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Use power series to solve the differential equation
\[y''+xy'+y=0\]
\[y=\sum_{n=0}^{\infty} a_n x^n\]
\[y'=\sum_{n=1}^{\infty}a_nnx^{n1}\]
\[y''=\sum_{n=2}^{\infty} n(n1)a_nx^{2}\]
\[xy'=\sum_{n=1}^{\infty}a_nnx^n\]
\[\sum_{n=2}^{\infty} n(n1)a_nx^{2}+\sum_{n=1}^{\infty}a_nnx^n+\sum_{n=0}^{\infty} a_n x^n=0\]
 one year ago
 one year ago
Use power series to solve the differential equation \[y''+xy'+y=0\] \[y=\sum_{n=0}^{\infty} a_n x^n\] \[y'=\sum_{n=1}^{\infty}a_nnx^{n1}\] \[y''=\sum_{n=2}^{\infty} n(n1)a_nx^{2}\] \[xy'=\sum_{n=1}^{\infty}a_nnx^n\] \[\sum_{n=2}^{\infty} n(n1)a_nx^{2}+\sum_{n=1}^{\infty}a_nnx^n+\sum_{n=0}^{\infty} a_n x^n=0\]
 one year ago
 one year ago

This Question is Closed

TuringTestBest ResponseYou've already chosen the best response.1
yep, now what's first?
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
changing the sums so that they all begin with n=2 changing the last sum from n=0 to n=1 would be \[a_1+\sum_{n=1}^\infty a_nx^n\]
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
no we don't strip out the terms yet, we just do an index shift
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
the first order of business is to get all the x's to be to the nth power through using index shifts
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
the only series that has x to some power other than n is the first one how can we change the index of the first series to make the exponent on x be n?
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
you mean on \[y''=\sum_{n=2}^{\infty} n(n1)a_nx^{n2}\]
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
yes, we need to start from a different n to get x^n through an index shift
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
so I'm shifting the starting point up by 2 so I have to subtract 2 from {n=22} \[\sum_{n=0}^\infty n(n1)a_nx^n\] negative to make up for the negative?
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
\[\sum_{n=0}^\infty n(n+1)a_nx^n\]
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
because that would give us zero
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
you didn't add 2 to every n
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
\[\sum_{n=0}^\infty (n+2)(n+1)a_nx^n\]
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
and I do mean \(every\) n (subscripts included)
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
again better, but...^
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
\[\sum_{n=0}^\infty (n+2)(n+1)a_{n+2}x^n \]
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
now that, I do believe, is correct :)
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
or just leave as is?
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
add what? oh that? I wouldn't yet
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
yeah leave as is asd now rewrite what you have
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
the whole expression please I'm lazy
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
\[\sum_{n=0}^{\infty} (n+2)(n1)a_{n+2}x^{n}+\sum_{n=1}^{\infty}a_nnx^n+\sum_{n=0}^{\infty} a_n x^n=0\]
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
\[\sum_{n=0}^{\infty} (n+2)(n+1)a_{n+2}x^{n}+\sum_{n=1}^{\infty}a_nnx^n+\sum_{n=0}^{\infty} a_n x^n=0\]okay next order of business is what you tried to do earlier; get all indices to start at the same n value the only problem child here is which term?
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
so what do we do to fix it?
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
you have to be a bit clever here...
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
\[ xa_n +\sum_{n=0}^\infty a_n(n+1)x^n \]
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
did you try to do an index shift?
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
it looks like you combined the idea of the index shift with stripping out the first term
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
oh my...let's try this again
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
it is a bit tricky so let me give you a hint...
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
you cannot do another index shift. If you do you will change the exponent on x, which means we are screwed again, so we need to do something else to get this to start from n=0
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
we can't really strip out a term because we don't have this series defined for n<1, but ask yourself, if it was defined at n=0, what would the first term be?
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
just have the addition of an a_n up front like in the last example?
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
no, answer my question above and you will see why...
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
no, be more careful
