I was looking at Wikipedia's page for nth root and there is a section given for representation of nth root with infinite series. I was trying to calculate the $$ \sqrt[3]{12} $$ with$$ (1+x)^{s/t}=\sum_{n=0}^\infty {\frac {\prod_{k=0}^{n-1} (s-kt)} {n!t^n} x^n}$$ and compare the result with results given by nth root algorithm. If I am correct the infinite serie would become something like this $$ (1+11)^{1/3}=\sum_{n=0}^\infty {\frac {\prod_{k=0}^{n-1} (1-3k)} {n!3^n} 11^n}$$ I was expecting a number close to 2.3 but it gave me a really large number for at least 7 steps. a sample of my steps

I don't know why this happened. I think that the way I am calculating might be wrong because I had to calculate each product manually for each step maybe.if this is the case I would like to know what were my mistakes.

footnote : My knowledge of math is quite limited.I also couldn't find any source for this infinite serie.

  • $\begingroup$ What you link to has the condition that $|x|<1$ I notice that this is not true in your case. $\endgroup$ – Kitter Catter May 30 '19 at 12:52
  • $\begingroup$ I guess I misunderstood it $\endgroup$ – superuzerneim May 30 '19 at 12:58

What you link to has the condition that $|x|<1$ I notice that this is not true in your case. You can manipulate what you are trying to solve to make this true which is likely to get you to where you want to be.

What are some cubes close to 12? If you can think of one we can then change your expression into: $$ \sqrt[3]{a^3 + (12-a^3)} = a \sqrt[3]{1+\frac{12-a^3}{a^3}}$$

If $|\frac{12-a^3}{a^3}| < 1$ you can then use the expression you derived.

  • $\begingroup$ Thank you. I misunderstood that condition, I guess for a>1.81 fits in this case $\endgroup$ – superuzerneim May 30 '19 at 13:22
  • $\begingroup$ @superuzerneim try $a^3=8$ since that's one you should know ;) $\endgroup$ – Kitter Catter May 30 '19 at 22:49
  • $\begingroup$ :D I don't why but for the inequality 2,3,4,5,6,.. fits, the smallest a I could calculate was 1.8175. however I kinda need to find some other way to simplify the sigma and product cause they are really time consuming. $\endgroup$ – superuzerneim May 31 '19 at 1:25
  • $\begingroup$ So I kinda did it but a weird thing happened. For $ \sqrt[3]{12} $ I don’t why but the result of the infinite serie gives a number say 0.25 for an “a” but I was expecting 1.25. if I multiply (1+0.25) with the respective “a” it gives me the answer to $ \sqrt[3]{12} $, and it seems that in this way it doesn’t matter what would be the “a” as long as it is any number above that makes the inequality to be true. Here are the results imgur.com/ZnvEwbd $\endgroup$ – superuzerneim May 31 '19 at 7:34
  • $\begingroup$ Random question but are you including the n=0 term? $\endgroup$ – Kitter Catter May 31 '19 at 16:58

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