Answer to Question #143376 in Quantitative Methods for Usman

Question #143376
Solve by iteration method:
(i) 1+logx=x
1
Expert's answer
2020-11-10T20:07:16-0500
"x=g(x)=1+\\log x"

"x_0=0.5"

"x_{i+1}=g(x_i)=1+\\log x_i, i=0, 1,2,..."

"\\def\\arraystretch{1.5}\n \\begin{array}{c:c:c}\n i & x_i & g(x_i) \\\\ \\hline\n 0 & 0.5 & 0.69897000 \\\\\n 1 & 0.69897000 & 0.84445854 \\\\\n 2 &0.84445854 & 0.92657833 \\\\\n 3 & 0.92657833 & 0.96688214\\\\\n 4 & 0.96688214 & 0.98537354 \\\\\n 5 & 0.98537354 & 0.99360090\\\\\n 6 & 0.99360090 & 0.99721198\\\\\n 7 & 0.99721198 & 0.99878749\\\\\n 8 & 0.99878749 & 0.99947309\\\\\n 9 & 0.99947309 & 0.99977111\\\\\n10 & 0.99977111 & 0.99990058\\\\\n11 & 0.99990058 & 0.99995682\\\\\n12 & 0.99995682 & 0.99998125\\\\\n13 & 0.99998125 & 0.99999186\\\\\n14 & 0.99999186 & 0.99999646\\\\\n15 & 0.99999646 & 0.99999846\\\\\n16 & 0.99999846 & 0.99999933\\\\\n17 & 0.99999933 & 0.99999971\\\\\n18 & 0.99999971 & 0.99999987\\\\\n19 & 0.99999987 & 0.99999994\\\\\n20 &0.99999994 & 0.99999997\\\\\n21 & 0.99999997 & 0.99999999\\\\\n22 &0.99999999 & 1.00000000\\\\\n\n\\hdashline\n23 &1.00000000 & \\\\\n\\end{array}"

"x\\approx1.00000000"



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