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numerical methods , fixed point iteration and initial guess ??


bimbo36

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i have this few doubts about taking an initial guess ... i am not sure how to do that when it comes to certain equations for solving them with numerical methods ...i dont know how to do that
you are supposed to take an initial guess when it comes to certain equations ...

is it about re arranging equations to fit certain formulas ... ???

do i have to take an initial guess of something in the equations when methods like these are involved ... ???

 

 

fixed point iteration
the bisection method
the newton raphson method
the secant method

 

few simple examples .. might look like this ...

 

 

that is fixed point iteration on a quadratic equation ...

they start by re arranging equations to apply the fixed point iteration ...

what is the initial guess there ??

and another example here ...

 

2ab997c.png

 

2e49ce0.png

:confused:

 

 

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One way of making an educated guess would be to sketch the function(s) you are studying. For example if you sketch the function f(x) = x^2 -x-1 you can see that one of the solutions to f(x) = 0 is a little below 2. My first guess here would be closer to 1.5, but that does not matter too much.

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2d97c96.png

 

 

 

 

3d99369.png

 

 

 

 

2ab997c.png

 

2e49ce0.png

first of all i am sorry for using few pictures... because i am completly lost without it ...

 

i am also so lost and confused between all these equations ... the terminologies and stuffs ..

 

we use numerical methods depending on the type of equations ...

 

the equations could be linear or non linear ... right ?

 

depending on the type of equation ...

 

we use the numerical methods .. mentioned below ....

 

 

 

 

 

 

When we know the degree we can also give the polynomial a name:
Degree Name Example
0 Constant 7
1 Linear 4x+3
2 Quadratic x2−3x+2
3 Cubic 2x3−5x2
4 Quartic x4+3x−2

 

then we have simulaneous equations ...which looks like these ...

 

 

x+2y-3z=10
2x-3y-4z=1
y-3x+z=-8

 

these two numerical methods can be applied to it if you have to deal with equations like these ...

 

 

http://www.ce.utexas.edu/prof/mckinney/ce311k/Overheads/12-LinEqs_Direct.pdf

 

http://www.ce.utexas.edu/prof/mckinney/ce311k/Overheads/13-LinEqs_Indirect.pdf

 

 

now for non linear equations ...

 

these methods can be applied ....

 

http://www.ce.utexas.edu/prof/mckinney/ce311k/Overheads/14-NonlinearEquations_1_FixedPoint.pdf

 

 

http://www.ce.utexas.edu/prof/mckinney/ce311k/Overheads/14-NonlinearEquations_2_Bisection.pdf

 

http://www.ce.utexas.edu/prof/mckinney/ce311k/Overheads/14-NonlinearEquations_3_Newton.pdf

 

 

http://www.maths.dit.ie/~dmackey/lectures/Roots.pdf

 

 

these all are non linear, fixed point iteration type numerical methods ...

 

 

 

 

fixed point iteration
the bisection method
the newton raphson method
the secant method

 

 

 

is that an alright graph ? :wacko:

 

so everytime i have to take an initial guess ? i have to plot its graph ...

 

i thought i almost had a grip on this subject ... now i lost it again ....

 

i am so lost ... why am i taking an initial guess for ? what is this initial guess consist of ??? :wacko:

Edited by bimbo36
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i am so lost ... why am i taking an initial guess for ? what is this initial guess consist of ??? :wacko:

Usually you are looking for f(x) =0, or maybe where the plot crosses one of the axes or meets another plot etc. It depends on the question. Sketching the functions maybe enough to give you an educated guess.

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Your last diagram seems okay to me. Sketching or carefully plotting the function will give you some reasonable starting point to try. It will also help you understand if there are multiple solutions, some of which maybe ruled out by further conditions (say +ive root required).

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sorry for making it look like never ending dumb questions ...

 

2ec01a2.png

 

do i keep typing that values in that plotter ??? anyway its giving me graph that looks like a straight forward line at different points?

 

and thats the approximate iterated answers ??

