By: Magnus - magnus-skog
RE: Error in "Reorthogonalising
a matrix"
2004-11-16 10:25 |
Seems we need a Taylor crash course :)
The general Taylor formula (around the point x = a) can be written as
follows:
f(x) = f(a) + (x - a) f'(a) + (x - a)^/2! f''(a) + (x - a)^3/3! f'''(a)
(i)
And the special case when x = 0 is called Maclaurins formula:
f(x) = f(0) + x f'(0) + x^2/2! f''(0)+ x^3/3! f'''(0)
So there is no difference between the two of them, we have been doing
Taylor
all along, but Maclaurin is just a special case of the more general Taylors
formula.
So in our example with f(x) = x^(-1/2) We get (just as you have gotten):
f(1) = 1
f'(1) = -1/2
f''(1) = 3/4
Substituting the values of the derivatives in (i) above yields:
f(x) = 1 - (x - 1)/2 + 3(x - 1)^2/8 - ...
So if we want to approximate x^(-1/2) with the first order taylor series,
we get:
x^(-1/2) ~~ 1 - (x - 1)/2 = (3 - x)/2 |
By: Magnus - magnus-skog
RE: Error in "Reorthogonalising
a matrix"
2004-11-16 17:54 |
Read through your derived expression again.
"f(x + h) = f(x) + h f'(x) + h2/2! f''(x)+ h3/3! f'''(x) ...
f(1 + h) = f(1) + h f'(1) + h2/2! f''(1)+ h3/3! f'''(1) ...
f(1 + h) = 1 - h /2 + h2 3/8 ...
where h = 1 - x ???"
Take a look at the last row with f(1 + h), it is the same expression as
mine, but you would want to evaluate it around h = 0. It gives the same
result as evaluating mine around x = 1.
Regards
Magnus |
By: Magnus - magnus-skog
RE: Error in "Reorthogonalising
a matrix"
2004-11-17 10:20 |
Of course it's ok :)
Just one minor correction left:
"f(1 + h) = 1 - h /2 + h2 3/8 ...
substituting h = 1 - x gives:"
It should be h = x - 1 =>
f(1 + h ) = f(1 + (x - 1)) = f(x) = 1 - (x - 1)/2 + ..
Otherwise it would be f(2 - x) = ..
Love your site and I will continue helping you if I can ! :)
Magnus |
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