Since 0^0 is undefined, assume none of the four digits is 0.

1^1 = 1
2^2 = 4
3^3 = 27
4^4 = 256
5^5 = 3125
6^6 = 46656
...

No digits can be 6 or above.

4^4 + 4^4 + 4^4 + 3^3 = 795 < 1000
4^4 + 4^4 + 4^4 + 4^4 = 1024
These are not solutions, so there is at least one digit 5.

5^5 + 5^5 + x^x + x^x > 6250
This number is either at least five digits or a four-digit number whose first digit is at least 6. Therefore, at most one digit 5.

5^5 + 1^1 + 1^1 + 1^1 = 3128
5^5 + 4^4 + 4^4 + 4^4 = 3893
3128 < 5^5 + x^x + x^x + x^x < 3893
First digit is always a 3.

With only sixteen choices for the remaining two digits, easiest to just brute force.

What is 0 to the 0 power?
This answer is adapted from an entry in the sci.math Frequently Asked Questions file, which is Copyright (c) 1994 Hans de Vreught (hdev@cp.tn.tudelft.nl).
According to some Calculus textbooks, 0^0 is an "indeterminate form." What mathematicians mean by "indeterminate form" is that in some cases we think about it as having one value, and in other cases we think about it as having another.

When evaluating a limit of the form 0^0, you need to know that limits of that form are "indeterminate forms," and that you need to use a special technique such as L'Hopital's rule to evaluate them. For instance, when evaluating the limit Sin[x]^x (which is 1 as x goes to 0), we say it is equal to x^x (since Sin[x] and x go to 0 at the same rate, i.e. limit as x->0 of Sin[x]/x is 1). Then we can see from the graph of x^x that its limit is 1.

Other than the times when we want it to be indeterminate, 0^0 = 1 seems to be the most useful choice for 0^0 . This convention allows us to extend definitions in different areas of mathematics that would otherwise require treating 0 as a special case. Notice that 0^0 is a discontinuity of the function f(x,y) = x^y, because no matter what number you assign to 0^0, you can't make x^y continuous at (0,0), since the limit along the line x=0 is 0, and the limit along the line y=0 is 1.

This means that depending on the context where 0^0 occurs, you might wish to substitute it with 1, indeterminate or undefined/nonexistent.

Some people feel that giving a value to a function with an essential discontinuity at a point, such as x^y at (0,0), is an inelegant patch and should not be done. Others point out correctly that in mathematics, usefulness and consistency are very important, and that under these parameters 0^0 = 1 is the natural choice.

The following is a list of reasons why 0^0 should be 1.


Rotando & Korn show that if f and g are real functions that vanish at the origin and are analytic at 0 (infinitely differentiable is not sufficient), then f(x)^g(x) approaches 1 as x approaches 0 from the right.
From Concrete Mathematics p.162 (R. Graham, D. Knuth, O. Patashnik):


Some textbooks leave the quantity 0^0 undefined, because the functions 0^x and x^0 have different limiting values when x decreases to 0. But this is a mistake. We must define x^0=1 for all x , if the binomial theorem is to be valid when x=0 , y=0 , and/or x=-y . The theorem is too important to be arbitrarily restricted! By contrast, the function 0^x is quite unimportant.
Published by Addison-Wesley, 2nd printing Dec, 1988.


As a rule of thumb, one can say that 0^0 = 1 , but 0.0^(0.0) is undefined, meaning that when approaching from a different direction there is no clearly predetermined value to assign to 0.0^(0.0) ; but Kahan has argued that 0.0^(0.0) should be 1, because if f(x), g(x) --> 0 as x approaches some limit, and f(x) and g(x) are analytic functions, then f(x)^g(x) --> 1 .
The discussion of 0^0 is very old. Euler argues for 0^0 = 1 since a^0 = 1 for a not equal to 0 . The controversy raged throughout the nineteenth century, but was mainly conducted in the pages of the lesser journals: Grunert's Archiv and Schlomilch's Zeitshrift. Consensus has recently been built around setting the value of 0^0 = 1 .


Magicwand think:
it is not undefined. it is "indeterminate form". they are different.
Mathmathic uses convention like 0! which make no sense but it fits other pattern.
0^0 is no different than 0!

if y * x = y then necessarily x = 1, isn't it ?
example 0 * 7 = 0 so 7 = 1 !

It is the oldest hamete in math and it is exactly the same reasoning the smart lady in the video is using.

Shouldn't that be "-1/0" on the second line I quoted? I don't see how you get from one of these to the other. Unless you're trying to assign some sort of literal value to x/0, in which case it should probably be ([-inf,inf]) and the solution from that point forward will look like a resolution for a quantum mechanics equation or something.

If it is secret I'll hide too. Trust Wikileaks!
If I understand you well you question my 5e line. At least its derivation. So I assume you buy my fourth line.
_ 4_ : 0 * ( -inf ) = 0 . ( it comes from 0^0 = 1, taking the log from 1, and applying the powerrule for logs )

If a * b = c then 1/a * 1/b = 1/c ( even in some sense if c equals zer0 : both sides of the second equation here are inf then.)
so from _ 4_ we get 1/0 * 1/(-inf) = 1/0 q.e.d.

About quantummechanic equations: Schöderinger's and Dirac's equations allow healthy solutions in physical cases. In the computing proces maybe distributions are handy but even these are respectable mathematical objects. Only with renormalisation it gets tricky, only actually to deal with infinities from the recursive aspects that comes along with general relativity. Disclaimer: long time ago I studied these things.

It's not the derivation of the 5e line I'm questioning... I'm pretty comfortable with distribution and commutation!

It's the derivation of the 6e line. How do you find this: -1 * (1/0) = inf ?

One of us has a blind spot. Only one-eyed I'm the suspect. Better to discuss string theory.
For the ease of reading I included some brackets.
LHS: push the minus sign to the front. RHS: substitute inf for 1/0.
Can't make it easier

Actually I'm not too proud of this whole "derivation".
I think there is a stronger case.
For example that for nice operators "op" we expect lim ( f(x) op g(x) ) = ( lim f(x)) op ( lim g(x) ) if RHS exists.
MW's definition would make ^ unnice.

Er, I understand *what* you did... I'm just saying it doesn't make sense to me, so I'm asking why you think it's possible. Could be I'm wrong.

Usually laws of equality and commutation work together to help solve equations so that you do something like this (same thing on each side of the equality symbol):
LHS: multiply every term by -1. RHS: multiply every term by -1

You seem to have done... well, what you say above, which doesn't make sense mathematically (unless I'm being very foolish somehow.) It looks like this to me:
LHS: multiply every term by -1. RHS: 1/0 ??? infinity ??? profit!

I'm not sure I see your problem, ethanb - what cyclops appears to have done is to simply rewrite the left-hand side, using that (-a)*b=a*(-b), and then write "inf" for 1/0 on the right, which at least seems intuitively plausible.

Of course, this is all horribly unrigorous, and makes especially little sense if you care about signs and are going to make a distinction between plus and minus infinity (as appears to be the case here). Saying 1/0 is infinity seems reasonable when we consider that 1/x tends to infinity as x tends to 0 from above, but if x tends to 0 from below then 1/x tends to minus infinity!

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D - the series of sums which contain the first n terms of the sequence doesn't converge, so the sum of the entire sequence is undefined. Specifically, the radius never gets smaller than 1 - if s_n =1, s_n+1 =0, and vice versa.

C, because it's the Cesaro sum for that series.