87 : Using Taylor's series, find √36.12.
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87 : Using Taylor's series, find √36.12.
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Answer:
defined :
f(x)=x−−√
Therefore :
f′(x)=12x−−√
f′′(x)=−14x3−−√
f′′′(x)=38x5−−√
for x=5 :
|R2(5)|=∣∣∣f(3)(c)3!∣∣∣=∣∣∣∣18c5√3!∣∣∣∣∣∣∣13!∗8c5−−√∣∣∣
I got stuck now, how can I evaluate:
???≥|3!∗8c5−−√|
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