6-digit arithmetic

(edited 12 years ago)

Let

f(x) = x ( \sqrt{x+1.42} - \sqrt{x} )

Rewrite

f(x)

in a form which reduces the errors arising from machine arithmetic and compute

f(10^5)

using 6-digit arithmetic.

If we let

\displaystyle \tilde{f}(x) = x ( \sqrt{x+1.42} - \sqrt{x}) \frac{\sqrt{x+1.42} + \sqrt{x}}{\sqrt{x+1.42} + \sqrt{x}} = \frac{x}{\sqrt{x+1.42} + \sqrt{x}}

Mistake here; the numerator of

\tilde{f}(x)

1.42x

(\sqrt{x+1.42} - \sqrt{x}) \times (\sqrt{x+1.42} + \sqrt{x}) = x + 1.42 - x = 1.42

(edited 12 years ago)

Reply 2

TheEd

Original post by james.h

Mistake here; the numerator of

\tilde{f}(x)

1.42x

(\sqrt{x+1.42} - \sqrt{x}) \times (\sqrt{x+1.42} + \sqrt{x}) = x + 1.42 - x = 1.42

But this then gives the answer 224.521 which is still out - am I doing the 6-digit arithmetic correctly?

Reply 3

james.h

Original post by TheEd

But this then gives the answer 224.521 which is still out - am I doing the 6-digit arithmetic correctly?

Good point. :erm:

I can't spot anything wrong with your method, but then I've only had fleeting encounters with this kind of maths. :dontknow:

Reply 4

nuodai

Original post by TheEd

But this then gives the answer 224.521 which is still out - am I doing the 6-digit arithmetic correctly?

I get 224.522; are you sure you rounded the final calculation correctly?

Reply 5

TheEd

Original post by nuodai

I get 224.522; are you sure you rounded the final calculation correctly?

\displaystyle \frac{142000}{632.457} = 224.5211927...

Reply 6

nuodai

Original post by TheEd

\displaystyle \frac{142000}{632.457} = 224.5211927...

Oh; multiplied by 1.42 right at the end, to get 158.114×1.42 = 224.52188

I suppose this is the price you pay by only using 6 digits :p:

Reply 7

TheEd

Original post by nuodai

Oh; multiplied by 1.42 right at the end, to get 158.114×1.42 = 224.52188

I suppose this is the price you pay by only using 6 digits :p:

Of course... :tongue:

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6-digit arithmetic

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