Given-that-y-1-x-a-Show-that-y-n-1-n-n-x-n-1-b-Find-an-expression-for-y-n-1-y-n-

Question Number 165641 by nadovic last updated on 05/Feb/22

Given that y = \frac{1}{x}

(a) Show that y^{(n)} = \frac{{(- 1)}^{n} n!}{x^{n + 1}}

(b) Find an expression for y^{(n - 1)} + y^{(n)}

Answered by aleks041103 last updated on 05/Feb/22

a)

y^{(0)} = \frac{{(- 1)}^{0} 0!}{x^{0 + 1}} = \frac{1}{x} = y ✓

s u p p o s e

y^{(k)} = \frac{{(- 1)}^{k} k!}{x^{k + 1}}

t h e n

Prime causes double exponent: use braces to clarify

= {(- 1)}^{k} k! {(x^{- k - 1})}^{'} =

= {(- 1)}^{k} k! (- k - 1) x^{- k - 2} =

= {(- 1)}^{k + 1} k! (k + 1) \frac{1}{x^{k + 2}} =

= \frac{{(- 1)}^{k + 1} (k + 1)!}{x^{(k + 1) + 1}} ✓

\Rightarrow b y i n d u c t i o n :

\forall n \in N^{0}, y^{(n)} = \frac{d^{n}}{{d x}^{n}} (\frac{1}{x}) = \frac{{(- 1)}^{n} n!}{x^{n + 1}}

Answered by aleks041103 last updated on 05/Feb/22

b)

y^{(n - 1)} = \frac{{(- 1)}^{n - 1} (n - 1)!}{x^{n}}

y^{(n)} = \frac{{(- 1)}^{n} n!}{x^{n + 1}}

\Rightarrow y^{(n - 1)} + y^{(n)} = \frac{{(- 1)}^{n} n!}{x^{n + 1}} (1 + \frac{(- 1) x}{n})

y^{(n - 1)} + y^{(n)} = \frac{{(- 1)}^{n} (n - 1)!}{x^{n + 1}} (n - x)