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 ∫


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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
$2) we have f(x)=x^n { (1+(1/x^2 ))^n −(1−(1/x^2 ))^n } but (1+(1/x^2 ))^n ∼ 1+(n/x^2 ) and (1−(1/x^2 ))^n ∼ 1−(n/x^2 ) (x→∞) ⇒ f(x) ∼ x^n { ((2n)/x^2 )} ⇒f(x)∼ 2n x^(n−2) n>2 ⇒lim_(x→+∞) f(x)=+∞ n=2 ⇒lim_(x→+∞) f(x)=2n =4 n=0 ⇒f(x)=0 n=1 ⇒ lim_(x→+∞) f(x)=0$
$2) w e h a v e f (x) = x^{n} {{(1 + \frac{1}{x^{2}})}^{n} - {(1 - \frac{1}{x^{2}})}^{n}} b u t$ ${(1 + \frac{1}{x^{2}})}^{n} \sim 1 + \frac{n}{x^{2}} a n d {(1 - \frac{1}{x^{2}})}^{n} \sim 1 - \frac{n}{x^{2}} (x \to \infty) \Rightarrow$ $f (x) \sim x^{n} {\frac{2 n}{x^{2}}} \Rightarrow f (x) \sim 2 n x^{n - 2}$ $n > 2 \Rightarrow {l i m}_{x \to + \infty} f (x) = + \infty$ $n = 2 \Rightarrow {l i m}_{x \to + \infty} f (x) = 2 n = 4$ $n = 0 \Rightarrow f (x) = 0$ $n = 1 \Rightarrow {l i m}_{x \to + \infty} f (x) = 0$
Commented by maxmathsup by imad last updated on 27/Oct/18
$3) we have f(x)=2 Σ_(k=0) ^([((n−1)/2)]) C_n ^(2k+1) x^(n−4k−2) ⇒ ∫_1 ^3 f(x)dx =2 Σ_(k=0) ^([((n−1)/2)]) C_n ^(2k+1) [(1/(n−4k−1)) x^(n−4k−1) ]_1 ^3 =2 Σ_(k=0) ^([((n−1)/2)]) (C_n ^(2k+1) /(n−4k−1)){3^(n−4k−1) −1} .$
$3) w e h a v e f (x) = 2 \sum_{k = 0}^{[\frac{n - 1}{2}]} C_{n}^{2 k + 1} x^{n - 4 k - 2} \Rightarrow$ $\int_{1}^{3} f (x) d x = 2 \sum_{k = 0}^{[\frac{n - 1}{2}]} C_{n}^{2 k + 1} {[\frac{1}{n - 4 k - 1} x^{n - 4 k - 1}]}_{1}^{3}$ $= 2 \sum_{k = 0}^{[\frac{n - 1}{2}]} \frac{C_{n}^{2 k + 1}}{n - 4 k - 1} {3^{n - 4 k - 1} - 1} .$
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