expression is very powerful because it allows the calculation of function values using a simple series formula. If the series converges, it can approximate functions more efficiently, avoiding complex calculations with derivatives.
Power Series
A power series is an infinite series of the form \( \sum_{n=0}^{\infty} a_n(x-c)^n \), where \( c \) is the center of the series and \( a_n \) are coefficients.
Power series are immensely useful in mathematics, especially for expanding functions in terms they can be approximated at different points.
In the context of Taylor polynomials, a power series allows us to express a function in a way that makes finding its derivatives straightforward. For functions like \( \frac{1}{1-x} \), we can expand them into a power series thus:
Power series are immensely useful in mathematics, especially for expanding functions in terms they can be approximated at different points.
In the context of Taylor polynomials, a power series allows us to express a function in a way that makes finding its derivatives straightforward. For functions like \( \frac{1}{1-x} \), we can expand them into a power series thus:
- \( \frac{1}{1-x} = 1 + x + x^2 + x^3 + \cdots \)
Quotient Rule
The quotient rule is a formula used to find the derivative of a quotient of two functions. It is given by:\[\left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2}\]where \( u \) and \( v \) are functions of \( x \). This rule is crucial when dealing with functions that can't be split into simpler terms.
For example, the function \( \frac{x}{1+x^2} \) doesn't lend itself to expansion like a geometric series. Instead, we use the quotient rule to find its derivative. The differentiation of \( \frac{x}{1+x^2} \) results in a complex expression, but the quotient rule streamlines the process by providing a systematic approach.
This rule allows us to tackle more complicated derivatives by breaking them into manageable parts, ensuring precise calculations without exhaustive manual manipulation.
For example, the function \( \frac{x}{1+x^2} \) doesn't lend itself to expansion like a geometric series. Instead, we use the quotient rule to find its derivative. The differentiation of \( \frac{x}{1+x^2} \) results in a complex expression, but the quotient rule streamlines the process by providing a systematic approach.
This rule allows us to tackle more complicated derivatives by breaking them into manageable parts, ensuring precise calculations without exhaustive manual manipulation.
Higher-Order Derivatives
Higher-order derivatives involve finding derivatives of derivatives, continuing as many times as needed. The notation \( f^{(n)}(x) \) represents the nth derivative of the function \( f(x) \).
This becomes particularly useful when constructing Taylor polynomials, as these require the evaluation of multiple derivatives at a point. Calculating higher-order derivatives helps determine the sensitivity of changes in a function.
This becomes particularly useful when constructing Taylor polynomials, as these require the evaluation of multiple derivatives at a point. Calculating higher-order derivatives helps determine the sensitivity of changes in a function.
- First derivative: Measures the rate of change or slope.
- Second derivative: Provides insights into concavity and acceleration.
- Third and higher derivatives: Reveal deeper aspects like the rate of change of the second derivative.