Estimating Exponents

Question

What's the best way to estimate exponents by hand? Say for example $(1.07)^{10}$ $\sim2$, or like $(1.07)^{15}$, or $(1.05)^{15}$. Is there any standard way of doing these calculations?

Answer 1

One popular method is using the $\textbf{rule of}$ $\mathbf{7 2}$ for estimating an investment's doubling time. It's a simple way to approximate the effect of compounding interest. According to this rule, you divide 72 by the annual interest rate to find how many years it will take for your investment to double. This method can be loosely adapted to estimate the growth of numbers slightly above 1 raised to high powers.

For a more general approach, especially for numbers that don't conveniently fit the doubling scenario, you can use the linear approximation method from calculus, which is essentially the first term of the Taylor series expansion. For small $x,(1+x)^n \approx 1+n x$, where $x$ is the rate of growth and $n$ is the number of periods.

Applying the Linear Approximation Method:

• For $(1.07)^{10}$, we approximate it as $1+10 \cdot 0.07=1.7$. This is an underestimate because it doesn't account for compounding beyond the first period.
• For $(1.07)^{15}$, it's $1+15 \cdot 0.07=2.05$. Again, a simple linear approximation.
• For $(1.05)^{15}$, it's $1+15 \cdot 0.05=1.75$.

Answer 2

I wanted to elaborate on my comments in the answer provided by @Cuteshrek in case it would be useful.

Let nonnegative integer $n$ be given and define $f(x)=(1+x)^n$. Then the $N$th order Maclaurin series expansion of $f(x)$ can be shown to be $$\sum_{i=0}^N {n \choose i} x^i.$$ Note that when $N \geq n$, $$\sum_{i=0}^N {n \choose i} x^i=\sum_{i=0}^n {n \choose i} x^i=(1+x)^n=f(x),$$ where we have used the binomial theorem $(y+x)^n = \sum_{i=0}^n {n \choose i} y^{n-i}x^i$ with $y=1$. This makes sense because $f$ is a $n$th degree polynomial.

Moreover, since $x>0$ in our context, the summand is always positive for all integers $0 \leq i \leq n$. Thus, for nonnegative integers $N<n$, the $N$th order Maclaurin series expansion for $f(x)$ can be seen as an underapproximation of $(1+x)^n$ that becomes increasingly more accurate as $N$ increases. The binomial approximation $(1+x)^n \approx 1+nx$ notably corresponds to the $N=1$ Maclaurin series expansion.

Since ${n \choose i}=\frac{n^{\underline{i}}}{i!} \leq \frac{n^i}{i!}$, we also have that $$\sum_{i=0}^N {n \choose i} x^i \leq \sum_{i=0}^N \frac{n^i}{i!} x^i = \sum_{i=0}^N \frac{(nx)^i}{i!},$$ which we recognize as the $N$th order Maclaurin series expansion for $e^{nx}$.

Putting everything together, we can write that for all nonnegative integers $N \leq n$, $$\sum_{i=0}^N {n \choose i} x^i \leq (1+x)^n \leq \sum_{i=0}^n \frac{(nx)^i}{i!} < e^{nx}, \tag{*}$$ where the lower bound has equality if and only if $N=n$ and the upper bound has equality if and only if $n \leq 1$. Increasing $N$ from $0$ to $n$ has the result of tightening the lower bound until it is an equality.

Answer 3

$(1+x)^{n}$ when x is very small: I subsitute $m = nx$ and then the starting expression is $(1+x)^{\frac{m}{x}}$ which is approximately $e^{m}$ so it is about $e^{nx}$.

This is very hard to use alone, merge it with the other methods