Question

\[\text{ Let } \\ u=1+\frac{x^3}{3!}+\frac{x^6}{6!}+\frac{x^9}{9!}+\cdots \\ v=x+\frac{x^4}{4!}+\frac{x^7}{7!}+\frac{x^{10}}{10!}+\cdots \\ w=\frac{x^2}{2!}+\frac{x^5}{5!}+\frac{x^8}{8!}+\cdots \\ \text{ Show that } u^3+v^3+w^3-3uvw=1.\] \[

Answer 1

First idea: \[\Large w'' = v' = u \\ \Large u+v+w = e^x\]

Answer 2

I guess also u' = w and others but yeah, I don't know if that's useful to us or not.

Answer 3

you can factor the left side of that equation

Answer 4

good idea, let's take (u+v+w)^3 and see what terms we can collect and throw around, maybe that helps too

Answer 5

(w+u+v)*(u^2-u*v-u*w+v^2-v*w+w^2)= 1

Answer 6

\[e^x(u^2+w^2+v^2-uv-uw-vw)\]

Answer 7

if you can show the latter term is equal to e^-x , you are done

Answer 8

but this could be a blind alley :)

Answer 9

\[(u+v+w)^3 -3u^2v-3u^2w-3uv^2-6uwv-3uw^2-3v^2w-3vw^2\]
just seeing what the other way looks like
I think I got rid of all the trash there

Answer 10

right that would be a better approach

Answer 11

i have no clue what the better approach is :p

Answer 12

maybe we can factor your other factor there @perl

Answer 13

i don't have you mad fancy factoring skills though

Answer 14

how did you factor that one thingy
or how did you see that

Answer 15

I used maple , a computer algebra system

Answer 16

ah i was trying to get wolfram to factor it for me but it told me to get lost

Answer 17

:)

Answer 18

Check this out:

w'=v, v'=u, u'=w so let's differentiate:

\[\large u^3+v^3+w^3 = 3uvw+1 \\ \large 3u^2 u' + 3v^2 v' + 3w^2 w' = 3u'vw + 3 uv'w + 3 uvw' \\ \large 3(u^2w+v^2 u+w^2v) = 3(wuw + uuw+ uvv)\]

Answer 19

I might have typed something wrong I don't know, there was a lot of stuff going on haha.

Answer 20

yes everything matches up except the w^2v=w^2u thing

Answer 21

*

Answer 22

I was trying it too...to see..I keep making a mistake I think...

Answer 23

I just worked it out on my paper, and it worked, the mistake is in the first term on the second half f the last equation, it should be w^2v, it's just the u and v look identical in that font lol.

\[\large u^3+v^3+w^3 = 3uvw+1 \\ \large 3u^2 u' + 3v^2 v' + 3w^2 w' = 3u'vw + 3 uv'w + 3 uvw' \\ \large 3(u^2w+v^2 u+w^2v) = 3(wuw + uuw+ uvv)\]

Answer 24

my v's and u's look the same in font and in my handwriting

Answer 25

omg @Kainui that is so freaking awesome
I thought this problem was going to have such terrible long solution

Answer 26

\[\Large v u \nu \mu\] Like all the same letter almost rofl.

Haha yeah, I hate these types of problems honestly, so I always try to find an easy way out. =P

Answer 27

the question about this question is the power series mentioned by @ganeshie8 last question
weird
it talks about the Fibonacci numbers being coefficients ...blah blah black sheep. If I was smart I could have cheated if I only had looked at my book right next to me.

Answer 28

Although I don't know if the question is completely done, when we took the derivative we lost that +1 term, I don't know if that matters or what exactly.

Answer 29

the question before this question*

Answer 30

No, this question lol.

Answer 31

sometimes I don't realize what I'm typing until its too late

Answer 32

Like, what I'm trying to say is we didn't show it is true because for all we know:

sin(x)=sin(x)+195

because they have the same derivative haha.

Answer 33

i have to type around a kitty all day
and it is tough

Answer 34

Yeah my old roommate had a kitty several years ago, and it loved keyboards idk why haha.

Answer 35

\[(f(x))'=0\]
f(x)=c
so we need to show c is 1

Answer 36

u(0)=1
v(0)=0
w(0)=0

Answer 37

Awesome! Plug that in!

Answer 38

1+0+0-3(0)=c

Answer 39

1=c

Answer 40

Ok, now that was satisfyingly easier than I thought it'd be. Wow, done. That was fun.

Answer 41

I like the last bit because I thought to do it
but now I will think to do the first part you mentioned hopefully later if I ever see a similar problem

Answer 42

Yeah I hope I do too. I wonder how often it will work for you though, since I don't think I've ever seen them defined like this before, kind of weird problem honestly.

Answer 43

well it was a weird problem
it was in the problem plus section

Answer 44

so all of those problems are weird and fun

Answer 45

I will probably most likely never do this again though
unless @xapproachesinfinity brings it up and @ganeshie8 says oh this is freckle's favorite problem :p

Answer 46

anyways goodnight!

Answer 47

lol don't burn the book, you might have to refer the same stuff again as xapproaches is following you like shadow :P gnite!

Answer 48

Haha well I think it's fun, these u,v, w functions are really the first power series I ever played with on my own outside of calculus 2 class. The thing that interested me was how e^x is its own first derivative, sinx is its own 4th derivative, and these are their own 3rd derivatives, etc... So I'll give you a challenge @freckles what is the power series representation of a function that is its own nth derivative?

Answer 49

and i guess you are looking for an answer other than e^x

Answer 50

I mean the power series representation for that function is:
\[1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\cdots \\ \text{ or in sigma notation } \\ \sum_{n=0}^{\infty} \frac{x^n}{n!}\]

Answer 51

Well, that is e^x, so maybe I should give some more examples of what I mean, e^-x and e^ix are their own 2nd and 4th derivatives respectively. \[\Large \frac{d}{dx} e^{-x} = -e^{-x} \\ \Large \frac{d^2}{dx^2} e^{-x} = e^{-x}\] \[\Large \frac{d}{dx} e^{ix} =ie^{ix} \\ \Large \frac{d^2}{dx^2} e^{ix} = -e^{ix} \\ \Large \frac{d^3}{dx^3} e^{ix} =-ie^{ix} \\ \Large \frac{d^4}{dx^4} e^{ix} = e^{ix}\]

So what function satisfies this: \[\Large \frac{d^n}{dx^n} f(x)=f(x)\] Even though e^x satisfies this for all n, I'm looking for the function that satisfies this as little as possible... lol

Answer 52

\[(r-1)(r^{n-1}+r^{n-2}+r^{n-3}+\cdots+r^3+r^2+r+1) \\ \text{ so I need to solve: } r^{n-1}+r^{n-2}+\cdots +1=0\]
?
So basically I need to solve this differential equation:
\[y^{(n-1)}+y^{(n-2)}+ \cdots +y^{(1)}+y=0\]
And I guess I could set my solution up as a power series...
which I actually have already done and have gotten something not so great looking...
But maybe I'm also choosing the wrong power series form...

Answer 53

Yeah, this is one way I think, but I think there's an easier way. Try to look at the examples I gave and see if you can find the common link between the examples I've given. I'll reveal the answer within an hour or two unless you really don't want me to tell you lol. =P

Answer 54

give me a day

Answer 55

post the answer in 24 hours or when you come back after 24 hours