Prove that (1+x+x^2 +...)(1+2x+3x^2 +...) =(1/2)(1.2+2.3x+3.4x^2 +...)


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Question Number 2795 by Rasheed Soomro last updated on 27/Nov/15

$P r o v e t h a t$ $(1 + x + x^{2} + . . .) (1 + 2 x + 3 x^{2} + . . .)$ $= \frac{1}{2} (1.2 + 2.3 x + 3.4 x^{2} + . . .)$
	Answered by prakash jain last updated on 27/Nov/15

	$A = (1 + x + x^{2} + . . .)$ $B = (1 + 2 x + 3 x^{2} + 4 x^{3})$ $Let us call coefficient of x^{n} i n A = a_{n} = 1$ $Let us call coefficient of x^{n} i n B = b_{n} = (n + 1)$ $c o e f f i c i e n t o f x_{n} i n p r o d u c t =$ $a_{0} b_{n} + a_{1} b_{n - 1} + a_{2} b_{n - 2} + . . . + a_{n} b_{0}$ $= (n + 1) + (n) + (n - 1) + . . . + 1$ $= \frac{(n + 1) (n + 2)}{2}$ $A \cdot B = \underset{i = 1}{\sum^{\infty}} \frac{(i + 1) (i + 2)}{2} x^{i} = \frac{1}{2} \underset{i = 1}{\sum^{\infty}} (i + 1) (i + 2) x^{i}$ $= \frac{1}{2} (1 \cdot 2 + 2 \cdot 3 x + 3 \cdot 4 x^{2} + 4 \cdot 5 x^{3} + . . .)$
	Answered by Yozzi last updated on 27/Nov/15
	$The Maclaurin series of (1/(1−x)) gives (1/(1−x))=1+x+x^2 +x^3 +...=Σ_(r=0) ^∞ x^r PROOF: Let f(x)=(1/(1−x)) ,x∈R−{1}. Differentiating we get f^((0)) (x)=(1−x)^(−1) f^((1)) (x)=(1−x)^(−2) f^((2)) (x)=2(1−x)^(−3) f^((3)) (x)=6(1−x)^(−4) . I conject that f^((n)) (x)=n!(1−x)^(−n−1) where f^((n)) (x) denotes the nth derivative of f(x)=(1−x)^(−1) (x≠1) and n∈N∪{0}. Let P(n) be the proposition that f^((n)) (x)=n!(1−x)^(−n−1) as conjectured. For n=0, P(0) gives f^((0)) =0!(1−x)^(−0−1) f^((0)) (x)=1×(1−x)^(−1) =(1−x)^(−1) which is f(x) differentiated zero times. Therefore, P(n) is true when n=0. Assume now that P(n) is true when n=k: f^((k)) (x)=k!(1−x)^(−k−1) . Fot n=k+1, f^((k+1)) (x)=(d/dx)(f^((k)) (x)) =(d/dx)(k!(1−x)^(−k−1) ) =(−k−1)(−1)k!(1−x)^(−k−2) =(k+1)k!(1−x)^(−k−2) f^((k+1)) (x)=(k+1)!(1−x)^(−(k+1)−1) So, P(k+1) is true, assuming P(k) is true. Since P(0) is true then,by the principle of mathematical induction, P(n) is true for all non−negative integers. Hence, f(x) is infinitely differentiable for x≠1 and f(x) with all its derivatives are defined at x=0. We can then obtain the Maclaurin series expansion of f(x)=(1/(1−x)). This is given by f(x)=Σ_(r=0) ^∞ ((x^r f^((r)) (0))/(r!)) Since f^((r)) (x)=r!(1−x)^(−r−1) f(x)=Σ_(r=0) ^∞ (x^r /(r!))×r!