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Old 07-08-2003, 07:59 PM   #1 (permalink)
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Cubic Root

Does anyone know how to determine cubic root without a calculator? Anyone who knows, I would appreciate it if you share the knowledge. Thanks.
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Old 07-08-2003, 08:43 PM   #2 (permalink)
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someone posted this before i think

http://mathforum.org/dr.math/faq/faq.sqrt.by.hand.html

search with google there are some more
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Old 07-08-2003, 08:52 PM   #3 (permalink)
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Old 07-08-2003, 08:52 PM   #4 (permalink)
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How about 5 easy steps?

To find a cube root by the "longhand" method (without a calculator), we would proceed similarly to how we would find a square root by hand.

Step 1. Draw a cube root symbol, or radical, with the number whose root you are seeking underneath. Start with the decimal point and mark off digits in both directions in groups of three. Put a decimal point above the radical, and directly above the other decimal point.

Step 2. Start with the first group of 1, 2, or 3 digits. Find the largest cube of a single-digit integer less than it. Write the single digit above the radical, and its cube under the first group. Draw a line under that cube, and subtract it from the first group.

Step 3. Bring down the next group below the last line drawn. This forms the current remainder. Draw a vertical line to the left of the resulting number, and to the left of that line put three hundred times the square of the number above the radical, a plus sign, thirty times the number above the radical, a multiplication sign, an underscore character, another plus sign, another underscore character, the exponent 2, an equals sign, and some blank space for the answer.

Step 4. Pick the biggest digit D that would fit into both underscore places, and give a number such that D times it is less than the current remainder. Put it above the radical above the last group of digits brought down, and put it in each of the blanks where the underscore characters are. Compute the number given by the expression, and put it after the equals sign. Multiply D times that number, and put that below the current remainder, draw a horizontal line below that, and subtract, to give a new current remainder.

Step 5. If the current answer, above the radical, has the desired accuracy, stop. Otherwise, go back to step 3.

<center>OR</center>

An approximation method:

If you had to solve x^3 = 86, you could proceed by Newton's Method as follows:

f(x) = x^3 - 86 = 0
f'(x) = 3x^2

Then we can use an iterative formula to improve on the accuracy of the root. If x0 is our first guess at the root, then a better approximation is given by:

x1 = x0 - f(x0)/f'(x0)
x1 = x0 - (x0^3-86)/(3x0^2)

Now 4^3 = 64 and 5^3 = 125, so we know the root lies between 4 and 5, closer to 4. So let us start with x = 4.3

x1 = 4.3 - (4.3^3 - 86)/[3(4.3)^2] = 4.417
x2 = 4.414
x3 = 4.414005 and x4 = 4.414004962

If you use a calculator to work out cube root(86) you get the same value as x4, so the method gets the right answer quite quickly.


Hope either is helpful....
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Old 07-09-2003, 12:40 AM   #5 (permalink)
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Wow thanks for the replies. JadziaDax, I see what you are doing in your post, but I'm completely ignorant to the f of x. How exactly does that work, if you don't mind explaining?
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Old 07-09-2003, 01:02 AM   #6 (permalink)
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f(x) is notation for a function. for example:
f(x) = x^2 could be an equation.

f(2) would be 2^2 = 4
you can write iterative equations like Jadz did above, with different X's (ie x1 x2 x3) usually in mathbooks the numbers are subscripted, which makes em a bit easier to read, but this'll work, for quick and dirty math example.

so, in her example, x0 is your initial guess.
x1 is the next iteration of the equation that will get closer and closer to the answer. so this:
x1 = x0 - f(x0)/f'(x0) in english, reads
the next iteration (x1) is equal to x0, minus a function using x0 as it's variable, divided by the derivative of that function, also using x0 as it's variable.

since f(x) = x^3 - 86 and
f'(x) (the derivative of f(x)) = x^2, that equation becomes:
x1 = x0 - ((x0^3 - 86)/x0^2)

it's 2 am so I hope that made sense/ was helpful!
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Old 07-09-2003, 01:06 AM   #7 (permalink)
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Quote:
Originally posted by KrazyKracka
Wow thanks for the replies. JadziaDax, I see what you are doing in your post, but I'm completely ignorant to the f of x. How exactly does that work, if you don't mind explaining?
I don't have to... cheerios did it

But if you need it explained further, let me know....

That's what I get paid to do.
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Old 07-09-2003, 10:41 AM   #8 (permalink)
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Quote:
Originally posted by JadziaDax
I don't have to... cheerios did it

But if you need it explained further, let me know....

That's what I get paid to do.
I was about to ask if you were a math teacher. Cheerios cleared it up for me a little bit, however it's been a long time since I've done math like that. I hate to sound stupid, but would you or cheerios be able to lay that out a little more in layman's terms? Thank you.
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Old 07-09-2003, 02:13 PM   #9 (permalink)
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Sure KrazyKracka, I can explain f(x) - aka f of x - a little more in layman's terms...

f(x) stands for function of x. You are probably more aware of equations in the form of y = ax^2 + bx + c (to use a general quadratic equation for an example). The variable "y" in this case is a "function of x". It directly depends on the value of x for its value. So, instead of introducing a new variable (y), we sometimes use the notation f(x). See the following example using a simple linear equation.

y = 2x + 7 <-- how you're used to seeing it
if x = 2, 2(2) + 7 = 11, then y = 11
if x = 5, 2(5) + 7 = 17, then y = 17

f(x) = 2x + 7 <-- stating the equation as a function of x
f(2) = 2(2) + 7 = 11
f(5) = 2(5) + 7 = 17

So, "f of 2" means evaluate the expression when x = 2, just like "f of 5" means evaluate the expression when x = 5. Just replace x with whatever value comes after the "f".

I hope that's better.
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Old 07-10-2003, 11:03 AM   #10 (permalink)
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Thanks, I get it now. I feel kind of stupid lol. Hit me like a ton of bricks.
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Old 07-11-2003, 01:31 PM   #11 (permalink)
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i think i just felt my brain Ooze from my head trying to figure out what you ppl were talking about
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Old 07-13-2003, 09:47 PM   #12 (permalink)
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Taylor Expansion
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Old 07-13-2003, 10:27 PM   #13 (permalink)
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now code them in C, and you'll be doing good.
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