By: Magnus  magnusskog
RE: Error in "Reorthogonalising
a matrix"
20041116 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  magnusskog
RE: Error in "Reorthogonalising
a matrix"
20041116 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  magnusskog
RE: Error in "Reorthogonalising
a matrix"
20041117 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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