(1−0)^(−r−1) f(x)=Σ_(r=0) ^∞ x^r Therefore, (1/(1−x))=Σ_(r=0) ^∞ x^r . □ Observe that 1+2x+3x^2 +4x^3 +...=Σ_(r=1) ^∞ rx^(r−1) R.H.S=Σ_(r=1) ^∞ (d/dx)(x^r ). The series is term by term differentiable. ∴ Σ_(r=1) ^∞ (d/dx)(x^r )=(d/dx)(Σ_(r=1) ^∞ x^r )=(d/dx)(xΣ_(r=1) ^∞ x^(r−1) ). Now,Σ_(r=1) ^∞ x^(r−1) =Σ_(r=0) ^∞ x^r =(1/(1−x)). ∴R.H.S=(d/dx)((x/(1−x))) =(((1−x)×1−(−1)(x))/((1−x)^2 )) (Quotient rule) =((1−x+x)/((1−x)^2 )) R.H.S=(1/((1−x)^2 )). ∴ Σ_(r=1) ^∞ rx^(r−1) =(1/((1−x)^2 )). The L.H.S product of the equation in question becomes (Σ_(r=0) ^∞ x^r )(Σ_(r=1) ^∞ rx^(r−1) )=(1/(1−x))×(1/((1−x)^2 )) L.H.S=(1/((1−x)^3 )). Let g(x)=(1−x)^(−3) . g^((1)) (x)=3(1−x)^(−4) =((3!)/(2!))(1−x)^(−4) g^((2)) (x)=3×4(1−x)^(−5) =((4!)/(2!))(1−x)^(−5) g^((3)) (x)=3×4×5(1−x)^(−6) =((5!)/(2!))(1−x)^(−6) I thus state without proof that the nth derivative$
	$T h e M a c l a u r i n s e r i e s o f \frac{1}{1 - x} g i v e s$ $\frac{1}{1 - x} = 1 + x + x^{2} + x^{3} + . . . = \underset{r = 0}{\sum^{\infty}} x^{r}$ $PROOF :$ $L e t f (x) = \frac{1}{1 - x}, x \in R - {1} .$ $D i f f e r e n t i a t i n g w e g e t$ $f^{(0)} (x) = {(1 - x)}^{- 1}$ $f^{(1)} (x) = {(1 - x)}^{- 2}$ $f^{(2)} (x) = 2 {(1 - x)}^{- 3}$ $f^{(3)} (x) = 6 {(1 - x)}^{- 4} .$ $I c o n j e c t t h a t f^{(n)} (x) = n! {(1 - x)}^{- n - 1}$ $w h e r e f^{(n)} (x) d e n o t e s t h e n t h d e r i v a t i v e$ $o f f (x) = {(1 - x)}^{- 1} (x \neq 1) a n d n \in N \cup {0} .$ $L e t P (n) b e t h e p r o p o s i t i o n t h a t$ $f^{(n)} (x) = n! {(1 - x)}^{- n - 1}$ $a s c o n j e c t u r e d .$ $F o r n = 0, P (0) g i v e s f^{(0)} = 0! {(1 - x)}^{- 0 - 1}$ $f^{(0)} (x) = 1 \times {(1 - x)}^{- 1} = {(1 - x)}^{- 1}$ $w h i c h i s f (x) d i f f e r e n t i a t e d z e r o t i m e s .$ $T h e r e f o r e, P (n) i s t r u e w h e n n = 0 .$ $A s s u m e n o w t h a t P (n) i s t r u e w h e n$ $n = k : f^{(k)} (x) = k! {(1 - x)}^{- k - 1} .$ $F o t n = k + 1,$ $f^{(k + 1)} (x) = \frac{d}{d x} (f^{(k)} (x))$ $= \frac{d}{d x} (k! {(1 - x)}^{- k - 1})$ $= (- k - 1) (- 1) k! {(1 - x)}^{- k - 2}$ $= (k + 1) k! {(1 - x)}^{- k - 2}$ $f^{(k + 1)} (x) = (k + 1)! {(1 - x)}^{- (k + 1) - 1}$ $S o, P (k + 1) i s t r u e, a s s u m i n g P (k) i s$ $t r u e . S i n c e P (0) i s t r u e t h e n, b y t h e$ $p r i n c i p l e o f m a t h e m a t i c a l i n d u c t i o n,$ $P (n) i s t r u e f o r a l l n o n - n e g a t i v e$ $i n t e g e r s .