What is the second approximation in GATE Civil 2026 Set-1 Q48?

The second approximation is 0.57. Using Newton-Raphson: f(x)=e^(-x)-x, f′(x)=-e^(-x)-1. At x0=0.5: f(0.5)=e^(-0.5)-0.5=0.1065, f′(0.5)=-e^(-0.5)-1=-1.6065. x1=0.5-0.1065/(-1.6065)=0.5+0.0663=0.5663≈0.57.

What is the derivative of f(x)=e^(-x)-x used in GATE Civil 2026 Set-1 Q48?

f′(x) = -e^(-x) - 1. This is always negative (f′(x) < 0 for all x), meaning the function is strictly decreasing. This guarantees the Newton-Raphson method converges to the unique root where e^(-x)=x, which is the Omega constant ≈ 0.5671.

What is the Newton-Raphson iteration formula used in GATE Civil 2026 Set-1 Q48?

The Newton-Raphson formula is x_{n+1} = x_n - f(x_n)/f′(x_n). Starting with x0=0.5, one iteration gives x1 = 0.5 - (e^(-0.5)-0.5)/(-e^(-0.5)-1) = 0.5 - (0.1065)/(-1.6065) = 0.5663 ≈ 0.57.

Newton-Raphson e^(-x)-x Second Approximation x=0.5 | GATE Civil 2026 Set-1 Q48

Q: What is the Newton-Raphson iteration formula used in GATE Civil 2026 Set-1 Q48?

The Newton-Raphson formula is x_{n+1} = x_n - f(x_n)/f′(x_n). Starting with x0=0.5, one iteration gives x1 = 0.5 - (e^(-0.5)-0.5)/(-e^(-0.5)-1) = 0.5 - (0.1065)/(-1.6065) = 0.5663 ≈ 0.57.

Civil Engineering > GATE 2026 SET-1 > Numerical Methods

Starting with the first approximation as x = 0.5, the second approximation for the root of the following function by the Newton-Raphson method is ______ (rounded off to two decimal places).

f(x) = e^−x − x

Correct : 0.57

f(x) = e^−x − x and f′(x) = −e^−x − 1.
The Newton-Raphson update formula is x₁ = x₀ − f(x₀)/f′(x₀).
Starting with x₀ = 0.5:
f(0.5) = e^−0.5 − 0.5 = 0.6065 − 0.5 = 0.1065
f′(0.5) = −e^−0.5 − 1 = −0.6065 − 1 = −1.6065
x₁ = 0.5 − (0.1065)/(−1.6065) = 0.5 + 0.0663 = 0.5663 ≈ 0.57
Correct answer: 0.57 ✓

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