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MathSofiya
I'm trying to find a pattern \[a_{n+2}=\frac{a_n}{(n+2)(n+1)}\] \[n=0:a_2=\frac{a_0}{2!}\] \[n=2:a_4=\frac{a_0}{4!}\] but that pattern changes because the next term is not \[\frac{a_0}{6!}\] \[n=4:a_6=\frac{a_0}{2\cdot 30}=\]
?? \[a _{4} = \frac{a _{0} }{ 4! } \]
this is the one we did yesterday? wasn't it\[a_{n+2}=-a_n\frac{1-n}{(n+1)(n+2)}\]?
no this is new problem \[y''=y\]
I'll write out what I have so far
I'll be able to able to help in about 15-20 min
\[\sum_{n=2}^\infty n(n-1)a_nx^{n-2}-\sum_{n=0}^\infty a_nx^n=0\]
@MathSofiya how exactly is the pattern changing?
let's establish that we have the right pattern first... I feel like you may have missed something continue plz @MathSofiya so far so good
\[\sum_{n=0}^\infty (n+2)(n+1)a_{n+2}x^n-\sum_{n=0}^\infty a_nx^n=0\]
so yeah, right pattern
\[\sum_{n=0}^\infty x^n\left[(n+2)(n+1)-a_n\right]=0\]
\[\sum_{n=0}^\infty x^n\left[(n+2)(n+1)a_{n+2}-a_n\right]=0\]
\[a_{n+2}=\frac{a_n}{(n+1)(n+2)}\]so you were right, just plug in each n and write out \(everything\) explicitly, even multiplication by 1
\[n=0:a_{0+2}=a_2=\frac {a_0}{2}\] \[n=1:a_{1+2}=a_3=\frac {a_1}{6}\] \[n=2:a_{2+2}=a_4=\frac {a_2}{12}\] \[n=3:a_{3+2}=a_5=\frac {a_3}{20}\] \[n=4:a_{4+2}=a_6=\frac {a_4}{30}\] \[n=5:a_{5+2}=a_7=\frac {a_5}{42}\]
so we established the relationship between \[a_2=\frac{a_0}{2}\] but how do I continue this to the next line? how do I change a_2 to a_0?
ooohh no need to because the next line is going to be zero
you are not writing it out explicitly enough to see the pattern reduce everything to terms of either a_0 or a_1 the two constants we can represent no more simply
now is the next line zero?
well it was yesterday LOL ....Ok I'll write everything out
\[n=0:a_{0+2}=a_2=\frac {a_0}{2\cdot1}\]\[n=1:a_3=\frac{a_1}{3\cdot2}\]write everything this way, with each number on the bottom specified
yesterday we had n-1 in the top, that's why
Haha I'm exercising while you crunch the numbers :P
\[n=0:a_{0+2}=a_2=\frac {a_0}{2}\] \[n=1:a_{1+2}=a_3=\frac {a_1}{6}=\frac{a_1}{2\cdot 3}\] \[n=2:a_{2+2}=a_4=\frac {a_2}{12}=\frac{a_0}{2\cdot2\cdot3}\] \[n=3:a_{3+2}=a_5=\frac {a_3}{20}=\frac{a_3}{10\cdot2}=\frac{a_1}{2\cdot2\cdot2\cdot3\cdot5}\] \[n=4:a_{4+2}=a_6=\frac {a_4}{30}=\frac{a_0}{2\cdot2\cdot2\cdot3\cdot5}\] \[n=5:a_{5+2}=a_7=\frac {a_5}{42}=\frac{a_1}{2\cdot2\cdot2\cdot3\cdot5\cdot42}\]
n=4 is wrong oh well...but I see a pattern now. How do you workout and do math?
push ups, then run over and check your work real quick :)
haha! That's what I call commitment!
