let f(x)=((2x+3)/(x^2 +1)) calculate f^((n)) (x) 2)find f^((10)) (x) and f^((15)) (x) 3)calculate f^((10)) (0) and f^((15)) (0) 4)developp f at integr serie 5)let g(x)=∫


Question and Answers Forum

All Questions Topic List
Relation and Functions Questions
Previous in All Question Next in All Question
Previous in Relation and Functions Next in Relation and Functions
Question Number 72260 by mathmax by abdo last updated on 26/Oct/19

$l e t f (x) = \frac{2 x + 3}{x^{2} + 1}$ $c a l c u l a t e f^{(n)} (x)$ $2) f i n d f^{(10)} (x) a n d f^{(15)} (x)$ $3) c a l c u l a t e f^{(10)} (0) a n d f^{(15)} (0)$ $4) d e v e l o p p f a t i n t e g r s e r i e$ $5) l e t g (x) = \int_{0}^{x} f (t) d t d e v e l o p p g a t i n t e g r s e r i e .$
Commented by mathmax by abdo last updated on 27/Oct/19
$1) f(x)=((2x+3)/(x^2 +1)) =((2x+3)/((x−i)(x+i))) =(a/(x−i)) +(b/(x+i)) a=((2i+3)/(2i)) =1+(3/(2i)) =1−((3i)/2) =((2−3i)/2) b=((−2i+3)/((−2i))) =((2i−3)/(2i)) =1−(3/(2i)) =1+((3i)/2) =((2+3i)/2) ⇒ f(x)=((2−3i)/(2(x−i))) +((2+3i)/(2(x+i))) ⇒f^((n)) (x)=((2−3i)/2)((1/(x−i)))^((n)) +((2+3i)/2)((1/(x+i)))^((n)) =((2−3i)/2) (((−1)^n n!)/((x−i)^(n+1) )) +((2+3i)/2) (((−1)^n n!)/((x+i)^(n+1) )) =(((−1)^n n!)/2){ ((2+3i)/((x+i)^(n+1) )) +((2−3i)/((x−i)^(n+1) ))} =(((−1)^n n!)/2){(((2+3i)(x−i)^(n+1) +(2−3i)(x+i)^(n+1) )/((x^2 +1)^(n+1) ))} =(((−1)^n n!)/((x^2 +1)^(n+1) ))Re{(2+3i)(x−i)^(n+1) } but 2+3i =(√(13))e^(iarctan((3/2))) and x−i =(√(x^2 +1))e^(−iarctan((1/x))) ⇒ (x−i)^(n+1) =(x^2 +1)^((n+1)/2) e^(−i(n+1)arctan((1/x))) ⇒ (2+3i)(x−i)^(n+1) =(√(13))(x^2 +1)^((n+1)/2) e^(i(arctan((3/2))−(n+1)arctan((1/x)))) ⇒ f^((n)) (x)=(((−1)^n n!)/((x^2 +1)^(n+1) ))(√(13))(x^2 +1)^((n+1)/2) cos(arctan((3/2))−(n+1)arctan((1/x))) f^((n)) (x)=(√(13))(−1)^n n!(x^2 +1)^(−((n+1)/2)) cos(arctan((3/2))−(n+1)arctan((1/x)))$
$1) f (x) = \frac{2 x + 3}{x^{2} + 1} = \frac{2 x + 3}{(x - i) (x + i)} = \frac{a}{x - i} + \frac{b}{x + i}$ $a = \frac{2 i + 3}{2 i} = 1 + \frac{3}{2 i} = 1 - \frac{3 i}{2} = \frac{2 - 3 i}{2}$ $b = \frac{- 2 i + 3}{(- 2 i)} = \frac{2 i - 3}{2 i} = 1 - \frac{3}{2 i} = 1 + \frac{3 i}{2} = \frac{2 + 3 i}{2} \Rightarrow$ $f (x) = \frac{2 - 3 i}{2 (x - i)} + \frac{2 + 3 i}{2 (x + i)} \Rightarrow f^{(n)} (x) = \frac{2 - 3 i}{2} {(\frac{1}{x - i})}^{(n)} + \frac{2 + 3 i}{2} {(\frac{1}{x + i})}^{(n)}$ $= \frac{2 - 3 i}{2} \frac{{(- 1)}^{n} n!}{{(x - i)}^{n + 1}} + \frac{2 + 3 i}{2} \frac{{(- 1)}^{n} n!}{{(x + i)}^{n + 1}}$ $= \frac{{(- 1)}^{n} n!}{2} {\frac{2 + 3 i}{{(x + i)}^{n + 1}} + \frac{2 - 3 i}{{(x - i)}^{n + 1}}}$ $= \frac{{(- 1)}^{n} n!}{2} {\frac{(2 + 3 i) {(x - i)}^{n + 1} + (2 - 3 i) {(x + i)}^{n + 1}}{{(x^{2} + 1)}^{n + 1}}}$ $= \frac{{(- 1)}^{n} n!}{{(x^{2} + 1)}^{n + 1}} R e {(2 + 3 i) {(x - i)}^{n + 1}} b u t$ $2 + 3 i = \sqrt{13} e^{i a r c t a n (\frac{3}{2})} a n d x - i = \sqrt{x^{2} + 1} e^{- i a r c t a n (\frac{1}{x})} \Rightarrow$ ${(x - i)}^{n + 1} = {(x^{2} + 1)}^{\frac{n + 1}{2}} e^{- i (n + 1) a r c t a n (\frac{1}{x})} \Rightarrow$ $(2 + 3 i) {(x - i)}^{n + 1} = \sqrt{13} {(x^{2} + 1)}^{\frac{n + 1}{2}} e^{i (a r c t a n (\frac{3}{2}) - (n + 1) a r c t a n (\frac{1}{x}))} \Rightarrow$ $f^{(n)} (x) = \frac{{(- 1)}^{n} n!}{{(x^{2} + 1)}^{n + 1}} \sqrt{13} {(x^{2} + 1)}^{\frac{n + 1}{2}} c o s (a r c t a n (\frac{3}{2}) - (n + 1) a r c t a n (\frac{1}{x}))$ $f^{(n)} (x) = \sqrt{13} {(- 1)}^{n} n! {(x^{2} + 1)}^{- \frac{n + 1}{2}} c o s (a r c t a n (\frac{3}{2}) - (n + 1) a r c t a n (\frac{1}{x}))$
Commented by mathmax by abdo last updated on 27/Oct/19

