Question

Solve this equation for x, (1 + x)^4=34.5 . I am interested in the math libraries you’d use.

the equation is MUCH SIMPLER (1 + x)^4=34.5

thanks

Answer 1

approximate x*(x+a)^b=c

You’ll need a more robust solution for more complex polynomials, but this may be good enough to finish your homework.

This algorithm uses Newton’s Method and is written in Ruby. You can verify that the derivative and answer is correct using wolfram|alpha.

def f(x,a,b,c)
 return x*(x+a)**b-c
end

def df(x,a,b,c)
 return (x+a)**b+b*x*(x+a)**(b-1)
end

def newton(a,b,c)
 xn=0   #initial seed for Newton's method
 
 while true
  xn2=xn-f(xn,a,b,c)/df(xn,a,b,c)  #Newton's method
  print "f(%.5f)=%.5f\n"%[xn,f(xn,a,b,c)]  
  break if (xn2*10000).to_i==(xn*10000).to_i #set desired precision here
  xn=xn2
 end
 print "root is %.5f"%[xn2]
end

newton(1,4,34.5)

this produces:

f(0.00000)=-34.50000
f(34.50000)=54793902.65625
f(27.44093)=17954483.09402
f(21.79391)=5883122.74717
f(17.27661)=1927672.51373
f(13.66318)=631598.66717
f(10.77301)=206926.07160
f(8.46171)=67782.26596
f(6.61400)=22194.34671
f(5.13819)=7259.61867
f(3.96214)=2367.67791
f(3.03097)=765.73665
f(2.30728)=241.54928
f(1.77466)=70.68568
f(1.43951)=16.48341
f(1.30101)=1.97186
f(1.27945)=0.04145
f(1.27897)=0.00002
root is 1.27897