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Question Number 14047 by FilupS last updated on 27/May/17

S=1+i−1−i+1+... (1/i)=−i S=i(−i+1+i−1−i+1+...) S=i(−i+S) S=1+iS S(1−i)=1 ∴ S=(1/(1−i)) a) Is this correct? b) Do there exist any other sequences in the form of: S=(a_1 +...+a_n )+(a_1 +...+a_n )+... S=(a_1 +...+a_n )(1+1+...+1_(m times) ) ⇒S=Σ_(i=1) ^(m→∞) Σ_(j=1) ^n a_j where a_(t+1) =ba_t , a_1 =ba_n I′m very interested in these sequences

$${S}=\mathrm{1}+{i}−\mathrm{1}−{i}+\mathrm{1}+... \\ $$$$\frac{\mathrm{1}}{{i}}=−{i} \\ $$$${S}={i}\left(−{i}+\mathrm{1}+{i}−\mathrm{1}−{i}+\mathrm{1}+...\right) \\ $$$${S}={i}\left(−{i}+{S}\right) \\ $$$${S}=\mathrm{1}+{iS} \\ $$$${S}\left(\mathrm{1}−{i}\right)=\mathrm{1} \\ $$$$\therefore\:{S}=\frac{\mathrm{1}}{\mathrm{1}−{i}} \\ $$$$\: \\ $$$$\left.\mathrm{a}\right)\:\mathrm{Is}\:\mathrm{this}\:\mathrm{correct}? \\ $$$$\left.\mathrm{b}\right)\:\mathrm{Do}\:\mathrm{there}\:\mathrm{exist}\:\mathrm{any}\:\mathrm{other}\:\mathrm{sequences} \\ $$$$\:\:\:\:\:\:\mathrm{in}\:\mathrm{the}\:\mathrm{form}\:\mathrm{of}: \\ $$$${S}=\left({a}_{\mathrm{1}} +...+{a}_{{n}} \right)+\left({a}_{\mathrm{1}} +...+{a}_{{n}} \right)+... \\ $$$${S}=\left({a}_{\mathrm{1}} +...+{a}_{{n}} \right)\left(\underset{{m}\:\mathrm{times}} {\mathrm{1}+\mathrm{1}+...+\mathrm{1}}\right) \\ $$$$\Rightarrow{S}=\underset{{i}=\mathrm{1}} {\overset{{m}\rightarrow\infty} {\sum}}\underset{{j}=\mathrm{1}} {\overset{{n}} {\sum}}{a}_{{j}} \\ $$$$\mathrm{where}\:{a}_{{t}+\mathrm{1}} ={ba}_{{t}} ,\:\:{a}_{\mathrm{1}} ={ba}_{{n}} \\ $$$$\: \\ $$$$\mathrm{I}'\mathrm{m}\:\mathrm{very}\:\mathrm{interested}\:\mathrm{in}\:\mathrm{these}\:\mathrm{sequences} \\ $$

Commented by prakash jain last updated on 27/May/17

This is geometric series with common ratio i. Since ∣i∣=1, by geometric series test the series diverges. S_(4n) =0 S_(4n+1) =1 S_(4n+2) =1+i S_(4n+3) =i S=((0+(1)+(1+i)+(i))/4)=((1+i)/2)

$$\mathrm{This}\:\mathrm{is}\:\mathrm{geometric}\:\mathrm{series}\:\mathrm{with} \\ $$$$\mathrm{common}\:\mathrm{ratio}\:{i}. \\ $$$$\mathrm{Since}\:\mid{i}\mid=\mathrm{1},\:\mathrm{by}\:\mathrm{geometric}\:\mathrm{series} \\ $$$$\mathrm{test}\:\mathrm{the}\:\mathrm{series}\:\mathrm{diverges}. \\ $$$${S}_{\mathrm{4}{n}} =\mathrm{0} \\ $$$${S}_{\mathrm{4}{n}+\mathrm{1}} =\mathrm{1} \\ $$$${S}_{\mathrm{4}{n}+\mathrm{2}} =\mathrm{1}+{i} \\ $$$${S}_{\mathrm{4}{n}+\mathrm{3}} ={i} \\ $$$${S}=\frac{\mathrm{0}+\left(\mathrm{1}\right)+\left(\mathrm{1}+{i}\right)+\left({i}\right)}{\mathrm{4}}=\frac{\mathrm{1}+{i}}{\mathrm{2}} \\ $$

