anonymous
  • anonymous
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 \[\frac{a_0}{6!}\] \[n=4:a_6=\frac{a_0}{2\cdot 30}=\]
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
?? \[a _{4} = \frac{a _{0} }{ 4! } \]
TuringTest
  • TuringTest
this is the one we did yesterday? wasn't it\[a_{n+2}=-a_n\frac{1-n}{(n+1)(n+2)}\]?
anonymous
  • anonymous
no this is new problem \[y''=y\]

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

TuringTest
  • TuringTest
oooooooh
anonymous
  • anonymous
I'll write out what I have so far
TuringTest
  • TuringTest
I'll be able to able to help in about 15-20 min
anonymous
  • anonymous
sure
anonymous
  • anonymous
\[\sum_{n=2}^\infty n(n-1)a_nx^{n-2}-\sum_{n=0}^\infty a_nx^n=0\]
nipunmalhotra93
  • nipunmalhotra93
@MathSofiya how exactly is the pattern changing?
TuringTest
  • TuringTest
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
anonymous
  • anonymous
\[\sum_{n=0}^\infty (n+2)(n+1)a_{n+2}x^n-\sum_{n=0}^\infty a_nx^n=0\]
TuringTest
  • TuringTest
so yeah, right pattern
anonymous
  • anonymous
\[\sum_{n=0}^\infty x^n\left[(n+2)(n+1)-a_n\right]=0\]
TuringTest
  • TuringTest
\[\sum_{n=0}^\infty x^n\left[(n+2)(n+1)a_{n+2}-a_n\right]=0\]
anonymous
  • anonymous
oops
TuringTest
  • TuringTest
\[a_{n+2}=\frac{a_n}{(n+1)(n+2)}\]so you were right, just plug in each n and write out \(everything\) explicitly, even multiplication by 1
anonymous
  • anonymous
ok
anonymous
  • anonymous
\[n=0:a_{0+2}=a_2=\frac {a_0}{2}\] \[n=1:a_{1+2}=a_3=\frac {a_1}{6}\] \[n=2:a_{2+2}=a_4=\frac {a_2}{12}\] \[n=3:a_{3+2}=a_5=\frac {a_3}{20}\] \[n=4:a_{4+2}=a_6=\frac {a_4}{30}\] \[n=5:a_{5+2}=a_7=\frac {a_5}{42}\]
anonymous
  • anonymous
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?
anonymous
  • anonymous
ooohh no need to because the next line is going to be zero
TuringTest
  • TuringTest
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
TuringTest
  • TuringTest
now is the next line zero?
TuringTest
  • TuringTest
how*
anonymous
  • anonymous
well it was yesterday LOL ....Ok I'll write everything out
TuringTest
  • TuringTest
\[n=0:a_{0+2}=a_2=\frac {a_0}{2\cdot1}\]\[n=1:a_3=\frac{a_1}{3\cdot2}\]write everything this way, with each number on the bottom specified
TuringTest
  • TuringTest
yesterday we had n-1 in the top, that's why
TuringTest
  • TuringTest
Haha I'm exercising while you crunch the numbers :P
anonymous
  • anonymous
\[n=0:a_{0+2}=a_2=\frac {a_0}{2}\] \[n=1:a_{1+2}=a_3=\frac {a_1}{6}=\frac{a_1}{2\cdot 3}\] \[n=2:a_{2+2}=a_4=\frac {a_2}{12}=\frac{a_0}{2\cdot2\cdot3}\] \[n=3:a_{3+2}=a_5=\frac {a_3}{20}=\frac{a_3}{10\cdot2}=\frac{a_1}{2\cdot2\cdot2\cdot3\cdot5}\] \[n=4:a_{4+2}=a_6=\frac {a_4}{30}=\frac{a_0}{2\cdot2\cdot2\cdot3\cdot5}\] \[n=5:a_{5+2}=a_7=\frac {a_5}{42}=\frac{a_1}{2\cdot2\cdot2\cdot3\cdot5\cdot42}\]
anonymous
  • anonymous
n=4 is wrong oh well...but I see a pattern now. How do you workout and do math?
TuringTest
  • TuringTest
push ups, then run over and check your work real quick :)
TuringTest
  • TuringTest
n=2 is wrong
TuringTest
  • TuringTest
12 is not 2x3
anonymous
  • anonymous
haha! That's what I call commitment!
