Skip to main content

N-R: sin x = e^-x with limitations

Short answer

With f(x)=sin x - e^{-x}, f'(x)=cos x + e^{-x}, from x0=0.6, the root is 0.5885. Limitations: requires derivative; may fail if f'≈0; needs close initial guess; no guaranteed convergence without bracketing; linear convergence at multiple roots; no error bound.

Find a real root of the equation sin x = e^(-x) correct up to four decimal places using N-R method. What are the limitations of this method?

Part 1: Root by Newton-Raphson

f(x)  = sin x − e^(−x)
f'(x) = cos x + e^(−x)
x(n+1) = x(n) − f(x(n)) / f'(x(n))     (x in radians)

f(0.5) = 0.4794 − 0.6065 = −0.1271 < 0 and f(1) = 0.8415 − 0.3679 = 0.4736 > 0, so a root lies in (0.5, 1). Take x0 = 0.6.

nx(n)f(x(n))f'(x(n))x(n+1)
00.60000.01581.37410.5885
10.5885-0.00011.38700.5885
20.5885-0.00001.38690.5885

Two successive iterates agree to four decimals.

Answer: root ≈ 0.5885

Part 2: Limitations of the method

  1. It needs f′(x), which must be derived and evaluated each step.
  2. If f′(x(n)) = 0 or is very small, the step is undefined or very large (here f′ = cos x + e^(−x) can become small, so a bad guess can fail).
  3. The initial guess must be close to the root; otherwise the iteration may diverge, oscillate or jump to another root (this function has many roots, such as 3.0964, 6.2850, ...).
  4. Convergence is guaranteed only when |f·f″/f′²| < 1 near the root.
  5. At a multiple root the convergence is only linear.
  6. There is no bracketing, so no error bound is available in advance.

✓ Answer checked by Claude Opus 5.5 (high effort)

Related questions

All Numerical Method chapter 2 questions →