taylor approximation near infinity

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December 24th 2009, 01:29 AM
Bryn

taylor approximation near infinity

Consider F(x)=[x^4 - x^2 + 1]/[x^4 + 2x^3 + x] for x not = 0

F(x) tends to 1 as x tends to infinity. When x is large, f(x) is approx but not exactly 1

Use change of variable y=1/x then do a maclaurin expansion in y to show that f(x) is approx 1-2x for large x and find the third term

Many thanks and happy christmas to anyone celebrating
December 24th 2009, 01:59 AM
HallsofIvy

Quote:

Originally Posted by Bryn

Consider F(x)=[x^4 - x^2 + 1]/[x^4 + 2x^3 + x] for x not = 0

F(x) tends to 1 as x tends to infinity. When x is large, f(x) is approx but not exactly 1

Use change of variable y=1/x then do a maclaurin expansion in y to show that f(x) is approx 1-2x for large x and find the third term

Excellent idea. Have you done that? To start with, what is f(1/x)?

Quote:

Many thanks and happy christmas to anyone celebrating
December 24th 2009, 12:57 PM
Bryn

No, I am finding every stage challenging, maybe I think I have been doing a little too much maths and have lost the ability to process, but any suggestions would be appreciated

thanks again
December 25th 2009, 03:47 AM
HallsofIvy

Quote:

Originally Posted by Bryn

Consider F(x)=[x^4 - x^2 + 1]/[x^4 + 2x^3 + x] for x not = 0

F(x) tends to 1 as x tends to infinity. When x is large, f(x) is approx but not exactly 1

Use change of variable y=1/x then do a maclaurin expansion in y to show that f(x) is approx 1-2x for large x and find the third term

Many thanks and happy christmas to anyone celebrating

Step one: find F(1/x) by replacing each "x" by "1/x".
$F(1/x)= \frac{\frac{1}{x^4}- \frac{1}{x^2}+ 1}{\frac{1}{x^4}+ \frac{2}{x^3}+ \frac{1}{x}}$

Get rid of the "little fractions" by multiplying numerator and denominator by $x^4$
$F(1/x)= \frac{1- x^2+ x^4}{1+ 2x+ x^3}$

Now find the MacLaurin series for that.