B Higher Roots of Positive Numbers

AI Thread Summary
Successive rooting operations on any positive non-zero number converge to one, as demonstrated by the limit expression $$\lim_{n\to \infty} a^{1/n} = 1$$ for all positive a. This can be proven using properties of the exponential function and logarithms, or through elementary methods showing the distance to one decreases significantly with each operation. The concept of an "infinite root" is not valid, and one cannot conclude that the infinite root of any positive number is zero. Instead, the correct interpretation is that the sequence defined by successive roots approaches one. Thus, the limit of the nth root of a positive number is always one.
dom_quixote
Messages
50
Reaction score
9
Playing around with my calculator, I realized that if I do successive rooting operations on any positive non-zero number, I always get the number one.
Can I conclude that the infinite root of any positive number will always be zero?
If the statement is true, is there any synthesized formula to prove this property?

P.S.:
Sorry if my way of expressing myself sounds strange, because I'm Brazilian and I use the automatic translator to communicate with all of you. It is a pleasure to observe the interest of the contributors to this Site, who strive to understand the questions I post on the Physics Forums.
 
Mathematics news on Phys.org
$$\lim_{n\to \infty} a^{1/n} = 1$$ for all positive a. You can show this e.g. using ##a^{1/n} = e^{\log(a)/n}## and continuity of the exponential function to bring the limit into the exponent, but you can also show it for successive square roots in more elementary ways by showing that e.g. the distance to 1 decreases at least by a factor 2 each time.
 
  • Like
  • Love
Likes mcastillo356, FactChecker, PeroK and 1 other person
dom_quixote said:
Playing around with my calculator, I realized that if I do successive rooting operations on any positive non-zero number, I always get the number one.
Can I conclude that the infinite root of any positive number will always be zero?
No, you cannot conclude anything from trying out a couple of numbers. And there is no such thing like an infinite root. However, you can pose the hypothesis that
$$
\lim_{n \to \infty} \sqrt[n]{x} = 1 \text{ for }x>0
$$
dom_quixote said:
If the statement is true, is there any synthesized formula to prove this property?
You can define a sequence ##a_n:=\sqrt[n]{x}## and prove that ##|a_n-1|## get's arbitrary small for large ##n##.
 
  • Like
Likes dom_quixote
mfb said:
$$\lim_{n\to \infty} a^{1/n} = 1$$ for all positive a. You can show this e.g. using ##a^{1/n} = e^{\log(a)/n}## and continuity of the exponential function to bring the limit into the exponent, but you can also show it for successive square roots in more elementary ways by showing that e.g. the distance to 1 decreases at least by a factor 2 each time.
You can also do an indirect proof. For any ##a > 1##, the sequence ##a^{1/n}## is monotone decreasing and bounded below by ##1##. Therefore, the limit exists.

However, for any number ##l > 1##, the sequence ##l^n## in unbounded; and so ##l \ne lim_{n \to \infty} a^{1/n}##. That leaves the only possibility that ##lim_{n \to \infty} a^{1/n} = 1##.
 
  • Like
Likes JFerreira, mfb and dom_quixote
Suppose ,instead of the usual x,y coordinate system with an I basis vector along the x -axis and a corresponding j basis vector along the y-axis we instead have a different pair of basis vectors ,call them e and f along their respective axes. I have seen that this is an important subject in maths My question is what physical applications does such a model apply to? I am asking here because I have devoted quite a lot of time in the past to understanding convectors and the dual...
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...
Back
Top