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

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 \[y''=y\]

oooooooh

I'll write out what I have so far

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

sure

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

@MathSofiya how exactly is the pattern changing?

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

so yeah, right pattern

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

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

oops

ok

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

now is the next line zero?

how*

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

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

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

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 :)

n=2 is wrong

12 is not 2x3

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

write out each number in the denom without simplifying every time

I cannot stress that enough to find patterns

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

ok I'm back

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

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

allow me to decipher it...

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

ok

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

by odd I mean 5! 3! and so on

exactly, and all even numbers can be written how?

2k

and odd?

n=2k

n=2k+1

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

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

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

picture*

ok bye

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

LOL I hope soo! see ya

where is Alan Turing?

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

right-o

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

ok found the laundry card brb

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

sounds good

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

is it?
what is\(a_3\) ?

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

yeah I'm wrong

3k?

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.

*patience

way easy

you forgot the factorials ;)

!!!!!!!!!!!!
there

LOL ok let me rewrite them

and you forgot that for odd subscripts it is a_1 in the numerator

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

oh my

haha :P

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

happy?

oh wait there is more...

you've got the right idea, but we need to use superposition to get the full answer

put them into one formula somehow

oh y and c_1 and so on and so forth...hold on

superposition...

adding them?

yep :)
c1=a0
c2=a1

\[c_1\sum_{k=0}^\infty\frac{x^{2k}}{(2k)!}+c_2\sum_{k=0}^\infty\frac{x^{2k+1}}{(2k+1)!}=0\]
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...

\[y=a_0\sum_{k=0}^\infty\frac{x^{2k}}{(2k)!}+a_1\sum_{k=0}^\infty\frac{x^{2k+1}}{(2k+1)!}=0\]

wait, why =0 ?

\[y=a_0\sum_{k=0}^\infty\frac{x^{2k}}{(2k)!}+a_1\sum_{k=0}^\infty\frac{x^{2k+1}}{(2k+1)!}\]

can I put my laundry in the dryer?

brb

2 mins

ok I'm back

yep makes sense

example 1 on*

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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