牛顿迭代法（MATLAB）求个问题

用牛顿迭代法求方程 X^x=10 的一个实根。精度要求为ε=〖10〗^(-6).

哪位达人帮忙给出这个问题的MATLAB求解的代码和结果
谢谢~~

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相关建议推荐于2017-11-26

x=10^(1/x),

{"代换次数", "x值"},
{1, 10.0000000000},
{2, 1.25892541179},
{3, 6.2277079027},
{4, 1.44734718383},
{5, 4.9081658429},
{6, 1.59860842235},
{7, 4.22225110674},
{8, 1.72520411363},
{9, 3.79875738892},
{10, 1.83334406211},
{11, 3.51116592745},
{12, 1.92666265812},
{13, 3.30394053156},
{14, 2.00756162740},
{15, 3.14859436240},
{16, 2.07779186863},
{17, 3.02886668640},
{18, 2.13873261347},
{19, 2.93471971603},
{20, 2.19153314320},
{21, 2.85957547453},
{22, 2.23718684977},
{23, 2.79891643939},
{24, 2.27657058379},
{25, 2.74952195010},
{26, 2.31046603117},
{27, 2.70902563673},
{28, 2.33957197630},
{29, 2.67564573598},
{30, 2.36451218146},
{31, 2.64801362230},
{32, 2.38584135304},
{33, 2.62506071440},
{34, 2.40405040970},
{35, 2.60594145898},
{36, 2.41957157870},
{37, 2.58997938804},
{38, 2.43278348440},
{39, 2.57662838939},
{40, 2.44401621907},
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{42, 2.45355631677},
{43, 2.55606357487},
{44, 2.46165153774},
{45, 2.54818725296},
{46, 2.46851538330},
{47, 2.54156832789},
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{49, 2.53600205167},
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{51, 2.53131819714},
{52, 2.48342520677},
{53, 2.52737489126},
{54, 2.48695231719},
{55, 2.52405365052},
{56, 2.48993547067},
{57, 2.52125535571},
{58, 2.49245779954},
{59, 2.51889696841},
{60, 2.49458994625},
{61, 2.51690883855},
{62, 2.49639187875},
{63, 2.51523248701},
{64, 2.49791445950},
{65, 2.51381877237},
{66, 2.49920079634},
{67, 2.51262637057},
{68, 2.50028740205},
{69, 2.51162051052},
{70, 2.50120518701},
{71, 2.51077192031},
{72, 2.50198030717},
{73, 2.51005594698},
{74, 2.50263488691},
{75, 2.50945182032},
{76, 2.50318763452},
{77, 2.50894203607},
{78, 2.50365436536},
{79, 2.50851183865},
{80, 2.50404844615},
{81, 2.50814878687},
{82, 2.50438117188},
{83, 2.50784238904},
{84, 2.50466208539},
{85, 2.50758379614},
{86, 2.50489924804},
{87, 2.50736554369},
{88, 2.50509946890},
{89, 2.50718133442},
{90, 2.50526849875},
{91, 2.50702585529},
{92, 2.50541119412},
{93, 2.50689462329},
{94, 2.50553165610},
{95, 2.50678385556},
{96, 2.50563334760},
{97, 2.50669035984},
{98, 2.50571919256},
{99, 2.50661144214},
{100, 2.50579165970},
{101, 2.50654482888},
{102, 2.50585283331},
{103, 2.50648860124},
{104, 2.50590447309},
{105, 2.50644113972},
{106, 2.50594806465},
{107, 2.50640107745},
{108, 2.50598486216},
{109, 2.50636726075},
{110, 2.50601592439},
{111, 2.50633871586},
{112, 2.50604214518},
{113, 2.50631462086},
{114, 2.50606427906},
{115, 2.50629428201},
{116, 2.50608296299},
{117, 2.50627711372},
{118, 2.50609873469},
{119, 2.50626262173},
{120, 2.50611204804},
{121, 2.50625038881},
{122, 2.50612328623},
{123, 2.50624006280},
{124, 2.50613277269},
{125, 2.50623134643},
{126, 2.50614078047},
{127, 2.50622398879},
{128, 2.50614754005},
{129, 2.50621777806},
{130, 2.50615324598},
{131, 2.50621253547},
{132, 2.50615806250},
{133, 2.50620811009},
{134, 2.50616212824},
{135, 2.50620437454},
{136, 2.50616556023},
{137, 2.50620122128},
{138, 2.50616845726},
{139, 2.50619855955},
{140, 2.50617090271},
{141, 2.50619631273},
{142, 2.50617296697},
{143, 2.50619441614},
{144, 2.50617470946},
{145, 2.50619281519},
{146, 2.50617618034},
{147, 2.50619146379},
{148, 2.50617742194},
{149, 2.50619032304},
{150, 2.50617847000},
{151, 2.50618936011},
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{187, 2.50618439248},
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{189, 2.50618435400},
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{194, 2.50618400915},

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其他看法

第1个回答 2009-01-02

syms x
f=x^x-10;
df=diff(f,x);

eps=1e-6;
x0=10;
cnt=0;
MAXCNT=200; %最大循环次数
while cnt<MAXCNT %防止无限循环
x1=x0-subs(f,x,x0)/subs(df,x,x0); %去掉这个分号,可以看到迭代过程.
if (abs(x1-x0)<eps)
break;
end
x0=x1;
cnt=cnt+1;
end
if cnt==MAXCNT
disp '不收敛'
else
vpa(x1,8)
end

得到结果：
x1 =

2.5062

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