decompose inside R(x) the fraction 1) F(x) =(1/(x^3 (x−2)^3 )) 2) F(x) =(1/((x+1)^4 (x−3)^4 ))


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Question Number 88032 by mathmax by abdo last updated on 07/Apr/20
$decompose inside R(x) the fraction 1) F(x) =(1/(x^3 (x−2)^3 )) 2) F(x) =(1/((x+1)^4 (x−3)^4 ))$
$d e c o m p o s e i n s i d e R (x) t h e f r a c t i o n$ $1) F (x) = \frac{1}{x^{3} {(x - 2)}^{3}}$ $2) F (x) = \frac{1}{{(x + 1)}^{4} {(x - 3)}^{4}}$
Commented by jagoll last updated on 08/Apr/20

$what is R (x) ? remainder ?$
Commented by abdomathmax last updated on 10/Apr/20
$space of rational fractions$
$s p a c e o f r a t i o n a l f r a c t i o n s$
Commented by abdomathmax last updated on 10/Apr/20
$1)F(x) =(1/8)(((x−(x−2))^3 )/(x^3 (x−2)^3 )) ⇒ 8F(x)?=((x^3 −3x^2 (x−2) +3x(x−2)^2 −(x−2)^3 )/(x^3 (x−2)^3 )) =((x^3 +3x(x−2)(−x+x−2) −(x−2)^3 )/(x^3 (x−2)^3 )) =((x^3 −6x(x−2)−(x−2)^3 )/(x^3 (x−2)^3 )) =(1/((x−2)^3 )) −(6/(x^2 (x−2)^2 ))−(1/x^3 ) =(1/((x−2)^3 ))−(1/x^3 ) −(3/2){(((x−(x−2))^2 )/(x^2 (x−2)^2 ))} =(1/((x−2)^3 ))−(1/x^3 )−(3/2){ ((x^2 −2x(x−2) +(x−2)^2 )/(x^2 (x−2)^2 ))} =(1/((x−2)^3 ))−(1/x^3 )−(3/(2(x−2)^2 )) +(6/(x(x−2))) +(3/(2x^2 )) =(1/((x−2)^3 ))−(1/x^3 )−(3/(2(x−2)^2 )) −3((1/x)−(1/(x−2))) +(3/(2x^2 )) =−(3/x) +(3/(x−2)) +(3/(2x^2 ))−(3/(2(x−2)^2 ))−(1/x^3 ) +(1/((x−2)^3 )) =8F(x)$
$1) F (x) = \frac{1}{8} \frac{{(x - (x - 2))}^{3}}{x^{3} {(x - 2)}^{3}} \Rightarrow$ $8 F (x) ? = \frac{x^{3} - 3 x^{2} (x - 2) + 3 x {(x - 2)}^{2} - {(x - 2)}^{3}}{x^{3} {(x - 2)}^{3}}$ $= \frac{x^{3} + 3 x (x - 2) (- x + x - 2) - {(x - 2)}^{3}}{x^{3} {(x - 2)}^{3}}$ $= \frac{x^{3} - 6 x (x - 2) - {(x - 2)}^{3}}{x^{3} {(x - 2)}^{3}}$ $= \frac{1}{{(x - 2)}^{3}} - \frac{6}{x^{2} {(x - 2)}^{2}} - \frac{1}{x^{3}}$ $= \frac{1}{{(x - 2)}^{3}} - \frac{1}{x^{3}} - \frac{3}{2} {\frac{{(x - (x - 2))}^{2}}{x^{2} {(x - 2)}^{2}}}$ $= \frac{1}{{(x - 2)}^{3}} - \frac{1}{x^{3}} - \frac{3}{2} {\frac{x^{2} - 2 x (x - 2) + {(x - 2)}^{2}}{x^{2} {(x - 2)}^{2}}}$ $= \frac{1}{{(x - 2)}^{3}} - \frac{1}{x^{3}} - \frac{3}{2 {(x - 2)}^{2}} + \frac{6}{x (x - 2)} + \frac{3}{2 x^{2}}$ $= \frac{1}{{(x - 2)}^{3}} - \frac{1}{x^{3}} - \frac{3}{2 {(x - 2)}^{2}} - 3 (\frac{1}{x} - \frac{1}{x - 2}) + \frac{3}{2 x^{2}}$ $= - \frac{3}{x} + \frac{3}{x - 2} + \frac{3}{2 x^{2}} - \frac{3}{2 {(x - 2)}^{2}} - \frac{1}{x^{3}} + \frac{1}{{(x - 2)}^{3}} = 8 F (x)$
