let-f-x-x-1-x-n-x-1-x-n-with-n-integr-natural-and-x-from-R-x-0-1-simplify-f-x-2-calculate-lim-x-f-x-3-calculate-1-3-f-x-dx-

Question Number 46421 by maxmathsup by imad last updated on 25/Oct/18

l e t f (x) = {(x + \frac{1}{x})}^{n} - {(x - \frac{1}{x})}^{n} w i t h n i n t e g r n a t u r a l a n d x f r o m R (x \neq 0)

1) s i m p l i f y f (x)

2) c a l c u l a t e {l i m}_{x \to + \infty} f (x)

3) c a l c u l a t e \int_{1}^{3} f (x) d x

Commented by maxmathsup by imad last updated on 27/Oct/18

1) t h e b i n o m e f o r m u l a e g i v e f (x) = \sum_{k = 0}^{n} C_{n}^{k} x^{k} \frac{1}{x^{n - k}} - \sum_{k = 0}^{n} C_{n}^{k} x^{k} \frac{{(- 1)}^{n - k}}{x^{n - k}}

= \sum_{k = 0}^{n} C_{n}^{k} x^{2 k - n} - \sum_{k . = 0}^{n} C_{n}^{k} {(- 1)}^{n - k} x^{2 k - n}

= \sum_{k = 0}^{n} C_{n}^{k} (1 - {(- 1)}^{n - k}) x^{2 k - n} = \sum_{k = 0}^{n} (1 - {(- 1)}^{n - k}) C_{n}^{k} x^{2 k - n}

=_{n - k = p} \sum_{p = 0}^{n} (1 - {(- 1)}^{p}) C_{n}^{n - p} x^{2 (n - p) - n}

= \sum_{p = 0}^{n} (1 - {(- 1)}^{p}) C_{n}^{p} x^{n - 2 p} = \sum_{k = 0}^{[\frac{n - 1}{2}]} 2 C_{n}^{2 k + 1} x^{n - 2 (2 k + 1)}

= 2 \sum_{k = 0}^{[\frac{n - 1}{2}]} C_{n}^{2 k + 1} x^{n - 4 k - 2} .

Commented by maxmathsup by imad last updated on 27/Oct/18

2) w e h a v e f (x) = x^{n} {{(1 + \frac{1}{x^{2}})}^{n} - {(1 - \frac{1}{x^{2}})}^{n}} b u t

{(1 + \frac{1}{x^{2}})}^{n} \sim 1 + \frac{n}{x^{2}} a n d {(1 - \frac{1}{x^{2}})}^{n} \sim 1 - \frac{n}{x^{2}} (x \to \infty) \Rightarrow

f (x) \sim x^{n} {\frac{2 n}{x^{2}}} \Rightarrow f (x) \sim 2 n x^{n - 2}

n > 2 \Rightarrow {l i m}_{x \to + \infty} f (x) = + \infty

n = 2 \Rightarrow {l i m}_{x \to + \infty} f (x) = 2 n = 4

n = 0 \Rightarrow f (x) = 0

n = 1 \Rightarrow {l i m}_{x \to + \infty} f (x) = 0

Commented by maxmathsup by imad last updated on 27/Oct/18

3) w e h a v e f (x) = 2 \sum_{k = 0}^{[\frac{n - 1}{2}]} C_{n}^{2 k + 1} x^{n - 4 k - 2} \Rightarrow

\int_{1}^{3} f (x) d x = 2 \sum_{k = 0}^{[\frac{n - 1}{2}]} C_{n}^{2 k + 1} {[\frac{1}{n - 4 k - 1} x^{n - 4 k - 1}]}_{1}^{3}

= 2 \sum_{k = 0}^{[\frac{n - 1}{2}]} \frac{C_{n}^{2 k + 1}}{n - 4 k - 1} {3^{n - 4 k - 1} - 1} .