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How to do decimal search?

How to do decimal search

Also, is there a quicker way to do it?
Reply 1
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Mesosleepy
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Reply 2
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Andy98
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Original post by Mesosleepy
How to do decimal search

Also, is there a quicker way to do it?


Not heard of that method before, could you give me a rough idea of what it is?
Reply 3
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Mesosleepy
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Original post by Andy98
Not heard of that method before, could you give me a rough idea of what it is?


I'm not sure, but it is something to do with locating roots.
Reply 4
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Andy98
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Original post by Mesosleepy
I'm not sure, but it is something to do with locating roots.


Ohhhh, that thing. I can't remember the details, but you basically find an iterative formula and spam equals. Although TeeEm would be better at explaining it than me.
Reply 5
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TeeEm
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Original post by Andy98
Ohhhh, that thing. I can't remember the details, but you basically find an iterative formula and spam equals. Although TeeEm would be better at explaining it than me.


sorry but I am teaching at the moment
Reply 6
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Andy98
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Original post by TeeEm
sorry but I am teaching at the moment


Fair enough, sorry to disturb :colondollar:
Is it like the bisection method, but splitting the interval into ten instead of two? If it is, just substitute values of x into the equation of the curve and then the root lies in the interval where it goes from negative to positive or positive to negative.
Original post by Mesosleepy
How to do decimal search

Also, is there a quicker way to do it?


A decimal search is a particular variation of the change of sign method. First assume there is a unique root. If f(0) and f(1) are of different sign, then you know there is a root between 0 and 1. Divide the interval [0,1] into ten and check the end-points of each of these sub intervals. The sub-interval that has a change of sign on its end-points contains the root. Now subdivide that interval. Continue until you reach a preset accuracy threshold. Deal with the case of multiple roots in the obvious fashion.

There a many ways of finding roots; the Wikipedia article is a pretty reasonable introduction.
Reply 9
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Mesosleepy
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Original post by Gregorius
A decimal search is a particular variation of the change of sign method. First assume there is a unique root. If f(0) and f(1) are of different sign, then you know there is a root between 0 and 1. Divide the interval [0,1] into ten and check the end-points of each of these sub intervals. The sub-interval that has a change of sign on its end-points contains the root. Now subdivide that interval. Continue until you reach a preset accuracy threshold. Deal with the case of multiple roots in the obvious fashion.

There a many ways of finding roots; the Wikipedia article is a pretty reasonable introduction.


Can you give an example?
Original post by Mesosleepy
Can you give an example?


Let f(x) = x - pi

Then f(3) < 0 & f(4) > 0. So there is a root in [3,4].

Divide up [3,4] into ten equal sub-intervals.
Then f(3.0) < 0, f(3.1) < 0 - so not this interval,
f(3.1) < 0 & f(3.2) > 0 - this interval.

So divide up [3.1, 3.2] into ten sub-intervals. Then
f(3.10) < 0 & f(3.11) < 0 - not this interval
f(3.11) < 0 & f(3.12) < 0 - not this interval
f(3.12) < 0 & f(3.13) < 0 - not this interval
f(3.13) < 0 & f(3.14) < 0 - not this interval
f(3.14) < 0 & f(3.15) > 0 -this interval. So, divide up [3.14, 3.15] into ten etc etc...
Reply 11
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Mesosleepy
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Original post by Gregorius
Let f(x) = x - pi

Then f(3) < 0 & f(4) > 0. So there is a root in [3,4].

Divide up [3,4] into ten equal sub-intervals.
Then f(3.0) < 0, f(3.1) < 0 - so not this interval,
f(3.1) < 0 & f(3.2) > 0 - this interval.

So divide up [3.1, 3.2] into ten sub-intervals. Then
f(3.10) < 0 & f(3.11) < 0 - not this interval
f(3.11) < 0 & f(3.12) < 0 - not this interval
f(3.12) < 0 & f(3.13) < 0 - not this interval
f(3.13) < 0 & f(3.14) < 0 - not this interval
f(3.14) < 0 & f(3.15) > 0 -this interval. So, divide up [3.14, 3.15] into ten etc etc...


When do I stop?
Original post by Mesosleepy
When do I stop?


When you reach a preset degree of accuracy. This method gives you an extra decimal place for each iteration.

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