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Taylor expansion in infinity?

  • Thread starter Vrbic
  • Start date
  • #1
381
11

Homework Statement


How to use Taylor series for condition x>>1? For example [itex]f(x)=x\sqrt{1+x^2}(2x^2/3-1)+\ln{(x+\sqrt{1+x^2})}[/itex]

Homework Equations




The Attempt at a Solution


I try to derived it and limit to infinity...for example first term [itex]\frac{x^4}{3\sqrt{1+x^2}}[/itex]. Limit this to infinity is obviously infinity. Any advice?
Thank you.
 

Answers and Replies

  • #2
34,361
10,432
Well, the function goes to infinity, so every approximation should do the same.
What exactly do you want to calculate or approximate where?

In general, to get good approximations for large x, you can calculate the taylor series of f(1/x) and evaluate it around 0. That won't work with divergent series, however, where you might need a Laurent series.
 
  • #3
381
11
Well, the function goes to infinity, so every approximation should do the same.
What exactly do you want to calculate or approximate where?

In general, to get good approximations for large x, you can calculate the taylor series of f(1/x) and evaluate it around 0. That won't work with divergent series, however, where you might need a Laurent series.
Nice trick, I tried it but for all derivative limit to 0 is divergent...for example 1.derivative of my f(1/x) is [itex] -\frac{1+1/x^2}{\pi x^4}[/itex]. Yes I would like to examine a behaviour of this function for large x and approximate it for this case.
 
  • #4
34,361
10,432
Well you'll need a Laurent series here, as I mentioned. If it stops, then you have a good approximation.
 

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