$2) f^{(10)} (x) = \sqrt{13} (10)! {(x^{2} + 1)}^{- \frac{11}{2}} c o s (a r c t a n (\frac{3}{2}) - 11 a r c t a n (\frac{1}{x})$ $f^{(15)} (x) = - \sqrt{13} (15)! {(x^{2} + 1)}^{- 8} c o s (a r c t a n (\frac{3}{2}) - 16 a r c t a n (\frac{1}{x}))$
Commented by mathmax by abdo last updated on 27/Oct/19

$4) w e h a v e f^{(n)} (x) =$ $\sqrt{13} {(- 1)}^{n} n! {(x^{2} + 1)}^{- \frac{n + 1}{2}} c o s (a r c t a n (\frac{3}{2}) - (n + 1) a r c t a n (\frac{1}{x})) \Rightarrow$ $f^{(n)} (0) = \sqrt{13} {(- 1)}^{n} n! c o s (a r c t a n (\frac{3}{2}) \bar{+} (n + 1) \frac{π}{2})$ $f (x) = \sum_{n = 0}^{\infty} \frac{f^{(n)} (x)}{n!} = \sum_{n = 0}^{\infty} \sqrt{13} {(- 1)}^{n} c o s (a r c t a n (\frac{3}{2}) \bar{+} (n + 1) \frac{π}{2}) x^{n}$
	Answered by mind is power last updated on 26/Oct/19
	$((2x+3)/(x^2 +1))=(a/(x+i))+(c/(x−i)) a+c=2,i(c−a)=3⇒c=((2−3i)/2),a=((2+3i)/2) f(x)=((2+3i)/(2(x+i)))+((2−3i)/(2(x−i))) f^n (x)=((2+3i)/2)(d^n x/dx^n ) (1/(x+i))+((2−3i)/2)(d^n x/dx^n ) (1/(x−i)) =((2+3i)/2) (((−1)^n n!)/((x+i)^(n+1) ))+((2−3i)/2).(((−1)^n n!)/((x−i)^(n+1) )) =(((−1)^n n!)/2)((((2+3i)(x−i)^(n+1) +(2−3i)(x+i)^(n+1) )/((x^2 +1)^(n+1) ))) =(((−1)^n n! (2.Re{ (2+3i)(x+i)^(n+1) }))/((x^2 +1)^(n+1) )) Re(2+3i)(x+i)^(n+1) (x+i)^(n+1) =ΣC_(n+1) ^k .i^k .(x)^(n+1−k) =Σ_(k=0) ^(E(((n+1)/2))) C_(n+1) ^(2k) .(−1)^k .X^(n+1−2k) +Σ_(k=0) ^(E((n/2))) C_(n+1) ^(2k+1) (−1)^k i.X^(n−2k) Re (2+3i).(x+i)^(n+1) = 2Σ_(k=0) ^(E(((n+1)/2))) C_(n+1) ^(2k) .(−1)^k .X^(n+1−2k) +3Σ_(k=0) ^(E((n/2))) C_(n+1) ^(2k+1) (−1)^(k+1) .X^(n−2k) f^n (x)=(((−1)^n n!)/((x^2 +1)^(n+1) )).2(2Σ_(k=0) ^(E(((n+1)/2))) C_(n+1) ^(2k) .(−1)^k .X^(n+1−2k) +3Σ_(k=0) ^(E((n/2))) C_(n+1) ^(2k+1) (−1)^(k+1) .X^(n−2k) ) f^((10)) (x)=2.((10!)/((x^2 +1)^(11) )).(2Σ_(k=0) ^5 C_(11) ^(2k) .