give me a second to do the problem on paper. It's kinda tedious to type it...one second
\[a_n=\frac{a_n}{(n+1)(n+2)}\]\[n=0:a_{0+2}=a_2=\frac {a_0}{2\cdot1}\]\[n=1:a_3=\frac{a_1}{3\cdot2}\]\[n=2:a_4=\frac {a_2}{4\cdot3}=\frac{a_0}{4\cdot3\cdot2\cdot1}\]\[n=3:a_5=\frac {a_3}{5\cdot4}=\frac{a_1}{5\cdot4\cdot3\cdot2}\]
write out each number in the denom without simplifying every time
I cannot stress that enough to find patterns
\[n=0:a_{0+2}=a_2=\frac {a_0}{2}\] \[n=1:a_{1+2}=a_3=\frac {a_1}{6}=\frac{a_1}{3\cdot 2}\] \[n=2:a_{2+2}=a_4=\frac {a_2}{12}=\frac{a_0}{4\cdot3\cdot2\cdot1}\] \[n=3:a_{3+2}=a_5=\frac {a_3}{20}=\frac{a_3}{10\cdot2}=\frac{a_1}{5\cdot4\cdot3\cdot2\cdot1}\] Ok I'm finally convinced...had to calculate like 5 times that the denominator is actually 120
good, but more importantly what is/are the general pattern(s)?
again, there are two; one for even n and one for odd n
sorry internet is super slow. I'm gonna restart my computer brb
I'm getting in the shower soon... so what are the two constants that every other constant \(a_n\) can be written in terms of ?
I wanna say \[(n+2)!\] and then sub in n=2k
wait...you go shower, and let me work on this
no, and that would only be evens, and needs a base constant like \(a_0\) write it as\[a_n=?\]and remember this time we do have odd terms
allow me to decipher it...
The denominator for both even and odds is the following...still working on the numerator \[\frac{?}{(n+2)!}\]
look at the difference between constants with an even and odd subscript what is the main difference you see>?
one will always be zero and the other will always be 1. I have come to that conclusion, but now I have to find a way to write it mathematically...I'm gonna cheat and look at yesterdays work
no wait, you can do it... let me give another hint...
odd are reducible to being in terms of the constant a_1 evens are reducible to being in terms of a_0 and for the denominator for any constant \(a_n\) what is it? look at the denom for a_2, a_3, a_4, what are they in terms of n ?
don't get confused with the n=(whatever) thing I mean what is the relation between the n in the subscript \(a_n\) and the number in the denominator?
hold on I didn't write anything yet
sorry It just took me 3 mins to load this page
no rush, still haven't showered yet
When the numerator is \[a_0\] the denominator is even When the numerator is \[a_1\] the denominator is odd
by odd I mean 5! 3! and so on
exactly, and all even numbers can be written how?
\[a_{2k}\] and \[a_{2k+1}\]
so then what is our formulas? we need one for all \(a_{2k}\) and another for all \(a_2k+1}\)
right, we need a formula for each one.. what are they ? that's probably the hardest part of the problem in my opinion is finding the pattern
I would like to put a_{2k} in the numerator and (n+2) in the numerator, but that wouldn't be right. So I have write it in one formula to account for both the even and odd?
I can't write two separate formulas correct? I've gotta write one?
everything should be written in terms of either a_0 and k, or a_1 and k n is now out of the poicture completely
gtg...I know this is the last part, and shouldn't take too long, but I gtg to the gym. i'll be back at 7:30 8pm
cool, health makes one think more clearly I believe see ya
LOL I hope soo! see ya
where is Alan Turing?