anonymous
  • anonymous
give me a second to do the problem on paper. It's kinda tedious to type it...one second
TuringTest
  • TuringTest
\[a_n=\frac{a_n}{(n+1)(n+2)}\]\[n=0:a_{0+2}=a_2=\frac {a_0}{2\cdot1}\]\[n=1:a_3=\frac{a_1}{3\cdot2}\]\[n=2:a_4=\frac {a_2}{4\cdot3}=\frac{a_0}{4\cdot3\cdot2\cdot1}\]\[n=3:a_5=\frac {a_3}{5\cdot4}=\frac{a_1}{5\cdot4\cdot3\cdot2}\]
TuringTest
  • TuringTest
write out each number in the denom without simplifying every time
TuringTest
  • TuringTest
I cannot stress that enough to find patterns
anonymous
  • anonymous
\[n=0:a_{0+2}=a_2=\frac {a_0}{2}\] \[n=1:a_{1+2}=a_3=\frac {a_1}{6}=\frac{a_1}{3\cdot 2}\] \[n=2:a_{2+2}=a_4=\frac {a_2}{12}=\frac{a_0}{4\cdot3\cdot2\cdot1}\] \[n=3:a_{3+2}=a_5=\frac {a_3}{20}=\frac{a_3}{10\cdot2}=\frac{a_1}{5\cdot4\cdot3\cdot2\cdot1}\] Ok I'm finally convinced...had to calculate like 5 times that the denominator is actually 120
TuringTest
  • TuringTest
good, but more importantly what is/are the general pattern(s)?
TuringTest
  • TuringTest
again, there are two; one for even n and one for odd n
anonymous
  • anonymous
sorry internet is super slow. I'm gonna restart my computer brb
anonymous
  • anonymous
ok I'm back
TuringTest
  • TuringTest
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 ?
anonymous
  • anonymous
I wanna say \[(n+2)!\] and then sub in n=2k
anonymous
  • anonymous
wait...you go shower, and let me work on this
TuringTest
  • TuringTest
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
anonymous
  • anonymous
allow me to decipher it...
anonymous
  • anonymous
The denominator for both even and odds is the following...still working on the numerator \[\frac{?}{(n+2)!}\]
TuringTest
  • TuringTest
look at the difference between constants with an even and odd subscript what is the main difference you see>?
anonymous
  • anonymous
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
TuringTest
  • TuringTest
no wait, you can do it... let me give another hint...
anonymous
  • anonymous
ok
TuringTest
  • TuringTest
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 ?
TuringTest
  • TuringTest
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?
anonymous
  • anonymous
hold on I didn't write anything yet
anonymous
  • anonymous
sorry It just took me 3 mins to load this page
TuringTest
  • TuringTest
no rush, still haven't showered yet
anonymous
  • anonymous
When the numerator is \[a_0\] the denominator is even When the numerator is \[a_1\] the denominator is odd
anonymous
  • anonymous
by odd I mean 5! 3! and so on
TuringTest
  • TuringTest
exactly, and all even numbers can be written how?
anonymous
  • anonymous
2k
TuringTest
  • TuringTest
and odd?
anonymous
  • anonymous
n=2k
anonymous
  • anonymous
n=2k+1
anonymous
  • anonymous
\[a_{2k}\] and \[a_{2k+1}\]
TuringTest
  • TuringTest
so then what is our formulas? we need one for all \(a_{2k}\) and another for all \(a_2k+1}\)
TuringTest
  • TuringTest
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
anonymous
  • anonymous
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?
anonymous
  • anonymous
I can't write two separate formulas correct? I've gotta write one?
TuringTest
  • TuringTest
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
TuringTest
  • TuringTest
picture*
anonymous
  • anonymous
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
anonymous
  • anonymous
ok bye
TuringTest
  • TuringTest
cool, health makes one think more clearly I believe see ya
anonymous
  • anonymous
LOL I hope soo! see ya
anonymous
  • anonymous
where is Alan Turing?
anonymous
  • anonymous
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
TuringTest
  • TuringTest
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 :)
anonymous
  • anonymous
I'll be back in like 10 mins, gotta start laundry
TuringTest
  • TuringTest
right-o
anonymous
  • anonymous
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)
anonymous
  • anonymous
Well it's what we called the pattern initially before writing it in terms of k's
anonymous
  • anonymous
ok found the laundry card brb
TuringTest
  • TuringTest
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\[n=5:a_6=\frac{a_4}6=\frac{a_0}{2\cdot4\cdot6}=\frac{a_0}{2^3(3!)}\]??
anonymous
  • anonymous
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?
TuringTest
  • TuringTest
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 \(a_{n+p}\)
TuringTest
  • TuringTest
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...
TuringTest
  • TuringTest
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
TuringTest
  • TuringTest
...and whatever base constant we can't determine, like \(a_0\) which we do not know since we are not given the initial conditions
anonymous
  • anonymous
ok that makes more sense now. Let's see if I can apply it to this problem now.
TuringTest
  • TuringTest
go for it, I'm gonna get a few empanadas for breakfast :) (sortof like Mexican hot-pockets :) brb
anonymous
  • anonymous
sounds good
anonymous
  • anonymous
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...
anonymous
  • anonymous
\[a_{2k+1}=\frac{a_1}{k!}\]
TuringTest
  • TuringTest
is it? what is\(a_3\) ?
anonymous
  • anonymous
oh my 2k+1=3 2k=2 k=1
anonymous
  • anonymous
yeah I'm wrong
anonymous
  • anonymous
3k?
anonymous
  • anonymous
your patients with me is truly astonishing. I would have strangled the student.
