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OpenStudy (anonymous):
\[\sum_{n=0}^{\infty}(x-2)^n/10^n\]
OpenStudy (anonymous):
heyy help
OpenStudy (anonymous):
i am confused about this radius
OpenStudy (anonymous):
when the radius is infinite
OpenStudy (anonymous):
radius of convergence is the value of x upto which the above series is convergent {i.e. Following values of the series are smaller than the initial values }
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OpenStudy (anonymous):
8
OpenStudy (anonymous):
book says 10
OpenStudy (anonymous):
any series is convergent if the term yields <1 value ..
hence (x-2)/10<1 ; X<12 hence 12 is radius of convergence
OpenStudy (anonymous):
you want:
\[\left| \frac{x-2}{10} \right|< 1\]
OpenStudy (anonymous):
yes joe
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OpenStudy (anonymous):
soory calc mistake
OpenStudy (anonymous):
do u find 10 phaniraj?
OpenStudy (anonymous):
so you get:
\[\left| \frac{x-2}{10} \right|<1 \Rightarrow -1< \frac{x-2}{10} < 1 \Rightarrow -10<x-2<10\]\[-8<x<12\]
OpenStudy (anonymous):
so x can be inbetween -8 and 12, giving an interval radius of 10. (the whole interval is 20 units long)