Commented by prakash jain last updated on 27/May/17

S=(1/(1+i)) is not correct. You can get 4 different answers for S. S=1+i−1−i+1+i−1−i =(1+i)−(1+i)+(1+i)+... =(1+i)−[(1+i)−(1+i)...] =(1+i)−S 2S=(1+i)⇒S=((1+i)/2)

$${S}=\frac{\mathrm{1}}{\mathrm{1}+{i}}\:\mathrm{is}\:\mathrm{not}\:\mathrm{correct}.\:\mathrm{You}\:\mathrm{can} \\ $$$$\mathrm{get}\:\mathrm{4}\:\mathrm{different}\:\mathrm{answers}\:\mathrm{for}\:{S}. \\ $$$${S}=\mathrm{1}+{i}−\mathrm{1}−{i}+\mathrm{1}+{i}−\mathrm{1}−{i} \\ $$$$\:\:\:\:=\left(\mathrm{1}+{i}\right)−\left(\mathrm{1}+{i}\right)+\left(\mathrm{1}+{i}\right)+... \\ $$$$\:\:\:\:=\left(\mathrm{1}+{i}\right)−\left[\left(\mathrm{1}+{i}\right)−\left(\mathrm{1}+{i}\right)...\right] \\ $$$$\:\:\:\:\:=\left(\mathrm{1}+{i}\right)−{S} \\ $$$$\mathrm{2}{S}=\left(\mathrm{1}+{i}\right)\Rightarrow{S}=\frac{\mathrm{1}+{i}}{\mathrm{2}} \\ $$

Commented by RasheedSindhi last updated on 27/May/17

S=1+iS ⇒^? S(1+i)=1

$${S}=\mathrm{1}+{iS} \\ $$$$\overset{?} {\Rightarrow}{S}\left(\mathrm{1}+{i}\right)=\mathrm{1} \\ $$$$ \\ $$

Commented by ajfour last updated on 27/May/17

S_n =((1/2)+(i/2))(1−i^n ) S_(4n) =0 S_(4n+1) =1 S_(4n+2) =1+i S_(4n+3) =i .

$${S}_{{n}} =\left(\frac{\mathrm{1}}{\mathrm{2}}+\frac{{i}}{\mathrm{2}}\right)\left(\mathrm{1}−{i}^{{n}} \right)\: \\ $$$${S}_{\mathrm{4}{n}} =\mathrm{0} \\ $$$${S}_{\mathrm{4}{n}+\mathrm{1}} =\mathrm{1} \\ $$$${S}_{\mathrm{4}{n}+\mathrm{2}} =\mathrm{1}+{i} \\ $$$${S}_{\mathrm{4}{n}+\mathrm{3}} ={i}\:\:\:. \\ $$

Commented by ajfour last updated on 27/May/17

Commented by prakash jain last updated on 27/May/17

S_n is fine geometric series and finite number of terms converges. Geometric series with ∣r∣≥1 diverges. For example S=1−1+1−1+... diverges and does not have a fixed sum. S=1 or 0 so S can be taken as (1/2) using analytical continuation.

$${S}_{{n}} \:\mathrm{is}\:\mathrm{fine}\:\mathrm{geometric}\:\mathrm{series} \\ $$$$\mathrm{and}\:\mathrm{finite}\:\mathrm{number}\:\mathrm{of}\:\mathrm{terms} \\ $$$$\mathrm{converges}. \\ $$$$\mathrm{Geometric}\:\mathrm{series}\:\mathrm{with}\:\mid{r}\mid\geqslant\mathrm{1} \\ $$$$\mathrm{diverges}.\: \\ $$$$\mathrm{For}\:\mathrm{example} \\ $$$$\mathrm{S}=\mathrm{1}−\mathrm{1}+\mathrm{1}−\mathrm{1}+... \\ $$$$\mathrm{diverges}\:\mathrm{and}\:\mathrm{does}\:\mathrm{not}\:\mathrm{have}\:\mathrm{a}\:\mathrm{fixed} \\ $$$$\mathrm{sum}. \\ $$$$\mathrm{S}=\mathrm{1}\:{or}\:\mathrm{0}\: \\ $$$$\mathrm{so}\:{S}\:\mathrm{can}\:\mathrm{be}\:\mathrm{taken}\:\mathrm{as}\:\frac{\mathrm{1}}{\mathrm{2}}\:\mathrm{using} \\ $$$$\mathrm{analytical}\:\mathrm{continuation}. \\ $$

Commented by FilupS last updated on 27/May/17

S=(1/(1+i)) has been corrected to S=(1/(1−i)) This is all interesting. I am aware of S=1−1+1−1+... These kinds if sequences are interesting

$${S}=\frac{\mathrm{1}}{\mathrm{1}+{i}}\:\mathrm{has}\:\mathrm{been}\:\mathrm{corrected}\:\mathrm{to}\:{S}=\frac{\mathrm{1}}{\mathrm{1}−{i}} \\ $$$$ \\ $$$$\mathrm{This}\:\mathrm{is}\:\mathrm{all}\:\mathrm{interesting}. \\ $$$$\mathrm{I}\:\mathrm{am}\:\mathrm{aware}\:\mathrm{of}\:{S}=\mathrm{1}−\mathrm{1}+\mathrm{1}−\mathrm{1}+... \\ $$$$\mathrm{These}\:\mathrm{kinds}\:\mathrm{if}\:\mathrm{sequences}\:\mathrm{are}\:\mathrm{interesting} \\ $$

Commented by prakash jain last updated on 27/May/17

For the sequence of the type. a_(t+1) =na_t ,a_1 =ba_n a_n =a_1 b^(n−1) a_1 =ba_n ⇒b^n =1 b^n =1=e^(2πi) b=e^(2πki/n) , k∈{0,1,2,...,n−1} b is n^(th) root of unity.

