Calculating the Riemann Sum of (cos1)^x

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The discussion centers around calculating the Riemann sum of the function (cos(1))^x, which participants clarify is actually a geometric series rather than a Riemann sum. The series converges due to the absolute value of cos(1) being less than 1, leading to the formula for the sum as cos(1)/(1-cos(1)). Participants debate the accuracy of numerical approximations, with one suggesting an approximate value of 1.175, while others emphasize the importance of providing exact expressions. Additionally, there is confusion regarding the interpretation of cos(1) in degrees versus radians, which affects the numerical results. The conversation highlights the need for clarity in mathematical terminology and precision in calculations.
  • #31
nick727kcin said:
so arctan2n doesn't have a sum right? because its divergent?

{arctan(2n)} is a convergent sequence, \lim_{n\rightarrow\infty}\arctan{2n}=\pi/2, not 1.57 as you said. 1.57 is an approximation to pi/2. If you are to learn one thing from this thread, make it to stop rounding and calling things equal. Or be prepared to have arildno point out this error everytime.

You mean to ask about the convergence of the series \sum_{n=1}^{\infty}\arctan{2n}, or possibly some other starting point (which won't affect convergence/divergence). This series is divergent, see d_leet's post.

It wasn't always clear whether you were asking about the convergence of the sequence or the corresponding series. You might want to make it more clear in the future by using latex to write the mathematical notation. If you click on the pretty graphics above, you can see how they were made. You don't have to know much about latex for stuff like this, and it let's you avoid using paint to post images.
 
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  • #32
finchi_88 said:
does it really matter

It may matter to professors who are looking for the best possible answer. You should always try and settle for the best answer if there is one :).
 
  • #33
finchie_88 said:
Yes it is, the method I used:
1. Put calculator in radian mode.
2. calculate cos1.
3. calculate 1-ans.
4. calculate cos1/previous ans.

That's not a good method to use. What would you have done if you had been given the following sum?

\sum_{n=0}^{\infty}(\cos(k))^n

(where k is in radians and is not equal to an integer multiple of \pi)
 
  • #34
One might, of course, object that it is too "trivial" to point out the error in calling a rounded number exact.
However, it is quite evident that for the OP at least, this error seems symptomatic of his confusion of terms, for example using the inappropriate term "Riemann sums" in the title, and mixing up the concepts of sequences and series.

My advice is that you start paying attention to definitions in maths, and try to understand them, rather than trying to memorize them.
 

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