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Numerical Solution of a System of Polynomial Parametric form Fuzzy Linear Equations 15

Example 8.4 Consider the following 2 by 2 system of equations



where

⎧

⎪

⎨

⎪

⎩

=(2 + ln{(e − 1)r + 1},4−ln{(e −1)r + 1}),

=(3 + ln(2r + 1),3+ 2ln3−ln(2r + 1)),

=(13 + ln{((e −1)r + 1)

(2r + 1)

},17+ 6ln3− ln{((e − 1)r + 1)

(2r + 1)

}),

=(12 + ln{((e −1)r + 1)

(2r + 1)

},18+ 4ln3− ln{((e − 1)r + 1)

(2r + 1)

}).

By choosing m

= 3, we have

⎧

⎪

⎨

⎪

⎩

= a

= 2.01846 + 1.48660r −0.64225r

+ 0.137185r

= a

= 3.98154 −1.48660r + 0.64225r

−0.13718r

= a

= 3.02605 + 1.66763r −0.75820r

+ 0.16313r

= a

= 5.17117 −1.66763r + 0.75820r

−0.16313r

⎧

⎪

⎨

⎪

⎩

= 13.11509 + 7.97609r −3.55909r

+ 0.76375r

= 23.47658 − 7.97609r + 3.55909r

−0.76375r

= 12.107499 + 7.79506r −3.44314r

+ 0.73781r

= 22.28695 − 7.79506r + 3.44314r

−0.73781r

and the 3

−DNLS solution of this system is x

= 2.000000 and x

= 3.000000.

Example 8.5 Consider the following 2 by 2 system of equations



where

⎧

⎪

⎨

⎪

⎩

=(ln(r + 1),ln(4 − 2r)),

=(ln(2r + 1),ln(4 − r)),

=(ln(r + 1),ln(5 − 3r)),

=(ln{(r + 1)

(2r + 1)},ln{(4 −2r)

(4 − r)}),

=(3ln(r + 1),ln{(4 −2r)

(5 − 3r)}).

the 1

−DNLS solution of this system is x

= 2.022850 and x

= 0.06491, meanwhile the

−DNLS solution of this system is x

= 2 + 0.0 × 10

−6

and x

= 1 + 0.0 × 10

−6

,alsothe

−DNLS solution of this system is x

= 2 + 0.0 ×10

−13

and x

= 1 + 0.0 ×10

−13

9. Conclusion

In this chapter a new method was proposed to solve a system of linear equations. If all

coefﬁcients are polynomial form fuzzy numbers then the system may have an exact solution

but otherwise we can approximate fuzzy coefﬁcients with m-degree polynomial-form fuzzy

numbers and ﬁnd an approximated solution of the system. Choosing m depends on the shape

of left and right spread functions L and R, and the derivation order of them. The presented

method can be applied on any system of equations with LR fuzzy number coefﬁcients.

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16 Ferroelectrics

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