Find a real root for x tan x + 1=0 using Newton Raphson method.
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given
f(x) = xtanx +1 =0
f'(x) = xsec²x + tan x
f(2) = 2 tan 2 +1 = -3.370079 < 0
f(3) = 2 tan 3+ 1 = -0.572370 > 0
the root is between 2 & 3.
taking
x₀ = (2+3)/2 = 2.5
according to the Newton Raphson method:
X i+1 = Xi - [f(xi)/f'(xi)]
X1 = X₀ - [f(X₀)/f'(X₀)]
X1 = 2.5 - (-0.86755)/3.14808
X1 = 2.77558
X₂ = X₁ - f(Xi)/f'(Xi)
f(X₁) = = – 0.06383, f'(x₁) = 2.80004
X₂ = 2.77558 - (-0.06383)/2.80004 = 2.798
similarly X₃ = 2.798= X₂
The real root of x tan x +1 = 2.798
I hope it works. If you have any confusion , comment below
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