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
what would be the first term of\[\sum_{n=0}^\infty a_n nx^n\]
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
\[0+\sum_{n=1}^\infty a_n nx^n=\sum_{n=0}^\infty a_n nx^n\]so we actually didn't need to change anything; the first term we "stripped out" was a zero anyway! It makes no difference.
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
so we can go ahead and just change the starting point of that series without further manipulation
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
always watch for that, it comes in handy
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
oh ok uhm.....I'll revisit the last example later, because the I'm not convinced that we needed to do a index shift (or was it stripping of a term)
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
first to get the x^n on all terms we did an index shift (where we add or subtract from each n in the summand) next to get all the starting points for each series the same, we strip out terms/or do the trick we just did they are different tricks, and the order in which you do them is imortant
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
\[\sum_{n=0}^{\infty} (n+2)(n+1)a_{n+2}x^{n}+\sum_{n=0}^{\infty}a_nnx^n+\sum_{n=0}^{\infty} a_n x^n=0\]
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
it is important for you to try to understand why the index shift requires you to add/subtract each n in the summand while stripping out a term does not
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
good, so now make it all into one summation...
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
\[\sum_{n=0}^\infty x^n[(n+2)(n+1)a_{n+2}+a_nn+a_n]=0\]
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
nice, so what will you set to zero to find the recurrence relation?
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
(I am going to type minimally now as I am eating dinner at the same time :)
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
food comes first in my book... :P Ok let's see here...
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
I can eat and respond with "yes" or "no" pretty well, no worries :)
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
\[(n+2)(n+1)a_{n+2}+a_nn+a_n=0\implies a_{n+2}=\frac{a_nna_n}{(n+2)(n+1)}\]
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
si, pero si quieras podrias factor la \(a_n\)
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
ok \[a_{n+2}=a_n\frac{n1}{(n+2)(n+1)}\]
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
and now comes the tedious part that you can do while I eat plug in n=0,1,2,3 and check what results from that recurrence relation until you notice a pattern like last time so just cranks it out; plug in n=0,n=1,n=2 and if a_2 can be written in terms of a_0 or something like last time, do it. That is how we find the pattern.
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
go ahead and list out the first 5 or so and let's see if we see a relation
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
\[n=1:a_3=0\] \[n=2:a_4=a_2\frac{1}{12}\] \[n=3:a_5=a_3\frac{2}{20}\] \[n=4:a_6=a_4\frac{3}{30}\] \[n=5:a_7=a_5\frac{4}{42}\]
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
what about n=0 ? we need to start from n=0 since our series starts from n=0 this time (last time it started from n=1)
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
\[n=0:a_2=a_0\frac{1}{2}\]
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
\[n=0:a_2=a_0\frac{1}{2}\]
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
torta de pollo=chicken sandwich on a french sort of bread
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
oh yeah...the pattern
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
so you did well in your subs I think, but now you must try to always sub all the way back as far as you can..
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
yeah, the pattern write each on as some finite number of unknowns, in this case everything in terms of \(a_0\) and \(a_3\)
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
in the previous example you seemed to have simply switched the \[a_{odd}\] to zero and the \[a_{even}\] to \[a_0\]
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
take you're time...i'm gonna watch a few more mins of my episode
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
it's not different, it's the same as last time odd subscript gives zero even has a pattern reducible to a_0
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
okay I got a fairly simple answer, let me see if I can confirm it...
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
stupid wolfram is of no help :/
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
\[n=0:a_2=a_0\frac{1}{2}=\frac{a_0}{2}\] \[n=1:a_3=0\] \[n=2:a_4=a_2\frac{1}{12}=\frac{a_0}{2(3!)}\] \[n=3:a_5=0\] \[n=4:a_6=a_4\frac{3}{30}=\frac{a_0}{5(3!)}\] \[n=5:a_7=0\]
