Is There a Simpler Method to Solve Complex Recurrence Relations?

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Homework Help Overview

The discussion revolves around finding a simpler method to solve complex recurrence relations, particularly in the context of polynomial coefficients that satisfy specific conditions. The original poster expresses a desire to avoid complicated nested loops in their approach.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the complexity of the recurrence relation and question whether it truly requires multiple nested loops. There are suggestions to calculate polynomial values theoretically and to express the problem as a system of linear equations.

Discussion Status

Participants are actively exploring different interpretations of the problem. Some have offered guidance on expressing the relations in matrix form, while others are questioning the necessity of the original complexity. There is acknowledgment of the potential for simpler approaches.

Contextual Notes

There is a mention of specific values needed for solving the equations, and some participants note the messy nature of deriving a general formula for the coefficients.

Jamin2112
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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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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??
 
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 "
 
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...
 
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
 
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...
 
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:
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
 
Yes, it would be something like that that you'll get.
 

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