difference equations?
amistre...hi
Step 1: What you want to do is drop the constant, the same as simply making the constant zero to get the homogeneous difference equation. Then you get a general solution to that homogenous difference equation. Step 2: Then go back and come up with a trial solution to the non-homogenous difference equation, the style of the trial solution being based on the kind of terms that are creating the non-homogeneous part (this will probably be the hardest part, so read that section of material carefully). Then put in particular values that you know solve the difference equation, specific values you got from working with the difference equation such as what xo is, what x1 is, allowing you to determine what the constants in your trial solution need to be. This gives you your particular solution. Step 3: As the last step you combine your general solution and your particular solution to get your complete solution for this specific problems, and of course verify that when applied it creates the correct values for x0, x1, .... xn. Overall you will have started with a "difference equation" expressing how to get x n+1 if you know x n, and you are ending with a formula that will calculate any x n by just knowing n. The formula allows you to calculate any x n term without working out all of the terms ahead of it.
isnt this from yesterday :)
yes need clear understanding for solutions starting with 2b
I have 0,0 1, -1 2, -1.5 , 3 -1.75 4 -1.875, 5 -1.9375 6, -1.96875 for a
this is on 2a right?
yes
you need to track these as community service hours ; )
add to resume
lol.... the problem or this session ;)
all of your time in OS
lol
just realized that was a joke, lol
homo diff is dropping the constant... :)
what does dropping the constant do for us?
gives us xsubn?
the good computers are all locked away today; this one hates openstudy. small font, no formating..... just awful
might go to McDs later and use the wifi.... if it aint busy/closed
yeah my laptop will not let me do anything in os anymore so have to sit at desktop
are you in school
so if we strip the constant; we are just left with: Xn+1 = (1/2)Xn which always equals itself, so I dont quite know why wed do that....
yeah, in the computer lab downstairs....
are you a student
yup
me to and unemployed since move
taught 8th grade Algebra back in SC
i cant get work since I hurt my back.
ima "liability"
oh no sorry to hear that...ever thought about starting online tutoring business..I can help you out
i thought about trying to do something like that; id be nice to have some sort of income other than PELL grant...
I'm studying off loans for now ; ( that is why I have to take this extra module that has nothing to do with my program of study...technically
send me your email address and I can send you a presentation I'm actually presenting on Tues. and the link to my website...I just started in March and have 3 students now but hope to advertise some more this summer...brishundra@yahoo.com
So the complete solution is jsut finding an equation that will sum up all the nth terms....
I have xn=(0.5)raised to n *C
keep looking back at table in a and messing me up bc different ?
couldnt we just solve for "Xn" and plug it in? maybe?...
y = .5(0) - 1 + .5(1) - 1 +.5(2)-1 + .5(3) -1 seems to be what they are getting at.
we can rearrange our equation to see it better: y - .5(0) + .5(1) + .5(n) - 1 - 1 - 1 - 1
we need ".5(n)" for any given interation of n right?
yes
and we need -(1n) for every iteration of n right?
whatis: .5(n) - (n) = ?
the constant
at =1 we get -1.5 right?
at n=2 we get: .5(2) - (2) = -2.5 right?
-1.5 maybe, gonna have to see this on a calculator.... terrible font
ok
whats yout table in a look like?
0,0 1, -1, 2, -1.5 3, -1.75...
4 -1.875 5 -1.9375 6 -1.96875
i get from teh original stuff, not my stuff: 0; -1; -1.5; -1.75; -1.875...
I had that when I first started this , lol
but looked at xo=0 and thought I was wrong
your "key returns" arent formating; they look like this:
the value of X0 = 0 then it iterates from there
ok so 6 -1.984375
we get add ons of 1/2; 3/4; 7/8; 15/16; 31/32; 63/64; 127/128; 511/512
(2^n)-1 ------ 2^n
[(2^n)-1] -1 - -------- = y 2^n
thats our formula/equation for this bit lol
n=[1,inf]
it fits right?
the bottom of the fractions ar plugging away at 2^n; and the tops are following one step behind at (2^n)-1
X{n+1} = (1/2) X{n} 2X{x+1} = X{n} I got no idea how they supposed to do this and come up with that lol
X{n+1} 1 ----- = --- X{n} 2 <-- 2^n ? if we borrow that "n"
any ideas?
lost...trying something
I got the right equation; just dont know how to get it from their stuff....
is that for c
thats for ..... D i think
you got a link to the "look over this material" part of it?
The style of the trial solution being based on the kind of terms that are creating the non-homogeneous part (this will probably be the hardest part, so read that section of material carefully).
got 10 minutes til the college closes....
and StLeo is a catholic college so that ones closed for today as well; my internets are getting scarce :)
email me the "material" section: tony031172@gmail.com
ok THANKS a bunch!
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