Some of friends are confused,regarding this solution as it wont solve recursion.NB:: Here I wont solve the recursion, instead I follow methods for solving it using series. From that I will get a condition,which helps to find the SUM. Instead of getting each term I prefer entire series sum.NB:: Here I wont solve the recursion, instead I follow methods for solving it using series. From that I will get a condition,which helps to find the SUM. Instead of getting each term I prefer entire series sum.

Your description of equation (6) is confusing. You should explicitly state equations (7) and (8) are the definition of $X$ and $Y$. Now here is the error or at least a leap of logic: how do you guarantee either $Y=0$ or $X=0$?

@Hansen I cant take the case Y=0 or X=0 . I can only take X=0 and Y=0 together to say X+Y = 0. You may say what about X=-Y? But if I ask since R.H.S =0,how can you say Y and X together as zero not a solution of X+Y=0? So what I did is, splited it as X,Y and considered the case both are zero. Please visit this [ link](ltcconline.net/greenl/courses/204/PowerLaplace/…), you can have exmaples of this method as I did. Since we have already one series as reference. We take $R_0,R_1$ as it is from recursion and treat SUM as variable

What is your "R.H.S."? Right now, you only know $X+Y=0$, not $X=0$ or $Y=0$. You can NOT assume the latter statement. $X=0 \wedge Y=0 \Rightarrow X+Y=0$ does not imply $X+Y=0 \Rightarrow X=0\wedge Y=0$. As a simple example, $2+(-2)=0$ does not imply $2=0$ and $-2=0$. Your derivation is wrong. The link you cite has nothing like what you are doing here. Point out which line if you think otherwise.

Please open the link cite. Then take the first example, y'' + xy' + y = 0 , you can see at some point we reach $ 2a_2+a_0+\sum_{n=1}^\infty\{ (n+2)(n+1)a_{n+2}+na_n+a_n\}=0$ What is the next step? Why it says $2a_2+a_0$ and $(n+2)(n+1)a_{n+2}+na_n+a_n =0$? It is same in next example too. The argument of 2+(-2) wont cross what I stated earlier. What I meant was if two variables X+Y=0, there exists so many solutions (0,0),(2,-2),(3,-3) etc. It doesnt say (0,0) is the only one. I took (0,0) for my case here.

Here I fixed $R_2,R_1$ and pulled down SUM as unknown variable..Would you explain why those links say like that or did I misinterpret ? Then it may clear my confusion. Another link see the equating step

I knew you were going to interpret it like that. In the example you cite from that link, have you faithfully copied the summand? Are these two versions the same? Have you understood the derivation in that solution or are you simply imitating the steps? Regarding $X+Y=0$, the problem is not asking for all $X$ and $Y$ such that the equation holds but merely$X$ and $Y$ satisfy the equation. The two statements are completely different.

I am bit confused or may be misinterpreted.Could you say why they made it zero?

I got up to that summand. But equating two parts to zero ,is bit confusing..It would have been a great help if you say what is that

Do you see the difference between your version and that on the site?

yes.. I have added SUM to it..

What do you mean? I am asking about the example in that link.

What could be the reason why they made two parts of it to 0 $ 2a_2+a_0+\sum_{n=1}^\infty\{ (n+2)(n+1)a_{n+2}+na_n+a_n\}=0$ in y'' + xy' + y = 0 first link I meant

I added -SUM coz of my misinterpretation

seond or first link?

ltcconline.net/greenl/courses/204/PowerLaplace/… let us go by this

I am asking you to check your left hand side again the one in the link ltcconline.net/greenl/courses/204/PowerLaplace/…

ok plz wait 2 min

yes it is $ 2a_2+a_0+\sum_{n=1}^\infty\{ (n+2)(n+1)a_{n+2}+na_n+a_n\}=0$

Are you simply copying what you have written before or really checking?

yes checking...no no I am not understanding which line of the link you are talking

there is an X^n there

I missed that

You just saw it now, after all this time...?

I kept asking you to check...

sorry for that some confusion

I was using x=1, that is why I omitted it

I used that for getting SUM hope u got what I meant

Alright, I hope we do waste too much time.

sorry..Did I waste ur time? sorry

what was the confusion with u...was it coz I used x as 1 and you didt see that? or any other issues

Well, I kept asking you to check one equation to see if they are the same version, it is a yes or no question. You kept not checking...

