B How many numbers can you make with four 4s?

[Greg Bernhardt]

Say you've got four 4s -- 4, 4, 4, 4 -- and you're allowed to place any normal math symbols around them. How many different numbers can you make? It's best to think of a number and then try to make it.

The origination of the puzzle and some examples are in this video

 

[Student100]

$4444\int dx =4444x +C$ I think that about covers it.

 

[jedishrfu]

Imagine what you could do with seven fours?

Spoiler

You could order Pizza Hut pizza! (From an old pizza commercial)

Nice video. I was waiting for Knuth arrow notation to come in but it never did.

 

[JonnyG]

$4444\int dx =4444x +C$ I think that about covers it.

I think that wraps things up!

 

[martinbn]

$4444\int dx =4444x +C$ I think that about covers it.

No, it doesn't. How do represent the number 10 for example?

 

[late347]

I think the log based proof was quite good for numberphile.

Also it seems you are allowed to put the minus sign in front of the log function which would extend the combos into the negative numbers for reals

 

[Greg Bernhardt]

Can anyone figure out Pi?

 

[late347]

0 could also be done with something log based like

lg(4/4)=0

 

[late347]

Can anyone figure out Pi?

I'm not quite certain if this is allowed by the problem such as using variable x, such as using standard symbols and four 4s as numbers...
But Pi can be gotten in certain ways

$lim_{x-> -1^{+}} =cos^{-1}(x)$

I'm not sure if cos^{-1} was defined exactluy at the value of (-1)

so that might be something like cos^{-4/4}(-4/4)= pi

But limit should still exist. I suck at trig.

that is in radians of course

 

[Stavros Kiri]

$4444\int dx =4444x +C$ I think that about covers it.

This is indefinite integral, i.e. function ...

Can anyone figure out Pi?

For any real number x: x = C/4444 . Just pick a C = x•4444 ...

Or x = C/(4+4+4+4), or x = C/(4•4•4•4) ...

or even (different form) x = C(4444) [C=x/4444], etc., or use other operations or operators ...

Thus trivial, unless you put more constraints in the problem.

 

[Student100]

No, it doesn't. How do represent the number 10 for example?

It wasn't meant to be taken seriously.

As for ten, x = 0, C = 10. There you go.

 

[Greg Bernhardt]

It wasn't meant to be taken seriously.

As for ten, x = 0, C = 10. There you go.

Assigning variables is kind of cheating :D

 

[mfb]

$$\pi = \sqrt{4}~ \arccos (\sqrt{4\cdot 4}-4) = \arccos \left(4-4-\frac 4 4 \right)$$
Similar expressions exist with the arcsine.
$$e=\exp\left(4-4+\frac 4 4 \right)$$
Well...

 

[PAllen]

No, it doesn't. How do represent the number 10 for example?

Let x=0, C=10

 

[Stavros Kiri]

Assigning variables is kind of cheating :D

It would appear so (and mine above could also be considered partially humour), but isn't solving an equation like inverting an operator or a function? Thus cosistent with:

It's best to think of a number and then try to make it.

And it is not limited by the video, or the original post [but since the OP sais now so we can now perhaps put it as a restriction ... (?)].

For example (aside/[or plus] the fact that functions can be considered as operarors, and operators are essentially functions):

If x is any real number (say π), and f any invertable real function (to avoid regular or perplexity form plain equations), and it happens that a number A which is directly/easily or else consructed by the 4 4s (say 4444) is related to x as A = f(x), then x = f^-1(A)
[e.g. x (or e.g. π) = f^-1(4444), and that's true for any [invertable] function f (providing it satisfies A = f(x) ) (e.g. a•x, sin, cos, exp, ln, log etc.) - just pick an appropriate one and be my guest! ...]

That's what I essentially did above, e.g.

x = C(4444)

where f(x) = (1/C)•x, A = 4444, and mfb did essentially the same with cos and arccos

π=√4 arccos(√4⋅4−4)=arccos(4−4−44)​

since arccos = cos^-1 ...

