Help with Solving 5a & Understanding Recursion Relations

[cloud360]

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-∫(-nx^n-1e^-x)

but then i have to keep integrating forever


how do i do 5a, any help, any online references to learn?

 

[tiny-tim]

hi cloud360!

(have an integral: ∫ )

integrate xⁿe^-x=-xⁿe^-x-|(-nx^n-1e^-x)

but then i have to keep integrating forever

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

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

 

[cloud360]

hi cloud360!

(have an integral: ∫ )


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

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

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)

 

[cloud360]

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

but don't really know what to do

 

[tiny-tim]

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 K_n = nK_n-1 …

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

 

[cloud360]

no, homogeneous functions are irrelevant …

your recurrence relation is k_n = nk_n-1 …

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

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

 

[tiny-tim]

That's right!

Similarly, if you know K_1/2, then you can find K_3/2, then you can find K_5/2, and so on.

 

[cloud360]

That's right!

Similarly, if you know K_1/2, then you can find K_3/2, then you can find K_5/2, and so on.

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?

K_3.5
K_3.5=3.5*2.5*1.5*k_0.5
K_3.5=3.5*2.5*1.5*0.5*(π^0.5)
K_3.5=11.6317

 

[cloud360]

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

 

[tiny-tim]

K_3.5=3.5*2.5*1.5*0.5*(π^0.5)
K_3.5=11.6317

Yes, or you can write it K_(2n+1)/2 = (2n+1)(2n-1)… 1/2ⁿ K_1/2

= (2n+1)(2n)… 1/2ⁿ2n(2n-2)… 2 K_1/2

= (2n+1)!/2²ⁿ(2n)! K_1/2

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

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.

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:

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?

 

[cloud360]

Yes, or you can write it K_(2n+1)/2 = (2n+1)(2n-1)… 1/2ⁿ K_1/2

= (2n+1)(2n)… 1/2ⁿ2n(2n-2)… 2 K_1/2

= (2n+1)!/2²ⁿ(2n)! K_1/2

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


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-∫(-nx^n-1e^-x)

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

 

[tiny-tim]

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

∫xⁿe^-x=-xⁿe^-x-∫(-nx^n-1e^-x)

i then said i would need to integrate forever because of this ∫(-nx^n-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

 

[cloud360]

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.

 

[cloud360]

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

 

[cloud360]

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

 

[tiny-tim]

… 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 lim_n->∞ in front of it

… 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!​

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.

 

[cloud360]

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 lim_n->∞ in front of it


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