((A and B) implies C) does not imply (A implies C).

oopz..what do you mean by same version.. I got confused there..sorry

Same version: is your equation exactly the same as the equation in that link?

not at all

method is same

in my case I didt want to solve recurrence

No, no, no...

You are driving me crazy.

My equation is not exactly same as the one in the link

I was about to say the difference

You wrote $ 2a_2+a_0+\sum_{n=1}^\infty\{ (n+2)(n+1)a_{n+2}+na_n+a_n\}=0$, the equation in that link is $ 2a_2+a_0+\sum_{n=1}^\infty[\{ (n+2)(n+1)a_{n+2}+na_n+a_n\}]x^n=0$

Agree?

yes... now i got where is the confusion...

I missed that x^n

sorry sorry

Are they the same or not? Yes or no?

they are same YES> but I made a typing mistake missing x^n are we clear

What? Missing a x^n makes the two equation the same?

2a_2+a_0+\sum_{n=1}^\infty[\{ (n+2)(n+1)a_{n+2}+na_n+a_n\}]x^n=0$ I meant this one,,.. as telling example ...COmparing this with my equation both are different

my equation means my question recurrence

I am not talking about your original equation, I am talking about what the equation you thought you copied from that link.

Forget about your original equation for now. Deal with one thing at a time.

yes let us talk about that.. You say my equation as $ 2a_2+a_0+\sum_{n=1}^\infty\{ (n+2)(n+1)a_{n+2}+na_n+a_n\}=0$

you *saw

actually I meant $ 2a_2+a_0+\sum_{n=1}^\infty[\{ (n+2)(n+1)a_{n+2}+na_n+a_n\}]x^n=0$ .. I missed that x^n

type error..sorry for that..are you clear now

i made mistake when I copied

hope ur ok now

OK. So answer my question are they the same or not, Yes or no?

Yes

With and without x^n are the same?

yes both same... actually I missed that x^n

dear friend I just copied that exmaple from that link I missed the x^n by typo error

But the error makes the two copies different, not the same, yes or no, my dear friend?

It is a logic question.

yes..

just talk about the one in the link ..the x^n one

Are you saying $2a_2+a_0+\sum_{n=1}^\infty\{ (n+2)(n+1)a_{n+2}+na_n+a_n\} = 2a_2+a_0+\sum_{n=1}^\infty[\{ (n+2)(n+1)a_{n+2}+na_n+a_n\}]x^n$?

No no ..dear friend..dont think mathematically now... I was trying to type 2a_2+a_0+\sum_{n=1}^\infty[\{ (n+2)(n+1)a_{n+2}+na_n+a_n\}]x^n$? but missed that x^n

I meant 2a_2+a_0+\sum_{n=1}^\infty[\{ (n+2)(n+1)a_{n+2}+na_n+a_n\}]x^n$ only

let us forget what we typed..start fresh one from link..we are in confusion now

just open the link

Alright, why can't you just say "no"? Any way,

coz I am getting confused what you are asking..let us start fresh

let us go to the link

ltcconline.net/greenl/courses/204/PowerLaplace/… ok?

Now that we are clear. Now how do you apply the logic of that link to your derivation?

now u can see $2a_2+a_0+\sum_{n=1}^\infty[\{ (n+2)(n+1)a_{n+2}+na_n+a_n\}]x^n$ =0

ok agreed?>

Yes.

ok ..finally we are on tack..sorry for making u confused.. I am proceeding

Go right ahead.

now see the next step he is saying $2a_2+a_0$ = 0

Agree. Then what?

and (n+2)(n+1)a_{n+2}+na_n+a_n =0 also ok?>

Agree. Then?

what is ur reasoning for writing like that?

What do you mean by "writing like that"?

I assumed that if we have two variables X+Y =0 ,x=0 and Y=0 is also a solution,..,means one of them

means they separated $2a_2+a_0$ and (n+2)(n+1)a_{n+2}+na_n+a_n =0

X+Y=0 have so many solutions... But among them I selected X=0 and Y=0 together..

Where do you get the idea that is the derivation step?

that was my reasoning

I just thought..if I am wrong could you say why it is liek that

Do you have a phone number? It will be easier and quicker to talk through orally.

Where are you?

I am in Delhi India.. you?

can u come in skype?

US.

Yes.

what is ur skype id ,,,I will add

now

hansennightrider

ok plz wait 2 min I will add u now