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Question Number 14113 by FilupS last updated on 28/May/17
Is this incorrect:  S=2×2×...  S=2(2×2×...)  S=2S  S=0     Please explain why
$$\mathrm{Is}\:\mathrm{this}\:\mathrm{incorrect}: \\ $$$${S}=\mathrm{2}×\mathrm{2}×… \\ $$$${S}=\mathrm{2}\left(\mathrm{2}×\mathrm{2}×…\right) \\ $$$${S}=\mathrm{2}{S} \\ $$$${S}=\mathrm{0} \\ $$$$\: \\ $$$$\mathrm{Please}\:\mathrm{explain}\:\mathrm{why} \\ $$
Commented by FilupS last updated on 28/May/17
the way i see it  S=2×2×2×...  S=2(1×2×2×...)  S≠1×2×2×...
$$\mathrm{the}\:\mathrm{way}\:\mathrm{i}\:\mathrm{see}\:\mathrm{it} \\ $$$${S}=\mathrm{2}×\mathrm{2}×\mathrm{2}×… \\ $$$${S}=\mathrm{2}\left(\mathrm{1}×\mathrm{2}×\mathrm{2}×…\right) \\ $$$${S}\neq\mathrm{1}×\mathrm{2}×\mathrm{2}×… \\ $$
Commented by RasheedSindhi last updated on 28/May/17
If  S=2×2×2×...  then 1×S=1×2×2×...           S=1×2×2×2×...
$$\mathrm{If}\:\:{S}=\mathrm{2}×\mathrm{2}×\mathrm{2}×… \\ $$$$\mathrm{then}\:\mathrm{1}×\mathrm{S}=\mathrm{1}×\mathrm{2}×\mathrm{2}×… \\ $$$$\:\:\:\:\:\:\:\:\:\mathrm{S}=\mathrm{1}×\mathrm{2}×\mathrm{2}×\mathrm{2}×… \\ $$$$ \\ $$
Commented by RasheedSindhi last updated on 28/May/17
S=2S⇒2S−S=∞−∞  ∞−∞  is indeterminte.  Hence S=0 is not necessarily  true.
$$\mathrm{S}=\mathrm{2S}\Rightarrow\mathrm{2S}−\mathrm{S}=\infty−\infty \\ $$$$\infty−\infty\:\:\mathrm{is}\:\mathrm{indeterminte}. \\ $$$$\mathrm{Hence}\:\mathrm{S}=\mathrm{0}\:\mathrm{is}\:\mathrm{not}\:\mathrm{necessarily} \\ $$$$\mathrm{true}. \\ $$
Commented by prakash jain last updated on 28/May/17
S=f(2)  f(x) be an analytical function  defined for some A⊂R.  S=0⇒f(x)=0∀x∈A  I think f(x)=0 is not analytic  so S=0 cannot be derived using  continuation of f(x)=0 for x=2
$${S}={f}\left(\mathrm{2}\right) \\ $$$${f}\left({x}\right)\:\mathrm{be}\:\mathrm{an}\:\mathrm{analytical}\:\mathrm{function} \\ $$$$\mathrm{defined}\:\mathrm{for}\:\mathrm{some}\:\mathrm{A}\subset\mathbb{R}. \\ $$$${S}=\mathrm{0}\Rightarrow{f}\left({x}\right)=\mathrm{0}\forall{x}\in\mathrm{A} \\ $$$$\mathrm{I}\:\mathrm{think}\:{f}\left({x}\right)=\mathrm{0}\:\mathrm{is}\:\mathrm{not}\:\mathrm{analytic} \\ $$$$\mathrm{so}\:{S}=\mathrm{0}\:\mathrm{cannot}\:\mathrm{be}\:\mathrm{derived}\:\mathrm{using} \\ $$$$\mathrm{continuation}\:\mathrm{of}\:{f}\left({x}\right)=\mathrm{0}\:\mathrm{for}\:{x}=\mathrm{2} \\ $$

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