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a reasonably challenging math problem
by MaxEntropy_Man Wed Sep 28, 2011 6:32 pm
show that if:
u(x) = 1+x^3/3! +x^6/6! + x^9/9!+….
v(x) = x+x^4/4!+x^7/7!+x^10/10!+…
w(x) = x^2/2!+x^5/5!+x^8/8!+x^11/11!+…
then
u^3+v^3+w^3-3uvw = 1
u(x) = 1+x^3/3! +x^6/6! + x^9/9!+….
v(x) = x+x^4/4!+x^7/7!+x^10/10!+…
w(x) = x^2/2!+x^5/5!+x^8/8!+x^11/11!+…
then
u^3+v^3+w^3-3uvw = 1
MaxEntropy_Man- Posts : 14702
Join date : 2011-04-28
Re: a reasonably challenging math problem
by MaxEntropy_Man Wed Sep 28, 2011 10:11 pm
nobody wants to give this a try? i'll wait a day and then post my solution. there are probably a few different ways to skin this cat.
MaxEntropy_Man- Posts : 14702
Join date : 2011-04-28
Re: a reasonably challenging math problem
by Guest Thu Sep 29, 2011 7:25 am
Just wondering - does the solution involve pure algebraic manipulation or does it involve calculus?
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Re: a reasonably challenging math problem
by MaxEntropy_Man Thu Sep 29, 2011 9:54 am
i'll post a bunch of hints when i have a bit more time later in the evening. rushed right now, so i'll just say that my own solution involved solving a second order ordinary diff equation and evaluating the two constants but i am sure there is a much easier solution which is escaping me. this is supposed to be solvable by high school students with some calculus background.
MaxEntropy_Man- Posts : 14702
Join date : 2011-04-28
Re: a reasonably challenging math problem
by Guest Thu Sep 29, 2011 10:18 am
I have the solution to this now but it is not my solution and so I won't post it.
Simple calculus is all that's required.
Simple calculus is all that's required.
Guest- Guest
Re: a reasonably challenging math problem
by Guest Thu Sep 29, 2011 10:44 am
MaxEntropy_Man wrote:this is supposed to be solvable by high school students with some calculus background.
really? integration of power series is not in the high school curriculum here and nor in the joint entrance syllabus. here is the aieee syllabus (look under "integral calculus").
Guest- Guest
Re: a reasonably challenging math problem
by Jeremiah Mburuburu Thu Sep 29, 2011 11:02 am
interesting! someone else has solved the problem; you saw or read his/her solution, and you want everyone here to know that! and is that last remark your contribution to the solution of the problem?blabberwock wrote:I have the solution to this now but it is not my solution and so I won't post it.
Simple calculus is all that's required.
Jeremiah Mburuburu- Posts : 1251
Join date : 2011-09-09
Re: a reasonably challenging math problem
by Guest Thu Sep 29, 2011 11:19 am
Jeremiah Mburuburu wrote:interesting! someone else has solved the problem; you saw or read his/her solution, and you want everyone here to know that! and is that last remark your contribution to the solution of the problem?blabberwock wrote:I have the solution to this now but it is not my solution and so I won't post it.
Simple calculus is all that's required.
VeeTee!
silly!
juvenile!
upset!
angry!
Use simple calculus and solve the problem, will ya?
Guest- Guest
Re: a reasonably challenging math problem
by Guest Thu Sep 29, 2011 11:21 am
chat wars come into the forums now!
Guest- Guest
Re: a reasonably challenging math problem
by Mosquito Thu Sep 29, 2011 2:38 pm
MaxEntropy_Man wrote:show that if:
u(x) = 1+x^3/3! +x^6/6! + x^9/9!+….
v(x) = x+x^4/4!+x^7/7!+x^10/10!+…
w(x) = x^2/2!+x^5/5!+x^8/8!+x^11/11!+…
then
u^3+v^3+w^3-3uvw = 1
You can solve this by differentiating it only once.
as
u=dw/dx
v=du/dx
w=dv/dx.
Mosquito- Posts : 706
Join date : 2011-04-28
Re: a reasonably challenging math problem
by Mosquito Thu Sep 29, 2011 3:15 pm
PseudoIntellectual wrote:MaxEntropy_Man wrote:show that if:
u(x) = 1+x^3/3! +x^6/6! + x^9/9!+….
v(x) = x+x^4/4!+x^7/7!+x^10/10!+…
w(x) = x^2/2!+x^5/5!+x^8/8!+x^11/11!+…
then
u^3+v^3+w^3-3uvw = 1
You can solve this by differentiating it only once.
as
u=dw/dx
v=du/dx
w=dv/dx.
d(u^3+v^3+w^3-3uvw)/dx=d(1)/dx
(3u^2v+3v^2w+3w^2u)-3(v^2w+w^2u+u^2v)=0
Mosquito- Posts : 706
Join date : 2011-04-28
Re: a reasonably challenging math problem
by MaxEntropy_Man Thu Sep 29, 2011 3:32 pm
i don't like the differentiating on both sides approach because it doesn't prove that the right hand side is 1, but just that it's a constant.
i give my approach below. those who still want to take a crack at this, read no further yet.
first i note that
u+v+w = Exp[x] (1)
then i note (by inspection of the power series) that
v=w'(x) and u = w''(x) (2)
substituting (2) in (1) gives me a second order ODE, i.e.
w''(x) + w'(x) + w(x) = Exp[x] (3)
i solved (3) to get an analytical solution with two constants C1 and C2 (i cheated and used mathematica although it's doable with some effort by hand).
the constants can be obtained by noting from the power series that
w(0) = 0 and w'(0) = v(0) =0.