So are we joking or serious? I am a bit confused.

My opinion is that the problem on the video etc. is either trivial or one would have to put explicitly more restrictions, namely to the basic functions and operations only, with no arbitrary constants (other than the 4 4s and their obvious basic step derivatives).
[In other words functions e.g. like a•x, or a•cosx ... are not allowed, unless a=1 ... etc., otherwise one can do anything - it's trivial]

So the problem is not the variables but the arbitrary constants.

 

[mfb]

I used common mathematical functions.
I didn't have to make up a function just for this exercise.

 

[Stavros Kiri]

I used common mathematical functions.
I didn't have to make up a function just for this exercise.

I know. Your's is ok

As long as the OP accepts inverse functions.

Mine has arbitrary constants, so may be not allowed, but the restrictions should be made explicit.

 

[mfb]

As long as the OP accepts inverse functions.

There is nothing inverse about it. The arccosine is a regular function (in more than one way). The cosine (restricted to [0,pi]) is the inverse function of the arccosine. And now?

 

[Stavros Kiri]

There is nothing inverse about it. The arccosine is a regular function (in more than one way). The cosine (restricted to [0,pi]) is the inverse function of the arccosine. And now?

But one has to define explicitly which functions are allowed to build from, and if inversing is allowed. (+see my edited posts above about arbitrariness of constants) But in any case I think your's is ok. I just wasn't sure if you were serious or not.

 

[Student100]

$$\pi = \sqrt{4}~ \arccos (\sqrt{4\cdot 4}-4) = \arccos \left(4-4-\frac 4 4 \right)$$
Similar expressions exist with the arcsine.
$$e=\exp\left(4-4+\frac 4 4 \right)$$
Well...

Trying to come up with a simple way to get closer...

Using factorial's and decimals you can get approximately close:

$$\sum_{n=1}^{n=4} .4! = 3.54905... $$

What about...

$$ .04! \times 4 - .4 -.4 = 3.113752 $$

 

[Stavros Kiri]

Trying to come up with a simple way to get closer...

Using factorial's and decimals you can get approximately close:

$$\sum_{n=1}^{n=4} .4! = 3.54905... $$

What about...

$$ .04! \times 4 - .4 -.4 = 3.113752 $$

A) You can just use the "Decimal Point Approximation Sequence", if you are familiar with it. You can get as close as you want [to any real (rational or irrational) number] ...

The formula is x_n = [10ⁿx]/10ⁿ, where [ ] notation is the "Integral Part" of a real number (e.g. [8.3] = 8, [11.9] = 11 , etc. ...), and x_n is the n th order approximation (of x).

But any such approximations are not acceptable solutions here, I think, unless you use somehow the fact that lim_n→∞x_n = x, and somehow involve the 4 4s ...

B) But I think there is also something wrong with your notation and/or understanding of sums and factorials ... (etc.)
[e.g. what is ".4!" ? ... . And the sum needs iteration index ...]

 

[Student100]

This avoids .04, but introduces logs, which I didn't want to do...

$$\log(\log((4!)!^{4!} \times .4)) +.4 = 3.156352..$$

This is a better use of log maybe, base 4!

$$log_4(44) +.4 = 3.12927...$$

Okay, I'm done.

 

[Student100]

B) But I think there is also something wrong with your notation and/or understanding of sums and factorials ... (etc.)
[e.g. what is ".4!" ? ... . And the sum needs iteration index ...]

Whats wrong with it? .4! is a gamma function. My sum reads... .4! + .4! + .4! + .4!. or .4!(4).

 

[Stavros Kiri]

This avoids .04, but introduces logs, which I didn't want to do...

$$\log(\log((4!)!^{4!} \times .4)) +.4 = 3.156352..$$

This is a correct notation (no factorials on a decimal, only for integers/naturals - unless you use advanced calculus ...), I haven't checked the result though, but still approximation, not full answer ...

Interesting as inventions though, that you get so close ...

