Question

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 \[\frac{a_0}{6!}\]
\[n=4:a_6=\frac{a_0}{2\cdot 30}=\]

Answer 1

??
\[a _{4} = \frac{a _{0} }{ 4! } \]

Answer 2

this is the one we did yesterday?
wasn't it\[a_{n+2}=-a_n\frac{1-n}{(n+1)(n+2)}\]?

Answer 3

no this is new problem \[y''=y\]

Answer 4

oooooooh

Answer 5

I'll write out what I have so far

Answer 6

I'll be able to able to help in about 15-20 min

Answer 7

sure

Answer 8

\[\sum_{n=2}^\infty n(n-1)a_nx^{n-2}-\sum_{n=0}^\infty a_nx^n=0\]

Answer 9

@MathSofiya how exactly is the pattern changing?

Answer 10

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

Answer 11

\[\sum_{n=0}^\infty (n+2)(n+1)a_{n+2}x^n-\sum_{n=0}^\infty a_nx^n=0\]

Answer 12

so yeah, right pattern

Answer 13

\[\sum_{n=0}^\infty x^n\left[(n+2)(n+1)-a_n\right]=0\]

Answer 14

\[\sum_{n=0}^\infty x^n\left[(n+2)(n+1)a_{n+2}-a_n\right]=0\]

Answer 15

oops

Answer 16

\[a_{n+2}=\frac{a_n}{(n+1)(n+2)}\]so you were right, just plug in each n and write out \(everything\) explicitly, even multiplication by 1

Answer 17

ok

Answer 18

\[n=0:a_{0+2}=a_2=\frac {a_0}{2}\]
\[n=1:a_{1+2}=a_3=\frac {a_1}{6}\]
\[n=2:a_{2+2}=a_4=\frac {a_2}{12}\]
\[n=3:a_{3+2}=a_5=\frac {a_3}{20}\]
\[n=4:a_{4+2}=a_6=\frac {a_4}{30}\]
\[n=5:a_{5+2}=a_7=\frac {a_5}{42}\]

Answer 19

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?

Answer 20

ooohh no need to because the next line is going to be zero

Answer 21

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

Answer 22

now is the next line zero?

Answer 23

how*

Answer 24

well it was yesterday LOL ....Ok I'll write everything out

Answer 25

\[n=0:a_{0+2}=a_2=\frac {a_0}{2\cdot1}\]\[n=1:a_3=\frac{a_1}{3\cdot2}\]write everything this way, with each number on the bottom specified

Answer 26

yesterday we had n-1 in the top, that's why

Answer 27

Haha I'm exercising while you crunch the numbers :P

Answer 28

\[n=0:a_{0+2}=a_2=\frac {a_0}{2}\]
\[n=1:a_{1+2}=a_3=\frac {a_1}{6}=\frac{a_1}{2\cdot 3}\]
\[n=2:a_{2+2}=a_4=\frac {a_2}{12}=\frac{a_0}{2\cdot2\cdot3}\]
\[n=3:a_{3+2}=a_5=\frac {a_3}{20}=\frac{a_3}{10\cdot2}=\frac{a_1}{2\cdot2\cdot2\cdot3\cdot5}\]
\[n=4:a_{4+2}=a_6=\frac {a_4}{30}=\frac{a_0}{2\cdot2\cdot2\cdot3\cdot5}\]
\[n=5:a_{5+2}=a_7=\frac {a_5}{42}=\frac{a_1}{2\cdot2\cdot2\cdot3\cdot5\cdot42}\]

Answer 29

n=4 is wrong oh well...but I see a pattern now. How do you workout and do math?

Answer 30

push ups, then run over and check your work real quick :)

Answer 31

n=2 is wrong

Answer 32

12 is not 2x3

Answer 33

haha! That's what I call commitment!

