simple exponential

Question

This is taken from an example given in Gilbert Strang's Linear Algebra. The topic is not relevant, but I don't understand the following:

$\left(1+\frac{0.06}{N}\right)^{5N} = e^{0.30}$

How is this derived?

Answer 1

Based on the definition of $\exp(x)$, we have: $$\lim_{n\to\infty}\left(1+\frac{x}{n}\right)^n = e^x$$

Thus, a useful approximation for some situations is (for large $N$): $$\left(1+\frac{x}{N}\right)^N \approx e^x$$

In your case (assuming $N$ is very large):

\begin{align} \left(1+\frac{0.06}{N}\right)^{5N}&= \left(1+\frac{5\cdot0.06}{\color{red}{5N}}\right)^{\color{red}{5N}}\\ &= \left(1+\frac{0.3}{5N}\right)^{5N} \\ &\approx e^{0.3} \end{align}

Answer 2

Another way to look at the problem. Let us define $$A=\left(1+\frac{0.06}{N}\right)^{5N} $$ Taking the logarithms of both sides gives $$\log(A)=5N\log\left(1+\frac{0.06}{N}\right)$$ where $N$ is large. Now, use the fact that if $x$ is small compared to $1$, $\log(1+x)\approx x - \frac {x^2}{2} +\cdots$. Replace $x$ by $\frac {0.06}{N}$ so $$ \log(A)\approx 5N \Big(\frac {0.06}{N}-\frac {0.0018}{N^2}\Big)=0.30 -\frac{0.009}{N}$$ where you can notice that the last term is quite small compared to the first.

If you do not require much accuracy, then, taking exponentials for both sides $$A=e^{0.30}$$ For more accuracy $$A=e^{0.30-\frac{0.009}{N}}$$

For illustration purposes, let us use the worst case corresponding to $N=1$. The exact value of $A$ is $1.33823$ while $e^{0.3}=1.34986$ (the second formula would give $e^{0.291}=1.33776$).