Commented by mathmax by abdo last updated on 10/Apr/20
$error of calculus 8F(x)=(1/((x−2)^3 ))−(1/x^3 )−(3/(2(x−2)^2 )) +(3/(x(x−2)))−(3/(2x^2 )) =(1/((x−2)^3 ))−(1/x^3 )−(3/(2(x−2)^2 )) −(3/2)((1/x)−(1/(x−2)))−(3/(2x^2 )) F(x)=(1/8){(1/((x−2)^3 ))−(1/x^3 )−(3/(2(x−2)^2 ))−(3/(2x)) +(3/(2(x−2)))−(3/(2x^2 ))}$
$e r r o r o f c a l c u l u s$ $8 F (x) = \frac{1}{{(x - 2)}^{3}} - \frac{1}{x^{3}} - \frac{3}{2 {(x - 2)}^{2}} + \frac{3}{x (x - 2)} - \frac{3}{2 x^{2}}$ $= \frac{1}{{(x - 2)}^{3}} - \frac{1}{x^{3}} - \frac{3}{2 {(x - 2)}^{2}} - \frac{3}{2} (\frac{1}{x} - \frac{1}{x - 2}) - \frac{3}{2 x^{2}}$ $F (x) = \frac{1}{8} {\frac{1}{{(x - 2)}^{3}} - \frac{1}{x^{3}} - \frac{3}{2 {(x - 2)}^{2}} - \frac{3}{2 x} + \frac{3}{2 (x - 2)} - \frac{3}{2 x^{2}}}$
Commented by mathmax by abdo last updated on 10/Apr/20
$2)F(x)=(1/4^4 )(({(x+1)−(x−3)}^4 )/((x+1)^4 (x−3)^4 )) ⇒4^4 F(x)=(({(x+1)^2 −2(x+1)(x−3)+(x−3)^2 }^2 )/((x+1)^4 (x−3)^4 )) we use the identity (a+b+c)^2 =a^2 +b^2 +c^2 +2(ab +bc +ca) ⇒ 4^4 F(x)=(((x+1)^4 +4(x+1)^2 (x−3)^2 +(x−3)^4 +2{ −2(x+1)^3 (x−3)+(x+1)^2 (x−3)^2 −2(x+1)(x−3)^3 })/((x+1)^4 (x−3)^4 )) =(1/((x−3)^4 )) +(4/((x+1)^2 (x−3)^2 )) +(1/((x+1)^4 )) −(4/((x+1)(x−3)^3 )) +(2/((x+1)^2 (x−3)^2 ))−(4/((x+1)^3 (x−3))) also we have (4/((x+1)^2 (x−3)^2 )) =(1/4){((((x+1)−(x−3))^2 )/((x+1)^2 (x−3)^2 ))} =(1/4){(((x+1)^2 −2(x+1)(x−3) +(x−3)^2 )/((x+1)^2 (x−3)^2 ))} =(1/4){ (1/((x−3)^2 )) −(2/((x+1)(x−3))) +(1/((x+1)^2 ))} =(1/(4(x−3)^2 ))+(1/8)((1/(x+1))−(1/(x−3)))+(1/(4(x+1)^2 )) =(1/(4(x−3)^2 )) +(1/(8(x+1)))−(1/(8(x−3)))+(1/(4(x+1)^2 )) −(4/((x+1)(x−3)^3 ))−(4/((x+1)^3 (x−3))) =−(4/((x+1)(x−3)))((1/((x−3)^2 ))+(1/((x+1)^2 ))) =((1/(x+1))−(1/(x−3)))((1/((x−3)^2 ))+(1/((x+1)^2 ))) =(1/((x+1)(x−3)^2 )) +(1/((x+1)^3 ))−(1/((x−3)^3 ))−(1/((x−3)(x+1)^2 )) =(1/((x+1)(x−3))){(1/(x−3))−(1/(x+1))}+(1/((x+1)^3 ))−(1/((x−3)^3 )) =−(1/4)((1/(x+1))−(1/(x−3)))((1/(x−3))−(1/(x+1)))+(1/((x+1)^3 ))−(1/((x−3)^3 )) =−(1/4)((1/((x+1)(x−3)))−(1/((x+1)^2 ))−(1/((x−3)^2 ))+(1/((x−3)(x+1))))+(1/((x+1)^3 ))−(1/((x−3)^3 )) ....$
