Use the Newton’s method to approximate the solution of the equation: 1-x-tan x = 0 starting at x0 = 0.5 (the initial value). i.e. Find x1,x2,x3.. .
+118703 Newton's method
You are probably supposed to know the formula but I can't remember formulas so I work it out ever time.
$Remember,thegradientofthetangentat$x=x1$isgivenby$f′(x0)sof′(x0)=f(x0)−0x0−x1f′(x0)=f(x0)x0−x1x0−x1=f(x0)f′(x0)−x1=−x0+f(x0)f′(x0)x1=x0−f(x0)f′(x0)$Nowyouhavetheformula$x1=x0−f(x0)f′(x0)
1-x-tan x = 0 starting at x0 = 0.5
Let
f(x)=1−x−tanxf′(x)=−1−sec2xf(0.5)=1−0.5−tan0.5f(0.5)=0.5−0.546f(0.5)=−0.046f′(0.5)=−1−sec2(0.5)f′(0.5)=−1−(1/cos2(0.5))f′(0.5)=−2.298sox1=0.5−f(0.5)f′(0.5)x1=0.5−−0.046−2.298x1=0.480
I have applied newtons method just once.
If you want more accuracy you can apply it again and again until you are happy. :)
+118703 Newton's method
You are probably supposed to know the formula but I can't remember formulas so I work it out ever time.
$Remember,thegradientofthetangentat$x=x1$isgivenby$f′(x0)sof′(x0)=f(x0)−0x0−x1f′(x0)=f(x0)x0−x1x0−x1=f(x0)f′(x0)−x1=−x0+f(x0)f′(x0)x1=x0−f(x0)f′(x0)$Nowyouhavetheformula$x1=x0−f(x0)f′(x0)
1-x-tan x = 0 starting at x0 = 0.5
Let
f(x)=1−x−tanxf′(x)=−1−sec2xf(0.5)=1−0.5−tan0.5f(0.5)=0.5−0.546f(0.5)=−0.046f′(0.5)=−1−sec2(0.5)f′(0.5)=−1−(1/cos2(0.5))f′(0.5)=−2.298sox1=0.5−f(0.5)f′(0.5)x1=0.5−−0.046−2.298x1=0.480
I have applied newtons method just once.
If you want more accuracy you can apply it again and again until you are happy. :)
