I couldn't really find even a bit of information regarding the type of problem I have. Problem states that I should use Taylor series (expansion) to calculate the given value with error maximum of 0,01.


My mentor gave me a little hint that I should somehow use Binomial Expansion and and take 243 as the nearest fifth root or something like that.

Looking forward to any help, thanks in advance

  • $\begingroup$ Did you expand $(243+7)^{1/5}$ using the general binomial expansion of $(x+y)^r$? It will give you an alternating series which will allow you to determine how many terms you must sum to obtain the accuracy required. $\endgroup$ – John Wayland Bales Mar 16 '17 at 19:28
  • $\begingroup$ Thank you a lot! That helped $\endgroup$ – Aidar Nugmanov Mar 16 '17 at 19:55

\begin{aligned}(x+y)^{r}&=\sum _{k=0}^{\infty }{r \choose k}x^{r-k}y^{k}\\&=x^{r}+rx^{r-1}y+{\frac {r(r-1)}{2!}}x^{r-2}y^{2}+{\frac {r(r-1)(r-2)}{3!}}x^{r-3}y^{3}+\cdots .\end{aligned}

\begin{eqnarray} {\displaystyle {\begin{aligned}(243+7)^{1/5}&=\sum _{k=0}^{\infty }{1/5 \choose k}243^{1/5-k}7^{k}\\&=3+\frac{1}{5}\cdot3^{-4}(7)-{\frac {4}{50}}3^{-9}\cdot7^{2}+{\frac {36}{750}}3^{-14}7^{3}+\cdots .\end{aligned}}} \end{eqnarray}

The value of the third term is $0.0002$ which means that the sum of the first two terms will give an accuracy $\pm0.0002$ which will satisfy the accuracy requirements.

So the answer is the sum of the first two terms, $3.02$ although you get a better answer if you use the three decimal place answer of $3.017$.

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  • $\begingroup$ what about an expansion when (x-y)^r? $\endgroup$ – Aidar Nugmanov Mar 16 '17 at 19:55
  • $\begingroup$ In that case you would have to expand $(1024-774)^{1/5}$ which would take many more terms to obtain the required precision. Also that series would not alternate. $\endgroup$ – John Wayland Bales Mar 16 '17 at 19:58

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