this then gives the complete solution for w(x) and therefore v(x) and u(x). these can be put back into the LHS of the identity to be proved and shown to be equal to 1. i'll give the gory details in a later post.
having done that, i am sure there is a much better solution.
i give my approach below. those who still want to take a crack at this, read no further yet.
first i note that
u+v+w = Exp[x] (1)
then i note (by inspection of the power series) that
v=w'(x) and u = w''(x) (2)
substituting (2) in (1) gives me a second order ODE, i.e.
w''(x) + w'(x) + w(x) = Exp[x] (3)
i solved (3) to get an analytical solution with two constants C1 and C2 (i cheated and used mathematica although it's doable with some effort by hand).
the constants can be obtained by noting from the power series that
w(0) = 0 and w'(0) = v(0) =0.
this then gives the complete solution for w(x) and therefore v(x) and u(x). these can be put back into the LHS of the identity to be proved and shown to be equal to 1. i'll give the gory details in a later post.
having done that, i am sure there is a much better solution.
MaxEntropy_Man- Posts : 14702
Join date : 2011-04-28
Re: a reasonably challenging math problem
by Guest Thu Sep 29, 2011 9:25 pm
MaxEntropy_Man wrote:i don't like the differentiating on both sides approach because it doesn't prove that the right hand side is 1, but just that it's a constant.
When x=0, u(x) is 1 and v(x)=w(x)=0. So u^3+v^3+w^3-3uvw is 1 when x=0.
Therefore the constant is 1.
Guest- Guest
Re: a reasonably challenging math problem
by MaxEntropy_Man Thu Sep 29, 2011 10:02 pm
fair enough. with that evaluation PI's proof is acceptable and far shorter than my brute force method. is that the proof you had also bw or was it different?
MaxEntropy_Man- Posts : 14702
Join date : 2011-04-28
Re: a reasonably challenging math problem
by Guest Thu Sep 29, 2011 10:16 pm
MaxEntropy_Man wrote:fair enough. with that evaluation PI's proof is acceptable and far shorter than my brute force method. is that the proof you had also bw or was it different?
It was the same. I threw this at my son last evening and he worked it out with his dad's help.
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Re: a reasonably challenging math problem
by Kris Fri Sep 30, 2011 3:57 am
blabberwock wrote:MaxEntropy_Man wrote:i don't like the differentiating on both sides approach because it doesn't prove that the right hand side is 1, but just that it's a constant.
When x=0, u(x) is 1 and v(x)=w(x)=0. So u^3+v^3+w^3-3uvw is 1 when x=0.
Therefore the constant is 1.
>>>> I vaguely thought along these lines, but was not sure. I'd better brush up on this. My son is in precalc now and honestly I am having trouble with even some of the inequality problems in his book.
Kris- Posts : 5460
Join date : 2011-04-28
Re: a reasonably challenging math problem
by Marathadi-Saamiyaar Fri Sep 30, 2011 4:48 am
Kris wrote:blabberwock wrote:MaxEntropy_Man wrote:i don't like the differentiating on both sides approach because it doesn't prove that the right hand side is 1, but just that it's a constant.
When x=0, u(x) is 1 and v(x)=w(x)=0. So u^3+v^3+w^3-3uvw is 1 when x=0.
Therefore the constant is 1.
>>>> I vaguely thought along these lines, but was not sure. I'd better brush up on this. My son is in precalc now and honestly I am having trouble with even some of the inequality problems in his book.
Funny....when your father daily asked you to study you never listened to him. Now, you are made to study on your own.
Isn't life cruel?....
P.S. I am back to renDon rendu; renduRendum Naalu.
Marathadi-Saamiyaar- Posts : 17675
Join date : 2011-04-30
Age : 110
Re: a reasonably challenging math problem
by Kris Fri Sep 30, 2011 10:53 am
>>>> I vaguely thought along these lines, but was not sure. I'd better brush up on this. My son is in precalc now and honestly I am having trouble with even some of the inequality problems in his book.[/quote]
Funny....when your father daily asked you to study you never listened to him. Now, you are made to study on your own.
Isn't life cruel?....
P.S. I am back to renDon rendu; renduRendum Naalu.[/quote]
>>>>> Its scary how much was missed. I was thinking yesterday about a grad school class in finance which was so heavily stats oriented and still can't figure out how I eked out a B.
Funny....when your father daily asked you to study you never listened to him. Now, you are made to study on your own.
Isn't life cruel?....
P.S. I am back to renDon rendu; renduRendum Naalu.[/quote]
>>>>> Its scary how much was missed. I was thinking yesterday about a grad school class in finance which was so heavily stats oriented and still can't figure out how I eked out a B.
Kris- Posts : 5460
Join date : 2011-04-28
Re: a reasonably challenging math problem
by Marathadi-Saamiyaar Fri Sep 30, 2011 11:40 am
Kris wrote:
>>>>> Its scary how much was missed. I was thinking yesterday about a grad school class in finance which was so heavily stats oriented and still can't figure out how I eked out a B.
Simple.... The entire educational system in the US (and much faster in India) has been dumbed down to show turnover of graduates and "productivity"
I recently heard that an Ivy league gave a PhD to a totally undeserving 50+ dude, who was a successful enterprenuer. He gave $20 mils, and the admin argument was giving him a PhD would not hurt as he was not likely to use it in any way other than to satisfy his ego.
(I heard this from one of the guys on his Dissertation committee).
Welcome to Reaganomics - which I blame for setting the downward motion of this country. Give Perry 4 years and watch it sink to the bottom.
Marathadi-Saamiyaar- Posts : 17675
Join date : 2011-04-30
Age : 110
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