1.8: Exercises
Exercise \(\PageIndex{1}\)
An alternative definition of the exponential function is the limiting expression \[\exp(x) \equiv \lim_{n\rightarrow\infty} \left(1+\frac{x}{n}\right)^n.\] Prove that this is equivalent to the definition in terms of an infinite series, \[\exp(x) \equiv 1 + \sum_{n=1}^\infty\frac{x^n}{n!}.\]
Exercise \(\PageIndex{2}\)
Prove that \[\exp(x+y) = \exp(x)\,\exp(y)\] using the definition of the exponential as an infinite series. Your proof must avoid using the concept of “raising to the power” of a non-natural number; this is to avoid circular logic, since this feature of the exponential function can be used in the generalized definition of the power operation (Section 1.4).
- Answer
-
To prove that \(\exp(x+y) = \exp(x)\,\exp(y)\) , we employ the infinite series formula \[\exp(x) = \sum_{n=0}^\infty\frac{x^n}{n!}.\] Here, for notational convenience, we let the sum start from \(n = 0\) , so that the leading term \(1\) in the definition of the exponential is grouped with the rest of the sum as its first term. This relies on the understanding that \(0! \equiv 1\) , and that \(x^0 = 1\) (the latter is consistent with the generalized definition of the power operation; but to avoid circular logic, treat this as the definition of \(x^0\) just for the sake of this proof). We begin by substituting the series formula into the right-hand side of our target equation: \[\exp(x)\exp(x) = \left(\sum_{n=0}^\infty\frac{x^n}{n!}\right)\;\left(\sum_{m=0}^\infty\frac{y^m}{m!}\right).\] Note that we use the symbol \(n\) for the first sum, and the symbol \(m\) for the second sum; \(n\) and \(m\) are bound variables, whose terms run over the values specified by the summation signs. The actual choice of symbol used in either sum is unimportant, except that we must not use the same symbol for both sums , because the two variables belong to distinct sums. In other words: \[\exp(x)\exp(x) \ne \left(\sum_{n=0}^\infty\frac{x^n}{n!}\right)\;\left(\sum_{n=0}^\infty\frac{y^n}{n!}\right). \quad(\text{Nonsense expression!})\] Next, we make use of the fact that the product of two series can be written as a double sum: \[\exp(x)\exp(x) = \sum_{n=0}^\infty \sum_{m=0}^\infty \frac{x^n}{n!} \frac{y^m}{m!}.\] Here, we are summing over all possible pair-wise combinations of \(n\) and \(m\) , which is precisely what happens when one expands the product of two series according to the usual rules of algebra. The next step is to perform a change of variables on \(m\) and \(n\) . In the above expression, we are summing over all non-negative integer \(m\) and \(n\) ; however, the bound variable \(n\) can be re-expressed in terms of a newly-defined variable, \[N = m + n.\] In the original double sum, \(n\) and \(m\) both run from \(0\) to \(+\infty\) , so it follows that their sum \(N\) runs from \(0\) to \(+\infty\) . For each given value of \(N\) , we can write \(n = N - m\) , and moreover the allowed values of \(m\) would only go from \(0\) to \(N\) (it can’t exceed \(N\) , otherwise \(n\) would be negative). In this way, the double sum is converted to \[\exp(x)\exp(x) = \sum_{N=0}^\infty \sum_{m=0}^N \frac{x^{N-m}}{(N-m)!} \frac{y^m}{m!}\] Note that after this change of variables, the two summation signs are no longer interchangeable. In the \(\sum_{m=0}^N\) sign, the variable \(N\) appears in the upper limit, so this needs to be written to the right of \(\sum_{N=0}^\infty\) . One sum is thus “encapsulated” inside the other; we could write the algebraic expression more rigorously like this: \[\exp(x)\exp(x) = \sum_{N=0}^\infty \left(\sum_{m=0}^N \frac{x^{N-m}}{(N-m)!} \frac{y^m}{m!}\right).\] Finally, we use the binomial theorem to simplify the inner sum: \[\exp(x)\exp(x) = \sum_{N=0}^\infty \frac{\left(x + y\right)^N}{N!}, \;\;\;\text{since} \;\; (x+y)^N = \sum_{m=0}^N \frac{N!}{m!(N-m)!} x^{N-m} \, y^m.\] Referring again to the series definition of the exponential, we obtain the desired result: \[\exp(x)\exp(x) = \exp(x+y)\]
Exercise \(\PageIndex{3}\)
One of the most important features of the exponential function \(\exp(x)\) is that it becomes large extremely quickly with increasing \(x\) . To illustrate this behavior, consider the graph shown in Section 1.2, which plots the exponential up to \(x = 4\) . On your screen or page, the height of the graph should be around 4 cm. Suppose we keep to the same resolution, and plot up to \(x = 10\) ; how high would the graph be? What if we plot up to \(x = 20\) ?
Exercise \(\PageIndex{4}\)
Prove that \(\displaystyle \exp(x) = e^x.\)
- Answer
-
The definition of non-natural powers is \[a^b = \exp[b\ln(a)].\] Let \(a = \exp(1) = e\) and \(b = x\) . Then \[^x = \exp\left[x\ln\Big(\exp(1)\Big)\right].\] Since the logarithm is the inverse of the exponential function, \(\ln(\exp(1)) = 1\) . Hence, \[e^x = \exp(x).\]
Exercise \(\PageIndex{5}\)
A “non-natural” logarithm of base \(c\) is defined as \[\log_c(x) = y \quad\mathrm{where}\;\; c^y = x.\] Derive an expression for the non-natural logarithm in terms of the natural logarithm.
Exercise \(\PageIndex{6}\)
Prove, using trigonometry, that \[\sin(\theta_1 + \theta_2) = \sin(\theta_1) \cos(\theta_2) + \cos(\theta_1)\sin(\theta_2).\] You may assume that \(\theta_1, \theta_2 \in [0, \pi/2].\)
Exercise \(\PageIndex{7}\)
Prove that \[\begin{align} \cos(3x) &= 4[\cos(x)]^3 -3\cos(x) \\ \sin(3x) &= 3\sin(x)-4[\sin(x)]^3. \end{align}\]