There is a famous story about a primary school teacher who wanted to occupy his

Question

There is a famous story about a primary school teacher who wanted to occupy his students' time by making the children compute the sum of 1 + 2 + 3 + ... + 100 by hand. As the story goes, the teacher was astounded when one of the children immediately produced the correct answer: 5050. The student, a child prodigy, was Carl Gauss, who grew up to be one of the most famous mathematicians of the eighteenth century. Repeat Gauss's remarkable calculation by writing a program with a loop that will compute and print the above sum. After you have the program working, rewrite it so you can compute 1 + 2 + ... + n where n is any positive integer. The Babylonian Algorithm for Square Root Approximation Perhaps the first algorithm used for approximating the square root of S is known as the "Babylonian method", named after the Babylonians, or "Heron's method", named after the first-century Greek mathematician Heron of Alexandria who gave the first explicit description of the method. It can be derived from (but predates by 16 centuries) Newton's method. The basic idea is that if x is an overestimate to the square root of a non-negative real number S then S/x will be an underestimate and so the average of these two numbers may reasonably be expected to provide a better approximation. More precisely, assuming S is a positive number: Make a guess at the answer (you can pick S/2 as your initial guess). Compute r = S / guess. Set new guess = (guess + r) / 2 Go back to step 2 until the last two guess values are within 1% of each other^1. Write a program that: Inputs the value of S from the user. Uses the Babylonian Algorithm to determine and output the square root of n.

Explanation / Answer

14.1 Gaussians sum =n(n+1)/2

int n = 100;

int sum=0;

for(i=0;i<=100;i++)

{

sum=sum+i;

}

system.out.println(sum);

}

////////////////////////////////////////////////////////////////////////////

public static void main(String[] args) {

int n=100;

sum=0;

sum=n*(n+1)/2;

system.out.println("sum is"+sum);

}

}

/////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////

14.2)

        The Baylonian Algorithm for square approximation

Babylonian Square Roots

Step 1: Make a guess.

Step 2: Divide your original number by your guess.

Step 3: Find the average of these numbers.

Step 4: Use this average as your next guess.

REPEAT THE PROCESS THREE TIMES.

For example, find sqrt 5

FIRST PROCESS

Step 1: Guess 2 (because 2*2=4, close to 5)

Step 2: Divide 5 by 2 = 2.5

Step 3: Find average of 2 and 2.5 = 2.25 (because (2+2.5)/2 = 2.25)

Step 4: Next guess is 2.25

SECOND PROCESS

Step 1: Guess 2.25

Step 2: Divide 5 by 2.25 = 2.22222222 (go 8 decimal places for accuracy)

Step 3: Find average of 2.25 and 2.22222222 = 2.23611111

Step 4: Next guess is 2.23611111

THIRD PROCESS

Step 1: Guess 2.236111111

Step 2: Divide 5 by 2.23611111 = 2.2360248

Step 3: Find average of 2.23611111 and 2.2360248 = 2.2360679

Step 4: FINAL guess is 2.236067

               

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