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**Subject Mathematics MATLAB for Mathematics**

Check the file: Questions.pdf

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1.

Matrix X starts as the identity matrix while matrix A is the operation of adding numbers according to the Fibonacci series; that is matrix X updates with the latest Fibonacci numbers. Every time enter is pressed we get the next number sequence in the Fibonacci series. n=0@X(2,2), n=1@X(2,1), n=2@X(1,1)=[X(2,2)+X(2,1)].

The process has to be repeated 1475 until X overflows, at the 1476th time the sum of numbers will be greater than realmax (=inf).

2.

e (expected) = 0 0 0 0 0 0 0 0 0 0

e (calculated) = 1.0e-15 *

0 0 -0.0555 0 0 -0.1110 -0.1110 0 0 0

t is a scalar. n is an array (size 10). Thus n determines the size of e, which is a size 10 array.

Each of n array values is similarly calculated, and should be zero (since: n/10 – n*0.1=0). This however is seen to not be the case.

MATLAB expresses divisions (1/10, etc.) as a binary approximation using an infinite series, up to a 52 bit term precision. In the above the decimal number 0.1 is developed into a series with a representation error, known as a round-off error. That is why, for example, the result of 3/10 differs slightly from 3*(1/10) and we are left with a non-zero result....

Matrix X starts as the identity matrix while matrix A is the operation of adding numbers according to the Fibonacci series; that is matrix X updates with the latest Fibonacci numbers. Every time enter is pressed we get the next number sequence in the Fibonacci series. n=0@X(2,2), n=1@X(2,1), n=2@X(1,1)=[X(2,2)+X(2,1)].

The process has to be repeated 1475 until X overflows, at the 1476th time the sum of numbers will be greater than realmax (=inf).

2.

e (expected) = 0 0 0 0 0 0 0 0 0 0

e (calculated) = 1.0e-15 *

0 0 -0.0555 0 0 -0.1110 -0.1110 0 0 0

t is a scalar. n is an array (size 10). Thus n determines the size of e, which is a size 10 array.

Each of n array values is similarly calculated, and should be zero (since: n/10 – n*0.1=0). This however is seen to not be the case.

MATLAB expresses divisions (1/10, etc.) as a binary approximation using an infinite series, up to a 52 bit term precision. In the above the decimal number 0.1 is developed into a series with a representation error, known as a round-off error. That is why, for example, the result of 3/10 differs slightly from 3*(1/10) and we are left with a non-zero result....

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