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\(S_0(x)=1+2x-x^3 \to 0 \leq x \leq 1\) \(S_1(x)=2+b(x-1)+c(x-1)^2+d(x-3)^3 \to 1 \leq x \leq 2\)
Sorry for the late reply. I was at class and then attended a seminar. Anyway: You know your three points. 0, 1 and 2. You have been given some of your coefficients ( specifically five of them) and need to find the other 3. So, you need 3 equations. Can you tell me which 3 equations you need to use? You're almost given one of them to start.
Yes I figured it out \(S_0'(1)=S_1'(1)\) \(S_0''(1)=S_1''(1)\) \(S_1''(2)=0\) THHAANKKSSSS :)
Um. Be careful there. S1''(2) = 0 implies that you have a natural spline. You don't know that. So, that doesn't work. But, you're almost there. Looking at the first two equations you gave me (which are correct), can you find a very similar equation that follows the same pattern as those two?
Oh i forgot to mention that it is a natural spline lol
Ah, okay. Well, in case you didn't know, you should still be able to solve it by using the fact that: S0(1) = S1(1), as they have the same y value.
Gotcha Can i dirve you crazy with one more question?
Go for it. I'm trying to avoid my own homework as it is.
hahaha I do the same. I sit on here answering questions and im like thinking y the hell am I on here when I gotta do my own
Construct a natural cubic spline to approximate f(x)=Cos(pi*x) by using the values given by f(x) at x=0, .25,.5, .75 and1 Integrate the spline over [0,1]
I think I am allowed to use a computer program but \i am wondering How can this be done by hand
Or is it wayyy too time consuming?
Well, you have 5 points. So, you'll end up having 16 coefficients to solve. You're better off solving the equations using a computer program. So, your points will be the x defined above and your five y values will be f(x) for each x. Then, just go back to making cubics on those intervals with unknown coefficients and do all the steps for defining all 16 equations.
A(I) B(I) C(I) D(I) 1.00000000 -0.7573593129 0. -6.627416998 0.7071067812 -2. -4.970562748 6.627416998 6.123233996*10^-17 -3.242640687 4.440892099*10^-16 6.627416998 -0.7071067812 -2. 4.970562748 -6.627416998
ohhhh it didnt work out uughhhhh
whtvrrrr ok once i get the 16 diff values then what?
It seems like I am suppossed to integrate but not sure y we are integrating
Don't worry about integrating right now. You want to find your coefficients first so you can find all 4 cubics. Then, after you know your cubics, you integrate them on their intervals. But, not right now. Worry about finding your coefficients using your equations.
ok I will write out all the equations then
\(S_0(x)=1-.7573593129x-6.627416998x^3\) \(S_1(x)=.707106812-2(x-.25)-4.970562748(x-.25)^2+6.627416998(x-.25)^3\) \(S_2(x)=-3.242640687(x-.5)+6.627416998(x-.5)^3\) \(S_3(x)=-.7071067812-2(x-.75)+4.970562748(x-.75)^2-6.627416998(x-.75)^3\)
So there are my four equations
Have you tried plotting them? If those are correct, they should match f(x) = cos(pi*x) pretty well.
hmm let me try
ummm pretty much sooo
Its weird that one of the equations has nothing to do with the graph at all
Yeah, only looks like one or two of them are correct. You may want to double check your equations.
ohhh I found my mistake mixed up signs Its only suppossed to match from [0,1] and it does
its annoying to work with sooooo mannnyy decimals
So now i integrate?
Yes. Now you can integrate using those cubics (so, from 0.25 use S1, 0.25 to 0.5 use S2, etc.).
I am just wondering y we are integrating lol I am just not seeing the logic behind that?
The idea is you're able to approximate the integral of any function in the real numbers by building a cubic spline and integrating the pieces of the spline.
ohhhhh I get it Ya that makes sense Like I am taking this course online and its def getting annoying wish I had a teacher to explain everything clearly THANKS YOU ARE GREATTTTTTTT
Do I add all the equations together and that will be my spline function?
It sounds like it's just asking you to find the derivative of f(x) = cos(pi*x) at 0.5.
No it asks me to compare both so it cant mean that
Compare both? Ah okay. So, yes. Find the derivative at 0.5 of the original function. Then, find the derivative of 0.5 using the cubic defined at 0.5 (there should be two of them, since it was one of your points you used when building your piecewise function). The values of both should be pretty close.
Ohhh what i did for a previous question gottcchhhaaaaaaa
I am seriously driving u crazy here. Thanks I really appreciate your help
Ha. That's okay.