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
almost, I actually realized that I made a mistake because I think I see a mistake from you so you did as well as me I'd say :D
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
let me try on paper again...
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
here I think I can demonstrate the benefit of not multiplying out terms and writing things overexplicitly to find patterns...
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
\[n=0:a_2=\frac{a_0}{2\cdot 1}\]\[n=2:a_4\frac{a_2}{4\cdot 3}=\frac{a_0}{4\cdot3\cdot2\cdot 1}\]now reevaluate the pattern
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
\[n=0:a_2=\frac{a_0}{2\cdot 1}=\frac{a_0}{2!}\] \[n=2:a_4\frac{a_2}{4\cdot 3}=\frac{a_0}{4\cdot3\cdot2\cdot 1}=\frac{a_0}{4!}\]
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
so try to do the same thing as last time name all even numbers and make a new series with index k
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
then represent the coefficients and plug that in for \(a_n\)
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
n=2k \[a_n=a_{2k}=\frac{a_0}{k!}\]
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
hhm...you know I thought about 2k!, but wouldn't that me too much? \[a_n=a_{2k}=\frac{a_0}{2k!}\]
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
well that is the answer I get... so unless you have a way of confirming it apart from this I say that is correct btw you should put the (2k)! with parentheses to indicate that it is not 2(k!)
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
\[a_n=a_{2k}=\frac{a_0}{(2k)!}\]
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
yeah, the sub it into the original guess for y
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
\[y=\sum_{n=0}^\infty a_nx^n=a_0\sum_{k=0}^\infty \frac{x^{2k}}{(2k)!}\]
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
...and that's what I got :D I sure think it's right as it comes out quite nicely, as opposed to wolf's answer http://www.wolframalpha.com/input/?i=y''%2Bxy'%2By%3D0&t=crmtb01
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
I forgot, what did you intend to study? Electrical Engineering or Mathematics?
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
it definitely has the e^(x^2) think in it though just by glancing, so I am even more confident now
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
Electrical engineering, though I admit I vacillate at times We'll seee how I feel after a year in uni
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
oh because the tayor expansion of e^(ax^2) always has x^(2k)/(2k)! in it (try the expansion yourself and see why, it's not that hard if you remember Taylor series)
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
don't have the brain cells left to do so, but I trust your reasoning
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
haha, well thanks. I hope I'm right. Nos vemos!
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
my spanish sucks as you might be able to tell
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
Nos vemos ~ we'll see each other
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
we've made a mistake
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
\[n=2:a_4\frac{a_2}{4\cdot 3}=\frac{a_0}{4\cdot3\cdot2\cdot 1}=\frac{a_0}{4!}\] is 24 not 12...so the pattern is probably incorrect too
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
and why should 24 be 12 exactly?
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
because we need 12 4*3 is 12
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
I was doing another problem an I got as far as the recurrent relationship and it was almost identical to this one
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
\[a_{n+2}=a_n\frac{n1}{(n+2)(n+1)}\]
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
you wrote\[n=0:a_2=a_0\frac{1}{2}=\frac{a_0}{2}\]\[n=2:a_4=a_2\frac{1}{12}=\frac{a_0}{2(3!)}\]but\[a_2=\frac{a_0}2\]and\[\frac1{12}=\frac1{2(3!)}\]so 1)\[a_4=a_2\frac1{2(3!)}=\frac{a_0}2\cdot\frac1{2(3!)}=\frac{a_0}{4!}\]and that minus sign you put that minus sign does not belong
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
I did not pick up on the\[ a_2=\frac{a_0}{2}\] idea...i just randomly subbed a_0 for a_2 :/
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
I'll finish y''=y tomorrow...I got it except for the parts stated above
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
once you find a base formula in terms of \(a_0\) or whatever the lowest you can express is, milk it to find the pattern
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
and like I say, sometime the writing out each term, even if it's just multiplying by one, helps make the pattern obvious g'night then!
 one year ago

