Answer to Question #114338 in Quantitative Methods for sana

Question #114338
solve for t in the following equation by means of linear interpolation
( 1+ t )^10 - 1/ t = 25
1
Expert's answer
2020-05-20T18:23:14-0400

(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

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