How to approximate $1.05^{50}$ by hand

Question

Is there some type of Taylor expansion or something which I could use to approximate quickly what for example $1.05^{50}$ is? or put bounds on that number? It's really annoying because I can't even guess what it would be because it's so close to 1. This calculation is from compound interest. Thanks!

Answer 1

The answer is not so close to $1$.

$1.05^{20}$ is approximately $e=\lim\limits_{n\to\infty}\left(1+\dfrac1n\right)^n$.

$1.05^{50}$ is approximately $e^{2.5}$, which is $\sqrt{e^2e^3}$.

$e\approx2.7$, so $e^2\approx 7.29\approx7$.

It helps to remember that $e^3\approx 20$.

Therefore, the answer is $\approx\sqrt{7\times20}=\sqrt{140}\approx12. $

My calculator says $11.467…$.

Answer 2

$\lim_\limits{n\to \infty} (1+\frac {1}{n})^n = e$

$(1+\frac {1}{n})^n < e$ for finite $n$ so this will give us an upper bound.

Let $n = 20$

$(1+\frac {1}{20})^{20} \approx e$

$1.05^{50} \approx e^{2.5}$

Now we could use the Taylor series for $e^x$ or do some more back-of-the-envelope calculations.

$e^{2.3} \approx 10$

$1.05^{50} \approx 10\cdot e^{0.2}$

And now use the Taylor series. $10(1+0.2 + \frac {0.04}{2} + \frac {0.008}{6})$

$12.2213$

Which compares to the $11.467$ I get on my calculator.

Can we do better?

$(1.05)^{50}$ is the future value of a dollar invested at $5\%$ for 50 years.

By the rule of $72$ we should expect our investment to double every 14.4 years.

$\frac {50}{14.4} = 3.47$

$2^{\frac {50}{144}} \approx 2^{3.5} = 8\sqrt 2 \approx 8(\frac {17}{12}) = \frac {34}{3} = 11.33$

Finally were do we get with the binomial theorem?

$(1 + 0.05)^{50} = 1 + \frac {50}{20} + \frac {50\cdot 49}{800} + \frac {50\cdot 49\cdot 48}{48000} + \cdots$

Let's make this simplification:

$(1 + 0.05)^{50} < 1 + \frac {50}{20} + (\frac 12)(\frac {5}{2})^2 + \frac 16(\frac 52)^3 + \frac 1{4!}(\frac 52)^4 + \cdots$

$1 + 2.5 + 3.125 + 1.3+ \cdots$ this is starting to converge, but seems to be the most work.

Answer 3

Here's bit more complicated approach if you want to improve the result. We have $$(1+x)^a = e^{a\log(1+x)}$$ The Taylor series of $\log(1+x)$ is: $$\log(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} -...$$ For our purpose, the second order is enough, then $$(1.05)^{50}\approx e^{50(0.05-0.05^2/2)}=e^{2.4375}$$ We know that $e^{2.3}\approx 10$, therefore $$e^{2.4375}=10\cdot e^{0.1375} \approx 10(1 + 0.1375+\frac{0.1375^2}{2})\approx11.47$$ Which is quite close to the answer $11.467...$

Answer 4

The rule of 70 says that an investment with an interest rate of $p$% takes about $70/p$ years to double. In this case an investment at APR 5% takes 70/5 = 14 years to double. 14 * 3.5 = 49, so after 50 years, it will have doubled a little bit more than 3.5 times, so the answer is a bit more than $2^{3.5} = (2^3)(2^.5) = 8 * \sqrt{2} \approx 8 * 1.4 = 8 + 3.2 = 11.2$.

Answer 5

At you mention compound interest, the rule of $72$ says the doubling time at $5\%$ interest is about $14$ years, so $50$ years is about $3 \frac 12$ doublings. The $3$ gives you a factor $8$ and the half gives a factor $\sqrt 2$, so the expression is about $8 \sqrt 2 \approx 8 \cdot 1.4 = 11.2$ This is rather close.

Answer 6

You can use the binomial theorem to give an approximation:

$(1.05)^{50}=(1+\frac{1}{20})^{50}\approx 1^{50}+50\times 1^{49}\times \frac{1}{20}+\ 1225\times 1^{48}\times(\frac{1}{20})^{2}...$

With the size of the numbers involved, this very quickly becomes an inefficient method though.