Is it possible to calculate exact value of cube root with only square root?

[geerky42]

For any real number a, is it possible to calculate exactly value of a^1/3 using basic calculator with square root?

(calculator with digit 0-9, any basic operations plus square root, and clear button.)

Explain why or why not?

I would say no, but I don't know how to show that.

 

[Ishuda]

For any real number a, is it possible to calculate exactly value of a^1/3 using basic calculator with square root?

(calculator with digit 0-9, any basic operations plus square root, and clear button.)

Explain why or why not?

I would say no, but I don't know how to show that.

When you say exact, I assume you mean to the limit of the calculators number of digits. Yes, you can use an iterterative procedure such as
\(\displaystyle x_{i+1} = \frac{2}{3} ( x_i + \frac{a}{2 x_i^2})\)
which will converge to the cube root of a fairly quickly. If fact,
\(\displaystyle x_{i+1} = \frac{n-1}{n} ( x_i + \frac{a}{(n-1) x_i^{n-1}})\)
will converge to the nth root of a fairly fast.

 

[lookagain]

When you say exact, I assume you mean to the limit of the calculators number of digits.
Yes, you can use an iterterative procedure such as
\(\displaystyle x_{i+1} = \frac{2}{3} ( x_i + \frac{a}{2 x_i^2})\)
which will converge to the cube root of a fairly quickly. If fact,
\(\displaystyle x_{i+1} = \frac{n-1}{n} ( x_i + \frac{a}{(n-1) x_i^{n-1}})\)
will converge to the nth root of a fairly fast.

For any real number a, is it possible to calculate exactly value of a^1/3 using basic calculator with square root?

geerky42 requires the use of the square root function for this.

 

[Ishuda]

geerky42 requires the use of the square root function for this.

O.K., take as an initial estimate sqrt(a).

 

[HallsofIvy]

geerky42 requires the use of the square root function for this.

He did not say that- he said, in the part you quoted, that he had to do it on a basic calculator that has only a square root function. I would interpret that as meaning, not that he had to use the square root but that he could use basic arithmetic operations, but that there was no "cube root" or "\(\displaystyle x^y\)" key.

 

[HallsofIvy]

SIMPLE: get another calculator:shock:

Can't argue with that!

 

[geerky42]

He did not say that- he said, in the part you quoted, that he had to do it on a basic calculator that has only a square root function. I would interpret that as meaning, not that he had to use the square root but that he could use basic arithmetic operations, but that there was no "cube root" or "\(\displaystyle x^y\)" key.

he's correct

Can't argue with that!

would be best solution if I always have access to another calculator haha...

 

[geerky42]

When you say exact, I assume you mean to the limit of the calculators number of digits. Yes, you can use an iterterative procedure such as
\(\displaystyle x_{i+1} = \frac{2}{3} ( x_i + \frac{a}{2 x_i^2})\)
which will converge to the cube root of a fairly quickly. If fact,
\(\displaystyle x_{i+1} = \frac{n-1}{n} ( x_i + \frac{a}{(n-1) x_i^{n-1}})\)
will converge to the nth root of a fairly fast.

ah right, I can just use Newton's method. Awesome! thanks.

 

[Deleted member 4993]

ah right, I can just use Newton's method. Awesome! thanks.

But you wanted exact solution...

Newton's method gives you approximate solution.

 

[Ishuda]

But you wanted exact solution...

Newton's method gives you approximate solution.

It's actually a repeated use of Newton's method. The trick is to get the right f(x). In this case you want
f(x) = 0 = xⁿ - a
Do a first order expansion to get the iterative scheme. You approximately double the accuracy each iterate.

 

[geerky42]

But you wanted exact solution...

Newton's method gives you approximate solution.

Well, see:

"When you say exact, I assume you mean to the limit of the calculators number of digits. Yes"

"it will converge fairly fast."

Plus calculator never really give us exactly solution lol. One example is pi key on Ti-84. Output for that is 3.141592654, which is not really pi, but very good approximation.

Sorry for poor worded problem thought.