# Maths - Infinite series - Forum

 By: Magnus - magnus-skog Error in "Reorthogonalising a matrix"   2004-11-15 13:56 On the page "https://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:  https://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:  https://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