$2) F (x) = \frac{1}{4^{4}} \frac{{(x + 1) - (x - 3)}^{4}}{{(x + 1)}^{4} {(x - 3)}^{4}}$ $\Rightarrow 4^{4} F (x) = \frac{{{(x + 1)}^{2} - 2 (x + 1) (x - 3) + {(x - 3)}^{2}}^{2}}{{(x + 1)}^{4} {(x - 3)}^{4}}$ $w e u s e t h e i d e n t i t y {(a + b + c)}^{2} = a^{2} + b^{2} + c^{2} + 2 (a b + b c + c a) \Rightarrow$ $4^{4} F (x) = \frac{{(x + 1)}^{4} + 4 {(x + 1)}^{2} {(x - 3)}^{2} + {(x - 3)}^{4} + 2 {- 2 {(x + 1)}^{3} (x - 3) + {(x + 1)}^{2} {(x - 3)}^{2} - 2 (x + 1) {(x - 3)}^{3}}}{{(x + 1)}^{4} {(x - 3)}^{4}}$ $= \frac{1}{{(x - 3)}^{4}} + \frac{4}{{(x + 1)}^{2} {(x - 3)}^{2}} + \frac{1}{{(x + 1)}^{4}} - \frac{4}{(x + 1) {(x - 3)}^{3}} + \frac{2}{{(x + 1)}^{2} {(x - 3)}^{2}} - \frac{4}{{(x + 1)}^{3} (x - 3)}$ $a l s o w e h a v e \frac{4}{{(x + 1)}^{2} {(x - 3)}^{2}} = \frac{1}{4} {\frac{{((x + 1) - (x - 3))}^{2}}{{(x + 1)}^{2} {(x - 3)}^{2}}}$ $= \frac{1}{4} {\frac{{(x + 1)}^{2} - 2 (x + 1) (x - 3) + {(x - 3)}^{2}}{{(x + 1)}^{2} {(x - 3)}^{2}}}$ $= \frac{1}{4} {\frac{1}{{(x - 3)}^{2}} - \frac{2}{(x + 1) (x - 3)} + \frac{1}{{(x + 1)}^{2}}}$ $= \frac{1}{4 {(x - 3)}^{2}} + \frac{1}{8} (\frac{1}{x + 1} - \frac{1}{x - 3}) + \frac{1}{4 {(x + 1)}^{2}}$ $= \frac{1}{4 {(x - 3)}^{2}} + \frac{1}{8 (x + 1)} - \frac{1}{8 (x - 3)} + \frac{1}{4 {(x + 1)}^{2}}$ $- \frac{4}{(x + 1) {(x - 3)}^{3}} - \frac{4}{{(x + 1)}^{3} (x - 3)} = - \frac{4}{(x + 1) (x - 3)} (\frac{1}{{(x - 3)}^{2}} + \frac{1}{{(x + 1)}^{2}})$ $= (\frac{1}{x + 1} - \frac{1}{x - 3}) (\frac{1}{{(x - 3)}^{2}} + \frac{1}{{(x + 1)}^{2}})$ $= \frac{1}{(x + 1) {(x - 3)}^{2}} + \frac{1}{{(x + 1)}^{3}} - \frac{1}{{(x - 3)}^{3}} - \frac{1}{(x - 3) {(x + 1)}^{2}}$ $= \frac{1}{(x + 1) (x - 3)} {\frac{1}{x - 3} - \frac{1}{x + 1}} + \frac{1}{{(x + 1)}^{3}} - \frac{1}{{(x - 3)}^{3}}$ $= - \frac{1}{4} (\frac{1}{x + 1} - \frac{1}{x - 3}) (\frac{1}{x - 3} - \frac{1}{x + 1}) + \frac{1}{{(x + 1)}^{3}} - \frac{1}{{(x - 3)}^{3}}$ $= - \frac{1}{4} (\frac{1}{(x + 1) (x - 3)} - \frac{1}{{(x + 1)}^{2}} - \frac{1}{{(x - 3)}^{2}} + \frac{1}{(x - 3) (x + 1)}) + \frac{1}{{(x + 1)}^{3}} - \frac{1}{{(x - 3)}^{3}}$ $. . . .$
Commented by abdomathmax last updated on 10/Apr/20
$1) another way we detemine D_2 f(0) with f(x)=(1/((x−2)^3 )) ⇒f(x)=f(0) +(x/(1!))f^((1)) (0)+(x^2 /(2!))f^((2)) (0) +x^3 ξ(x) f(0)=−(1/8) f(x)=(x−2)^(−3) ⇒f^′ (x)=−3(x−2)^(−4) ⇒ f^′ (0)=−3(−2)^(−4) =((−3)/2^4 ) =−(3/(16)) f^((2)) (0) =12(x−2)^(−5) ⇒f^((2)) (0) =((12)/((−2)^5 )) =−((12)/2^5 ) =−((12)/(32)) =−(3/8) decomposition of F is F(x)=Σ_(i=1) ^3 (a_i /x^i ) +Σ_(i=1) ^3 (b_i /((x−2)^i )) we have F(x)=(1/x^3 )(−(1/8)−(3/(16))x−(3/(16))x^2 +x^3 ξ(x)) =−(1/(8x^3 ))−(3/(16x^2 ))−(3/(16x)) +ξ(x) ⇒a_1 =−(3/(16)) a_2 =−(3/(16)) and a_3 =−(1/8) let find the b_i we do the