Assist with answers and working out.

Question

a) Show that 1/r(r+1) =1/r - 1/r+1

 

hence, show that (the sum from r=1 to n)  ∑ 1/r(r+!) = 1 - 1/n+1

                       

 

deduce that S_n the sum to infinity, if it exists.

 

 

b) The function f is given by f(x)) = 3 - x - x² show that;

 

    i) the equation of f(x) = 0 has a real root α in the interval [1,2]

    ii) by choosing X₁ = 1 as an initial approximation to the root α. Use Newton Raphson method to show that a better approximation of α is x₂ = 1.33

Answer 1

a) The easiest way is in reverse:

 

1/r - 1/(r+1)

 

= (r+1)/[r(r+1)] - r/[r(r+1)]

 

= [(r+1) - r] / [r(r+1)]

 

= 1/[r(r+1)]

 

This is a telescoping series:

 

 ∑_r=1...n 1/[r(r+1)]

 

= ∑_r=1...n [1/r - 1/(r+1)]

 

= ∑_r=1...n 1/r - ∑_r=1...n 1/(r+1)

 

= ∑_r=1...n 1/r - ∑_r=2...n+1 1/r

 

= 1 - 1/(n+1)

 

The infinite series is:

 

∑_r=1...∞ 1/[r(r+1)] = lim_n→∞ [1 - 1/(n+1)] = 1

 

b) Answers about f(x) = 3 - x - x²:

 

i) f(1) = 3 - 1 - 1² = 1 > 0 ; f(2) = 3 - 2 - 2² = -3 < 0 ; f(x) continuous.

 

     Thus by the intermediate value theorem, f(x) has a root α ∈ [1,2].

 

ii) Newton's method says that x_n+1 = x_n - f(x_n) / f'(x_n)

 

We have f'(x) = -1 - 2x

 

Therefore, x_n+1 = x_n + (3 - x_n - x_n²) / (1 + 2x_n).

 

x₂ = 1 + (3 - 1 - 1²) / (1+2·1) = 1 + 1/3 ≈ 1.33