/sci/ - Science & Math

>>10058744
Way to contribute to a conversation, recursively speaking.

>>10058742
Other communicating entities fascinate me
>Schizoid Handbook 101; Make Space For Friends

>>2168833
yeah I haven't had any formal education on series yet for some reason, You'd think in a Calculus 2 class I'd learn it... but apparently I learn that next semester or something, which is unfortunate when my discrete teacher expects me to know it all already.

>>2168768
>>2168768

sadly... no?
I see how the

r^n + n[r^(n+1) -1/(r-1) -r^n]

works. but as far as deducing the closed form solution, I would have never thought of that, and I'm sure if my teacher asks me about it, I won't be able to figure it out, which is why I was trying to get the geometric series one.
but,

r^n + n * (1-r^n)/(1-r)
if I plug in 2... r^2 + 2* (1-r^2)/(1-r) yields r^2 + 2 -2r unless im doing my math wrong, which seems weird because it should be working,..
then that third one.... I'm still trying to understand

>>2168763
k, in this part
<span class="math"> \sum_{n=0}^{\infty }\frac{t_{n+1}x^n}{n!} -r\sum_{n=0}^{\infty }\frac{t_nx^n}{n!}=\sum_{n=0}^{\infty }\frac{(n+1)x^n}{n!} \;\; \;(2). [/spoiler]
you're just removing all the terms before the one that appeared first kinda?

uhm the solution to your first order differential didnt appear, something bout a close brace?

I haven't learned series so I don't understand how to do pretty much anything involving them, the only reason I know geometric series is becuase our teacher goes over it when he was showing us other recurrences like this one...

uhm and the last bit, I can'teven begin to see how plugging in 1 or 2 into n could be simplified to become the terms like r+1 and r^2 + r + 2

>>2168746

<span class="math">T(1)=r^{(1)}+(1)(\frac{(r^{1+(1)}-1)}{r-1} - r^1)=r+\frac{(r^2-1)}{r-1} - r=r+\frac{(r+1)(r-1)}{r-1}-r= r+1[\math][/spoiler]

T(1)=r^{(1)}+(1)(\frac{(r^{1+(1)}-1)}{r-1} - r^1)=r+\frac{(r^2-1)}{r-1} - r=r+\frac{(r+1)(r-1)}{r-1}-r= r+1[/math]

>>2168696
>>2168732

I don't think subtracting is actually necessary, becuase sigma 0 to n-1 of r^i is in fact the geometric series

r^n + n* (1-r^n)/(1-r)
which does seem to give the correct answers,
t(1) = r^1 + 1* 1-r/1-r
t(2) = r^2 + 2* (1-r^2)/1-r which... I don't think simplifies correctly, or im doing my math wrong.

>>2168696
r^n + n *[ (r^(n+1)-1)/r-1 -r^n]
wait a minute, if I plug in 1 to this
I get

r&1 + 1 * r^2-1/r-1 -r^1
simplifying,
r(r-1) = r^2 - r
r^2-1 - (r^2-r)
r^2s cancel, then you get r-1r-1 .. which equals one!
ok, that worked, when I did it the first time on my whiteboard I got -1... probably didn't subtract right

>>2168696
ok
so to change n-1 to n
we added in a r^n term into the summation, to keep things balanced we subtracted it as well.
k

so then, once we do all the steps and such again,
r^n + n * (r^(n+1)-1)/r-1 -r^n

is the last term also being multiplied by n? it must be right? becuase it was involved in manipulating the summation, so
r^n + n *[ (r^(n+1)-1)/r-1 -r^n] yes?
>>2168586

still waiting in case you are writing out your steps, I am interesting

>>2168651

wait... so I was right on.. the n-1 thing? k..

[Sum (from i = 0 to n) { r^(i) }] - r^n
we're subtracting r^n because
ugh im still confused,

>>2168586

I think that would be really really helpful if you could. thank you. It really doesn't help me that my math professor in my calculus classes haven't gotten to series I think

ok... so the sum got manipulated wrong
wait a minute.
sigma 1 to n i^i-1 to add an i term in, would you have to increase n by 1? so it would be sigma 1 to n+1 ? or am i wrong again?
I hate working with sigma....
but making it n+1 is the same as decreasing teh lower bound by one... ugh

>>2168564
so ...

[Sum (from i = 0 to n) { r^(i) }] <=> (r^(n+1) - 1) / (r - 1)

By induction:

Base Case: r^0 = 1 and (r^(0+1) - 1) / (r - 1) = (r-1)/(r-1) =1

So we assume this shit works for 0 to some n. We'll prove it for n+1.

[Sum (from i = 0 to n+1) { r^(i) }] <=>
[Sum (from i = 0 to n) { r^(i) }] + r^(n+1) <=>
(r^(n+1) - 1) / (r - 1) + r^(n+1) // This is the induction step.

Simplifying gives you:

[ r^(n + 1) - 1 + r^(n+2) - r^(n+1) ] / (r - 1) <=>

(r^((n+2) - 1) / (r - 1) []


is that all wrong? because it seems solid...

ugh

so can you not pull it to become sigma 0 to n of r^i from sigma 1 to n i^i-1 ?

>>2168549
ok? Sooo it's a method to solve them. What kinda classes did you learn it in?
I know wikipedia was talking about linear algebra being able to solve recurrences easily or something?

T(n) = (r^n)T(0) + (n)[Sum (from i = 1 to n) { r^(i-1) }]

if I plug in 1 into this, it does come out to r+1
but yeah,
1 into r^n + n*[(r^n+1)-1]/r-1
gets me...
r + 1* r^2-1/r-1 which is.. not right

>>2168527

like as in... some random website or something?
or is this some type of algorithm I haven't learned yet?

>>2168503
should be r^(n+1) - 1 not just r^n+1

>>2168497
>>2168409

aw damn, sooo
how did you get to this answer?
While I want the answer, I want to know how to get to it as well, or else it's pretty meaningless

>>2168481
ah ok, I must have missed it when I read it.
ugh

<span class="math"> r^{n} + n* (r^{n+1}) / ( r-1) [/spoiler]

hopefully that turns out alright if not:
r^n + n* (r^(n+1)) / ( r-1)

thanks.

>>2168450

where is this n coming from!?
ebfore we do anything we've got

r^n + sigma 1 to n r^i-1
then we work our induction magic on the sigma part and

r^n + r^n+1 / r-1

so where does the n come from?

>>2168409
wait.. what?

>>2168389
ok

I do know how to do PBI but I didn't know why you choose r^(n+1)-1 / r-1 or where you got it from.
I don't have anything like that in my book, and couldn't find it in any summation tables. So I've been keeping with what the book has had me doing, which is using the geometric series for most things.

so then, you've got (r^((n+2) - 1) / (r - 1) .. right!
ok, so then the answer to the whole problem is
r^n + (r^((n+1) / (r - 1)
alright.

thank you very much for your help! I'm still new to solving recurrences, and I've mainly been using the theorems given in the book becuase the logic and methods behind their guessing the formulas is lost to me

>>2168362
oooooooooooooooo
i just thought you forgot n-1 one or something
ok that makes sense.

right so, now its r^n + sigma 0 to n r^i
not: r^n n*sigma 0 to n r^i right?

or should be taking an n out so we can get a geometric series?