Help with Solving 5a & Understanding Recursion Relations

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Homework Help Overview

The discussion revolves around understanding recursion relations and solving a specific problem labeled as 5a, which involves integration of the form ∫x^n e^(-x) dx. Participants express confusion about the concept of recursion relations, the integration process, and the application of limits in the context of the problem.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants attempt to integrate the expression and discuss the implications of integrating "forever." There are questions about the notation used for representing the integration process and the relevance of the gamma function. Some participants express uncertainty about the use of homogeneous functions and their connection to the problem. Others explore the relationship between different sequences defined by the same recursion relation.

Discussion Status

The discussion is active, with participants sharing insights and questioning various aspects of the problem. Some have offered partial guidance on the integration process and recursion relations, while others are seeking clarification on specific steps and concepts. There is a recognition that understanding the limits and the behavior of the recursion relation is crucial, but no consensus has been reached on how to approach the problem fully.

Contextual Notes

Participants note the lack of prior exposure to certain concepts in class, which adds to their confusion. There are references to specific values and limits in the integration process, as well as the need to understand the implications of these limits in the context of recursion relations.

cloud360
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Would be really grateful if someone helped me with 5a, and explained the ideas behind recursion relations. I don't know what a recursion relation is, and how to apply for forum ales given

Homework Statement


[PLAIN]http://img839.imageshack.us/img839/2301/recurrencerelation.gif
http://img839.imageshack.us/img839/2301/recurrencerelation.gif


Homework Equations



none

The Attempt at a Solution




∫xne-x=-xne-x-∫(-nxn-1e-x)

but then i have to keep integrating forever


how do i do 5a, any help, any online references to learn?
 
Last edited by a moderator:
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hi cloud360! :smile:

(have an integral: ∫ :wink:)
cloud360 said:
integrate xne-x=-xne-x-|(-nxn-1e-x)

but then i have to keep integrating forever


but you have proved the main part of 5(a) ! :smile:

(and what's "forever" about going from n down to zero? :confused:)
 


tiny-tim said:
hi cloud360! :smile:

(have an integral: ∫ :wink:)


but you have proved the main part of 5(a) ! :smile:

(and what's "forever" about going from n down to zero? :confused:)


ok, so if i have to integrate something forever. what notation can i use to represent this.

i keep seeing the gamme function, but don't know what its used for

(my teachers like putting things on hwk that we never did in class)
 


ok, i heard soemthing about homegenous function which go down to 0?

but don't really know what to do
 
cloud360 said:
ok, i heard soemthing about homegenous function which go down to 0?

but don't really know what to do

no, homogeneous functions are irrelevant …

your recurrence relation is Kn = nKn-1 …

so for example K7 = 7K6 = 7*6K5, and so on down to K0, where it stops! :wink:
 


tiny-tim said:
no, homogeneous functions are irrelevant …

your recurrence relation is kn = nkn-1 …

so for example k7 = 7k6 = 7*6k5, and so on down to k0, where it stops! :wink:

k7 = 7k6 = 7*6k5
=7*6*5k4=7*6*5*4k3 = 7*6*5*4*3k2= 7*6*5*4*3*2k1= 7*6*5*4*3*2*1k0


I know very litle about recurssion relations. But know, if i have 2 squences with same recurssion relations. then sequence 1 - sqeuence 2 = 0

I have the solutions to the question i posted. but don't know why its the answer

[PLAIN]http://img231.imageshack.us/img231/3002/recussion2.gif
 
Last edited by a moderator:
That's right! :smile:

Similarly, if you know K1/2, then you can find K3/2, then you can find K5/2, and so on. :wink:
 


tiny-tim said:
That's right! :smile:

Similarly, if you know K1/2, then you can find K3/2, then you can find K5/2, and so on. :wink:


Question: is this factorial llike thing... where you keep multiplying downwards e.g 5*4*3..e.t.c (does this happen for all recursion relations), or only this. does it depend on forumale they give us?

K3.5
K3.5=3.5*2.5*1.5*k0.5
K3.5=3.5*2.5*1.5*0.5*(π0.5)
K3.5=11.6317
 
Last edited:


Still. please can you tell me what i would do to work out 5a. i got an exam on this tomorrow. and 5a, and one like it are only things i don't know how to do
 
  • #10
cloud360 said:
K3.5=3.5*2.5*1.5*0.5*(π0.5)
K3.5=11.6317

Yes, or you can write it K(2n+1)/2 = (2n+1)(2n-1)… 1/2n K1/2

= (2n+1)(2n)… 1/2n2n(2n-2)… 2 K1/2

= (2n+1)!/22n(2n)! K1/2 :smile:

(which is easier for large n because you can look up factorials in a book or on your calculator :wink:)
Question: is this factorial liek this where you keep mutipying downards e.g 5*4*3..e.t.c (does this happen for all recursion relations)?

