Home

 Solve any problem (step by step solutions) Input table (Matrix, Statistics)
Mode :
SolutionHelp
Solution will be displayed step by step (In 2 parts)
Solution
 Problem: muller method 2x^3-2x-5, a=0, b=1, c=2 [ Calculator, Method and examples ]Solution:Your problem -> muller method 2x^3-2x-5, a=0, b=1, c=2Here 2x^3-2x-5=0Let f(x) = 2x^3-2x-5x_0 = 0x_1 = 1x_2 = 21^(st) iteration :f(x_0)=f(0)=2*(0)^3-2*(0)-5=-5f(x_1)=f(1)=2*1^3-2*1-5=-5f(x_2)=f(2)=2*(2)^3-2*(2)-5=7h_1=x_1-x_0=1-0=1h_2=x_2-x_1=2-1=1delta_1=(f(x_1)-f(x_0))/h_1=(-5--5)/1=0delta_2=(f(x_2)-f(x_1))/h_2=(7--5)/1=12a=(delta_2-delta_1)/(h_2+h_1)=(12-0)/(1+1)=6b=a xx h_2 + d_2=6xx1+12=18c=f(x_2)=7x_3=x_2+(-2c)/(b +- sqrt(b^2-4ac))x_3=x_2+(-2c)/(b +sign(b) sqrt(b^2-4ac))=2+(-2 xx 7)/(18 + sqrt(18^2 - 4xx 6 xx 7))=2+(-14)/(18 + sqrt(156))=2+(-14)/(18 + 12.49)=1.5408Relative percent errorvarepsilon_(a^1)=|(x_3-x_2)/x_3| xx 100%=|(1.5408-2)/1.5408| xx 100%=29.7999%Now,x_0=x_1=1x_1=x_2=2x_2=x_3=1.54082^(nd) iteration :f(x_0)=f(1)=2*1^3-2*1-5=-5f(x_1)=f(2)=2*(2)^3-2*(2)-5=7f(x_2)=f(1.5408)=2*(1.5408)^3-2*(1.5408)-5=-0.7653h_1=x_1-x_0=2-1=1h_2=x_2-x_1=1.5408-2=-0.4592delta_1=(f(x_1)-f(x_0))/h_1=(7--5)/1=12delta_2=(f(x_2)-f(x_1))/h_2=(-0.7653-7)/-0.4592=16.9117a=(delta_2-delta_1)/(h_2+h_1)=(16.9117-12)/(-0.4592+1)=9.0817b=a xx h_2 + d_2=9.0817xx-0.4592+16.9117=12.7417c=f(x_2)=-0.7653x_3=x_2+(-2c)/(b +- sqrt(b^2-4ac))x_3=x_2+(-2c)/(b +sign(b) sqrt(b^2-4ac))=1.5408+(-2 xx -0.7653)/(12.7417 + sqrt(12.7417^2 - 4xx 9.0817 xx -0.7653))=1.5408+(1.5306)/(12.7417 + sqrt(190.15))=1.5408+(1.5306)/(12.7417 + 13.7895)=1.5985Relative percent errorvarepsilon_(a^2)=|(x_3-x_2)/x_3| xx 100%=|(1.5985-1.5408)/1.5985| xx 100%=3.6089%Now,

Solution provided by AtoZmath.com
Any wrong solution, solution improvement, feedback then Submit Here
Want to know about AtoZmath.com and me
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "http://atozmath.com/Default.aspx?q1=muller%20method%202x%5E3-2x-5%2C%20a%3D0%2C%20b%3D1%2C%20c%3D2%60883&do=1", "fetch_time": 1566052396000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2019-35/segments/1566027313428.28/warc/CC-MAIN-20190817143039-20190817165039-00234.warc.gz", "warc_record_offset": 19992712, "warc_record_length": 143348, "token_count": 1010, "char_count": 1839, "score": 4.03125, "int_score": 4, "crawl": "CC-MAIN-2019-35", "snapshot_type": "longest", "language": "en", "language_score": 0.31596264243125916, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00014-of-00064.parquet"}