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**Relentless****Member**- Registered: 2015-12-15
- Posts: 624

I don't have the time to figure out how to do it right now, so I thought I might post it while I'm here.

It gives the solutions for the discount rates per annum that set two engineering options equal in terms of present financial value. It is not an assessment question, it is just something I encountered. Exact solutions are sought.

It is already simplified a little, but I am a bit lost now.

I am concerned more with answers than explanations at the moment. (:

*Last edited by Relentless (2016-04-13 15:51:05)*

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**Relentless****Member**- Registered: 2015-12-15
- Posts: 624

Apparently it is equivalent to...

*Last edited by Relentless (2016-04-13 17:05:50)*

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

Without a CAS?

There appears to be only two real roots. Finding analytical forms for them might be difficult or impossible.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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**Relentless****Member**- Registered: 2015-12-15
- Posts: 624

It does not have to be done by hand, but I would like to know something about the two analytical forms (or determining that they cannot be found). If you cannot find them, can you elaborate on how this might be investigated?

It is not a practical problem since it is relatively easy to approximate the answers to a desired degree. I am just curious (and pedantic)

*Last edited by Relentless (2016-04-13 21:28:40)*

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

Hi;

The form you produced in post #2 is not the same as post #1.

The two real roots are:

x = .046648023052674644930849814289785515159221216598276...

x = .31595555552492574948785882734255958583342547212420...

To get them obviously requires a CAS.

The procedure is to first establish a bound on the real roots using a Cauchy bound and or graphing, with a further refinement using Sturm sequences.. Then some iterative scheme has to be used to zero in on the roots once they have been isolated approximately.

To get closed forms requires the use of a PSLQ. But since most numbers can not be expressed in terms of known constants, this will often not be possible.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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**Relentless****Member**- Registered: 2015-12-15
- Posts: 624

Hi bobbym,

The only difference between the forms in post #2 and #1 is that #2 is continuous at x=0.

Thank you for bringing integer relation algorithms to my attention!

It has honestly never occurred to me before that there might be numbers that can't be expressed in terms of known constants. It sounded counter-intuitive to me, until I realised that it does not imply that there are numbers which cannot be expressed in terms of defined operations. It merely means that it is sometimes necessary to expand our collection of "known constants".

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Relentless wrote:

Apparently it is equivalent to...

For linear, quadratic, cubic and quartic polynomials, finding the roots is an easy task, because there are formulae for those roots in terms of the co-efficients of the polynomial and radicals. But this is in fact not true for polynomials of degree 5 or higher in general: the reasoning behind this is answered by Galois theory.

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**Relentless****Member**- Registered: 2015-12-15
- Posts: 624

It would be interesting to be an expert in the representation of solutions with no analytic expression. Surely there is in principle some expression with no self-reference, if everything including limits, integrals, functions etc. are included?

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

Solving equations, except for the simplest polynomial requires numerical analysis and with it computers.

The only difference between the forms in post #2 and #1 is that #2 is continuous at x=0.

That form has an additional real zero. Also, when you plug in numbers you will see the two forms are not equal.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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