Before getting into the course, let's look at 'what numerical methods' really mean.

### What are numerical methods?

Numerical methods are techniques by which mathematical problems are formulated so that they can be solved with arithmetic operations yielding numerical results instead of closed-form results. Most numerical methods involve large amounts of arithmetic calculations; most of them are *trials and erros*.

Sometimes, engineering problems cannot be efficiently solved for exact answer. But instead with a large number of engineering problems being solved using trials and errors (due to the development of fast and advanced computers).

let's look at an example :)

##### Finding irrational square root

We continue if x

^{2}< 6. But if it still exceeds, we move on to the next decimal point. BUT we have to stop somewhere, and it's when the result isclose enough; which we get from approximating. (But we have to be careful aboutaccuracyandprecision.

A picture to demonstrate what are accuracy and precision:

Let's move on to see how we solve a problem..

### ERROR

But hey, if there's something that comes with an approximation, it's an *error*. There are 2 types of errors; *true error* and *approximation error*.

##### True error

True error (E

_{t}) = True value - Approximation

True relation error (Ɛ^{t}) =^{True value - Approximation}⁄_{True value}x 100

But the problem with true error is that we don't know what the exact *true value* is. So instead we use *Approximation error*.

##### Approximation error

Ɛ

_{α}=^{Approximation error}⁄_{Approximation}x 100

Ɛ_{α}=^{Present approximation - Previous approximation}⁄_{Present approximation}x 100

And there are 2 types of approximation error:

1) **Truncation error** is the error made by truncating an infinite sum and approximating it by a finite sum.

2) **Round-off error** is the difference between the calculated approximation of a number and its exact mathematical value due to rounding.

Ok, so back to the 'when to stop' thing and approximation. How will we be *confident* enough to stop the calculation?

The answer is (not really, but what we use for now) when.. |Ɛ_{α}| < Ɛ_{s}

Ɛ

_{α}is approximation error

Ɛ_{s}= (0.5 x 10^{2-n})%

- Ɛ
_{s}is already in percentage- n is a number of significant figures we want

And below is an octave (similar to matlab, but a free, open-source one) code for calculating the √6.

```
% take squareroot of 6
m = 6;
true_value = 2.44948974270319;
% nf = the number of the significant figures
nf = 4;
es = 0.5 * 10^(2-nf);
printf('es = %f\n', es);
% set x = the approximation answer
% initialize x = 0
previous_result = 0;
x = 0;
% outer loop to loop for the number of digit
% we want to do this for ndigits
n = 6;
%for d = 0:n;
ea = 100;
d = 0;
while (abs(ea) > es)
previous_result = x;
for i = 0:10
x = previous_result + i/(10^d);
if (x*x > m)
break;
endif
endfor
x = x - 1/(10^d);
ea = ((x - previous_result)/x)*100;
et = ((true_value - x)/true_value) * 100;
printf('x = %f, ea = %f, et = %f\n', x, ea, et);
d++;
endwhile
%printf('x = %f.\n', x);
```

And below is the result of the above code:

```
es = 0.005000
x = 2.000000, ea = 100.000000, et = 18.350342
x = 2.400000, ea = 16.666667, et = 2.020410
x = 2.440000, ea = 1.639344, et = 0.387417
x = 2.449000, ea = 0.367497, et = 0.019994
x = 2.449400, ea = 0.016331, et = 0.003664
x = 2.449480, ea = 0.003266, et = 0.000398
```

Ok, I'm going to stop here for today, see you in the next part! xD

Posted on by:

### rinzzzz

Hey ya! Just another girl who codes ;) Still new and learning (hopefully I'll be getting better and learning soon)

## Discussion

One semester about approximations in computer science didn't help me understand them and this post did. Thank you for the post!

Thank you! I'm so glad it helps

^{^}I'll be writing the next lesson next, I hope it also helps!!Thank you for the post. You saved my life.