written 7.7 years ago by | modified 2.1 years ago by |
Mumbai University > First Year Engineering > Sem 1 > Applied Maths 1
Marks : 8 M
Year : Dec 2014
written 7.7 years ago by | modified 2.1 years ago by |
Mumbai University > First Year Engineering > Sem 1 > Applied Maths 1
Marks : 8 M
Year : Dec 2014
written 7.7 years ago by | • modified 7.7 years ago |
The given equations can be written as follows:
$ x= \dfrac{-6}{27} y + \dfrac{1}{27} z+ \dfrac{85}{27} \\ \; \\ y= -\dfrac{6}{15} x- \dfrac{2}{15} z+ \dfrac{72}{15} \\ \; \\ z=-\dfrac{1}{54} x- \dfrac{1}{54} y+ \dfrac{110}{54} \\ \; \\ \; \\ \; \\ $
We will make use of following Iteration formulae :
$ x^{(k+1)}= \dfrac{-6}{27} y^{(k)} + \dfrac{1}{27} z^{(k)} + \dfrac{85}{27} \\ \; \\ y^{(k+1)}= -\dfrac{6}{15} x^{(k+1)}- \dfrac{2}{15} z^{(k)} + \dfrac{72}{15} \\ \; \\ z^{(k+1)}=-\dfrac{1}{54} x^{(k+1)}- \dfrac{1}{54} y^{(k+1)}+ \dfrac{110}{54} \\ \; \\ \; \\ \; \\ $
We will start from k=0. Hence, initially x, y and z values are equal to zero. Now, we will go on incrementing k value by 1.
$ \therefore x^{1} \;=\; \dfrac{85}{27} \;=\; 3.148 \\ \; \\ \; \\ y^{1} \;=\; -\dfrac{6}{15} \times (3.148) \;-\; \dfrac{2}{15}(0) + \dfrac{72}{15} \;=\; 3.541 \\ \; \\ \; \\ z^{1} \;=\; -\dfrac{1}{54} (3.148)- \dfrac{1}{54} (3.541)+ \dfrac{110}{54} \;=\; 1.913 \\ \; \\ \; \\ \; \\ \; \\ $
Likewise, we will calculate till values of x, y, z become unchanged for further iterations.
k | x | y | z |
---|---|---|---|
1 | 3.148 | 3.541 | 1.913 |
2 | 2.432 | 3.572 | 1.926 |
3 | 2.426 | 3.573 | 1.926 |
4 | 2.426 | 3.573 | 1.926 |
$ \\ \; \\ \; \\ $
Hence, x=2.426, y=3.573 and z=1.926