$$\mathrm{For}\:\mathrm{the}\:\mathrm{sequence}\:\mathrm{of}\:\mathrm{the}\:\mathrm{type}. \\ $$$${a}_{{t}+\mathrm{1}} ={na}_{{t}} ,{a}_{\mathrm{1}} ={ba}_{{n}} \\ $$$${a}_{{n}} ={a}_{\mathrm{1}} {b}^{{n}−\mathrm{1}} \\ $$$${a}_{\mathrm{1}} ={ba}_{{n}} \\ $$$$\Rightarrow{b}^{{n}} =\mathrm{1} \\ $$$${b}^{{n}} =\mathrm{1}={e}^{\mathrm{2}\pi{i}} \\ $$$${b}={e}^{\mathrm{2}\pi{ki}/{n}} ,\:{k}\in\left\{\mathrm{0},\mathrm{1},\mathrm{2},...,{n}−\mathrm{1}\right\} \\ $$$${b}\:\mathrm{is}\:{n}^{{th}} \:\mathrm{root}\:\mathrm{of}\:\mathrm{unity}. \\ $$

Commented by ajfour last updated on 27/May/17

if S=1−1+1−1+... then even we have a formula for S_n : S_n =(1/2)(1−i^(2n) ) .

$${if}\:\:\:\mathrm{S}=\mathrm{1}−\mathrm{1}+\mathrm{1}−\mathrm{1}+... \\ $$$${then}\:{even}\:{we}\:{have}\:{a}\:{formula} \\ $$$${for}\:\mathrm{S}_{{n}} : \\ $$$${S}_{{n}} =\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}−{i}^{\mathrm{2}{n}} \right)\:. \\ $$

Commented by prakash jain last updated on 27/May/17

(1/(1−i))=((1+i)/2) is correct. This value is correct with analytic continuation. This series itself is divergent. S=1+2+2^2 +.... S=1+2(1+2+2^2 +..) S=−1. S=−1 is derived from analytical continuation of f(x)=(1/(1−x)) =1+x+x^2 +...

$$\frac{\mathrm{1}}{\mathrm{1}−{i}}=\frac{\mathrm{1}+{i}}{\mathrm{2}}\:\mathrm{is}\:\mathrm{correct}.\: \\ $$$$\mathrm{This}\:\mathrm{value}\:\mathrm{is}\:\mathrm{correct}\:\mathrm{with}\:\mathrm{analytic} \\ $$$$\mathrm{continuation}.\:\mathrm{This}\:\mathrm{series}\:\mathrm{itself} \\ $$$$\mathrm{is}\:\mathrm{divergent}. \\ $$$$\mathrm{S}=\mathrm{1}+\mathrm{2}+\mathrm{2}^{\mathrm{2}} +.... \\ $$$$\mathrm{S}=\mathrm{1}+\mathrm{2}\left(\mathrm{1}+\mathrm{2}+\mathrm{2}^{\mathrm{2}} +..\right) \\ $$$$\mathrm{S}=−\mathrm{1}. \\ $$$$\mathrm{S}=−\mathrm{1}\:\mathrm{is}\:\mathrm{derived}\:\mathrm{from}\:\mathrm{analytical} \\ $$$$\mathrm{continuation}\:\mathrm{of}\:{f}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{1}−{x}} \\ $$$$=\mathrm{1}+{x}+{x}^{\mathrm{2}} +... \\ $$

Commented by ajfour last updated on 27/May/17

incomprehensible indeed !

$${incomprehensible}\:{indeed}\:! \\ $$

Commented by prakash jain last updated on 27/May/17

Analytical continuation is interesting for example using analytical continuation it can be proved that 1+2+3+4+...to infinity=−(1/(12))

$$\mathrm{Analytical}\:\mathrm{continuation}\:\mathrm{is}\:\mathrm{interesting} \\ $$$$\mathrm{for}\:\mathrm{example}\:\mathrm{using}\:\mathrm{analytical} \\ $$$$\mathrm{continuation}\:\mathrm{it}\:\mathrm{can}\:\mathrm{be}\:\mathrm{proved} \\ $$$$\mathrm{that} \\ $$$$\mathrm{1}+\mathrm{2}+\mathrm{3}+\mathrm{4}+...\mathrm{to}\:\mathrm{infinity}=−\frac{\mathrm{1}}{\mathrm{12}} \\ $$

Commented by FilupS last updated on 27/May/17

Yeah, i know a bit about the topic. I don′t really see ′how′ half of the solutions are logical, but i love it

$$\mathrm{Yeah},\:\mathrm{i}\:\mathrm{know}\:\mathrm{a}\:\mathrm{bit}\:\mathrm{about}\:\mathrm{the}\:\mathrm{topic}. \\ $$$$\mathrm{I}\:\mathrm{don}'\mathrm{t}\:\mathrm{really}\:\mathrm{see}\:'{how}'\:\mathrm{half}\:\mathrm{of}\:\mathrm{the}\:\mathrm{solutions} \\ $$$$\mathrm{are}\:\mathrm{logical},\:\mathrm{but}\:\mathrm{i}\:\mathrm{love}\:\mathrm{it} \\ $$

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