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

Mathematics
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yep, now what's first?
changing the sums so that they all begin with n=2 changing the last sum from n=0 to n=1 would be \[a_1+\sum_{n=1}^\infty a_nx^n\]
no we don't strip out the terms yet, we just do an index shift

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the first order of business is to get all the x's to be to the nth power through using index shifts
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?
you mean on \[y''=\sum_{n=2}^{\infty} n(n-1)a_nx^{n-2}\]
yes, we need to start from a different n to get x^n through an index shift
so I'm shifting the starting point up by 2 so I have to subtract 2 from {n=2-2} \[-\sum_{n=0}^\infty n(n-1)a_nx^n\] negative to make up for the negative?
nooooo
hold on
\[\sum_{n=0}^\infty n(n+1)a_nx^n\]
nice :)
still not right
because that would give us zero
no but closer
you didn't add 2 to every n
\[\sum_{n=0}^\infty (n+2)(n+1)a_nx^n\]
and I do mean \(every\) n (subscripts included)
again better, but...^
\[\sum_{n=0}^\infty (n+2)(n+1)a_{n+2}x^n \]
can I add them?
now that, I do believe, is correct :)
or multiply
or just leave as is?
add what? oh that? I wouldn't yet
ok
yeah leave as is asd now rewrite what you have
and*
the whole expression please I'm lazy
\[\sum_{n=0}^{\infty} (n+2)(n-1)a_{n+2}x^{n}+\sum_{n=1}^{\infty}a_nnx^n+\sum_{n=0}^{\infty} a_n x^n=0\]
\[\sum_{n=0}^{\infty} (n+2)(n+1)a_{n+2}x^{n}+\sum_{n=1}^{\infty}a_nnx^n+\sum_{n=0}^{\infty} a_n x^n=0\]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?
the middle
so what do we do to fix it?
you have to be a bit clever here...
\[ xa_n +\sum_{n=0}^\infty a_n(n+1)x^n \]
did you try to do an index shift?
it looks like you combined the idea of the index shift with stripping out the first term
oh my...let's try this again
it is a bit tricky so let me give you a hint...
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
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?
just have the addition of an a_n up front like in the last example?
no, answer my question above and you will see why...
a 1
times a_n
no, be more careful
what would be the first term of\[\sum_{n=0}^\infty a_n nx^n\]
zero
right, so...
\[0+\sum_{n=1}^\infty a_n nx^n=\sum_{n=0}^\infty a_n nx^n\]so we actually didn't need to change anything; the first term we "stripped out" was a zero anyway! It makes no difference.
so we can go ahead and just change the starting point of that series without further manipulation
always watch for that, it comes in handy
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)
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
ok
\[\sum_{n=0}^{\infty} (n+2)(n+1)a_{n+2}x^{n}+\sum_{n=0}^{\infty}a_nnx^n+\sum_{n=0}^{\infty} a_n x^n=0\]
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
good, so now make it all into one summation...
\[\sum_{n=0}^\infty x^n[(n+2)(n+1)a_{n+2}+a_nn+a_n]=0\]
nice, so what will you set to zero to find the recurrence relation?
(I am going to type minimally now as I am eating dinner at the same time :)
food comes first in my book... :P Ok let's see here...
I can eat and respond with "yes" or "no" pretty well, no worries :)
\[(n+2)(n+1)a_{n+2}+a_nn+a_n=0\implies a_{n+2}=-\frac{a_nn-a_n}{(n+2)(n+1)}\]
si?
si, pero si quieras podrias factor la \(-a_n\)
ok \[a_{n+2}=-a_n\frac{n-1}{(n+2)(n+1)}\]
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.
go ahead and list out the first 5 or so and let's see if we see a relation
\[n=1:a_3=0\] \[n=2:a_4=-a_2\frac{1}{12}\] \[n=3:a_5=-a_3\frac{2}{20}\] \[n=4:a_6=-a_4\frac{3}{30}\] \[n=5:a_7=-a_5\frac{4}{42}\]
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)
\[n=0:a_2=-a_0\frac{-1}{2}\]
\[n=0:a_2=a_0\frac{1}{2}\]
whatcha eating?
torta de pollo=chicken sandwich on a french sort of bread
con chipotle
nice
oh yeah...the pattern
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..
yeah, the pattern write each on as some finite number of unknowns, in this case everything in terms of \(a_0\) and \(a_3\)
each one*
in the previous example you seemed to have simply switched the \[a_{odd}\] to zero and the \[a_{even}\] to \[a_0\]
take you're time...i'm gonna watch a few more mins of my episode
it's not different, it's the same as last time odd subscript gives zero even has a pattern reducible to a_0
okay I got a fairly simple answer, let me see if I can confirm it...
stupid wolfram is of no help :/
\[n=0:a_2=a_0\frac{1}{2}=\frac{a_0}{2}\] \[n=1:a_3=0\] \[n=2:a_4=-a_2\frac{1}{12}=\frac{a_0}{2(3!)}\] \[n=3:a_5=0\] \[n=4:a_6=-a_4\frac{3}{30}=\frac{a_0}{5(3!)}\] \[n=5:a_7=0\]
how did I do?
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
let me try on paper again...
here I think I can demonstrate the benefit of not multiplying out terms and writing things over-explicitly to find patterns...
ok
\[n=0:a_2=\frac{a_0}{2\cdot 1}\]\[n=2:a_4\frac{a_2}{4\cdot 3}=\frac{a_0}{4\cdot3\cdot2\cdot 1}\]now reevaluate the pattern
\[n=0:a_2=\frac{a_0}{2\cdot 1}=\frac{a_0}{2!}\] \[n=2:a_4\frac{a_2}{4\cdot 3}=\frac{a_0}{4\cdot3\cdot2\cdot 1}=\frac{a_0}{4!}\]