 

do i call it the possible roots of the equations ??

 

:confused:

Edited by bimbo36
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and thats the approximate iterated answers ??

So x = 1.618 is an approximation for one of the roots. Meaning that f(1.618) is close to zero. How close you can check yourself.

 

If you look at the graph of the function you see that near x= 1.618 the graph crosses the x-axis. Notice it also crosses some where near x = -0.5. You can try your same method near this negative value to see what you get. Then substitute that value back into the original equation to check how close it really is to a root.

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so this is what its mostly about .. f(1.618) being close to zero .. therefore a root of the equation ...

 

what practical value has this value has ???

 

772af83.jpg

 

i am trying to think if it has any practical application ... that value being there ???

 

:confused:

 

is it like .. at f(1.618 ) .. the ball touches the ground ???

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so this is what its mostly about .. f(1.618) being close to zero .. therefore a root of the equation ...

 

what practical value has this value has ???

You would need to provide some context to the equations you are solving before anyone can answer that in general.

 

However, for the specific quadratic you have given, the positive root is known as the golden ratio. (This you can get an exact closed expression for using 'The Formula')

 

In more generality, it may not be possible to find a closed expression or exact value for f(x)=0 (for any f) and one will have to use approximations or numerical methods.

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thanks a lot for the answers ....

 

yea i almost forgot that , the value will be used to fit certain formulas ...

 

and the formulas should be able to give you more points in the graph after applying the appropriate numerical methods ...

 

then it becomes like possible points on that particular space .... ??? to do something ... ??

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i have few more questions ... i am almost exhausted trying to ask proper questions ... because its almost like i was trying to learn this from scratch ... it was a bit hard to find simple examples too ...

 

anyway the good thing is i am starting to understand this numerical methods types problems... i now sort of understand what others in this forum were trying to tell me about .. exact values and approximate values ..

 

i have few more doubts about some basic concepts ...

 


 

 

i think the one below all are the fixed point iteration type questions too...

 

 

 

 

fixed point iteration
the bisection method
the newton raphson method
the secant method

 

let me again start with .. with this simple example ...

 

i was trying to get a better picture with few images like these ...

 

it was helping me to take an initial guess ....

 

 

 


 

06f2c9d.jpg

 

772af83.jpg

 

 

 

 

2ab997c.png

2e49ce0.png

 

 

 

i think i figured it out that the initial guess ,(the initial root )... the answer of the question ... is like f(1.618) ... which makes the equation close to zero ... ??

this positive root is called a golden ratio ... right ??

is this like an answer to the question already ???

should i be looking for more values which makes the equation closer to zero ?

is that why we use numerical methods such as fixed point iteration ... ??

anyway then you put the values to fit the formulas of iteration methods ...

to get more approximate values which makes the equation closer to zero ???

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i think i figured it out that the initial guess ,(the initial root )... the answer of the question ... is like f(1.618) ... which makes the equation close to zero ... ??

Yes

 

this positive root is called a golden ratio ... right ??

For this particular example the positive root is known as the golden ratio.

 

is this like an answer to the question already ???

Yes, for this example you can find a closed expression. In general for more complicated functions this is not always possible.

 

should i be looking for more values which makes the equation closer to zero ?

Your numerical approximation for the root can be made more and more accurate making the value of f closer and closer to zero.

 

is that why we use numerical methods such as fixed point iteration ... ??

In this case each iteration should get us closer to the actual root. The reason for numerical methods is that it can be impossible to find exact expressions as solutions.

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thanks a lot .. these replies means a lot to me ... i thought i was never going to learn these things properly ... i have understood a lot of basic stuffs ... thanks to this forum and its members ....

 

now i have something to think about and pursue at free times ...

 

i have lot more to learn ...but now atleast i know where this is all heading ...

 

i was unable to get a bigger picture of this subject for a very long time ...

 

but after these discussions and replies .. i should be able to follow this a bit more deeply ...maybe not today or tommorow ... but someday its going to happen ....

 

 

^_^

Edited by bimbo36
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