(−1)^k X^(11−2k) +3Σ_(k=0) ^5 C_(11) ^(2k+1) (−1)^(k+1) X^(10−2k) ) f^(15) (x) same idee 3) f^(10) (x)=2.10!.(3.)=6.10! f^(15) (0)=2(−1)^(15) .15!.(2.)=−4.15! 4) f^n (0) if n=2s f^n (0)=(−1)^(2s) .(2s)!.2.(3.(−1)^(s+1) )=(((−1)^(s+1) .6.(2s)!)/) n=2s+1⇒f^n (0)=(((−1)^(2s+1) .(2s+1)!.2(2.(−1)^(s+1) ))/)=(−1)^s .4(2s+1)! f(x)=Σ_(n≥0) ((f^n (0).x^n )/(n!))=Σ((f^(2s) (0)x^(2s) )/((2s)!))+Σ((f^(2s+1) (0)x^(2s+1) )/((2s+1)!)) =Σ_(s=0) ^(+∞) (((f^(2s) (0)x^(2s) )/((2s)!))+((f^(2s) (0)x^(2s+1) )/((2s+1)!)))=Σ_(s=0) ^(+∞) (6(−1)^(s+1) x^(2s) +4(−1)^s x^(2s+1) ) =Σ_(s≥0) (−1)^s x^(2s) (−6+4x) 5)gx=∫_0 ^x f g(x)=Σ_(s≥0) (∫_0 ^x 6(−1)^(s+1) t^(2s) +4(−1)^s t^(2s+1) ) =Σ_(s≥0) .(6(−1)^(s+1) .(x^(2s+1) /(2s+1))+((2(−1)^s x^(2s+2) )/(s+1)))$
	$\frac{2 x + 3}{x^{2} + 1} = \frac{a}{x + i} + \frac{c}{x - i}$ $a + c = 2, i (c - a) = 3 \Rightarrow c = \frac{2 - 3 i}{2}, a = \frac{2 + 3 i}{2}$ $f (x) = \frac{2 + 3 i}{2 (x + i)} + \frac{2 - 3 i}{2 (x - i)}$ $f^{n} (x) = \frac{2 + 3 i}{2} \frac{d^{n} x}{{dx}^{n}} \frac{1}{x + i} + \frac{2 - 3 i}{2} \frac{d^{n} x}{{dx}^{n}} \frac{1}{x - i}$ $= \frac{2 + 3 i}{2} \frac{{(- 1)}^{n} n!}{{(x + i)}^{n + 1}} + \frac{2 - 3 i}{2} . \frac{{(- 1)}^{n} n!}{{(x - i)}^{n + 1}}$ $= \frac{{(- 1)}^{n} n!}{2} (\frac{(2 + 3 i) {(x - i)}^{n + 1} + (2 - 3 i) {(x + i)}^{n + 1}}{{(x^{2} + 1)}^{n + 1}})$ $= \frac{{(- 1)}^{n} n! (2 . Re {(2 + 3 i) {(x + i)}^{n + 1}})}{{(x^{2} + 1)}^{n + 1}}$ $Re (2 + 3 i) {(x + i)}^{n + 1}$ ${(x + i)}^{n + 1} = Σ C_{n + 1}^{k} . i^{k} . {(x)}^{n + 1 - k}$ $= \underset{k = 0}{\sum^{E (\frac{n + 1}{2})}} C_{n + 1}^{2 k} . {(- 1)}^{k} . X^{n + 1 - 2 k} + \underset{k = 0}{\sum^{E (\frac{n}{2})}} C_{n + 1}^{2 k + 1} {(- 1)}^{k} i . X^{n - 2 k}$ $Re (2 + 3 i) . {(x + i)}^{n + 1} =$ $2 \underset{k = 0}{\sum^{E (\frac{n + 1}{2})}} C_{n + 1}^{2 k} . {(- 1)}^{k} . X^{n + 1 - 2 k} + 3 \underset{k = 0}{\sum^{E (\frac{n}{2})}} C_{n + 1}^{2 k + 1} {(- 1)}^{k + 1} . X^{n - 2 k}$ $f^{n} (x) = \frac{{(- 1)}^{n} n!