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

This Question is Closed

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1yep, now what's first?

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

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1no we don't strip out the terms yet, we just do an index shift

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

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1the 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?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0you mean on \[y''=\sum_{n=2}^{\infty} n(n1)a_nx^{n2}\]

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so 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?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\sum_{n=0}^\infty n(n+1)a_nx^n\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0because that would give us zero

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1you didn't add 2 to every n

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\sum_{n=0}^\infty (n+2)(n+1)a_nx^n\]

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1and I do mean \(every\) n (subscripts included)

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1again better, but...^

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

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1now that, I do believe, is correct :)

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1add what? oh that? I wouldn't yet

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1yeah leave as is asd now rewrite what you have

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1the whole expression please I'm lazy

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

TuringTest
 4 years ago
Best 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?

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1so what do we do to fix it?

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1you have to be a bit clever here...

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

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1did you try to do an index shift?

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1it looks like you combined the idea of the index shift with stripping out the first term

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0oh my...let's try this again

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1it is a bit tricky so let me give you a hint...

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1you 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

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1we 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?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0just have the addition of an a_n up front like in the last example?

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1no, answer my question above and you will see why...

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

TuringTest
 4 years ago
Best 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.

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1so we can go ahead and just change the starting point of that series without further manipulation

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1always watch for that, it comes in handy

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0oh 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)

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1first 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

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

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1it 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

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1good, so now make it all into one summation...

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

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1nice, so what will you set to zero to find the recurrence relation?

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1(I am going to type minimally now as I am eating dinner at the same time :)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0food comes first in my book... :P Ok let's see here...

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1I can eat and respond with "yes" or "no" pretty well, no worries :)

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

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1si, pero si quieras podrias factor la \(a_n\)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ok \[a_{n+2}=a_n\frac{n1}{(n+2)(n+1)}\]

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1and 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.

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1go ahead and list out the first 5 or so and let's see if we see a relation

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

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1what 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)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[n=0:a_2=a_0\frac{1}{2}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[n=0:a_2=a_0\frac{1}{2}\]

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1torta de pollo=chicken sandwich on a french sort of bread

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0oh yeah...the pattern

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1so 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..

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0in the previous example you seemed to have simply switched the \[a_{odd}\] to zero and the \[a_{even}\] to \[a_0\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0take you're time...i'm gonna watch a few more mins of my episode

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

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1okay I got a fairly simple answer, let me see if I can confirm it...

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1stupid wolfram is of no help :/

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

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1almost, 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

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1let me try on paper again...

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

TuringTest
 4 years ago
Best 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

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

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

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1then represent the coefficients and plug that in for \(a_n\)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0n=2k \[a_n=a_{2k}=\frac{a_0}{k!}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0hhm...you know I thought about 2k!, but wouldn't that me too much? \[a_n=a_{2k}=\frac{a_0}{2k!}\]

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1well 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!)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[a_n=a_{2k}=\frac{a_0}{(2k)!}\]

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1yeah, the sub it into the original guess for y

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[y=\sum_{n=0}^\infty a_nx^n=a_0\sum_{k=0}^\infty \frac{x^{2k}}{(2k)!}\]

TuringTest
 4 years ago
Best 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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I forgot, what did you intend to study? Electrical Engineering or Mathematics?

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

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

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1oh 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)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0don't have the brain cells left to do so, but I trust your reasoning

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1haha, well thanks. I hope I'm right. Nos vemos!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0my spanish sucks as you might be able to tell

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1Nos vemos ~ we'll see each other

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[n=2:a_4\frac{a_2}{4\cdot 3}=\frac{a_0}{4\cdot3\cdot2\cdot 1}=\frac{a_0}{4!}\] is 24 not 12...so the pattern is probably incorrect too

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1and why should 24 be 12 exactly?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0because we need 12 4*3 is 12

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I was doing another problem an I got as far as the recurrent relationship and it was almost identical to this one

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[a_{n+2}=a_n\frac{n1}{(n+2)(n+1)}\]

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1you 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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I did not pick up on the\[ a_2=\frac{a_0}{2}\] idea...i just randomly subbed a_0 for a_2 :/

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I'll finish y''=y tomorrow...I got it except for the parts stated above

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

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1and 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!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1348252850986:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1348252918564:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1348253019736:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1348253097421:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1348253165705:dw

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

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1oh no, you dropped an x

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1at n=1 the exponent on x is 1

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

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

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1thought so, probably should have done that a while ago, hehe :P

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Makes perfect sense now, no doubt in my mind anymore. Haha, the brain exercise was good though

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1Awesome, always a pleasure :D

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@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 \[a_4=a_2\frac{1}{12}=\frac{a_0}{24}\]

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1hm... I guess you may be right

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0^^^^my way of saying: yes!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so it's an alternating series? for 2k+1?

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0let's see if I remember, can I cheat?

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1cheating 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 :)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0cheating would be using a the google search engine, but let's see if I can figure it out on my own without google.com

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

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1what can we do to 1 to make it even?

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1very awesome!!!!!!!!!!!!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0hey I got 240 for a_6

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1you love multiplying the numbers out, I just call it 6!

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1haha 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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0sounds good... sorry don't mean to be nitpicking :S

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1oh no, I;m glad you pointed it out, but.... must... eat

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dude go for it LOL I'll see if I can figure this pattern out

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I 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 \[a_n\frac{(n+1)}{(n+2)(n+1)}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[a_{n+2}=\frac{a_n}{n+2}\]

TuringTest
 4 years ago
Best 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...

TuringTest
 4 years ago
Best 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

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1well, 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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Yeah, let's see if I can figure the pattern out... (btw I did yoga today...feeling more alert, or it could be plain coincidence =)

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1well you're catching all my mistakes, so it seems to be working :)

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1the pattern seems harder to write concisely for me now... though it seems obvious

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[a_2=\frac{a_0}{2}=\frac{a_0}{2!}\] \[a_3=\frac{a_1}{3}\] \[a_4=\frac{a_2}{4}=\frac{a_0}{3!}\] \[a_5=\frac{a_3}{5}\] \[a_6=\frac{a_4}{6}=\frac{a_0}{36}\] what's withe the 36?

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1it'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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yeah it's quite strange, but the formula is correct, right? could we have still made an error there?

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1a 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...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0sure I'll start from the beginning and post it in a new post...this one is getting quite long

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0oh ok, I'll see ya trmw then

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0hasta luego mi amigo!
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