 

[Stavros Kiri]

Whats wrong with it? .4! is a gamma function. My sum reads... .4! + .4! + .4! + .4!. or .4!(4).

Ok, I didn't know you knew advanced calculus, but still the sum could be fixed, as a notation. (e.g. ∑x_i , and defining the limits of i on the sum ...) [unless you just want to say 4 times (.4!) , in this case, or give the explicit sum, as you mentioned ..]

Interesting as inventions though, that you get so close ...

 

[jbriggs444]

If you allow both factorials (as extended using the gamma function) and square roots it should be possible to get arbitrarily close to $\pi$ or any other positive real number by simply applying both operations a large number of times in an appropriately interleaved pattern such as:

$\sqrt{\sqrt{\sqrt{\frac{4+4+4}{4}!}!}}!$

Edit: while this approach will (I think) directly give you an approximation for any x > 1, one would need to apply a minor manipulation to get approximations for 0 < x < 1.

 

[Stavros Kiri]

But back on track with the exact solution (not approximation), for any real number x (e.g. π) [integer, rational, or irrational ...].

Another, perhaps better, and more direct way (than what I attempted earlier above etc. - e.g. see/cf. post #15) is to in fact begin with the actual question: x = f(A), where A is [the only constant allowed (?), besides the 4s themselves and their as stated (in the problem) derivative numbers...] a number like 4444, 4+4+4+4, and the like, as stated in the problem ..., and f an appropriate real function or operator, or combination or sequence of operations (etc.).

x = f(A) is also already the answer, as well as the method of construction, by choosing f appropriately, but the restrictions on the functions and operations (etc.) must be made explicit and clear, otherwise there are plenty of choices for f (and the problem becomes easy and trivial), if one allows arbitrariness of constants and functions [as well as inverting] (etc.). [See/cf. my other posts #15, #17, #19 (as well as #10) earlier above ...] (mainly for the restrictions)

 

[martinbn]

Let x=0, C=10

This way you use four 4's, a 0 and a 10. The problem says only four 4's.

 

[Stavros Kiri]

This way you use four 4's, a 0 and a 10. The problem says only four 4's.

But 0 and 10 can be expressed primatively in terms of 4 4s (see the video etc.), so I think it is allowed. But arbitrary constants C, A .. etc. aren't ... , if Greg has agreed on this (etc.) ...

 

[mfb]

But 0 and 10 can be expressed primatively in terms of 4 4s (see the video etc.), so I think it is allowed.

Both at the same time with a total of 4 4s? Yes they can be expressed like that, but then there is no point in introducing integrals and whatever because you have expressed 10 via 4s already.

 

[Stavros Kiri]

Both at the same time with a total of 4 4s? Yes they can be expressed like that, but then there is no point in introducing integrals and whatever because you have expressed 10 via 4s already.

Thus one has to pick which method they follow ... (The primitive one is already in the video, so the question about 10 shouldn't have been asked in the first place ! ...)

 

[jbriggs444]

But back on track with the exact solution (not approximation), for any real number x (e.g. π) [integer, rational, or irrational ...].

A simple counting argument shows that this is not possible. The cardinality of the reals is larger than the cardinality of the set of finite expressions.

 

[Stavros Kiri]

A simple counting argument shows that this is not possible. The cardinality of the reals is larger than the cardinality of the set of finite expressions.

Unless you use continues functions (in the variables and/or constants) to invert from the 4 4s numbers business. Then the cardinality can even be |R|, or |RxR| etc., if you fix appropriately variables and constants for the functions ... . But the point is that we say that this is not actually allowed (see previous posts about restrictions on arbitrariness and inversion, as well as the examples of continues functions there).

So basically, yes, I agree. Do you?

 

[jbriggs444]

Unless you use continues functions (in the variables and/or constants) to invert from the 4 4s numbers business. Then the cardinality can even be |R|, or |RxR| etc., if you fix appropriately variables and constants for the functions ... . But the point is that we say that this is not actually allowed (see previous posts about restrictions on arbitrariness and inversion, as well as the examples of continues functions there).

