tidaltinkerer

[tidaltinkerer]

Pure Mathematics. Topic Numerical Techniques.

Since I have a test on Monday I thought I would use this blog for its intended purpose.

When the normal techniques do not work, we use numerical Techniques to find the root of a function. The normal techniques are factorization, completing the square, and using the quadratic formula x= -b +- square root (b2-4ac )  /2a. 

    Firstly, we NEED to establish that there is a root, to do so we use the intermediate value theorem (IMVT). The formal definition of the IMVT is that

IF y=f(x) is a continuous function in the interval a<x<b and if “k” is a number which lies in between f(a)<k<f(b) or f(b)<k<f(a) then there exists at least one number “c” such that “c” lies between a<c<b and f(c)=k. 

BUT there is no need to commit the definition to memory 

Going back to establishing the root. 

Eg. Show that x^2 -3x - 1 = 0 has a root in [3,4]      

         The [3,4] in this case means interval,  3 < root < 4 or the root lies in between 3 and 4.

         For a root to exist there MUST be a change in sign either (-ve to +ve) or (+ve to -ve).

f(3) = 3^2 - 3(3) -1 

      = -1

f(4) = 4^2 - 3(4) - 1

      = 3

          What I just did is called substitution. I replaced the value of x with the values in the given interval.  

    Now we write a STATEMENT, the change in sign indicates that there is a root, and a statement is needed.

Since f(x) is a continuous function in [3,4] and f(3) x f(4) <0 then by IMVT there exits a root such that 3 <root< 4

3< root< 4 This means that ← the root lies in between the interval                

f(3) x f(4) <0 this means multiplied it produces a negative value which is less than 0  

This needs to be memorised. 

There are four methods in this topic:

Interval bisection

Liner Interpolation

Newton-Raphson

and

Fixed Pt Iteration 

In this post, I'll be covering Interval Bisection and Liner Interpolation.

Interval Bisection

First Prove that there is a root

f(x) ; x^2 - 4x + 1 = 0,  [3,4]

f(3) = (3)^2 - 4(3) + 1

      = -2

f(4) = 1

Since f(x) is a continuous function in [3,4] and f(3) x f(4) <0 then by IMVT there exists a root such that 3 < root < 4. 

Now using the interval bisection method

WE FIND THE MIDPOINT

3+4/2 = 7/2 = 3.5

Consider f(3.5) = (3.5)2 - 4(3.5) +1 = 0

f(3.5)= -0.75 

Now we change the interval 

Since f(3.5) x f(4) <0

Then 3.5 < root < 4   or [3.5, 4]

Keep in mind that the root EXISTS where there is a change of sign 

3.5 +4/2 = 3.75

Consider f(3.75)= 0.0625

Since f(3.5) x f(3.75) < 0

    Then 3.5 < root < 3.75   or [3.5, 3.75]

3.75 + 3.5 /2  = 3.625

Consider f(3.625) = -0.359

Since f(3.625) x f( 3.75)

Then 3.625 < root < 3.75

After a few more rounds you should get 

3.719 < root < 3.734    identical to 1 decimal place

As you can see this method can be lengthy

A question you might have is How do you know when to stop?

Well 2 ways:

When they are identical to the asked decimal places

eg. 3.7 < root < 3.7 which is identical to 1 decimal places

or after how many iterations they ask you for.

eg. Do interval bisection 3 times. With reference to the example we worked out above the answer here would be 3.625 < root < 3.75

 LINEAR INTERPOLATION 

 There are two proofs included in this method that I’ll attach a picture of below. You must memorise these proofs

In this method, we assign the interval value as either a or b. A is normally used for the smaller interval value and b for the larger interval value. 

We then find f(a) and f(b) and then substitute into any one of the formulas in the above proof.  An example of this is the picture below

There was no need to continue since 0.2 < root < 0.2

As stated we only need to put the statement: since f(a) x f(b)< 0, a< root< b if we are doing another interpolation which is why there is an absence of that statement after working out x2.

Again we stop either when the interval is identical to the stated decimal points or after how many iterations.

Hope you found this helpful, I'll post the second part... sometime😅 but for now I have to revise those other two methods for my test tomorrow.

[tidaltinkerer]

My oc made using different picrews... really wished more Picrews had more options for braided hair, but they're cool either way.