nonlinear reading

It’s easier in some formats (book) than others (web). Also,]]>nobody told meI could/should skip around in maths books. Took me years (which includes long enough to understand standard definitions, and to understand that they are standard and won’t differ in ways that will make me really misunderstand the later parts) to get to the point where Icanflip around in a definitions-first maths book.It’s hard to remember being a novice, but for example looking at Birkhoff/Mac Lane’s Algebra. I tried flipping around and it was clear I was missing

Also, it can be beneficial for the stodgy to feel they’ve been “verbally permitted” to skip around (like in a foreword or preface, the author says “Feel free to skip around–that’s how I designed this book”.

As long as there’s

somepathway the reader can take where they will always be able to encounter interesting, comprehensible text, then we can have a eunoia.

*Thanks — I’ve put that right now.*

I presume the obvious pattern doesn’t continue. In other words, I presume that it is *not* true that . Indeed, it can’t hold because the left-hand side is even.

One could also ask the much more general question of whether there are other examples of identities of the form

For that to hold, we need , which means that must be small enough that the powers are still increasing pretty fast. So we need to be of a similar order of magnitude to . This order of magnitude should be around , so that needs to be around . A back-of-envelope calculation then tells us that will have to be of similar magnitude to . But whether there are any more solutions I don’t know: I wouldn’t expect an infinite family of solutions.

In fact, I think there could be a heuristic argument that suggests that there are only finitely many solutions. Roughly speaking, for each there will be only a small number of s that could possibly work. If you now test each one, it will have a probability of … this is the bit I’m not quite sure about, but I’m pretty confident that the answer is small … something like of actually working. Summing over all gives a finite result.

]]>$$3^2 + 4^2 = 5^2$$

$$3^3 + 4^3 + 5^3 = 6^3$$

which is not only more related to the post than what I wrote, but also more true. ]]>

$$3^2 + 4^2 = 5^2$$

$$3^3 + 3^4 + 3^5 = 3^6$$

?

]]>and since most of the LHS sum of cancels against most of the RHS sum of , we can rearrange trivially to get a formula for the next term down, namely :

This approach generalises to give a formula for each in terms of all the previous ones:

and so on in an obvious pattern.

]]>Then and each have size . The intersection of any two of them has size . And finally the intersection of all three of them has size . So by the inclusion-exclusion formula we get

That gives us that , which equals .

In general, if we do things this way, we get that

For example, , so .

This formula is perhaps simpler conceptually than the one in the post, but requires more calculation because you have to calculate all the earlier .

]]>Not sure if that’s significant or obvious.

]]>*Thanks — I’ve corrected that now.*

Very nice proof for the sum of kth powers. Recommended for reading! ]]>

2) From the finite difference considerations it is clear that the formula for the sum of the k-th powers is a polinomial of degree k+1 in n. Actually we can just guess that from examples. Then we can figure out the coefficients of this polynomial by fitting the small n data, and then finish the proof by induction.

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