I'm reviewing the previous problems we did...and there might be some errors. I not convinced with the way that the final answer was written in k's
Allow me to restore your faith, though there are other manners of making the notation I'm sure... Paul to the rescue again: http://tutorial.math.lamar.edu/Classes/DE/SeriesSolutions.aspx Notice that since the constants become different depending on even-odd subscripts for many answers we divide the subscripts in the recursion formula for them into terms of 2k and 2k+1. This particular problem you are doing is almost identical to example 1 one this page, with the exception of the minus between y'' and y exchanged for a plus (reading this example you will probably discover the pattern I was asking you to look for). Notice that it is in fact essential that we change the terms from n to k, since we can get no single summation formula in terms of n since the coefficients depend on whether n is odd n. This is not true for all problems with series solutions, but when we need to distinguish between even and odd n the only way to do this is to put them as two separate series in terms of k. Read example 1 carefully to understand the intricacies of the process, and the why we put n in terms of k. Sorry I never returned last night, I was teaching political philosophy. Look forward to finishing the problem with you today if possible. Tag me when you are ready to continue :)
I'll be back in like 10 mins, gotta start laundry
I can't find my laundry card! Ok so I was trying to rewrite the first problem, and got to the part where 1. all even n#'s =0, or all odd "a" subscripts=0 2. We've gotten a pattern that we called \[a_n=\frac{a_0}{2^n(n!)}\] but if we call them "n" for now wouldn't that conflict because Let's say for n=5 \[\frac{a_0}{2^3(3!)}\] which is not true (I'm referring to y'=xy, the first problem)
Well it's what we called the pattern initially before writing it in terms of k's
ok found the laundry card brb
The pattern is not in terms of n like that, if it were it would be\[a_n=\frac{a_0}{2^{n/2}(\frac n2!)}\]for even n subscripts of a, and\[a_n=0\] for all odd n subscripts of a. I think a major problem here is that we are getting confused with the n in the \(n=...\) that we are using to plug in to find the particular constant \(a_n\), and the n in the \(a_n\) in the constant. Those two n's are not equal! What exactly is your argument that in the last problem you are referring too we do no not have that\[n=5:a_6=\frac{a_4}6=\frac{a_0}{2\cdot4\cdot6}=\frac{a_0}{2^3(3!)}\]??
Ok that makes sense. I thought the n's were the same. So the n's are just a variable to hold place until we change the argument to k's I guess. wait, what do you mean by the last sentence? so the n=5 statement above is not accurate?
yes it is accurate, plugging in n=5 into the recursion formula\[a_{n+1}=\frac{a_{n-1}}{n+1}\]gives\[n=5:a_6=\frac{a_4}6=\frac{a_0}{2\cdot4\cdot6}=\frac{a_0}{2^3(3!)}\]the \(n=...\) is just what we use to get an expression relating two other constants \(a_{n+ p}\)and \(a_{n+ q}\) where p and q are some number (in this case above p=-1 and q=1) so the n we plug in to test is \(not\) the same as the n in the subscripts, because the subscripts in our recursion relation are usually not just \(a_n\) (the same n as we are plugging in for the test) but really \(a_{n+p}\)
We want our expression for each constant in terms of its \(subscript\), not the \(n=...\) we use to test the relation, hence we divide the subscripts into even and odd to make the pattern more clear. Once we notice that by doing that we can get meaningful expressions for each constant a in terms of the subscript (however we write) that is what we use...
in this case by dividing the subscripts into those of the form 2k and those of the form 2k+1 we get a new way to write each constant in terms of k
...and whatever base constant we can't determine, like \(a_0\) which we do not know since we are not given the initial conditions
ok that makes more sense now. Let's see if I can apply it to this problem now.
go for it, I'm gonna get a few empanadas for breakfast :) (sortof like Mexican hot-pockets :) brb
Let's see here.... \[a_2=\frac{a_0}{2!}\] \[a_3=\frac{a_1}{3!}\] \[a_4=\frac{a_0}{4!}\] \[a_5=\frac{a_1}{5!}\] \[a_2k=\frac{a_0}{k!}\] my way of saying if the subscript is even...
\[a_{2k+1}=\frac{a_1}{k!}\]
is it? what is\(a_3\) ?
oh my 2k+1=3 2k=2 k=1
your patients with me is truly astonishing. I would have strangled the student.
haha, but I haven't done this in a long time either, so it's a good review for me too.