}{{(x^{2} + 1)}^{n + 1}} . 2 (2 \underset{k = 0}{\sum^{E (\frac{n + 1}{2})}} C_{n + 1}^{2 k} . {(- 1)}^{k} . X^{n + 1 - 2 k} + 3 \underset{k = 0}{\sum^{E (\frac{n}{2})}} C_{n + 1}^{2 k + 1} {(- 1)}^{k + 1} . X^{n - 2 k})$ $f^{(10)} (x) = 2 . \frac{10!}{{(x^{2} + 1)}^{11}} . (2 \underset{k = 0}{\sum^{5}} C_{11}^{2 k} . {(- 1)}^{k} X^{11 - 2 k} + 3 \underset{k = 0}{\sum^{5}} C_{11}^{2 k + 1} {(- 1)}^{k + 1} X^{10 - 2 k})$ $f^{15} (x) same idee$ $3)$ $f^{10} (x) = 2.10! . (3 .) = 6.10!$ $f^{15} (0) = 2 {(- 1)}^{15} . 15! . (2 .) = - 4.15!$ $4) f^{n} (0) if n = 2 s$ $f^{n} (0) = {(- 1)}^{2 s} . (2 s)! . 2 . (3 . {(- 1)}^{s + 1}) = \frac{{(- 1)}^{s + 1} . 6 . (2 s)!}{}$ $n = 2 s + 1 \Rightarrow f^{n} (0) = \frac{{(- 1)}^{2 s + 1} . (2 s + 1)! . 2 (2 . {(- 1)}^{s + 1})}{} = {(- 1)}^{s} . 4 (2 s + 1)!$ $f (x) = \sum_{n ⩾ 0} \frac{f^{n} (0) . x^{n}}{n!} = Σ \frac{f^{2 s} (0) x^{2 s}}{(2 s)!} + Σ \frac{f^{2 s + 1} (0) x^{2 s + 1}}{(2 s + 1)!}$ $= \underset{s = 0}{\sum^{+ \infty}} (\frac{f^{2 s} (0) x^{2 s}}{(2 s)!} + \frac{f^{2 s} (0) x^{2 s + 1}}{(2 s + 1)!}) = \underset{s = 0}{\sum^{+ \infty}} (6 {(- 1)}^{s + 1} x^{2 s} + 4 {(- 1)}^{s} x^{2 s + 1})$ $= \sum_{s ⩾ 0} {(- 1)}^{s} x^{2 s} (- 6 + 4 x)$ $5) gx = \int_{0}^{x} f$ $g (x) = \sum_{s ⩾ 0} (\int_{0}^{x} 6 {(- 1)}^{s + 1} t^{2 s} + 4 {(- 1)}^{s} t^{2 s + 1})$ $= \sum_{s ⩾ 0} . (6 {(- 1)}^{s + 1} . \frac{x^{2 s + 1}}{2 s + 1} + \frac{2 {(- 1)}^{s} x^{2 s + 2}}{s + 1})$
	Commented by mathmax by abdo last updated on 27/Oct/19

	$t h a n k y o u s i r .$
	Commented by mind is power last updated on 27/Oct/19

	$y^{'} re welcom$
	Commented by mind is power last updated on 27/Oct/19

	$y^{'} re welcom$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum