Series / remainder term?

[HappyCalculusStudent]

Hi. I have two questions. I think the fist one is pretty straight forward, but for some reason I'm having trouble with it. I'd appreciate some help.

Use series to evaluate the limit:
lim x?0 (1-cos x)/(1+x-(e^x))

I know the answer is -1 (if I use L' Hopital's Rule), but I'm not sure how the terms cancel when using series.
I'm left with - (-x^2/2! + x^4/4! - x^6/6! +...)/ - (x^2/2! + x^3/3! + x^4/4! +...)
I was able to do other similar problems, for this one, I'm confused about what remains and what cancels. Could someone please show me how it's done?

Also, when dealing with Taylor's Inequality, I don't really understand problems with the "Remainder term".
For example,
"Approximate lnx by a 5th degree polynomial in x; for x on [1,1.2]"
My book says "the remainder term is given by R5(x)=f(6)(z)/6! *(x-1)^6 = -((x-1)^6)/(6z^6)
(where the first 6 is a superscript)
And one row down it has abs[R5(x)] < (0.2)^6/6 < 0.000011

???I'm not sure what they are doing here. I've been working from several books and it still isn't clear to me...
If possible, I'd like to check out some more examples. If there's a good website, please let me know. Thanks!

 

[Deleted member 4993]

Hi. I have two questions. I think the fist one is pretty straight forward, but for some reason I'm having trouble with it. I'd appreciate some help.

Use series to evaluate the limit:
lim x?0 (1-cos x)/(1+x-(e^x))

I know the answer is -1 (if I use L' Hopital's Rule), but I'm not sure how the terms cancel when using series.
I'm left with - (-x^2/2! + x^4/4! - x^6/6! +...)/ - (x^2/2! + x^3/3! + x^4/4! +...)

What happens when you factor out (x^2/2!) - from numerator and denominator - and take the limit?

I was able to do other similar problems, for this one, I'm confused about what remains and what cancels. Could someone please show me how it's done?

Also, when dealing with Taylor's Inequality, I don't really understand problems with the "Remainder term".
For example,
"Approximate lnx by a 5th degree polynomial in x; for x on [1,1.2]"
My book says "the remainder term is given by R5(x)=f(6)(z)/6! *(x-1)^6 = -((x-1)^6)/(6z^6)
(where the first 6 is a superscript)
And one row down it has abs[R5(x)] < (0.2)^6/6 < 0.000011

???I'm not sure what they are doing here. I've been working from several books and it still isn't clear to me...
If possible, I'd like to check out some more examples. If there's a good website, please let me know. Thanks!

 

[BigGlenntheHeavy]

Second problem:

\(\displaystyle We \ see \ that \ the \ (n+1)st \ derivative \ for \ f(x) \ = \ ln|x|, \ centered \ at \ one \ is \ f^{n+1}(x) \ = \ \frac{(-1)^{n}n!}{x^{n+1}}.\)

\(\displaystyle Taylor's \ theorem \ states \ that \ R_n(x) \ = \ \frac{f^{n+1}(z)}{(n+1)!}(x-c)^{n+1}.\)

\(\displaystyle Hence, \ R_5(1.2) \ = \ \frac{f^{6}(z)}{6!}(1.2-1)^{6} \ = \ \frac{(-1)^{5}5!(.2)^{6}}{z^{6}6!} \ = \ \frac{(-1)^{5}(.2)^{6}}{6z^{6}}\)

\(\displaystyle Now \ 1 \ < \ z \ < \ 1.2 \ and \ in \ this \ interval, \ |R_5(1.2)| \ is \ largest \ when \ z \ = \ 1.\)

\(\displaystyle Therefore |R_5(1.2)| = \frac{.2^{6}}{6} \ < \ .000011\)

\(\displaystyle Now \ P_5(1.2) \ = \ .182330666667.\)

\(\displaystyle Hence, \ .182330666667 \ - \ R_5(1.2) \ < \ ln(1.2) \ < \ .182330666667\)

\(\displaystyle .182330666667 \ - \ .000011 \ < \ ln(1.2) \ < \ .182330666667\)

\(\displaystyle Ergo, \ .182319666667 \ < \ ln(1.2) \ < \ .182330666667\)

\(\displaystyle Actual \ value \ of \ ln(1.2) \ is \ .182321556794.\)

Note: For an even closer approximation, take the average, viz.,

\(\displaystyle \frac{.182319666667+.182330666667}{2} \ = \ .182325166667\)

\(\displaystyle Now, \ .182325166667-.182321556794 \ = \ .000003609873, \ good \ enough \ for \ government \ work.\)

 

[HappyCalculusStudent]

Much better. Thank you!
I think I just need more practice now.
I may post another thing or two this week...I still seem to have difficulty with some tests for series.
Thanks very much!