\[a_n=\frac{a_n}{(n+1)(n+2)}\]\[n=0:a_2=\frac {a_0}{2}=\frac{a_0}{2!}\]\[n=1:a_3=\frac {a_1}{6}=\frac{a_1}{3\cdot 2}=\frac{a_1}{3!}\]\[n=2:a_4=\frac {a_2}{12}=\frac{a_0}{4\cdot3\cdot2\cdot1}=\frac{a_0}{4!}\]\[n=3:a_5=\frac {a_3}{20}=\frac{a_3}{10\cdot2}=\frac{a_1}{5\cdot4\cdot3\cdot2\cdot1}=\frac{a_1}{5!}\]now hopefully you have already noticed that the denominator on each \(a_n\) is \(n!\)\[a_n=\frac{a_0}{n!}\]if n is even, and\[a_n=\frac{a_1}{n!}\]if n is odd given that, simply rewrite the subscripts as 2k or 2k+1 depending on whether they are even or odd, and then sub in for n in the general expressions
yep, I honestly saw that relationship all along, I just didn't know how to express that in k's \[a_{2k}=\frac{a_0}{2k}\] \[a_{2k+1}=\frac{a_0}{2k+1}\] sorry I kinda feel silly now
you forgot the factorials ;)
LOL ok let me rewrite them
and you forgot that for odd subscripts it is a_1 in the numerator
\[a_{2k}=\frac{a_0}{(2k)!}\] \a_{2k+1}=\frac{a_1}{2k+1}[\] I didn't forgot...I just checking if you were paying attention :P
\[a_{2k+1}=\frac{a_1}{2k+1}\]
\[a_{2k+1}=\frac{a_1}{(2k+1)!}\]
oh wait there is more...
Much better :) I wanna make a subtle point about the possible values of k we are talking about, but I will save it till the end so as not to confuse you.
\[y=\sum_{n=0}^\infty a_nx^n=a_0\sum_{k=0}^\infty\frac{x^{2k}}{(2k)!} \] \[y=\sum_{n=0}^\infty a_nx^n=a_1\sum_{k=0}^\infty\frac{x^{2k+1}}{(2k+1)!} \]
you've got the right idea, but we need to use superposition to get the full answer
if we accept only one answer or the other we are excluding either all the even or odd exponents, so how can we fix that?
put them into one formula somehow
oh y and c_1 and so on and so forth...hold on
\[c_1\sum_{k=0}^\infty\frac{x^{2k}}{(2k)!}+c_2\sum_{k=0}^\infty\frac{x^{2k+1}}{(2k+1)!}=0\] huh?
gosh the sums probably shouldn't be there...
no I was just saying that the c1 you are used to is called a0 in this case, so...
\[y=a_0\sum_{k=0}^\infty\frac{x^{2k}}{(2k)!}+a_1\sum_{k=0}^\infty\frac{x^{2k+1}}{(2k+1)!}=0\]
\[y=a_0\sum_{k=0}^\infty\frac{x^{2k}}{(2k)!}+a_1\sum_{k=0}^\infty\frac{x^{2k+1}}{(2k+1)!}\]
I just looked at the example in the book Uhm Ohhh,,, but they don't have to be written in c_1 and c_2 right?
can I put my laundry in the dryer?
nah, constants are constants, you can call them whatever you want but I just wanted to keep our notation consistent, and since they are equivalent to c1 and c2 (constants that depend on the initial conditions) and yes, you have my permission to do your laundry :P
so review what we did, look again at example one one here: http://tutorial.math.lamar.edu/Classes/DE/SeriesSolutions.aspx and tell me if you have any questions
I read it, I don't think I have any questions. 22 hrs later I finally understand it (hopefully I'll be able to recreate it)
awesome! there is, as I said, a subtle point to be made about the possible values of k (as paul mentions) but since it really doesn't affect this problem I'll leave it for you to read, or we'll cross that bridge when we come to it :)
sounds good...I'll do 2 problems on my own from beginning to end and we'll see if I make error or still struggle on some points here and there...
I do encourage you to read the whole page I linked you to though, to see some of the differences encountered in other problems. Good luck!
I will. Thanks! quick question, did you make your own empanadas?
nope, bought 'em at the corner I only cook hamburgers and pasta
what? you've been in Mexica for a little while now! Gotta learn to cook like the locals!
well I make quesadillas, I guess that's a start lol
yep, work your way up from here :P
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