Question

Find the first few terms of the Maclaurin series for each of the following functions and check your results by computer. $$\cos (e^x-1)$$

Short answer

The first few terms are $1-\frac{x^2}{2}-\frac{x^3}{2}$.

Step-by-step solution

Understand Maclaurin Series

The Maclaurin series for a function can be expressed as: $$f(x)=f(0)+f^{\prime }(0)x+\frac{f^{″}(0)}{2!}{x}^2+\frac{f^{‴}(0)}{3!}{x}^3+\cdots$$ where $f^{(n)}(0)$ represents the $n$-th derivative of the function evaluated at 0.

Define the Function

Let $f(x)=\cos (e^x-1)$. To find the Maclaurin series, we need the derivatives of $f(x)$ evaluated at $x=0$.

Evaluate $f(0)$

First, evaluate the function at $x=0$:$$f(0)=\cos (e^0-1)=\cos (0)=1$$.

Find the First Derivative

Calculate $f^{\prime }(x)$:Using the chain rule, $f^{\prime }(x)=-\sin (e^x-1)\cdot \frac{d}{dx}(e^x-1)=-\sin (e^x-1)e^x$.Evaluate $f^{\prime }(0)$: $f^{\prime }(0)=-\sin (e^0-1)e^0=-\sin (0)=0$.

Find the Second Derivative

Calculate $f^{″}(x)$:$$f^{″}(x)=-\cos (e^x-1)e^{2x}-\sin (e^x-1)e^x$$.Evaluate $f^{″}(0)$: $f^{″}(0)=-\cos (0)e^0-\sin (0)e^0=-1$.

Find the Third Derivative

Calculate $f^{‴}(x)$:$$f^{‴}(x)=(-\sin (e^x-1)e^{2x}-\cos (e^x-1)e^{4x})-(\cos (e^x-1)e^{2x}+\sin (e^x-1)e^x)e^x$$.Evaluate $f^{‴}(0)$: $f^{‴}(0)=\sin (0)e^0-3\cos (0)e^0=-3$.

Construct the Series

Substitute the values into the Maclaurin series formula:$$f(x)\approx 1+0\cdot x+\frac{-1}{2!}{x}^2+\frac{-3}{3!}{x}^3=1-\frac{x^2}{2}-\frac{x^3}{2}$$.

Key concepts

Taylor Series

The Taylor series is a way to represent a function as an infinite sum of terms that are calculated from the values of its derivatives at a single point. It generalizes the concept by expanding the function around any point, not just zero. The Maclaurin series is a special case of the Taylor series where the expansion is around 0. The general form of the Taylor series is:
$$f(x)=f(a)+f^{\prime }(a)(x-a)+\frac{f^{″}(a)}{2!}(x-a{)}^2+\frac{f^{‴}(a)}{3!}(x-a{)}^3+\cdots$$
Here, $$f^{(n)}(a)$$ represents the $n$-th derivative of the function evaluated at point $a$. By using the series, we make complex functions much more manageable.
For example, in the given exercise, we are dealing with the function $\cos (e^x-1)$ and we use its derivatives to construct the Maclaurin series.

Derivatives

The concept of derivatives is fundamental to calculus. A derivative represents the rate at which a function is changing at any given point and is a measure of the function's sensitivity to change in its input. For instance, the first derivative of $f(x)$, denoted as $f^{\prime }(x)$, provides the slope of the tangent to the function at any point x. Higher-order derivatives, like the second derivative $f^{″}(x)$, provide further information about the function's curvature and rate of change of the slope.
In the solution provided:

• The first derivative, calculated as $f^{\prime }(x)=-\sin (e^x-1)e^x$, tells us how the function $\cos (e^x-1)$ changes as $x$ changes.
• The second derivative $f^{″}(x)=-\cos (e^x-1)e^{2x}-\sin (e^x-1)e^x$, gives us information about the concavity of the function.
• The third derivative reveals even more intricate details about the function’s behavior as x increases or decreases.

These derivatives are essential for constructing a series expansion.

Series Expansion

Series expansion is a way to write a function as a sum of simpler terms. This is incredibly useful in mathematics because it allows one to work with complex functions in a more manageable form. By finding the first few terms of the series, we can approximate how a function behaves around a certain point.
Take for instance the expansion of $\cos (e^x-1)$ in the original exercise. By using the derivatives of the function and evaluating them at 0, the Maclaurin series is constructed: $$f(x)\approx 1+0\cdot x+\frac{-1}{2!}{x}^2+\frac{-3}{3!}{x}^3=1-\frac{x^2}{2}-\frac{x^3}{6}$$
This series is an infinite sum that approximates $\cos (e^x-1)$ around $x=0$. The first few terms give you a good approximation close to $x=0$.

Trigonometric Functions

Trigonometric functions, like $\cos$ and $\sin$, describe the relationships between angles and lengths in right-angled triangles. They are periodic and have specific patterns that make them particularly interesting in calculus and series expansion.
For example, the function $\cos (x)$ tells us the horizontal coordinate of a point on a unit circle as it sweeps around the circle from an angle $x$.
In our exercise, the function $\cos (e^x-1)$ is a bit more complex as it combines the exponential function $e^x$ with the trigonometric function $\cos$. To find its Maclaurin series, we treat $e^x-1$ as the argument of $\cos$ and find the derivatives accordingly.

Calculus

Calculus is the branch of mathematics that studies how things change. It is divided into two main parts: differential calculus, which deals with the concept of derivatives, and integral calculus, which deals with the concept of integration.
Differential calculus focuses on the idea of a derivative, which represents the rate of change of a function. By finding the derivatives of a function, we can analyze its behavior deeply. For instance:

• The first derivative tells us the slope of the tangent at any point.
• The second derivative gives us information about the concavity of the function.
• Higher-order derivatives can tell us about even more subtle aspects of the function's graph.

In the problem given, we used derivatives to find the Maclaurin series for $\cos (e^x-1)$. This exercise highlights how calculus allows us to approximate and understand complex functions more intuitively.