$ $H e n c e, f (x) i s i n f i n i t e l y d i f f e r e n t i a b l e$ $f o r x \neq 1 a n d f (x) w i t h a l l i t s d e r i v a t i v e s a r e$ $d e f i n e d a t x = 0 . W e c a n t h e n o b t a i n$ $t h e M a c l a u r i n s e r i e s e x p a n s i o n o f$ $f (x) = \frac{1}{1 - x} . T h i s i s g i v e n b y$ $f (x) = \underset{r = 0}{\sum^{\infty}} \frac{x^{r} f^{(r)} (0)}{r!}$ $S i n c e f^{(r)} (x) = r! {(1 - x)}^{- r - 1}$ $f (x) = \underset{r = 0}{\sum^{\infty}} \frac{x^{r}}{r!} \times r! {(1 - 0)}^{- r - 1}$ $f (x) = \underset{r = 0}{\sum^{\infty}} x^{r}$ $T h e r e f o r e, \frac{1}{1 - x} = \underset{r = 0}{\sum^{\infty}} x^{r} . ◻$ $O b s e r v e t h a t$ $1 + 2 x + 3 x^{2} + 4 x^{3} + . . . = \underset{r = 1}{\sum^{\infty}} {r x}^{r - 1}$ $R . H . S = \underset{r = 1}{\sum^{\infty}} \frac{d}{d x} (x^{r}) .$ $T h e s e r i e s i s t e r m b y t e r m d i f f e r e n t i a b l e .$ $∴ \underset{r = 1}{\sum^{\infty}} \frac{d}{d x} (x^{r}) = \frac{d}{d x} (\underset{r = 1}{\sum^{\infty}} x^{r}) = \frac{d}{d x} (x \underset{r = 1}{\sum^{\infty}} x^{r - 1}) .$ $N o w, \underset{r = 1}{\sum^{\infty}} x^{r - 1} = \underset{r = 0}{\sum^{\infty}} x^{r} = \frac{1}{1 - x} .$ $∴ R . H . S = \frac{d}{d x} (\frac{x}{1 - x})$ $= \frac{(1 - x) \times 1 - (- 1) (x)}{{(1 - x)}^{2}} (Q u o t i e n t r u l e)$ $= \frac{1 - x + x}{{(1 - x)}^{2}}$ $R . H . S = \frac{1}{{(1 - x)}^{2}} . ∴ \underset{r = 1}{\sum^{\infty}} {r x}^{r - 1} = \frac{1}{{(1 - x)}^{2}} .$ $T h e L . H . S p r o d u c t o f t h e e q u a t i o n$ $i n q u e s t i o n b e c o m e s$ $(\underset{r = 0}{\sum^{\infty}} x^{r}) (\underset{r = 1}{\sum^{\infty}} {r x}^{r - 1}) = \frac{1}{1 - x} \times \frac{1}{{(1 - x)}^{2}}$ $L . H . S = \frac{1}{{(1 - x)}^{3}} .$ $L e t g (x) = {(1 - x)}^{- 3} .$ $g^{(1)} (x) = 3 {(1 - x)}^{- 4} = \frac{3!}{2!} {(1 - x)}^{- 4}$ $g^{(2)} (x) = 3 \times 4 {(1 - x)}^{- 5} = \frac{4!}{2!} {(1 - x)}^{- 5}$ $g^{(3)} (x) = 3 \times 4 \times 5 {(1 - x)}^{- 6} = \frac{5!}{2!} {(1 - x)}^{- 6}$ $I t h u s s t a t e w i t h o u t p r o o f t h a t t h e n t h d e r i v a t i v e$
	Commented by Yozzi last updated on 27/Nov/15

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