so try to do the same thing as last time name all even numbers and make a new series with index k
then represent the coefficients and plug that in for \(a_n\)
n=2k \[a_n=a_{2k}=\frac{a_0}{k!}\]
is it over k! ?
hhm...you know I thought about 2k!, but wouldn't that me too much? \[a_n=a_{2k}=\frac{a_0}{2k!}\]
*be
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!)
\[a_n=a_{2k}=\frac{a_0}{(2k)!}\]
yeah, the sub it into the original guess for y
then*
\[y=\sum_{n=0}^\infty a_nx^n=a_0\sum_{k=0}^\infty \frac{x^{2k}}{(2k)!}\]
...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
I forgot, what did you intend to study? Electrical Engineering or Mathematics?
it definitely has the e^(-x^2) think in it though just by glancing, so I am even more confident now
why the e^?
Electrical engineering, though I admit I vacillate at times We'll seee how I feel after a year in uni
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)
don't have the brain cells left to do so, but I trust your reasoning
haha, well thanks. I hope I'm right. Nos vemos!
see ya!
my spanish sucks as you might be able to tell
Nos vemos ~ we'll see each other
I was right!
sweet :)
we've made a mistake
\[n=2:a_4\frac{a_2}{4\cdot 3}=\frac{a_0}{4\cdot3\cdot2\cdot 1}=\frac{a_0}{4!}\] is 24 not 12...so the pattern is probably incorrect too
let's see...
and why should 24 be 12 exactly?
because we need 12 4*3 is 12
I was doing another problem an I got as far as the recurrent relationship and it was almost identical to this one
you dropped a 2...
where?
\[a_{n+2}=-a_n\frac{n-1}{(n+2)(n+1)}\]
oh I see
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
I did not pick up on the\[ a_2=\frac{a_0}{2}\] idea...i just randomly subbed a_0 for a_2 :/
I'll finish y''=y tomorrow...I got it except for the parts stated above
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
ok
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!
g'night
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@TuringTest would this statement about striping a term be true (generally speaking) \[0+a_1+\sum_{n=2}^\infty a_n nx^n\]
yep, exactly
oh no, you dropped an x
at n=1 the exponent on x is 1
\[\sum_{n=0}^\infty a_nnx^n=a_0(0)x^0+\sum_{n=1}^\infty a_nnx^n=0+a_1(1)x^1+\sum_{n=2}^\infty a_nx^n\]\[=a_1x+\sum_{n=2}^\infty a_nx^n\]
easier example, stripping out terms for the sum of the first n integers\[\sum_{i=1}^n i=1+\sum_{i=2}^n=1+2+\sum_{i=3}^n i=1+2+3+\sum_{i=4}^n i\]
much better
thought so, probably should have done that a while ago, hehe :P
Makes perfect sense now, no doubt in my mind anymore. Haha, the brain exercise was good though
cool thanks !
Awesome, always a pleasure :D
@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 \[a_4=-a_2\frac{1}{12}=-\frac{a_0}{24}\]
hm... I guess you may be right
yuuss!!!!
^^^^my way of saying: yes!
so it's an alternating series? for 2k+1?
I mean 2k
sorry I was double-checking this whole time yeah it alternates, so you know the only change we're going to make?
let's see if I remember, can I cheat?
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 :)
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
good idea brb
k
argh! \[\sum_{n=0}^{\infty}x^n=0,x,x^2,etc.\] \[\sum_{n=0}^{\infty}(-1)x^n=0,-x,-x^2\]
what can we do to -1 to make it even?
sorry, positive*
\[(-1)^n\]
yes
AWESOME!
very awesome!!!!!!!!!!!!
hey I got 240 for a_6
you love multiplying the numbers out, I just call it 6!
no that's 720
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
sounds good... sorry don't mean to be nitpicking :S
oh no, I;m glad you pointed it out, but.... must... eat
dude go for it LOL I'll see if I can figure this pattern out
I think this is incorrect \[a_{n+2}=-a_n\frac{n-1}{(n+2)(n+1)}\] because \[a_{n+2}=\frac{-a_nn-1}{(n+2)(n+1)}\neq-a_n\frac{n-1}{(n+2)(n+1)}\] It should be \[-a_n\frac{(n+1)}{(n+2)(n+1)}\]
\[-\frac{a_n}{n+2}\]
\[a_{n+2}=-\frac{a_n}{n+2}\]
\[(n+2)(n+1)a_{n+2}+a_nn+a_n=0\implies a_{n+2}=-\frac{a_nn-a_n}{(n+2)(n+1)}\]is what you wrote, let's see if that is correct...
\[(n+2)(n+1)a_{n+2}+a_n(n+1)=0\implies a_{n+2}=-a_n\frac{n+1}{(n+2)(n+1)}\]aw crap you're right again
yuuus!!!
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
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 =)
well you're catching all my mistakes, so it seems to be working :)
the pattern seems harder to write concisely for me now... though it seems obvious
\[a_2=-\frac{a_0}{2}=-\frac{a_0}{2!}\] \[a_3=-\frac{a_1}{3}\] \[a_4=-\frac{a_2}{4}=\frac{a_0}{3!}\] \[a_5=-\frac{a_3}{5}\] \[a_6=-\frac{a_4}{6}=-\frac{a_0}{36}\] what's withe the 36?
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
yeah it's quite strange, but the formula is correct, right? could we have still made an error there?
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...
sure I'll start from the beginning and post it in a new post...this one is getting quite long
good idea I am thinking\[a_{2k}=(-1)^k\frac{a_0}{2^kk!}\]and do not yet have an expression for the odds
oh ok, I'll see ya trmw then
hasta luego
hasta luego mi amigo!

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