Riemann sums

  • #26
finchie_88
If you want the sum to infinity, is it not simply [tex] \frac{\cos1}{1 - \cos1} [/tex], which equals 1.1753426496700214107767867596561. Also, 1-co1 is the same as [itex] 2(sin0.5)^2 [/itex], if that helps.
 
  • #27
arildno
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finchie_88 said:
If you want the sum to infinity, is it not simply [tex] \frac{\cos1}{1 - \cos1} [/tex], which equals 1.1753426496700214107767867596561
No, it doesn't, at least not with radians or degrees as your angular measure.
 
  • #28
finchie_88
arildno said:
No, it doesn't, at least not with radians or degrees as your angular measure.

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.

Overall answer = 1.175 to 3 d.p.
 
  • #29
VietDao29
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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.

Overall answer = 1.175 to 3 d.p.
finchie_88, exact solutions are always the best. So giving the answer as:
[tex]
\frac{\cos 1}{1 - \cos 1}[/tex] is more preferable than giving it in the form: 1.175 to 3 d.p..
No? :)
This is arildno's point, I believe.
 
  • #30
arildno
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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..
No it isn't.
Learn the difference between exact and approximate answers.
What do think your calculator answer is in this case?
 
  • #31
shmoe
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nick727kcin said:
so arctan2n doesnt have a sum right? because its divergent?

{arctan(2n)} is a convergent sequence, [tex]\lim_{n\rightarrow\infty}\arctan{2n}=\pi/2[/tex], 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 [itex]\sum_{n=1}^{\infty}\arctan{2n}[/itex], 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 lets you avoid using paint to post images.
 
  • #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
Tom Mattson
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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?

[tex]\sum_{n=0}^{\infty}(\cos(k))^n[/tex]

(where [itex]k[/itex] is in radians and is not equal to an integer multiple of [itex]\pi[/itex])
 
  • #34
arildno
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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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