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Question-5843




Question Number 5843 by Rojaye Shegz last updated on 31/May/16
Commented by Rojaye Shegz last updated on 31/May/16
What is wrong here?
$${What}\:{is}\:{wrong}\:{here}? \\ $$
Commented by Yozzii last updated on 31/May/16
∞≠∞
$$\infty\neq\infty\: \\ $$
Commented by prakash jain last updated on 01/Jun/16
1+2+4+8+... is divergent series.  Compare it with   1+x+x^2 +....to ∞  S=f(x)=(1/(1−x)) for ∣x∣<1  Analytical continuation of f(x) to x∈R  f(2)=(1/(1−2))=−1  So the result  1+2+2^2 +...+to ∞=−1 is valid using analytical  continuity.  Another way  S=1+2+2^2 +...=1+2(1+2+2^2 +..)=1+2S  S=1+2S⇒S=−1
$$\mathrm{1}+\mathrm{2}+\mathrm{4}+\mathrm{8}+…\:\mathrm{is}\:\mathrm{divergent}\:\mathrm{series}. \\ $$$$\mathrm{Compare}\:\mathrm{it}\:\mathrm{with}\: \\ $$$$\mathrm{1}+{x}+{x}^{\mathrm{2}} +….\mathrm{to}\:\infty \\ $$$$\mathrm{S}={f}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{1}−{x}}\:\mathrm{for}\:\mid{x}\mid<\mathrm{1} \\ $$$$\mathrm{Analytical}\:\mathrm{continuation}\:\mathrm{of}\:{f}\left({x}\right)\:\mathrm{to}\:{x}\in\mathbb{R} \\ $$$${f}\left(\mathrm{2}\right)=\frac{\mathrm{1}}{\mathrm{1}−\mathrm{2}}=−\mathrm{1} \\ $$$$\mathrm{So}\:\mathrm{the}\:\mathrm{result} \\ $$$$\mathrm{1}+\mathrm{2}+\mathrm{2}^{\mathrm{2}} +…+\mathrm{to}\:\infty=−\mathrm{1}\:\boldsymbol{\mathrm{is}}\:\boldsymbol{\mathrm{valid}}\:\mathrm{using}\:\mathrm{analytical} \\ $$$$\mathrm{continuity}. \\ $$$$\mathrm{Another}\:\mathrm{way} \\ $$$$\mathrm{S}=\mathrm{1}+\mathrm{2}+\mathrm{2}^{\mathrm{2}} +…=\mathrm{1}+\mathrm{2}\left(\mathrm{1}+\mathrm{2}+\mathrm{2}^{\mathrm{2}} +..\right)=\mathrm{1}+\mathrm{2S} \\ $$$$\mathrm{S}=\mathrm{1}+\mathrm{2S}\Rightarrow\mathrm{S}=−\mathrm{1} \\ $$
Commented by prakash jain last updated on 01/Jun/16
The result S=−1 is correct.
$$\mathrm{The}\:\mathrm{result}\:\mathrm{S}=−\mathrm{1}\:\mathrm{is}\:\mathrm{correct}. \\ $$
Commented by FilupSmith last updated on 01/Jun/16
S=−1  is one answer  just to make it clear,  lim S = ∞  but through analytical continuation  S=−1 is correct
$${S}=−\mathrm{1}\:\:\mathrm{is}\:\boldsymbol{{one}}\:\boldsymbol{{answer}} \\ $$$$\mathrm{just}\:\mathrm{to}\:\mathrm{make}\:\mathrm{it}\:\mathrm{clear}, \\ $$$$\mathrm{lim}\:{S}\:=\:\infty \\ $$$$\mathrm{but}\:\mathrm{through}\:\mathrm{analytical}\:\mathrm{continuation} \\ $$$${S}=−\mathrm{1}\:\mathrm{is}\:\mathrm{correct} \\ $$
Commented by Rojaye Shegz last updated on 01/Jun/16
but why is it analytically correct?  how can the fact tht 1+ Σ_0 ^∞ 2k=−1 be put to use?  how cn i conceptualize it?
