Recent content by dobry_den

In the temperate latitudes of the northern hemisphere, the sun is due south when it reaches the highest point in the sky (see for example http://www.solarplots.info/pages/definitions.aspx" ). What about latitudes between the Tropic of Cancer and the equator? Is the sun due north when it reaches...

i made a mistake in the first post, the limit comparison test is applicable only to non-negative series. then the limit should be like: \lim_{n\rightarrow \infty} \frac{\left|\ln \left(1+\frac{(-1)^n}{n^p}\right)\right|}{\frac{1}{n^p}} which is equal to zero when p is positive. However, the...

Homework Statement \sum_{n=2}^{\infty} \ln \left(1+\frac{(-1)^n}{n^p}\right) p is a real parameter, determine when the series converges absolutely/non-absolutely The Attempt at a Solution I tried to do the limit \lim_{n\rightarrow \infty} \frac{\ln...

the same with me, that's why I'm confused about it.. but still, thanks a lot!

it's an indefinite integral

Homework Statement \int \left|\cos t\right| \ dtThe Attempt at a Solution I divided the interval (-infinity, +infinity) into 2 types of subintervals where cosine is positive and negative respectively. But I'm not sure how to combine these two integrals to get one formula for the whole...

Well, I've finally found a different way how to solve it... I post it here since it can be useful to someone.. I substituted x with y = x-1, and then after some steps I've arrived at the following limit: \lim_{y \rightarrow...

Homework Statement Solve the following limit: \lim_{x \rightarrow 1}(1-x)\tan{\left(\frac{\pi x}{2}\right)} The Attempt at a Solution I solved it using L'Hospital rule, it's equal to 2/pi, but is there any other way how to solve it? thanks a lot! The same question would apply to \lim_{x...

that's exactly what i don't get... shouldn't the result of \mathcal{O}(x^8) - \mathcal{O}(x^{10}) + \mathcal{O}(x^{12}) \right) be O(x^12) since that is the largest term?

\biggl(-\frac{x^2}2 + \frac{x^4}{24} - \frac{x^6}{720} +\mathcal{O}(x^8)\biggr)-\frac12\biggl(-\frac{x^2}2+\frac{x^4}{24}+\mathcal{O}(x^6)\biggr)^2+\frac13\biggl(-\frac{x^2}2+\mathcal{O}(x^4)\biggr)^3 + \mathcal{O}(x^8)\\ & =-\frac{x^2}2 + \frac{x^4}{24}-\frac{x^6}{720} - \frac{x^4}8 +...

Ok, thanks you very much, it helped me a lot... ad "o(1/n^3)".. my book says it should be o(1/n^2) since we're using the small o notation, not the big o, so that's may be the difference.

Could you just check if this is correct? I'm not quite sure how to work with the small oh notation: First I get this: e^{1}e^{(-n)\log{(1+\frac{1}{n}}) Then I express the logarithm as a Taylor series: log(1+\frac{1}{n})= \frac{1}{n}-\frac{1}{2n^2}+o(\frac{1}{n^2}) This is multiplied with...

thanks a lot...! and btw, by expansion parameter do you mean the degree to which I expand the Taylor polynomial?

but by combining the nth powers (i.e. (n^n)/(n+1)^n), wouldn't you get rather (1-1/(n+1))^n than (1+1/n)^n?

Could you please elaborate a little bit on that? Do you mean to rewrite this part: (\left n+1 \right)^{n} as e^{n\log{(n+1)}} and then to do a Taylor expansion of this expression?