Maths - Infinite series - Forum

By: Magnus - magnus-skog
Error in "Reorthogonalising a matrix"  
2004-11-15 13:56

On the page "http://www.euclideanspace.com/maths/algebra/matrix/orthogonal/index.htm"; 
 
x^(-1/2) is approximated with the taylor series of the first order: (1 - x)/2, which is not correct. 
 
The value x = 1 should give the same result but it's 1 and 0 (and a value of x = 1.01 even gives a negative approximation). The correct first order approximation should be 1 - (x - 1)/2 = (3 - x)/2 
 
Regards 
Magnus Skog

By: Martin Baker - martinbaker
RE: Error in "Reorthogonalising a matrix"  
2004-11-16 09:10

Magnus, 
 
Thanks very much, this is very useful. 
I am not very familiar with taylors series, Since your message I have tried to derive the expression here: 
http://www.euclideanspace.com/maths/algebra/polynomial/series/ 
this seems to work for f(1 + h) where h = 1 - x 
Is this the correct way to do it? 
 
Martin 

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: Martin Baker - martinbaker
RE: Error in "Reorthogonalising a matrix"  
2004-11-17 08:36

Magnus, 
 
Many thanks again, I have updated both pages as you suggested.  
I have also put examples here to check that it works: 
http://www.euclideanspace.com/maths/algebra/matrix/orthogonal/reorthogonalising/ 
 
I still have some concerns about the derivation of [C] such as deciding that it is symmetric, but at least the expression appears to be correct. 
 
I have also linked the pages to this thread if that is OK with you. 
 
Martin 

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