TuringTest
  • TuringTest
haha, but I haven't done this in a long time either, so it's a good review for me too.
anonymous
  • anonymous
*patience
TuringTest
  • TuringTest
\[a_n=\frac{a_n}{(n+1)(n+2)}\]\[n=0:a_2=\frac {a_0}{2}=\frac{a_0}{2!}\]\[n=1:a_3=\frac {a_1}{6}=\frac{a_1}{3\cdot 2}=\frac{a_1}{3!}\]\[n=2:a_4=\frac {a_2}{12}=\frac{a_0}{4\cdot3\cdot2\cdot1}=\frac{a_0}{4!}\]\[n=3:a_5=\frac {a_3}{20}=\frac{a_3}{10\cdot2}=\frac{a_1}{5\cdot4\cdot3\cdot2\cdot1}=\frac{a_1}{5!}\]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
anonymous
  • anonymous
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
anonymous
  • anonymous
way easy
TuringTest
  • TuringTest
you forgot the factorials ;)
anonymous
  • anonymous
!!!!!!!!!!!! there
anonymous
  • anonymous
LOL ok let me rewrite them
TuringTest
  • TuringTest
and you forgot that for odd subscripts it is a_1 in the numerator
anonymous
  • anonymous
\[a_{2k}=\frac{a_0}{(2k)!}\] \a_{2k+1}=\frac{a_1}{2k+1}[\] I didn't forgot...I just checking if you were paying attention :P
anonymous
  • anonymous
\[a_{2k+1}=\frac{a_1}{2k+1}\]
anonymous
  • anonymous
oh my
TuringTest
  • TuringTest
haha :P
anonymous
  • anonymous
\[a_{2k+1}=\frac{a_1}{(2k+1)!}\]
anonymous
  • anonymous
happy?
anonymous
  • anonymous
oh wait there is more...
TuringTest
  • TuringTest
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.
anonymous
  • anonymous
\[y=\sum_{n=0}^\infty a_nx^n=a_0\sum_{k=0}^\infty\frac{x^{2k}}{(2k)!} \] \[y=\sum_{n=0}^\infty a_nx^n=a_1\sum_{k=0}^\infty\frac{x^{2k+1}}{(2k+1)!} \]
TuringTest
  • TuringTest
you've got the right idea, but we need to use superposition to get the full answer
TuringTest
  • TuringTest
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?
anonymous
  • anonymous
put them into one formula somehow
anonymous
  • anonymous
oh y and c_1 and so on and so forth...hold on
TuringTest
  • TuringTest
superposition...
anonymous
  • anonymous
adding them?
TuringTest
  • TuringTest
yep :) c1=a0 c2=a1
anonymous
  • anonymous
\[c_1\sum_{k=0}^\infty\frac{x^{2k}}{(2k)!}+c_2\sum_{k=0}^\infty\frac{x^{2k+1}}{(2k+1)!}=0\] huh?
anonymous
  • anonymous
gosh the sums probably shouldn't be there...
TuringTest
  • TuringTest
no I was just saying that the c1 you are used to is called a0 in this case, so...
TuringTest
  • TuringTest
\[y=a_0\sum_{k=0}^\infty\frac{x^{2k}}{(2k)!}+a_1\sum_{k=0}^\infty\frac{x^{2k+1}}{(2k+1)!}=0\]
TuringTest
  • TuringTest
wait, why =0 ?
TuringTest
  • TuringTest
\[y=a_0\sum_{k=0}^\infty\frac{x^{2k}}{(2k)!}+a_1\sum_{k=0}^\infty\frac{x^{2k+1}}{(2k+1)!}\]
anonymous
  • anonymous
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?
anonymous
  • anonymous
can I put my laundry in the dryer?
anonymous
  • anonymous
brb
anonymous
  • anonymous
2 mins
TuringTest
  • TuringTest
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
anonymous
  • anonymous
ok I'm back
anonymous
  • anonymous
yep makes sense
TuringTest
  • TuringTest
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
TuringTest
  • TuringTest
example 1 on*
anonymous
  • anonymous
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)
TuringTest
  • TuringTest
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 :)
anonymous
  • anonymous
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...
TuringTest
  • TuringTest
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!
anonymous
  • anonymous
I will. Thanks! quick question, did you make your own empanadas?
TuringTest
  • TuringTest
nope, bought 'em at the corner I only cook hamburgers and pasta
anonymous
  • anonymous
what? you've been in Mexica for a little while now! Gotta learn to cook like the locals!
TuringTest
  • TuringTest
well I make quesadillas, I guess that's a start lol
anonymous
  • anonymous
yep, work your way up from here :P
TuringTest
  • TuringTest
:)
anonymous
  • anonymous
|dw:1348252365896:dw|
anonymous
  • anonymous
|dw:1348252603116:dw|
anonymous
  • anonymous
|dw:1348252694415:dw|
anonymous
  • anonymous
|dw:1348252778070:dw|

Looking for something else?

Not the answer you are looking for? Search for more explanations.