So basically, yes, I agree. Do you?

I am not sure that I know what you mean by "use continuous functions". But if one is going to use a function there is a requirement that one identify the function. That is why the general approach given in post #2 (just use an arbitrary constant C that gives the desired result) is not acceptable.

It does not matter how rich the specification language is. The number of things that you can specify cannot exceed the number of valid specifications.

It does not matter that the set of not-necessarily-continuous functions has cardinality $|\mathbb{R}^{|\mathbb{R}|}|$. The set of the ones you can finitely specify has cardinality no more than |N|.

 

[Stavros Kiri]

I am not sure that I know what you mean by "use continuous functions"

Real continuous functions on the [also continuous] variables and constants, e.g. say (x, A) [continuous quantities (i.e. their domain is a continuous subset of R)], plus f(x;A) [or notation like f_A(x)], for one variable and one constant etc.], continuous functions on their domain, a requirement for simplicity to give you easily every choice possible, that would make the puzzle trivial.

But as mentioned, and mostly agreed, none of that should be allowed, according to the video and OP. (Just the 4 4s, used only once [each], and basic step operations, basic functions [inverses I think are allowed*, if there are no arbitrary constants (that you can covenienty fix etc.)], mathematical symbols etc. ... ; no other numbers or arbitrary constants etc.)

So I think overall we probably agree, and for the rest of your post I completely agree and liked them both [posts of yours].

But although we cannot get all real numbers, we can try to get as many as we can, following:

How many different numbers can you make? It's best to think of a number and then try to make it

Thus the puzzle and challenge is not trivial, but a real and valid one (Note that in the video he deals only about 0, 1, 2 ... n ... ∞ , with "Dirac's busting" at the end ... , but the OP actually gave more challenge ... (as many numbers as we can, i.e. reals ...; π was one new accomplished etc.) ).

* e.g. ~as in mfb's solution for π (earlier), as both cos and arccos I think should be allowed (one is the inverse of the other, both basic functions ...).

 

[jbriggs444]

Real continuous functions on the [also continuous] variables and constants, e.g. say (x, A) [continuous quantities (i.e. their domain is a continuous subset of R)], plus f(x;A) [or notation like fA(x)], for one variable and one constant etc.], continuous functions on their domain, a requirement for simplicity to give you easily every choice possible, that would make the puzzle trivial.

I still do not understand this.

A constant function is continuous. But if you allow for arbitrary constant functions to be invoked (specified how?) then the problem becomes trivial. So one must have a limited roster of functions.

If you are allowed to include functions with no arguments (also known as zero-place functions or constants) then the problem becomes trivial. Given one such function (call it k), one can write $\frac{k+k+k+k}{k+k+k}$ or similar to encode any rational number. So we cannot allow zero-place functions . (Not even zero, see below).

If one is given a one-place monotone increasing function f and an unrelated one-place monotone decreasing function g then one can write expressions like f(f(g(g(f(g(x)))))) and use the pattern of f's and g's to encode values. So we must exercise care with the set of one-place functions that are permitted, otherwise the problem becomes trivial. [That was the point of post #26].

For two-place or greater functions, the fact that no constants (not even zero) are allowed is helpful. With only four constant terms (the four fours) to use, a maximum of three two-place functions can possibly be used. e.g. f(4,g(4,h(4,4))). Similarly, at most two three place functions or at most one four-place function. The point is that the user is prevented from chaining together an arbitrarily long string of function compositions.

So. Which functions do you have in mind, consistent with the above guidelines?

Plus, minus, times, divide, exponentiation and unary negation?

 

[Stavros Kiri]

then the problem becomes trivial. So one must have a limited roster of functions.

That's what I also said.+Many interesting points in your reply.

+

Plus, minus, times, divide, exponentiation and unary negation?

In the video he uses more, e.g. concatenation, percentage and a bunch of special tricks ...