Answer to Question #114338 in Quantitative Methods for sana

Question

solve for t in the following equation by means of linear interpolation
( 1+ t )^10 - 1/ t = 25

Answer 1

(1+ t )^10 - 1/t = 25

(1+ t )^10 - 1/t - 25 = 0

f(t) = (1+ t )^10 - 1/t - 25 = 0

The equation has three real solutions.

The first root of f(t₁) is between 0.3 and 0.4.

f(0.3) = -14.54

f(0.4) = 1.42

In liner interpolation two points are joined by a straight line and the t-vallue of the t-intercept of this line is calculated.

The first two points will be (0.3, -14.54) and (0.4, 1.42).

Using simslar triangles, with the root at t₁ = 0.3 + p₁, you have

p₁/14.54 = (0.1 - p₁)/1.42

p₁ = 0.09

A better approximation of the root f(t₁) = 0 is 0.39, (0.3 + p₁), which gives f(0.39) = -0.63

The root is between 0.39 and 0.4.

Next approximation is based on using similar trianles:

p₂/0.63 = (0.01 - p₂)/1.42

p₂ = 0.003

The second approximation of the root of f(t₁) = 0 is 0.393, (0.39 + 0.003), which gives

f(0.393) = -0.0332

The root is between 0.393 and 0.4.

Repeating the procedure again, you obtain

p₃/0.0332 = (0.007 - p₃)/1.42

p₃ = 0.00016

The third approximation of the root is 0.39316, which gives

f(0.39316) = -0.00057

t₁ = 0.39316

The second root of f(t₂) is between -0.05 and -0.03.

f(-0.05) = -4.40

f(-0.03) = 9.07

The first two points will be (-0.05, -4.4012) and (-0.03, 9.0707).

Using simslar triangles, with the root at t₂ = (-0.03 - p₄), you have

p₄/9.0707 = (0.02 - p₄)/4.4012

p₄ = 0.01346

A better approximation of the root f(t₂) = 0 is -0.04346, (-0.03 - 0.01346), which gives f(-0.04346) = -1.3491

The root is between -0.04346 and -0.03.

Next approximation is based on using similar trianles:

p₅/9.0707 = (0.01346 - p₅)/1.3491

p₅ = 0.01171

The second approximation of the root of f(t₂) = 0 is -0.04171, (-0.03 - 0.01171), which gives

f(-0.04171) = -0.3718

The root is between -0.04171 and -0.03.

Repeating the procedure again, you obtain

p₆/9.0707 = (0.01171 - p₆)/0.3718

p₆ = 0.0112

The third approximation of the root is -0.0412, which gives

f(-0.0412) = -0.0715

t₂ = -0.0412

The third root of f(t₃) is between -2.4 and -2.2.

f(-2.4) = 4.34

f(-2.2) = -18.35

The first two points will be (-2.4, 4.34) and (-2.2, -18.35).

Using simslar triangles, with the root at t₃ = (-2.4 + p₇), you have

p₇/4.34 = (0.2 - p₇)/18.35

p₇ = 0.04

A better approximation of the root f(t₃) = 0 is -2.36, (-2.4 + 0.04), which gives f(-2.36) = -2.929

The root is between -2.4 and -2.36.

Next approximation is based on using similar trianles:

p₈/4.34 = (0.04 - p₈)/2.93

p₈ = 0.024

The second approximation of the root of f(t₃) = 0 is -2.376, (-2.4 + 0.024), which gives

f(-2.376) = -0.2467

The root is between -2.4 and -2.376.

Repeating the procedure again, you obtain

p₉/4.34 = (0.024 - p₉)/0.2467

p₉ = 0.0227

The third approximation of the root is -2.3773, which gives

f(-2.3773) = -0.016

t₃ = -2.3773

Answer: t₁ = 0.39316, t₂ = -0.0412, t₃ = -2.3773