Question

When is this statement true?

Answer 1

\[\Large (x+y)^n=x^n+y^n\]

Answer 2

Hmmm, when you think about the binomial theorem...

Answer 3

It's only really true when \(n=1\) or when \(x\) or \(y\) equals \(0\)... maybe a few other special cases

Answer 4

It's sort of a trick question. How can this statement be true if n is greater than 1 and x and y are arbitrary nonnegative integers?

Answer 5

\[
(x+y)^n = \sum_{k=0}^n {n\choose k}x^ky^{n-k}
\]

Answer 6

\[
(x+y)^n = \sum_{k=0}^n {n\choose k}x^ky^{n-k} = nx^n+ny^n +\sum_{k=1}^{n-1} {n\choose k}x^ky^{n-k}
\]

Answer 7

You're not wrong, but you've not solved it yet either. There's a particular set of limitations you have to put on it. =P

Answer 8

\[
\frac 1n\sum_{k=1}^{n-1} {n\choose k}x^ky^{n-k} = 0
\]

Answer 9

Alright should I reveal the answer?

Answer 10

When n is prime, \[\Large (x+y)^n = x^n+y^n \mod(n)\] for x and y arbitrary non negative integers.

Answer 11

Trick question indeed, you're changing the question!

Answer 12

@mathmate No I'm not, I'm just asking when this is true. And it happens to be true when you're in modular arithmetic! =D

Answer 13

:)

Answer 14

http://en.wikipedia.org/wiki/Freshman%27s_dream

Answer 15

I guess I'll give the medal to someone who can explain why n has to be prime.

Answer 16

It's already explained in the reference, that's why you got a medal! :)

Answer 17

Fine haha. But what's even more interesting is if you follow the link to Sophomore's dream...

http://en.wikipedia.org/wiki/Sophomore%27s_dream

Answer 18

Just nitpicking on notation :
\[\Large (x+y)^p \ne x^p+y^p \pmod p\]
\[\Large (x+y)^p \equiv x^p+y^p \pmod p\]

Answer 19

What's the difference?

Answer 20

"congrunce" is not same "identically equal"

Answer 21

@ganeshie8 agree!

Answer 22

you say two triangles are similar to each other
but replacing "similar" with "equal" is not acceptable "strictly speaking"

Answer 23

@Kainui Yes, sophomore is even more beautiful! :)
Apart from the two dreams, I learned about the Bernoulli family.

Answer 24

i remember seeing sophomore dream recently
http://openstudy.com/study#/updates/54579995e4b0a717ff61b10b

@Princer_Jones

Answer 25

Interesting, I wonder what equation satisfies this: \[\Large \int\limits_{f(x)}^1x^{-x}dx = \sum_{n=1}^{1/f(n)}n^{-n}\] If you catch my drift.

Answer 26

don't the limits of integration have to be constant with respect to the variable of integration?

Answer 27

I guess I have written it incorrectly, Interesting, I wonder what equation satisfies this: \[\Large \int\limits_{f(t)}^1x^{-x}dx = \sum_{n=1}^{1/f(t)}n^{-n}\] If you catch my drift.

Answer 28

\[
\int_a^xdx
\]Doesn't make sense because you're saying \(a<x<x\)

Answer 29

I would put:\[\Large \int\limits_{1/c}^1x^{-x}dx = \sum_{n=1}^{c}n^{-n}\]

Answer 30

Sure, I think the main problem with this is we might end up with infinity/2 as the halfway point, which is pretty much nonsense currently.

Answer 31

\[-\Large \int\limits_{1}^{1/c}x^{-x}dx = \sum_{n=1}^{c}n^{-n}\]

Answer 32

\[-\Large \sum_{i=0}^{m} {x_i^*}^{-x_i^*} \frac{1/c-1}{m} = \sum_{n=1}^{c}n^{-n}\]

Answer 33

\(m\to \infty\)

Answer 34

I wonder if this is at all related to: \[\LARGE f(x) = e^{|lnx|}\]

Answer 35

I guess I have some bit of weird intuition in me that says we can rotate the graph \[\large y=x^{-x}\] around the point (1,1) and then stretch the points from 0 to 1 out to 1 to infinity and vice versa to have a sort of "odd" function if that makes sense.

Answer 36

I'm sort of just making stuff up as I go, but it is sort of graphically apparent what it looks like if I do an "even" reflection, the graph changes from this to this:\[\Large x^{-x} \rightarrow x^{1/x}\]

Answer 37

What I'm looking for is some kind of heuristic argument that says something similar to the proof that there are just as many numbers between 0 and 1 as there are from 1 to infinity. But it has to expand upon this and say that we can turn sums into integrals in a similar sort of manner by squeezing the sum into an integral in a sense. That's kind of my motivation right now, since it seems like an intuitive sort of thing going on here that I don't quite understand yet but would like to.