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 Problem: iteration method 2x^3-2x-5, a=1, b=4 [ Calculator, Method and examples ]Solution:Your problem -> iteration method 2x^3-2x-5, a=1, b=4Method-1Let f(x) = 2x^3-2x-5Here 2x^3-2x-5=0:.2x^3-2x=5:.2x(x^2-1)=5:.2x=(5)/(x^2-1):.x=((5)/(x^2-1))/2:.phi(x)=((5)/(x^2-1))/2Here f(1) = -5 < 0 and f(4) = 115 > 0:. Root lies between 1 and 4x_0 = (1 + 4)/2 = 2.5x_1 = phi(x_0) = phi(2.5) = 0.4762x_2 = phi(x_1) = phi(0.4762) = -3.2331x_3 = phi(x_2) = phi(-3.2331) = 0.2645x_4 = phi(x_3) = phi(0.2645) = -2.688x_5 = phi(x_4) = phi(-2.688) = 0.4016x_6 = phi(x_5) = phi(0.4016) = -2.9807x_7 = phi(x_6) = phi(-2.9807) = 0.3171x_8 = phi(x_7) = phi(0.3171) = -2.7794x_9 = phi(x_8) = phi(-2.7794) = 0.3717x_10 = phi(x_9) = phi(0.3717) = -2.9009x_11 = phi(x_10) = phi(-2.9009) = 0.3372x_12 = phi(x_11) = phi(0.3372) = -2.8206x_13 = phi(x_12) = phi(-2.8206) = 0.3594x_14 = phi(x_13) = phi(0.3594) = -2.8708x_15 = phi(x_14) = phi(-2.8708) = 0.3452x_16 = phi(x_15) = phi(0.3452) = -2.8383x_17 = phi(x_16) = phi(-2.8383) = 0.3543x_18 = phi(x_17) = phi(0.3543) = -2.8589x_19 = phi(x_18) = phi(-2.8589) = 0.3485x_20 = phi(x_19) = phi(0.3485) = -2.8456x_21 = phi(x_20) = phi(-2.8456) = 0.3522x_22 = phi(x_21) = phi(0.3522) = -2.8541x_23 = phi(x_22) = phi(-2.8541) = 0.3499x_24 = phi(x_23) = phi(0.3499) = -2.8487x_25 = phi(x_24) = phi(-2.8487) = 0.3514x_26 = phi(x_25) = phi(0.3514) = -2.8521x_27 = phi(x_26) = phi(-2.8521) = 0.3504x_28 = phi(x_27) = phi(0.3504) = -2.8499x_29 = phi(x_28) = phi(-2.8499) = 0.351x_30 = phi(x_29) = phi(0.351) = -2.8513x_31 = phi(x_30) = phi(-2.8513) = 0.3506

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