Function that looks a lot like exponential, but isn't

Question

I'm looking for a continuous function f(x) with the following properties. I've been playing with exponentials, but that doesn't seem to be the answer, although my high school mathematics is a bit rusty, I must admit.

• $f(1) = 2$
• $f(2) = 6 = 2+4$
• $f(3) = 14 = 2+4+8$
• $f(4) = 30 = 2+4+8+16$
• And so on

I'm looking for a continuous function, so something with a meaningful answer for $f(2.5)$, which is less then $6 + \frac{14 - 6}{2} = 10$.

My simple high school math made me look at something like $f(x) = a^x$, but that doesn't seem to be the answer.

Any better ideas?

Answer 1

We have for the sequence $f(n)$

$$\begin{align} f(n)&=\sum_{k=1}^n2^k\\\\ &=2^{n+1}-2\tag 1 \end{align}$$

where we summed a Geometric Progression to arrive at $(1)$.

Thus, the continuation of $f$ for real arguments is

$$\bbox[5px,border:2px solid #C0A000]{f(x)=2^{x+1}-2}$$

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NOTE:

We remark that the continuation is not unique inasmuch as we can add any continuous functions that has zeros for each integer. As example, the function $g(x)=2^{x+1}-2+C\,\sin ( \pi x)$, where $C$ is any constant, is continuous and for all integer-valued arguments $n$, is given by $g(n)=2^{n+1}-2$ since $\sin (n\pi)=0$ for all integer values of $n$.

Answer 2

HINT: $$\sum_{i=1}^n 2^i=2(2^n-1)$$

Answer 3

The other answers show you how to derive the function $f(x) = 2^{x+1} - 2$, but I'm just going to show how the problem is approached (in complete unrigorous terms). I know from experience how magical it seems that the above answers just managed to come up with the right function all of a sudden.

So, we look at the problem and we can see that it's forming a pattern $2, 6, 14, 30, \ldots$ - now this reminds us of something, all the numbers seem to be even, but if you look hard enough they are all $2$ less than powers of $2$. So we can see that really, the sequence is actually $$2^2 - 2, 2^3 - 2, 2^4 - 2, 2^5 - 2, \ldots$$

From this, it's pretty obvious to us that we should start thinking that this problem is related to something like $2^{\text{something}} - 2$ and once you know what to arrive at it makes getting there a whole lot easier, which is where the other answers come in.

Answer 4

If you're computer-minded, it might seem natural to express these numbers in binary, since they're sums of powers of two!

Then you see that they're:

$$f(1)=10_2$$

$$f(2)=110_2$$

$$f(3)=1110_2$$

$$f(4)=11110_2$$

and it's immediately apparent that adding $10_2$ to these will give powers of two.