Trying to find an easy way to solve this

  • Thread starter Jamin2112
  • Start date
In summary, The conversation is about finding a better way to solve for the coefficients in a polynomial that satisfies certain conditions. The person suggests using a system of linear equations and matrices, and another person suggests using MATLAB to solve it. The conversation also touches on the complexity of using a recurrence relation with nested for loops to solve the problem.
  • #1
Jamin2112
986
12

Homework Statement



screen-capture-38.png


Homework Equations



n/a

The Attempt at a Solution




I understand that I could make a REALLY complicated recurrence relation, one involving like 5 nested for loops, but I would prefer a better way of finding d1, ..., dn.


Is there a better way?
 
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  • #2
Uuh, what is the complete question?? All I can see is the question "Compute coefficients so that the polynomial..."

Maybe attach your pdf to here??
 
  • #3
Jamin2112 said:

Homework Statement



screen-capture-38.png


Homework Equations



n/a

The Attempt at a Solution




I understand that I could make a REALLY complicated recurrence relation, one involving like 5 nested for loops, but I would prefer a better way of finding d1, ..., dn.


Is there a better way?



Whoops!


" ... so that the polynomial satisfies the n conditions

p(t1) = b1 , ... , p(tn) = bn "
 
  • #4
Is the recurrence THAT complicated?? I'm really surprised how you can find 5 nested loops for this problem. Care to explain your reasoning?

What you could do is calculate (theoretically) p(t1),...,p(tn), and see what you get. It shouldn't be that hard...
 
  • #5
micromass said:
Is the recurrence THAT complicated?? I'm really surprised how you can find 5 nested loops for this problem. Care to explain your reasoning?

What you could do is calculate (theoretically) p(t1),...,p(tn), and see what you get. It shouldn't be that hard...

I still think it's complicated.

We have

b1 = d1
b2 = d1 + d2(t2 - t1)
b3 = d1 + d2(t3 - t1) + d3(t3 - t1)(t3 - t2)
b4 = d1 + d2(t4 - t1) + d3(t4 - t1)(t4 - t2) + d4(t4 - t1)(t4 - t2)(t4 - t3)
.
.
.

And it got really messy when I tried to find a general formula for di in terms of di-1
 
  • #6
Ah, yes, I see the problem

Jamin2112 said:
I still think it's complicated.

We have

b1 = d1
b2 = d1 + d2(t2 - t1)
b3 = d1 + d2(t3 - t1) + d3(t3 - t1)(t3 - t2)
b4 = d1 + d2(t4 - t1) + d3(t4 - t1)(t4 - t2) + d4(t4 - t1)(t4 - t2)(t4 - t3)
.
Express this as a system of linear equations, and thus as matrices. MATLAB is really good in solving these!

And it got really messy when I tried to find a general formula for di in terms of di-1

Don't worry with a general formula. I suppose you'll only need to solve it for specific values...
 
  • #7
micromass said:
Ah, yes, I see the problem


.
Express this as a system of linear equations, and thus as matrices.



Ooooooooooohhhhhhh! HOW DID I NOT SEE THAT BEFORE?


[URL]http://www.gifbin.com/bin/1233445870_ae19b02.gif[/URL]
 
Last edited by a moderator:
  • #8
micromass said:
Ah, yes, I see the problem


.
Express this as a system of linear equations, and thus as matrices. MATLAB is really good in solving these!



Don't worry with a general formula. I suppose you'll only need to solve it for specific values...

Is this the sort of set-up I'll have?


screen-capture-39.png
 
  • #9
Yes, it would be something like that that you'll get.
 

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