CS101: Numerical Computing 2

Abhiram Ranade

Topics

More examples of for statement Evaluating ex. Hemachandra Numbers Newton Rhapson method for finding roots

Defining Functions

ex

We use the series ex = 1 + x + x2/2! + x3/3! + ...

We will evaluate the series to some 10 terms.

ith term is xi/i!. How can we calculate it quickly?

Direct method: write a loop to calculate xi and also i!, then divide. loops within loop.

Better idea

ith term

= xi/i! = xi-1/(i-1)! * (x/i)

= i-1th term * (x/i)

So we calculate ith term from the i-1th term calculated earlier, rather than directly.

Program

float x; cin >> x;

float term = 1.0; // 0th term of the series.

float answer = 0.0;

for( int i=1; i <= 10; i++){

answer += term;

term = term * x / i;}

cout << answer;

Notes

You must put the following comment to your program:

/* After ith iteration, i terms of the series have been summed up */

Check that this is true!

- by inspection,

- Add a statement to print after each iteration and check.

Program 2

float x; cin >> x;

float term = 1; // 0th term of the series.

float answer = 0;

for( int i=1; term > 0.0001; i++){

answer += term;

term = term * x / i;}

cout << answer;

Remarks

If we calculate xi/i! independently in each iteration, we will need about 2i arithmetic operations in each iteration.

In our program, we just need 2 in each iteration. Idea of reusing what was calculated in the

previous iteration is very important, and also very commonly useful.

Hemachandra’s Problem (12th century AD)

Suppose I have to build a wall of length 8 feet. I have bricks 2 feet long and also 1 foot long. In how many ways I can lay the bricks so that I fill the 8 feet?

Possiblities: 2,2,2,2; 1,1,1,1,1,1,1,1; 2,2,2,1,1

....

Hemachandra’s Actual Problem

Suppose I am designing a poetic meter with 8 beats. The meter is made of short syllables and long syllables. Short syllable = 1 beat, Long syllable = 2 beats. How many ways are there of filling 8 beats?

Example of a poetic meter

Aa ji chya Ja wa lii gha dya l ka sa lay aa he ch mat kaa ri k

ya kun den du tu sha r ha r dha va la ya shubh r vas tra vru ta

l l l s s l s l s s s l l l s l l s s

Hemachandra’s Solution

“By the method of Pingala, it is enough to observe that the last beat is long or short”

Pingala: mathematician/poet from 500 A.D. Hemachandra is giving credit to someone who

lived 1500 years before him!! Copy but give credit..

Hemachandra’s solution contd.

S : Class of 8 beat patterns with short last beat.

L : Class of 8 beat patterns with long last beat.

Each 8 beat pattern is in class L or class S

S = all 7 beat patterns + long beat appended.

| class S | = Number of patterns with 7 beats

| class L | = Number of patterns with 6 beats

|8 beat patterns| = |class S| + |class L|

= |7 beat patterns| + |6 beat patterns|

Algebraically..

Hn = number of patterns with n beats

H8 = H7 + H6

In general Hn = Hn-1 + Hn-2

Does this help us to compute H8?

We need to know H7, H6, for which we need H5, ...

Algorithm Idea

H1 = number of patterns with 1 beat = 1 H2 = Number with 2 beats = 2 ...{SS, L} H3 = H2 + H1 = 2 + 1 = 3 H4 = H3 + H2 = 3 + 2 = 5 H5 = H4 + H3 = 5 + 3 = 8 H6 = H5 + H4 = 8 + 5 = 13 H7 = H6 + H5 = 13 + 8 = 21 H8 = H7 + H6 = 21 + 13 = 34 ...

Program to compute Hn

int n; cin >> n; // which number to compute

int hprev = 1, hcurrent = 2;

for(int i=3; i <= n; i++){

hnext = hprev + hcurrent;

hprev = hcurrent; // prepare for next iteration

hcurrent = hnext; }

cout << hnext;

Code is tricky!

Need a comment:

/* At the begining of an iteration

hcurrent = Hi-1 write h[i] if you like.

hprev = Hi-2

where i is the value of variable i */ Can you prove this? Will mathematical induction help? Proving this is enough -- hnext = hprev + hcurrent --

hence correct answer will be generated.

Proof by induction

Base case: At the beginning of the first iteration is this true? Yes, i will have value 3, and hprev = 1 = H1, hcurrent = 2 = H2

Suppose it is true at some later iteration, when i has value v >= 3. By induction hypothesis hprev, hcurrent have values Hv-1,Hv-2.

