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Help with difference equation (hard undergrad calculus)

Suppose that the sequence pi1, pi2, pi3,....... Satisfies the recurrence relation

Pi(sub n+1) * n/(n+1) = Pi(sub n)/n for n=1,2,3,....

Along with the condition sumof(Pi(sub n)) = 1 from n>=1.

What is the value of Pi(sub 1)?

Sub = subscript

Now I don't know where to start.

Thanks for help!
(edited 8 years ago)
Reply 1
Original post by mathsRus
Suppose that the sequence pi1, pi2, pi3,....... Satisfies the recurrence relation

Pi(sub n+1) * n/(n+1) = Pi(sub n)/n for n=1,2,3,....

Along with the condition sumof(Pi(sub n)) = 1 from n>=1.

What is the value of Pi(sub 1)?

Sub = subscript

Now I don't know where to start.

Thanks for help!



please post a photo of the question
Lol
Sub=subscript
Wtf does that mean.


Posted from TSR Mobile
Reply 3
Original post by TeeEm
please post a photo of the question


For some reason it is saying that your post it edited but cannot be seen. Please refresh the page.
Reply 4
Original post by physicsmaths
Lol
Sub=subscript
Wtf does that mean.


Posted from TSR Mobile


Subscript meaning the little wordings below the notation on the right. Get it? Using iPhone so I can't write the question properly image.jpeg
Reply 5
Original post by TeeEm
please post a photo of the question


image.jpeg

Managed to upload it. :smile:
Reply 6
Original post by mathsRus
image.jpeg

Managed to upload it. :smile:


will look at it
Original post by mathsRus
image.jpeg

Managed to upload it. :smile:


You're given two pieces of information about the sequence. It might be a good idea to use them together.
Reply 8
Original post by mathsRus
For some reason it is saying that your post it edited but cannot be seen. Please refresh the page.


I just looked at it over dinner and I got it as 1/(2e-1)
which is not there (clearly I am making an error somewhere)

I am desperate to get this right as I want it for my own library of resources
(edited 8 years ago)
Reply 9
Original post by TeeEm
I just looked at it over dinner and I got it as 1/(2e-1)
which is not there (clearly I am making an error somewhere)

I am desperate to get this right as I want it for my own library of resources


I have some more super tricky questions coming up!
Reply 10
Original post by TeeEm
I just looked at it over dinner and I got it as 1/(2e-1)
which is not there (clearly I am making an error somewhere)

I am desperate to get this right as I want it for my own library of resources


I think it's 0.5e^-1
Reply 11
Original post by mathsRus
I have some more super tricky questions coming up!


I will post a solution in around 1/2 hour but I am not getting your answer

maybe you can check my solution
Reply 12
Original post by Gome44
I think it's 0.5e^-1


check my work in 1/2 hour
Reply 13
Original post by Gome44
I think it's 0.5e^-1


Original post by mathsRus
I have some more super tricky questions coming up!


this is my solution... I cannot get any of the 4 answers there, but someone can check where I have gone wrong.
I hope I have not made any embarrassing error for which I am notorious

corrected the error in the adding now
(edited 8 years ago)
Original post by TeeEm
this is my solution... I cannot get any of the 4 answers there, but someone can check where I have gone wrong.
I hope I have not made any embarrassing error for which I am notorious
Your infinite sum starts at P2P_2 (i.e. you are finding 2Pk\sum_2^\infty P_k when it should be 1Pk\sum_1^\infty P_k).
Reply 15
Original post by DFranklin
Your infinite sum starts at P2P_2 (i.e. you are finding 2Pk\sum_2^\infty P_k when it should be 1Pk\sum_1^\infty P_k).


Just saw it before your post
I was considering tippex but I really need to rewrite

thanks
Reply 16
Original post by TeeEm
this is my solution... I cannot get any of the 4 answers there, but someone can check where I have gone wrong.
I hope I have not made any embarrassing error for which I am notorious


Gome44 is correct. The answer is 0.5e^-1
Reply 17
Original post by mathsRus
Gome44 is correct. The answer is 0.5e^-1


see the method I posted
there is a tiny error in the adding (see if you can spot it)
then there is your question ...
Reply 18
Original post by TeeEm
see the method I posted
there is a tiny error in the adding (see if you can spot it)
then there is your question ...


haha thanks!

some tough question to understand
Reply 19
Original post by mathsRus
haha thanks!

some tough question to understand


a little bit perhaps

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