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How to evaluate 55^(1/4)?
- Thread starter rbcc
- Start date
How do you evaluate 55^(1/4) without a calculator? right now I just estimate, is there an actual way to do this?
55^(1/4) = (55^(1/2))^(1/2)
Find the "inside" square root first, then repeat the process for the "outer" square root.
There are multiple methods that have been developed for approximating square roots. Here are a couple:
http://www.math.com/school/subject1/les ... 1L9DP.html
http://math.arizona.edu/~kerl/doc/square-root.html
How about
55[sup:3fg1ffkd]1/4[/sup:3fg1ffkd]=x so x[sup:3fg1ffkd]4[/sup:3fg1ffkd]=55 so x[sup:3fg1ffkd]4[/sup:3fg1ffkd]-55 = 0.
Where this graph crosses the x axis give the positive and negative solutions.
You can just find the positive one and you'll know the other solution also.
2[sup:3fg1ffkd]4[/sup:3fg1ffkd]=16 and 3[sup:3fg1ffkd]4[/sup:3fg1ffkd]=81
therefore the answer lies between 2 and 3.
If you locate f(3) and draw a tangent to the curve at that point,
the tangent crosses the x axis much nearer the solution you are looking for.
Now, find f(this new x) and draw a new tangent and you zig-zag in on the solution.
It is a mathematical "locking device".
If you keep this up, you keep drawing tangents that home in on the answer.
Won't take long to get an extremely accurate answer.
It's called the Newton-Raphson method for discovering polynomial solutions.
If you draw the first triangle, you get f(x[sub:3fg1ffkd]1[/sub:3fg1ffkd])/{x[sub:3fg1ffkd]1[/sub:3fg1ffkd]-x[sub:3fg1ffkd]2[/sub:3fg1ffkd]} = f[sup:3fg1ffkd]'[/sup:3fg1ffkd](x[sub:3fg1ffkd]1[/sub:3fg1ffkd])
Rearranging that you get in general "next x" = x[sub:3fg1ffkd]n[/sub:3fg1ffkd] = x[sub:3fg1ffkd]n-1[/sub:3fg1ffkd] - f(x[sub:3fg1ffkd]n-1[/sub:3fg1ffkd])/f[sup:3fg1ffkd]'[/sup:3fg1ffkd](x[sub:3fg1ffkd]n-1[/sub:3fg1ffkd])
If you try one iteration of that procedure, you get x = 2.76 about.
(2.76)[sup:3fg1ffkd]2[/sup:3fg1ffkd] is 57.97
The second iteration will bring you much closer to the true answer
55[sup:3fg1ffkd]1/4[/sup:3fg1ffkd]=x so x[sup:3fg1ffkd]4[/sup:3fg1ffkd]=55 so x[sup:3fg1ffkd]4[/sup:3fg1ffkd]-55 = 0.
Where this graph crosses the x axis give the positive and negative solutions.
You can just find the positive one and you'll know the other solution also.
2[sup:3fg1ffkd]4[/sup:3fg1ffkd]=16 and 3[sup:3fg1ffkd]4[/sup:3fg1ffkd]=81
therefore the answer lies between 2 and 3.
If you locate f(3) and draw a tangent to the curve at that point,
the tangent crosses the x axis much nearer the solution you are looking for.
Now, find f(this new x) and draw a new tangent and you zig-zag in on the solution.
It is a mathematical "locking device".
If you keep this up, you keep drawing tangents that home in on the answer.
Won't take long to get an extremely accurate answer.
It's called the Newton-Raphson method for discovering polynomial solutions.
If you draw the first triangle, you get f(x[sub:3fg1ffkd]1[/sub:3fg1ffkd])/{x[sub:3fg1ffkd]1[/sub:3fg1ffkd]-x[sub:3fg1ffkd]2[/sub:3fg1ffkd]} = f[sup:3fg1ffkd]'[/sup:3fg1ffkd](x[sub:3fg1ffkd]1[/sub:3fg1ffkd])
Rearranging that you get in general "next x" = x[sub:3fg1ffkd]n[/sub:3fg1ffkd] = x[sub:3fg1ffkd]n-1[/sub:3fg1ffkd] - f(x[sub:3fg1ffkd]n-1[/sub:3fg1ffkd])/f[sup:3fg1ffkd]'[/sup:3fg1ffkd](x[sub:3fg1ffkd]n-1[/sub:3fg1ffkd])
If you try one iteration of that procedure, you get x = 2.76 about.
(2.76)[sup:3fg1ffkd]2[/sup:3fg1ffkd] is 57.97
The second iteration will bring you much closer to the true answer