How do you find the zeroes for #f(x)=x^5+3x^3-x+6#?

2 Answers
mason m
Dec 6, 2015

Use a graphing calculator.

Explanation:

Use a graphing calculator.

graph{x^5+3x^3-x+6 [-15.09, 16.95, -5.13, 10.89]}

There's only one zero: #(-1.178,0)#

George C.
Dec 6, 2015

Use Newton's method to find Real root:

#x ~~ -1.178259848#

and Complex roots:

#x ~~ -0.20255440359096+-1.893126314652i#

#x ~~ 0.79168432752598+-0.882050595027i#

Explanation:

#f(x) = x^5+3x^3-x+6#

#f'(x) = 5x^4+9x^2-1#

Starting with a first approximation #a_0#, iterate using the formula:

#a_(i+1) = a_i - (f(a_i))/(f'(a_i))#

Putting this into a spreadsheet, choosing #a_0 = -1#, we find:

#a_0 = -1#

#a_1 = -1.230769231#

#a_2 = -1.181552182#

#a_3 = -1.178273619#

#a_4 = -1.178259848#

#a_5 = -1.178259848#

The same method can be applied to find the #4# Complex roots, using Complex arithmetic and Complex initial approximations.

Since my spreadsheet application does not support Complex arithmetic directly, I created columns for #Re(z)#, #Im(z)#, #Re(z^2)#, #Im(z^2)#,.., #Re(z^5)#, #Im(z^5)#, #Re(f(z))#, #Im(f(z))#, #Re(f'(z))#, #Im(f'(z))#, #Re(f(z)bar(f'(z)))#, #Im(f(z)bar(f'(z)))#, #Re(f'(z)bar(f'(z)))#, #Im(f'(z)bar(f'(z)))#.

With an initial approximation of #0+2i# I saw the following approximations:

#a_0 = 0+2i#

#a_1 = -0.13953488372093+1.860465116279i#

#a_2 = -0.21027440054555+1.889317276079i#

#a_3 = -0.20248503148821+1.893034069471i#

#a_4 = -0.20255442138413+1.893126325251i#

#a_5 = -0.20255440359096+1.893126314652i#

#a_6 = -0.20255440359096+1.893126314652i#

With an initial approximation of #1+i# I saw the following approximations:

#a_0 = 1+i#

#a_1 = 0.83921568627451+0.909803921569i#

#a_2 = 0.79458495418768+0.883839764146i#

#a_3 = 0.79169561531938+0.882058297006i#

#a_4 = 0.79168432769581+0.882050595168i#

#a_5 = 0.79168432752598+0.882050595027i#

#a_6 = 0.79168432752598+0.882050595027i#

Note that Complex roots occur in conjugate pairs so the #4# Complex roots are approximately:

#-0.20255440359096+-1.893126314652i#

#0.79168432752598+-0.882050595027i#

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