No, it just happens to work for this one.
cloud360 said:
Still. please can you tell me what i would do to work out 5a. i got an exam on this tomorrow. and 5a, and one like it are only things i don't know how to do

ah, when I posted my last answer, I hadn't see this, which you edited in later:
cloud360 said:
I have the solutions to the question i posted. but don't know why its the answer

[PLAIN]http://img231.imageshack.us/img231/3002/recussion2.gif[/QUOTE]

which part of that is worrying you?
 
Last edited by a moderator:
  • #11


tiny-tim said:
Yes, or you can write it K(2n+1)/2 = (2n+1)(2n-1)… 1/2n K1/2

= (2n+1)(2n)… 1/2n2n(2n-2)… 2 K1/2

= (2n+1)!/22n(2n)! K1/2 :smile:

(which is easier for large n because you can look up factorials in a book or on your calculator :wink:)


No, it just happens to work for this one.


ah, when I posted my last answer, I hadn't see this, which you edited in later:


which part of that is worrying you?
[PLAIN]http://img839.imageshack.us/img839/2301/recurrencerelation.gif

the first question


5a...(i)

∫uv'=uv-u'v
u=xn
dv=e-x

1. i don't understand why they are using limits to show the recursion relation holds for the given equation?
The way i would do it is by working out K0 , K1 , K2...And find a nth term formulae?
2. I don't know how to get K0 , K1 , K2 by direct integration> the question as for K0 by direct integration, what does this even mean
[PLAIN]http://img231.imageshack.us/img231/3002/recussion2.gif

in fact you said that i did the main part of the question when i posted this

∫xne-x=-xne-x-∫(-nxn-1e-x)

i then said i would need to integrate forever because of this ∫(-nxn-1e-x)
 
Last edited by a moderator:
  • #12
cloud360 said:
in fact you said that i did the main part of the question when i posted this

∫xne-x=-xne-x-∫(-nxn-1e-x)

i then said i would need to integrate forever because of this ∫(-nxn-1e-x)

but you don't have to integrate forever, because the value of n keeps falling each time …

after n integrations, you get down to ∫e-x dx, which of course you can do immediately
 
  • #13


tiny-tim said:
but you don't have to integrate forever, because the value of n keeps falling each time …

after n integrations, you get down to ∫e-x dx, which of course you can do immediately

So you get e^-x by using limits.

I don't understand the second line of the solutions. please can you explain

[PLAIN]http://img231.imageshack.us/img231/3002/recussion2.gif


why are they subbing infinity into the equation. and why sub infinity into :

-xne-x

where did this come from?

were did they get this from.
 
Last edited by a moderator:
  • #14


this is the solution to part b (see first post and check black area)

[PLAIN]http://img709.imageshack.us/img709/7861/ans2k.gif


I don't understand why they don't use limits in this question to prove the recursion relations holds for the given forumale? but did for part a

i can't see any consistency. therefore i don't know the general way of solving/proving a recursion relation holds for a given equation
 
Last edited by a moderator:
  • #15


2. I don't know how to get K0 for part a

can you help me with this. i don't even know what it means "by direct integration". i have seen the soluton but don't understand why they sub "0" into x, and not n?


[PLAIN]http://img231.imageshack.us/img231/3002/recussion2.gif
 
Last edited by a moderator:
  • #16
cloud360 said:
… I don't understand the second line of the solutions. please can you explain
…
why are they subbing infinity into the equation. and why sub infinity into :

-xne-x

where did this come from?

were did they get this from.

oh i see … it's just the [xn(-e-x)]0∞ that's worrying you …

ok, that's ∞n(-e-∞) - 0n(-e-0), wchih is 0 - 0

except of course we can't really write the first part like that, we have to put a limn->∞ in front of it :wink:
cloud360 said:
… I don't understand why they don't use limits in this question to prove the recursion relations holds for the given forumale? but did for part a

i can't see any consistency. therefore i don't know the general way of solving/proving a recursion relation holds for a given equation

because the limits were there only because ∞ was one of the bounds of the integration, ∫0∞, so you got a […]0∞, which involved taking a limit

in most recurrence relations, there's either no integral at all, or there's an integral with finite bounds …

you do not normally get limits in a recurrence relation!​

cloud360 said:
2. I don't know how to get K0 for part a

can you help me with this. i don't even know what it means "by direct integration". i have seen the soluton but don't understand why they sub "0" into x, and not n?

K0 by definition is just ∫0∞ e-x dx, = [-e-x]0∞ = 1.
 
  • #17


tiny-tim said:
oh i see … it's just the [xn(-e-x)]0∞ that's worrying you …

ok, that's ∞n(-e-∞) - 0n(-e-0), wchih is 0 - 0

except of course we can't really write the first part like that, we have to put a limn->∞ in front of it :wink:


because the limits were there only because ∞ was one of the bounds of the integration, ∫0∞, so you got a […]0∞, which involved taking a limit

in most recurrence relations, there's either no integral at all, or there's an integral with finite bounds …

you do not normally get limits in a recurrence relation!​



K0 by definition is just ∫0∞ e-x dx, = [-e-x]0∞ = 1.

i am so grateful for your help. thanks a lot !

the 0-0 part was the thing i couldn't picture
 

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