The first statement hnext = hprev + hcurrent makes hnext = Hv-1 + Hv-2 = Hv. After this the statement hprev = hcurrent makes hprev = Hv-1. The next statement hcurrent = hnext makes hcurrent=Hv.

In the next iteration i will have value v+1. But hprev,hcurrent will have exactly the right values!

On Hemachandra Numbers

Series is very interesting. Number of petals in many flowers. Ratio of consecutive terms tends to a limit.

What are these numbers more commonly known as?

Hemachandra lived before Fibonacci. Mathematics from poetry!

Newton Raphson method

Method to find the root of f(x), i.e. x s.t. f(x)=0.

Method works if:

f(x) and f '(x) can be easily calculated.

A good initial guess is available.

Example: To find square root of k.

use f(x) = x2 - k. f ‘ (x) = 2x.

f(x), f ‘ (x) can be calculated easily. 2,3 arithmetic ops.

Initial guess x0 = 1 always works! can be proved.

How to get better xi+1 given xi

f(x)x

ix

i+1

Point A =(xi,0) known.

A

B

C

f ‘ (xi) = AB/AC = f(xi)/(xi - xi+1) xi+1

= (xi- f(xi)/f ‘ (xi))

Calculate f(xi ).

Point B=(xi,f(x

i)) known

Approximate f by tangentC= intercept on x axis C=(x

i+1,0)

Square root of k

xi+1

= (xi- f(xi)/f ‘ (xi))

f(x) = x2 - k, f ‘ (x) = 2x

xi+1 = xi - (xi2 - k)/(2xi) = (xi + k/xi)/2

Starting with x0=1, we compute x1, then x2, then...

We can get as close to sqrt(k) as required.

Proof not part of the course.

Code

float k;

cin >> k;

float xi=1; // Initial guess. Known to work.

for(int i=0; i < 10; i++){ // 10 iterations

xi = (xi + k/xi)/2;

}

cout << xi;

Another wayfloat xi, k; cin >> k;

for( xi = 1 ; // Initial guess. Known to work.

xi*xi – k > 0.001 || k - xi*xi > 0.001 ;

// until error in the square is at most 0.001

xi = (xi + k/xi)/2);

cout << xi;

Yet Another way

float k; cin >> k;

float xi=1;

while(xi*xi – k > 0.001 || k - xi*xi > 0.001){

xi = (xi + k/xi)/2 ;

}

cout << xi;

}

While statement while (condition) { loop body}

check condition, then execute loop body if true.

Repeat.

If loop body is a single statement, then need not

use { }. Always putting braces is

recommended; if you insert a statement, you

may forget to put them, so do it at the

beginning.

True for other statements also: for/repeat/if.

For vs. while

If there is a “control” variable with initial value,

update rule, and whose value distinctly defines

each loop iteration, use for.

If loop executes fixed number of times, use for.

Personal taste.

“break” and “continue” : read from your book.

Procedures which return values

Polygon(sides, length) draws. But it will be nice to have a procedure sqrt

which can be called as

y = sqrt(2.56); // should return 1.6

which would suspend the current program, compute sqrt(k), send back the value, and resume.

This is possible!

Procedure for sqrt

float sqrt(float k){

float xi=1;

while(xi*xi – k > 0.001 || k - xi*xi > 0.001){

xi = (xi + k/xi)/2 ;

}

return xi;

}

Notes

First word tells the type of the value being returned.

Note the return statement: this says what value is to be sent back.

Other nomenclature is same. Parameters (k), argument (2.56 in our example).

“y = sqrt(2.56);” is a “call”.

Execution Mechanism

Almost identical to what was described earlier.

Calling program suspends.

Area constructed for sqrt procedure. Argument value (2.56) copied to k from main program.

Sqrt program executes in its own area.

After it finishes, result (1.6 if it works) is copied back to calling program. The result replaces the call. Area in which sqrt executed is destroyed.

Calling program resumes.

Notes

In C++ procedures are called functions. Uniform syntax for functions that do not return

values. So far we have used:

“ procedure Polygon(...)...”

Instead in C++ it is more common to write

“ void Polygon(...) ...” “Procedure” was a keyword defined only for this

course, stop using it from now on.

Math library

You didnt actually need to write the sqrt

function. Just add the line:

#include <cmath>

at the top of your file. This allows you to use in

your program the mathematical functions in the

cmath library. sqrt is in the library. So are

functiones such as sine, cos, tan, ...

Homework

Write a program to calculate the cube root

using Newton Raphson method.

Check how many iterations are needed to get

good answers. Should be very few.

Use sqrt function to draw a right angled

isoceles triangle.