changement x−2 =t ⇒ F(x)=(1/(t^3 (t+2)^3 )) and find D_2 (0) for g(t)=(t+2)^(−3) g(t)=g(0)+t g^′ (0) +(t^2 /2)g^((2)) (0) +t^3 δ(x) g(0)=(1/8) , g^′ (t)=−3(t+2)^(−4) ⇒g^′ (0)=((−3)/(16)) g^((2)) (0) =12(t+2)^(−5) ⇒g^((2)) (0)=((12)/2^5 ) =((12)/(32)) =(3/8) F(x)=(1/t^3 ){(1/8)−(3/(16))t +(3/(16))t^2 +t^3 δ(t)} =(1/(8t^3 )) −(3/(16t^2 )) +(3/(16t)) +δ(t) =(1/(8(x−2)^3 ))−(3/(16(x−2)^2 ))+(3/(16(x−2))) +δ(t) ⇒ b_1 =(3/(16)) , b_2 =−(3/(26)) and b_3 =(1/8) tbe coefficients are found$
$1) a n o t h e r w a y w e d e t e m i n e D_{2} f (0) w i t h$ $f (x) = \frac{1}{{(x - 2)}^{3}} \Rightarrow f (x) = f (0) + \frac{x}{1!} f^{(1)} (0) + \frac{x^{2}}{2!} f^{(2)} (0)$ $+ x^{3} ξ (x)$ $f (0) = - \frac{1}{8}$ $f (x) = {(x - 2)}^{- 3} \Rightarrow f^{^{'}} (x) = - 3 {(x - 2)}^{- 4} \Rightarrow$ $f^{^{'}} (0) = - 3 {(- 2)}^{- 4} = \frac{- 3}{2^{4}} = - \frac{3}{16}$ $f^{(2)} (0) = 12 {(x - 2)}^{- 5} \Rightarrow f^{(2)} (0) = \frac{12}{{(- 2)}^{5}} = - \frac{12}{2^{5}}$ $= - \frac{12}{32} = - \frac{3}{8} d e c o m p o s i t i o n o f F i s$ $F (x) = \sum_{i = 1}^{3} \frac{a_{i}}{x^{i}} + \sum_{i = 1}^{3} \frac{b_{i}}{{(x - 2)}^{i}} w e h a v e$ $F (x) = \frac{1}{x^{3}} (- \frac{1}{8} - \frac{3}{16} x - \frac{3}{16} x^{2} + x^{3} ξ (x))$ $= - \frac{1}{8 x^{3}} - \frac{3}{16 x^{2}} - \frac{3}{16 x} + ξ (x) \Rightarrow a_{1} = - \frac{3}{16}$ $a_{2} = - \frac{3}{16} a n d a_{3} = - \frac{1}{8} l e t f i n d t h e b_{i} w e d o t h e$ $c h a n g e m e n t x - 2 = t \Rightarrow$ $F (x) = \frac{1}{t^{3} {(t + 2)}^{3}} a n d f i n d D_{2} (0) f o r g (t) = {(t + 2)}^{- 3}$ $g (t) = g (0) + t g^{^{'}} (0) + \frac{t^{2}}{2} g^{(2)} (0) + t^{3} δ (x)$ $g (0) = \frac{1}{8}, g^{^{'}} (t) = - 3 {(t + 2)}^{- 4} \Rightarrow g^{^{'}} (0) = \frac{- 3}{16}$ $g^{(2)} (0) = 12 {(t + 2)}^{- 5} \Rightarrow g^{(2)} (0) = \frac{12}{2^{5}} = \frac{12}{32} = \frac{3}{8}$ $F (x) = \frac{1}{t^{3}} {\frac{1}{8} - \frac{3}{16} t + \frac{3}{16} t^{2} + t^{3} δ (t)}$ $= \frac{1}{8 t^{3}} - \frac{3}{16 t^{2}} + \frac{3}{16 t} + δ (t)$ $= \frac{1}{8 {(x - 2)}^{3}} - \frac{3}{16 {(x - 2)}^{2}} + \frac{3}{16 (x - 2)} + δ (t) \Rightarrow$ $b_{1} = \frac{3}{16}, b_{2} = - \frac{3}{26} a n d b_{3} = \frac{1}{8}$ $t b e c o e f f i c i e n t s a r e f o u n d$
Commented by abdomathmax last updated on 10/Apr/20

$b_{2} = - \frac{3}{16}$
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