mahmit2012Best ResponseYou've already chosen the best response.0
dw:1348252850986:dw
 one year ago

mahmit2012Best ResponseYou've already chosen the best response.0
dw:1348252918564:dw
 one year ago

mahmit2012Best ResponseYou've already chosen the best response.0
dw:1348253019736:dw
 one year ago

mahmit2012Best ResponseYou've already chosen the best response.0
dw:1348253097421:dw
 one year ago

mahmit2012Best ResponseYou've already chosen the best response.0
dw:1348253165705:dw
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
@TuringTest would this statement about striping a term be true (generally speaking) \[0+a_1+\sum_{n=2}^\infty a_n nx^n\]
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
oh no, you dropped an x
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
at n=1 the exponent on x is 1
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
\[\sum_{n=0}^\infty a_nnx^n=a_0(0)x^0+\sum_{n=1}^\infty a_nnx^n=0+a_1(1)x^1+\sum_{n=2}^\infty a_nx^n\]\[=a_1x+\sum_{n=2}^\infty a_nx^n\]
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
easier example, stripping out terms for the sum of the first n integers\[\sum_{i=1}^n i=1+\sum_{i=2}^n=1+2+\sum_{i=3}^n i=1+2+3+\sum_{i=4}^n i\]
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
thought so, probably should have done that a while ago, hehe :P
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
Makes perfect sense now, no doubt in my mind anymore. Haha, the brain exercise was good though
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
Awesome, always a pleasure :D
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
@TuringTest shouldn't \[a_4 \] have a minus sign? I know you said it shouldn't but \[a_2=\frac{a_0}{2}\] and now we have \[a_4=a_2\frac{1}{12}=\frac{a_0}{24}\]
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
hm... I guess you may be right
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
^^^^my way of saying: yes!
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
so it's an alternating series? for 2k+1?
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
sorry I was doublechecking this whole time yeah it alternates, so you know the only change we're going to make?
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
let's see if I remember, can I cheat?
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
cheating would be having me tell you, any way you wanna figure out how to make a series change sign every other term is up to you :)
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
cheating would be using a the google search engine, but let's see if I can figure it out on my own without google.com
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
argh! \[\sum_{n=0}^{\infty}x^n=0,x,x^2,etc.\] \[\sum_{n=0}^{\infty}(1)x^n=0,x,x^2\]
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
what can we do to 1 to make it even?
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
very awesome!!!!!!!!!!!!
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
hey I got 240 for a_6
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
you love multiplying the numbers out, I just call it 6!
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
haha whatever I'm not even checking if you are having a problem with the constants again give me a min and I'll brb again to help
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
sounds good... sorry don't mean to be nitpicking :S
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
oh no, I;m glad you pointed it out, but.... must... eat
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
dude go for it LOL I'll see if I can figure this pattern out
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
I think this is incorrect \[a_{n+2}=a_n\frac{n1}{(n+2)(n+1)}\] because \[a_{n+2}=\frac{a_nn1}{(n+2)(n+1)}\neqa_n\frac{n1}{(n+2)(n+1)}\] It should be \[a_n\frac{(n+1)}{(n+2)(n+1)}\]
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
\[\frac{a_n}{n+2}\]
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
\[a_{n+2}=\frac{a_n}{n+2}\]
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
\[(n+2)(n+1)a_{n+2}+a_nn+a_n=0\implies a_{n+2}=\frac{a_nna_n}{(n+2)(n+1)}\]is what you wrote, let's see if that is correct...
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
\[(n+2)(n+1)a_{n+2}+a_n(n+1)=0\implies a_{n+2}=a_n\frac{n+1}{(n+2)(n+1)}\]aw crap you're right again
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
well, being right that we were wrong is always a little bittersweet, haha gotta do the pattern over, but at least it's only\[a_{n+2}=\frac{a_n}{n+2}\]which should be easier I would think
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
Yeah, let's see if I can figure the pattern out... (btw I did yoga today...feeling more alert, or it could be plain coincidence =)
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
well you're catching all my mistakes, so it seems to be working :)
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
the pattern seems harder to write concisely for me now... though it seems obvious
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
\[a_2=\frac{a_0}{2}=\frac{a_0}{2!}\] \[a_3=\frac{a_1}{3}\] \[a_4=\frac{a_2}{4}=\frac{a_0}{3!}\] \[a_5=\frac{a_3}{5}\] \[a_6=\frac{a_4}{6}=\frac{a_0}{36}\] what's withe the 36?
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
it's like the odds get cut out of the factorials and I'm trying to figure out how to write that same for the evens
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
yeah it's quite strange, but the formula is correct, right? could we have still made an error there?
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
a mistake in the recursion relation? we could have... to be safe we ought to take it from the top, but I can't do that, I have to leave...
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
sure I'll start from the beginning and post it in a new post...this one is getting quite long
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
good idea I am thinking\[a_{2k}=(1)^k\frac{a_0}{2^kk!}\]and do not yet have an expression for the odds
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
oh ok, I'll see ya trmw then
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
hasta luego mi amigo!
 one year ago
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