Calculating Limits with Taylor Series: Tips and Guidelines for Accurate Results

[Azael]

Taylor rule of thumb??

When calculating limits by using taylor series is there any easy way to know how many elements that should be included in the taylor series?

if I have

\lim_{x\rightarrow\zero} \frac{exp(x-x^2)-Cos2x-Ln(1+x+2x^2)}{x^3}


How do I know many terms to include in the taylors series for the exponent, cos and logaritm functions?? Should I include all terms that contains a x^3??

 

[Azael]

x->0 but for some reason the latex won't include the 0.

 

[Azael]

When I solve things like this I always notice at the end that I have used to many terms in the taylor series or to few. To many isn't much of a problem obviously except that it clutters up the calculations. But very anoying when I end up with to few.

 

[matt grime]

How many you 'should' include depends on what the question asks for, surely. There is no universal 'correct amount'. If you want to know what the error is for any truncation then you should look up Talyor Remainder and other such things which should be in your book, assuming you have one.

 

[Azael]

yes I have one, not the best to be honest, but it usualy works pretty good. The answere it gave to this particular question is that is trial and error.

This question doesn't involve error estimations in the series. But rather how to get a "feel" for how many terms to include to make limit problems like the one I postes as easily solvable as possible. If you solve that one how many terms would you include??
I first tried by using the first 2 terms of each taylor series but it couldn't be solved that way. So I tried with the first 3 terms of each taylor series and that way I got the correct answere.

When I look at the solution they used the first 3 terms from the taylor series for the exponent, the first 2 for the cos function and the first 3 for the logaritm. How can I "tell" right away that I need no more then 2 terms from the cos to solve it?

 

[HallsofIvy]

Approximating a function by a Taylor's polynomial is just that- an approximation. No approximation, no matter how close it is well tell you exactly what a limit is. In this case, however, since the denominator is x³, writing out the entire Taylor's series, then dividing by x³, you know that all terms in the series past x³ will have an x in them and will go to 0. That's not a "rule", that's basic algebra.

 

[shmoe]

When I look at the solution they used the first 3 terms from the taylor series for the exponent, the first 2 for the cos function and the first 3 for the logaritm. How can I "tell" right away that I need no more then 2 terms from the cos to solve it?

When you're counting the number of terms they used, remember that the x^3 term in the taylor series for cos has a 0 coefficient. The actual number of terms isn't really what's important, it's the size of the error between your taylor polynomial and your function.

In this question, you are dividing by an x^3. Your book should give some explanation about finding how many terms needed so the error/x^3 goes to zero (maybe in terms of big-Oh notation, but possibly not), then the limit will be determined by your taylor polynomials to whatever degree was needed for this size error term.

More precisely, if P_n(x) is the nth degree taylor polynomial of f(x) and R_n(x) is the error, so f(x)=P_n(x)+R_n(x), you want to be able to determine the value of n so that \lim_{x\rightarrow 0 }R_n(x)/x^3=0. Then \lim_{x\rightarrow 0}f(x)/x^3=\lim_{x\rightarrow 0}P_n(x)/x^3. You can replace x^3 with any power x^m and it's the same idea (the required n may change of course).

 

[Azael]

thanks shmoe that clears it up very good :)

Yeah we are taught to use the big-Oh notation(or well we call it big Ordo but I guess its the same thing)in problems like these.