# Why does the approximation for exponents $(a+b)^c \approx a^{c-bc} (a+1)^{cb}$ work?

I was working with some code involving exponents in an environment where exponents can only be calculated if the base of the exponent is an integer. I needed a good fast way to approximate this without causing overflow issues. I accidentally stumbled upon an incredible approximation method and I'm not sure why it works.

Suppose you have an exponent in the form of $x^y$ where $x$ is not an integer and you want to approximate the value using only exponents which have integers for their base-values.

Break $x$ into two parts, an integer part, and an additive. For example $3.7\to 3 + 0.7$.

Therefor, $x\to(a+b)$ where $a$ is an integer part.

The approximation formula is:

$$(a+b)^c \approx a^{c-bc} (a+1)^{cb}$$

Or in my original form:

$$(a+b)^c \approx ((a+b)-b)^{c(1-b)} ((a+b)+(1-b))^{cb}$$

It's remarkably close to the right solution seemingly every time. Granted I've only been able to check about 100 cases, but I'm fascinated.

For example:

$$37.5^{28} ≈ 37^{14}\cdot38^{14}$$

And sure enough, if we divide both parts, the ratio is 1.002 which is very close.

Edit: Thanks to RayDansh pointing out in the comments, this is accurate IFF $a+b$ is big. In fact, the larger $a$ gets the more accurate this approximation seems to get.

Can anyone shed some light as to why this approximation method I've stumbled upon works?

• Well, here's your intuition. The exponent terms you've discovered are such that whichever of $a$ and $a+1$ are closer to $a+b$, those will be weighted more than the other term. And both happen to be remarkably close to $a+b$. Hence your discovery. Aug 14, 2018 at 23:48
• What about $1.5^{100}≈1^{50}*2^{50}$? Quite inaccurate. Aug 14, 2018 at 23:48
• Just trying to show how the approximation loses accuracy as $x$ decreases and $y$ increases. Aug 14, 2018 at 23:49
• @RayDansh I think the point is that it improves for large $a$.
– Ian
Aug 14, 2018 at 23:49
• @RayDansh Right you are; I was only testing with large numbers. Seems to get 100% accurate as a approached infinity. How odd Aug 14, 2018 at 23:50

Write your approximation $$a^{c-bc} \cdot (a+1)^{cb}=a^c\left(1+\frac 1a\right)^{bc}$$ and your approximation is $$\left(1+\frac 1a\right)^{b}\approx 1+\frac ba$$ Which is the first two terms of the binomial expansion. It will be reasonably accurate when $\frac ba \ll 1$ The next term is $\frac {b(b-1)}{2a^2}$

• Doesn't get clearer than this +1 Aug 15, 2018 at 0:31

As a general rule, if you see a lot of products and exponents it may clear things up to take logs. In your case,

$$(a+b)^c ≈ a^{c-bc} \cdot (a+1)^{cb}$$

becomes

$$c \log(a+b) \approx c(1-b) \log(a) + cb \log(a+1)$$

or just

$$\log(a+b) \approx (1-b) \log(a) + b \log(a+1).$$

This is equivalent to doing a linear interpolation of $\log x$ between the points $a$ and $a+1$. This will be pretty accurate when $a$ is large because $\log x$ will be close to linear between $a$ and $a+1$.

The answer is obvious. If we expand both sides of the formula by the binomial theorem, we get for the first two: $(a+b)^c=a^c+cba^{c-1}+ ...$ for both forms of the equation. Now, it is immediately apparent that this approximation will improve as a gets larger because the discarded terms become less significant. That is, for larger a, $a^n>>a^{n-1}$.

• Of course it improves, but why is it significantly better than just keeping $a^c$?
– Ian
Aug 14, 2018 at 23:58
• Look at the rest of the binomial expansion terms that are omitted above
– user553944
Aug 15, 2018 at 0:00