solve for a

Onyx · 2010-12-30 05:18:43

Hi, can anybody help me solve this for a:

I'm really stuck

bobbym · 2010-12-30 06:39:47

Hi Onyx;

Who would not be stuck with that? If you are looking for a hand method then you have a major problem. Can be done by numerical methods with great care and great skill and lots of computing.

You should recognize the above answer. It comes with a caveat. An important one.

Took a while to bound the error somewhat. That answer is correct to at least 1000 digits.
If you look at it you will see why that answer is very close. It is even possible to come up with another possible root.
If you want we could look for more and I will show you how to find them using the new "experimental math."

soroban · 2010-12-30 06:49:56

. . .

. . . . . . . . . .

. . . . . . . . . . . . .

. . I'll wait in the car . . .

.

Onyx · 2010-12-31 00:17:49

haha thanks both!

I beleive you are both right, since I have |a|<1, so there are two approximate solutions....thats good enough for me.

(Application is design of a discrete time LTI filter for an audio DSP)

bobbym · 2010-12-31 01:15:51

Hi Onyx;

I do not agree, there is one root extremely close to

The exact root differs from this starting at the 1671 decimal place!
There is also another root at 1.000337250317441004525771... I am sorry you did not want to see how these are bounded. derived and proved through the methods of Numerical Analysis.

tobleronebeater · 2010-12-31 04:38:15

hi, what do u mean by methods of numerical analysis? do u mean with a computer program, or is there an actual method to do it by hand that doesnt include trial and error?

bobbym · 2010-12-31 05:02:36

Hi tobleronebeater;

do u mean with a computer program,

Not exactly, more accurate to say an algorithm to find the roots.

or is there an actual method to do it by hand that doesnt include trial and error

Hand methods are extremely limited. They are okay for polynomials of the first and second degree after that they become difficult or impossible. For the general case of a polynomial of higher than the 4th degree it is not possible. For non polynomials it is even more difficult. Unless you are dealing with a book problem designed to illustrate some hand technique in the real world you are stuck with numerical solutions.

Iteration is not trial and error. Newton's iteration is a common iterative technique.

Welcome to the forum and Happy New Year!

Onyx · 2010-12-31 05:45:46

Hi bobbym, actually as I said there are two approximate roots of the equation, given the constraint |a|<1. They are:

But only,

satisfies the constraint |a|<1, since,

Also, I know that these are only approximations, and for my engineering design application, they are extremely accurate approximations, and will suffice for the limited precision required.

I'm also familiar with iterative methods such as Newton or Newton-Raphson in numerical analysis, and am a proficient programmer, and would be able to implement such algorithms myself....Excel could also be used. I was just interested to see if the problem had a solution using algebraic manipulation, since my maths is a bit rusty in some areas, and this one really stumped me. I appreciate the offer though, but the solution is good enough for my problem.

Thank you

bobbym · 2010-12-31 06:08:07

Hi Onyx;

That is the Golden Ratio and it is not a root of that polynomial, nor is it an approximate root. That is easy to prove.

My point is this, I found two roots one near the golden ratio - 1 and the other close to 1.000333... That is an 8001th degree polynomial! There could be many roots close by to those two.

The problem is hardly solved. It needs to have some analytical work done on it. If you use Descartes rule of signs it can have at most 2 positve roots ( they are accounted for. See my earlier post ) and at most 3 negative roots. What about these 3 possible negative roots? They could be in the interval |a|<1.

I'm also familiar with iterative methods such as Newton or Newton-Raphson in numerical analysis, and am a proficient programmer,

If you attempt to utilize newtons iteration on this problem in say C++ you might underflow or overflow. Just look at those exponents. You might bounce around and never converge. Round off error might kill those 17 digit precision long doubles. Newtons requires one to be close to the roots. Currently we have no idea where they are at.

Onyx · 2011-01-01 04:49:56

Why is it not possible to do this?

Then:

Then by the quadratic formula:

but,

bobbym · 2011-01-01 04:54:26

Hi Onyx;

It is not impossible, I used that to get the estimate for the first root. It is just not exhaustive! Also it does not rule out

nor does it find 1.000333... Notice that it is larger than one but still converges.

Besides any of that there are either 1 or 3 negative roots that could have an absolute value < 1.

aleclarsen12 · 2011-01-02 16:50:53

The only solution such that |a| < 1 is:
a ~~ 0.6180339887498948482045868343656381177203...

(obtained using Mathematica command FindRoot[100a^8001 - 100a^7999 - a^2 - a + 1 == 0, {a,-1, 1,WorkingPrecision -> 39] )

bobbym · 2011-01-02 20:40:46

Hi aleclarsen12;

Using the FindRoot command like that is misleading. What happened is that it converged to .618.... That does not mean that is the only root in the interval [-1,1]. FindRoot finds one root, it does not find them all.

Ex:

FindRoot[(x-1)(x-2)(x-3)==0,{x,0,4},WorkingPrecision->50]

{x->2.0000000000000000000000000000000000000000000000000}

That polynomial obviously has roots of 1,2,3 in the interval of [0,4]. FindRoot finds only one. This is typical of iterative solvers.
When trying to get them all you have to use Solve,NSolve,Reduce, etc.

Math Is Fun Forum

#1 2010-12-30 05:18:43

solve for a

#2 2010-12-30 06:39:47

Re: solve for a

#3 2010-12-30 06:49:56

Re: solve for a

#4 2010-12-31 00:17:49

Re: solve for a

#5 2010-12-31 01:15:51

Re: solve for a

#6 2010-12-31 04:38:15

Re: solve for a

#7 2010-12-31 05:02:36

Re: solve for a

#8 2010-12-31 05:45:46

Re: solve for a

#9 2010-12-31 06:08:07

Re: solve for a

#10 2011-01-01 04:49:56

Re: solve for a

#11 2011-01-01 04:54:26

Re: solve for a

#12 2011-01-02 16:50:53

Re: solve for a

#13 2011-01-02 20:40:46

Re: solve for a

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