# Math Q&A challenge (Part 11)

#### eddybob123

So Hoempa gets the next Q.

#### Hoempa

Math Team
This think kept me busy for a while, wonder what you guys make of it:
$$\displaystyle u_{n+1} = \left(\sum_{k=0}^{m} {m + 2\choose 2k} u_n^{m+2-2k} \cdot x^k\right) \cdot \left( \sum_{k=0}^{m-1}{m+2\choose 2k+1} u_n^{m+1-2k} \cdot x^{k} \right)^{-1}$$ for even positive m.
What value does it get when n->infty? I have an idea, but can't explicitely prove it.

#### icemanfan

What is the value of $$\displaystyle u_0$$? Is x just a variable or does it have a specific value?

#### Hoempa

Math Team
Oh, eh, x>0, u_0 <> 0. I slipped, it's supposed to be. The answer only depends on x.

$$\displaystyle u_{n+1}=\left(\sum_{k=0}^{0.5m} {m \choose 2k}\cdot u_n^{m-2k}\cdot x^k\right) \cdot \left(\sum_{k=0}^{0.5m-1} {m \choose 2k+1}\cdot u_n^{m-2k-1}\cdot x^k\right)^{-1}$$ for even positive m.

#### agentredlum

Math Team
I ask the honorable gentleman from UTC-5 to post his solutions to his questions please.

#### CRGreathouse

Forum Staff
agentredlum said:
I ask the honorable gentleman from UTC-5 to post his solutions to his questions please.

I have no time to explain* ( :cry: ) but
http://oeis.org/A000028
and the Lambek-Moser paper linked therein should get you there.

* Personal matters: got in a terrible car wreck, now dealing with fallout.

#### agentredlum

Math Team
Hoempa said:
No worries, agent, I'm happy to read CRG is fine. Hope you guys like the Q.
I like the question but as usual am having trouble understanding the notation. It appears m is arbitrary even number > 0 and determines the number of terms in the summation , is that right?

Would you be kind enough to give numerical values of $$\displaystyle \ \ \ u_{0 + 1} \ , \ u_{1 + 1} \ , \ and \ u_{2 + 1} \ \ \$$ so that I may work backwards and make sure that I have not misinterpreted the notation ?

#### Hoempa

Math Team
agentredlum said:
I like the question but as usual am having trouble understanding the notation. It appears m is arbitrary even number > 0 and determines the number of terms in the summation , is that right?
Yes, that's right.
agentredlum said:
Would you be kind enough to give numerical values of $$\displaystyle \ \ \ u_{0 + 1} \ , \ u_{1 + 1} \ , \ and \ u_{2 + 1} \ \ \$$ so that I may work backwards and make sure that I have not misinterpreted the notation ?
Yes. I'll chose x = 100, u_0 = 10000, m = 2.
u_1 = 1000001/200 = 5000.005
u_2 = 1000006000001/400000400 ~= 2500.0125
u_3 = 1000028000070000028000001/800005600005600005600000800 ~= 1250.026.

#### Hoempa

Math Team
Could you work backwards agent? I'll admit that my values are tricky, the next is about half the previous but that isn't the general case.

#### agentredlum

Math Team
Hoempa said:
Could you work backwards agent? I'll admit that my values are tricky, the next is about half the previous but that isn't the general case.
This problem appears to require programming capability using Computer Algebra Systems in order to determine trends. Right now i don't have a computer ... unless you know how to program the free online version of Wolfram Alpha ? I don't think WIA can run program loops but i'm not sure ... i suppose i could input each value separately and write down each result but it's so tedious ...

Do you know of any online CAS that does not require the user to download onto the hard drive? My PS3 hard drive won't accept any download unless it comes from Sony as an app and Sony hasn't made a CAS app as far as i know. The PS3 has very powerful computing power (in relation to the cost) but unless we hack it we can't use that power for anything other than gaming. I remember reading an article about a university that hacked together a bunch of PS3's and turned them into 1 supercomputer.