Answer 34

give me a second to do the problem on paper. It's kinda tedious to type it...one second

Answer 35

\[a_n=\frac{a_n}{(n+1)(n+2)}\]\[n=0:a_{0+2}=a_2=\frac {a_0}{2\cdot1}\]\[n=1:a_3=\frac{a_1}{3\cdot2}\]\[n=2:a_4=\frac {a_2}{4\cdot3}=\frac{a_0}{4\cdot3\cdot2\cdot1}\]\[n=3:a_5=\frac {a_3}{5\cdot4}=\frac{a_1}{5\cdot4\cdot3\cdot2}\]

Answer 36

write out each number in the denom without simplifying every time

Answer 37

I cannot stress that enough to find patterns

Answer 38

\[n=0:a_{0+2}=a_2=\frac {a_0}{2}\]
\[n=1:a_{1+2}=a_3=\frac {a_1}{6}=\frac{a_1}{3\cdot 2}\]
\[n=2:a_{2+2}=a_4=\frac {a_2}{12}=\frac{a_0}{4\cdot3\cdot2\cdot1}\]
\[n=3:a_{3+2}=a_5=\frac {a_3}{20}=\frac{a_3}{10\cdot2}=\frac{a_1}{5\cdot4\cdot3\cdot2\cdot1}\]
Ok I'm finally convinced...had to calculate like 5 times that the denominator is actually 120

Answer 39

good, but more importantly what is/are the general pattern(s)?

Answer 40

again, there are two; one for even n and one for odd n

Answer 41

sorry internet is super slow. I'm gonna restart my computer brb

Answer 42

ok I'm back

Answer 43

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 ?

Answer 44

I wanna say \[(n+2)!\] and then sub in n=2k

Answer 45

wait...you go shower, and let me work on this

Answer 46

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

Answer 47

allow me to decipher it...

Answer 48

The denominator for both even and odds is the following...still working on the numerator
\[\frac{?}{(n+2)!}\]

Answer 49

look at the difference between constants with an even and odd subscript
what is the main difference you see>?

Answer 50

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

Answer 51

no wait, you can do it... let me give another hint...

Answer 52

ok

Answer 53

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 ?

Answer 54

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?

Answer 55

hold on I didn't write anything yet

Answer 56

sorry It just took me 3 mins to load this page

Answer 57

no rush, still haven't showered yet

Answer 58

When the numerator is \[a_0\] the denominator is even
When the numerator is \[a_1\] the denominator is odd

Answer 59

by odd I mean 5! 3! and so on

Answer 60

exactly, and all even numbers can be written how?

Answer 61

2k

Answer 62

and odd?

Answer 63

n=2k

Answer 64

n=2k+1

Answer 65

\[a_{2k}\]
and
\[a_{2k+1}\]

Answer 66

so then what is our formulas?
we need one for all \(a_{2k}\) and another for all \(a_2k+1}\)

Answer 67

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

Answer 68

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?

Answer 69

I can't write two separate formulas correct? I've gotta write one?

Answer 70

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

Answer 71

picture*

Answer 72

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

Answer 73

ok bye

Answer 74

cool, health makes one think more clearly I believe
see ya

Answer 75

LOL I hope soo! see ya

Answer 76

where is Alan Turing?

Answer 77

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

Answer 78

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 :)

Answer 79

I'll be back in like 10 mins, gotta start laundry

Answer 80

right-o

Answer 81

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)

Answer 82

Well it's what we called the pattern initially before writing it in terms of k's

Answer 83

ok found the laundry card brb

Answer 84

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\[n=5:a_6=\frac{a_4}6=\frac{a_0}{2\cdot4\cdot6}=\frac{a_0}{2^3(3!)}\]??

Answer 85

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?

Answer 86

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 \(a_{n+p}\)

Answer 87

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...

Answer 88

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

Answer 89

...and whatever base constant we can't determine, like \(a_0\) which we do not know since we are not given the initial conditions

Answer 90

ok that makes more sense now. Let's see if I can apply it to this problem now.

Answer 91

go for it, I'm gonna get a few empanadas for breakfast :)
(sortof like Mexican hot-pockets :) brb

Answer 92

sounds good

Answer 93

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...

Answer 94

\[a_{2k+1}=\frac{a_1}{k!}\]

Answer 95

is it?
what is\(a_3\) ?

Answer 96

oh my
2k+1=3
2k=2
k=1

Answer 97

yeah I'm wrong

Answer 98

3k?