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Find the Taylor series for \(e^{x}\) about a general center \(c\)

Short Answer

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The Taylor series for \(e^{x}\) centered at a general center \(c\) is \(e^{x} = \sum_{n=0}^{\infty} \frac{e^{c}}{n!}(x - c)^n\).

Step by step solution

01

Calculating the nth derivative

The first step is to note that the nth derivative of \(e^{x}\) is always equal to \(e^{x}\), therefore, \(f^{(n)}(c) = e^{c}\). This fact will greatly simplify the process of expanding the Taylor series.
02

Applying the Taylor series formula

Using the formula for Taylor series, we know that the nth term is given by \(\frac{f^{(n)}(a)}{n!}(x - a)^n\). From the previous step, we know \(f^{(n)}(c) = e^{c}\), so substituting this into the formula we get that the nth term is \(\frac{e^{c}}{n!}(x - c)^n\).
03

Writing out the series

Now we can write out the Taylor series for \(e^{x}\) centered at \(c\). The series is \(e^{x} = \sum_{n=0}^{\infty} \frac{e^{c}}{n!}(x - c)^n\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Taylor Series Expansion
Imagine you want to understand the essence of a function, like peering into the soul of its mathematical being. That's where the Taylor series expansion comes in. This powerful tool unravels any continuous function into an infinite sum of terms calculated from the function's derivatives at a single point. This point is the 'center' of the series, often denoted as 'a' or 'c' in mathematical notation.

The allure of the Taylor series lies in its ability to approximate functions using polynomials—those simple algebraic equations we're more comfortable handling. It’s like crafting a mathematical mosaic that gets increasingly detailed as you add more pieces, or in this case, terms. For functions that behave nicely, adding more terms makes the approximation more precise, giving us insights into function behavior even far from the center point.

The formula for the nth term in a Taylor series about the center 'c' is: \[\frac{f^{(n)}(c)}{n!}(x - c)^n\]. Here, \(f^{(n)}(c)\) is the nth derivative evaluated at 'c', \(n!\) is n factorial, and \(x - c\) raised to the nth power enables the series to capture the curvature and nuances of the original function around the center. This series can be thought of as an infinite polynomial, gracefully tracking the shape of \(e^{x}\) as closely as one desires by including more terms.
nth Derivative of Exponential Function
Some functions change dramatically with each twist and turn of calculus. Others, like the exponential function \(e^{x}\), are the serene 'Steady Eddies' of the mathematical world. Every derivative you take of \(e^{x}\) looks just like the original—it's a rare and convenient characteristic.

What this means is that when you calculate the nth derivative, denoted as \(f^{(n)}(x)\), it's the same as the function itself. That's right, no matter how many times you differentiate \(e^{x}\), the result is always \(e^{x}\). This simplicity plays a starring role when creating a Taylor series for \(e^{x}\). It's like a special function filter that means every term in the series includes a factor of \(e^{c}\), the exponential function evaluated at the center 'c'. This feature is a mathematician's dream, drastically simplifying computations and providing clear insights into the behavior of the series.
Mathematical Series
A mathematical series is akin to a choo-choo train of numbers, each car hitched to the next by a specific rule or formula. When we talk about converging series, like the Taylor series for certain functions, imagine that this train is headed towards a scenic destination—the actual value of the function we're trying to approximate.

Each sequence of numbers (the train cars) adds up to form the series (the entire train). When it comes to the Taylor series, the numbers in our sequence emerge from the function's derivatives at a point, coupled with powers of \(x - c\) and divided by factorial terms. If the terms in the sequence get smaller quickly enough, the series will converge to a specific value, providing a precise estimate for the function over a range of 'x' values.

Many mathematical series can be summed up to reveal elegant relationships and behaviors across mathematics and physics. It’s a way of distilling endless possibilities into concrete values and can reveal patterns and connections that measurements and discrete calculations simply cannot capture.
Calculus
Calculus might seem daunting, with its limits, derivatives, and integrals. But at its heart, calculus is the mathematics of change—how quantities evolve and how they accumulate. It's the language we use to describe the dynamic world around us, from the motion of planets to the growth of populations.

In the context of the Taylor series, calculus shows how a function can be approximated by a polynomial. Derivatives, which give us information about a function's rate of change, become the building blocks for constructing the series. The act of taking a derivative, integration, working with limits—they all converge in the Taylor series, providing a symphony of insights into the function we're studying.

Understanding calculus is a cornerstone for analyzing and predicting the behavior of systems. The principles of calculus are pervasive, supporting fields as diverse as engineering, economics, and ecology. It gives us the superpower to dissect the universe's continuous processes into understandable and manageable parts.

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Most popular questions from this chapter

Use a Taylor series to verify the given formula. $$\sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k}=\ln 2$$

Find the first five terms in the Taylor series about \(c=0\) for \(f(x)=e^{x} \sin x\) and compare to the product of the Taylor polynomials about \(c=0\) of \(e^{x}\) and \(\sin x\)

The function \(\sin 8 \pi t\) represents a \(4-\mathrm{Hz}\) signal \((1 \mathrm{Hz} \text { equals } 1\) cycle per second) if \(t\) is measured in seconds. If you received this signal, your task might be to take your measurements of the signal and try to reconstruct the function. For example, if you measured three samples per second, you would have the data \(f(0)=0, f(1 / 3)=\sqrt{3} / 2, f(2 / 3)=-\sqrt{3} / 2\) and \(f(1)=0\) Knowing the signal is of the form \(A\) sin \(B t,\) you would use the data to try to solve for \(A\) and \(B\). In this case, you don't have enough information to guarantee getting the right values for A and \(B\). Prove this by finding several values of \(A\) and \(B\) with \(B \neq 8 \pi\) that match the data. A famous result of \(\mathrm{H}\). Nyquist from 1928 states that to reconstruct a signal of frequency \(f\) you need at least \(2 f\) samples.

An electric dipole consists of a charge \(q\) at \(x=1\) and a charge \(-q\) at \(x=-1 .\) The electric field at any \(x > 1\) is given by \(E(x)=\frac{k q}{(x-1)^{2}}-\frac{k q}{(x+1)^{2}},\) for some constant \(k .\) Find a power series representation for \(E(x)\)

(a) use a Taylor polynomial of degree 4 to approximate the given number, (b) estimate the error in the approximation and (c) estimate the number of terms needed in a Taylor polynomial to guarantee an accuracy of \(10^{-10}\) $$\sqrt{1.2}$$

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