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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?

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    $\begingroup$ $f(4)=2^1+2^2+2^3+2^4$ $\endgroup$ – Dr. Sonnhard Graubner Jul 19 '15 at 15:59
  • $\begingroup$ what about $\sum_{i=1}^{n}2^i$ $\endgroup$ – Cardinal Jul 19 '15 at 16:00
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    $\begingroup$ Did you try $f(n) = 2^{n+1}-2$? $\endgroup$ – jbuddenh Jul 19 '15 at 16:05
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    $\begingroup$ it must be $$f(n)=2^{n+1}-2$$ $\endgroup$ – Dr. Sonnhard Graubner Jul 19 '15 at 16:06
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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

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

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 $n$.

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    $\begingroup$ Ok, this was exactly what I was looking for. Feeling a little ashamed right now that I didn't come up with this myself. Thanks! $\endgroup$ – Rolf van de Krol Jul 19 '15 at 16:06
  • $\begingroup$ You're welcome! My pleasure. $\endgroup$ – Mark Viola Jul 19 '15 at 16:07
  • $\begingroup$ @Dr.MV I changed the link so that it goes directly to the relevant formula. I hope this is agreeable. $\endgroup$ – Ben Grossmann Jul 19 '15 at 16:09
  • $\begingroup$ @Omnomnomnom Absolutely! No worry. I just didn't see any part of the text that was edited recognizably and was baffled. $\endgroup$ – Mark Viola Jul 19 '15 at 16:10
  • $\begingroup$ Of course if you add $g(x)=\sin (2\pi x)$ or a similar function to your solution, that would be a new "solution" (at least in the sense that it is continuous (even analytic) and has the right values at the natural numbers). $\endgroup$ – Jeppe Stig Nielsen Jul 20 '15 at 14:50
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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.

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

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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.

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    $\begingroup$ "Looking for patterns" in this way makes it easy to fall victim to the "strong law of small numbers." That being said, it sometimes helps us make an educated guess. $\endgroup$ – Ben Grossmann Jul 19 '15 at 16:13
  • $\begingroup$ @Omnomnomnom, Like I said, this is more just to provide a direction of where to start looking in, so you can see that the problem is related to $2^x$ so you can start going in that direction. If that direction works, perfect! If not, at least you've learnt something and you can try some other tactic. $\endgroup$ – Zain Patel Jul 19 '15 at 16:15
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    $\begingroup$ Still, I object to the phrasing that it's "pretty obvious" that the formula holds because it happens to fit the first 5 values. $\endgroup$ – Ben Grossmann Jul 19 '15 at 16:18
  • $\begingroup$ @Omnomnomnom, fair enough. I've edited my answer. :-) $\endgroup$ – Zain Patel Jul 19 '15 at 16:19

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