Use Newton's method to find all the roots of the equation correct to eight decimal places. Use Newton's method to find all the roots of the...

Use Newton's method to find all the roots of the equation correct to eight decimal places. 

Use Newton's method to find all the roots of the equation correct to eight decimal places. Start by drawing a graph to find initial approximations. (Enter your answers as a comma-separated list.)

 

x6 − x5 − 7x4 − x2 + x + 8 = 0

x=_____________________?

Asked on by gulusoy

1 Answer

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embizze | High School Teacher | (Level 2) Educator Emeritus

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Using a graph we get approximations for the 4 real zeros: -2.2,-1,1,3.2

Note that `f'(x)=6x^5-5x^4-28x^3-2x+1`

Newtons method begins with a "guess", `x_1` , and then generates a new "guess" by `x_n=x_(n-1)-(f(x_n))/(f'(x_n))`

(1) Let `x_1=-2.2`

`x_2=-2.2-(1.897024)/(-122.80192)=-2.184552163`

`x_3=-2.184552163-.059782097192/-115.10915551=-2.184032812`

`x_4=-2.184032812-.0000659453474/-114.85541759=-2.184032238`

`x_5=-2.184032238-.0000000184267/-114.85513732=-2.184032238`

Thus the first zero is at x=-2.18403224

(2) Let `x_1=-1`

`x_2=-1-1/20=-1.05`

`x_3=-1.05-(-.04466654688)/21.778279375=-1.047949032`

`x_4=-1.047949032-.000115474486/21.7061461225=-1.047954352`

`x_5=-1.047954352-(-.00018911458)/21.7066379578=-1.04794563994`

`x_6=-1.04794563994-(-.000000001355)/21.7063325891=-1.04794564`

Thus the second zero is at x=-1.04794564

(3) Let `x_1=1`

` `Then `x_2=1-1/(-28)=1.0357142857`

`x_3=1.0357142857-(-.04928418522)/(-30.782575596)=1.034113244`

`x_4=1.034113244-(-.00010221366)/(-30.654936565)=1.0341099097`

`x_5=1.0341099097-(-.000000000438)/(-30.654671025)=1.03410991`

The third zero is x=1.03410991

(4) Let `x_1=3.2`

`x_2=3.2-5.154304/566.07392=3.190894644996`

`x_3=3.190894644996-.067146424201/551.363362448=3.190772862468`

`x_4=3.190772862468-.0000118863551/551.16816262=3.1907728409`

`x_5=3.1907728409-.0000000000624/551.168128057=3.190772841`

So the 4th zero is x=3.19077284