$$\mathrm{but}\:\mathrm{why}\:\mathrm{is}\:\mathrm{it}\:{analytically}\:\mathrm{correct}? \\ $$$$\mathrm{how}\:\mathrm{can}\:\mathrm{the}\:\mathrm{fact}\:\mathrm{tht}\:\mathrm{1}+\:\underset{\mathrm{0}} {\overset{\infty} {\sum}}\mathrm{2}{k}=−\mathrm{1}\:\mathrm{be}\:\mathrm{put}\:\mathrm{to}\:\mathrm{use}? \\ $$$$\mathrm{how}\:\mathrm{cn}\:\mathrm{i}\:\mathrm{conceptualize}\:\mathrm{it}? \\ $$$$ \\ $$$$ \\ $$
Commented by FilupSmith last updated on 02/Jun/16
There are aplications in more complex  physics for such things.  A famous example is: 1+2+3+...=−(1/(12))  This has applications in astrophysics.  I don′t have any better examples or specifics.    Sometimes people refer to the analytical  solution as an algebraic solution.    e.g.  S=2+4+8+...  2S=4+8+16+...  ∴2S+2=S  S=−2  ∴ 2+4+8+16+...=−2 is an algebraic solution  through analytical continuation.    or, another method  S=2+4+8+...  S=2(1+2+4+...)  S=2(1+S)  S=2+2S  S=−2
$$\mathrm{There}\:\mathrm{are}\:\mathrm{aplications}\:\mathrm{in}\:\mathrm{more}\:\mathrm{complex} \\ $$$$\mathrm{physics}\:\mathrm{for}\:\mathrm{such}\:\mathrm{things}. \\ $$$$\mathrm{A}\:\mathrm{famous}\:\mathrm{example}\:\mathrm{is}:\:\mathrm{1}+\mathrm{2}+\mathrm{3}+…=−\frac{\mathrm{1}}{\mathrm{12}} \\ $$$$\mathrm{This}\:\mathrm{has}\:\mathrm{applications}\:\mathrm{in}\:\mathrm{astrophysics}. \\ $$$$\mathrm{I}\:\mathrm{don}'\mathrm{t}\:\mathrm{have}\:\mathrm{any}\:\mathrm{better}\:\mathrm{examples}\:\mathrm{or}\:\mathrm{specifics}. \\ $$$$ \\ $$$$\mathrm{Sometimes}\:\mathrm{people}\:\mathrm{refer}\:\mathrm{to}\:\mathrm{the}\:\mathrm{analytical} \\ $$$$\mathrm{solution}\:\mathrm{as}\:\mathrm{an}\:\mathrm{algebraic}\:\mathrm{solution}. \\ $$$$ \\ $$$$\mathrm{e}.\mathrm{g}. \\ $$$${S}=\mathrm{2}+\mathrm{4}+\mathrm{8}+… \\ $$$$\mathrm{2}{S}=\mathrm{4}+\mathrm{8}+\mathrm{16}+… \\ $$$$\therefore\mathrm{2}{S}+\mathrm{2}={S} \\ $$$${S}=−\mathrm{2} \\ $$$$\therefore\:\mathrm{2}+\mathrm{4}+\mathrm{8}+\mathrm{16}+…=−\mathrm{2}\:\mathrm{is}\:\mathrm{an}\:\mathrm{algebraic}\:\mathrm{solution} \\ $$$$\mathrm{through}\:\mathrm{analytical}\:\mathrm{continuation}. \\ $$$$ \\ $$$${or},\:\mathrm{another}\:\mathrm{method} \\ $$$${S}=\mathrm{2}+\mathrm{4}+\mathrm{8}+… \\ $$$${S}=\mathrm{2}\left(\mathrm{1}+\mathrm{2}+\mathrm{4}+…\right) \\ $$$${S}=\mathrm{2}\left(\mathrm{1}+{S}\right) \\ $$$${S}=\mathrm{2}+\mathrm{2}{S} \\ $$$${S}=−\mathrm{2} \\ $$

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