Calculating the Riemann Sum of (cos1)^x

[pureouchies4717]

hi, is it possible to find the riemann sum of (cos1)^x?

it looks divergent to me

can someome please help me... even if it is convergent, i don't know how to find the sum of a trigonometric function

 

[arildno]

Eeh, do you mean (cos(1))^x?

 

[d_leet]

hi, is it possible to find the riemann sum of (cos1)^x?

it looks divergent to me

can someome please help me... even if it is convergent, i don't know how to find the sum of a trigonometric function

Did you really mean the riemann sum? Or did you mean the sum of the infinite series? Well since cos(0) = 1 and cos(pi) = -1 etc.. Then for any x that is not a multiple of pi |cos(x)| will be less than 1. Since this absolute value is less than 1 the geometric series will converge, specifically it will converge to a/(1-r) where a is the first term, and r is the common ratio.

 

[pureouchies4717]

thanks a lot for the input guys

can someone please tell me if 1.175 is right? thanks

 

[d_leet]

thanks a lot for the input guys

can someone please tell me if 1.175 is right? thanks

Where is the sum from? From 0 to infinity or what?

 

[arildno]

1.175 is definitely not "right". What you've got is a geometric series, and you should be able to get a simple, exact expression for the sum, rather than an approximation.

 

[nrqed]

thanks a lot for the input guys

can someone please tell me if 1.175 is right? thanks

yes, that is correct.

 

[pureouchies4717]

ah i think i got it

i think the sum would be .540

 

[nrqed]

1.175 is definitely not "right". What you've got is a geometric series, and you should be able to get a simple, exact expression for the sum, rather than an approximation.

well, you are right. But his answer is correct within 4 sig figs. I assumed he wanted a numerical, approximate value.

 

[pureouchies4717]

thanks so much for the help guys :!)


but now I am kinda confused

 

[arildno]

Since there isn't said anywhere whether cos(1) is the cosine to 1 degree or to 1 radian, for example, the numerical answers are just meaningless.

 

[pureouchies4717]

Since there isn't said anywhere whether cos(1) is the cosine to 1 degree or to 1 radian, for example, the numerical answers are just meaningless.

o, its to the radians

 

[arildno]

And why is it "right" just to keep 4 significant figures? Why not 15, or just 1?
Nothing in what you posted gives any suggestion of what appoximation scheme you ought to use.

 

[pureouchies4717]

ok here's what i did to get the 1.7... figure:

you said that it converges to a/(1-r)

A. a1= cos1
r= cos1 (right?)

so then: cos1/(1-cos1)

this comes out to 1.7



B. this is how i got the .5... answer:
a/(1-r)
a1= cos1
r= (cos1)^k

cos1/(1-(cos1)^k)

then i took the limits (cos1)^k becomes 0

so its: cos1/(1)= cos1 or .5...

 

[arildno]

To be to the point:
Why are you so reluctant to give your answer as the exact expression
cos(1)/(1-cos(1)) ??

 

[pureouchies4717]

o ok

hmm i just felt like rounding. thanks guys/girls

 

[d_leet]

ok here's what i did to get the 1.7... figure:

you said that it converges to a/(1-r)

A. a1= cos1
r= cos1 (right?)

so then: cos1/(1-cos1)

this comes out to 1.7



B. this is how i got the .5... answer:
a/(1-r)
a1= cos1
r= (cos1)^k

cos1/(1-(cos1)^k)

then i took the limits (cos1)^k becomes 0

so its: cos1/(1)= cos1 or .5...

Well for case B, if you even just write out the first few terms of the eries you should see that that definitely cannot be teh answer because you have

cos(1) + cos²(1) + cos³(1) + ...

As you can see the first term in that series is about .54... and since none of those terms are ever negative the sum of the series must be larger than the first term about .54...

Case A gives you an approximation of the correct sum of this series.

 

[arildno]

I'm glad you finally posted this, since finally, we can get rid of some misconceptions you show here.

B. this is how i got the .5... answer:
a/(1-r)
a1= cos1
r= (cos1)^k

cos1/(1-(cos1)^k)

then i took the limits (cos1)^k becomes 0

so its: cos1/(1)= cos1 or .5...

A geometric series is of the form:
S=a*r^{0}+ar^{2}+++ar^{k}+
where a and r are real numbers, and |r|<1
In your case, your series is:
\cos(1)*(\cos(1))^{0}+\cos(1)*(\cos(1))^{1}+++\cos(1)*(\cos(1))^{k}+++
Thus, a=r=cos(1)

 

[pureouchies4717]

thanks everyone :)

and also,

i don't want to flood the board with topics, so can you guys please help with something else?

how is arctan(2n) divergent when, if you make n a big number, it clearly has a limit of 1.57?

 

[d_leet]

thanks everyone :)

and also,

i don't want to flood the board with topics, so can you guys please help with something else?

how is arctan(2n) divergent when, if you make n a big number, the answer comes out to 90?

ex:

a50= 89.4
a1000000=90

I'm assuming you mean the infinite series with a general term of arctan(2n). Well it's usually the first test for convergence/divergence that you learn: that if the series converges then the limit of the terms as n goes to infinity must be 0, thus if the terms don't go to 0 then the series diverges.

 

[arildno]

1. It doesn't have a limit at 1.57.

2. It isn't divergent.

 

[arildno]

Just a question:
Whatever did this thread have to do with RIEMANN sums??

 

[pureouchies4717]

o I am sorry for the confusion... i meant normal sums

so arctan2n doesn't have a sum right? because its divergent?

 

[nrqed]

ok here's what i did to get the 1.7... figure:

you said that it converges to a/(1-r)

A. a1= cos1
r= cos1 (right?)

so then: cos1/(1-cos1)

That's the answer. There is nothing else to do..

(that comes out to 1.17534265...but I have been chastised for writing an approcimate numerical value so I will say that the answer is cos(1)/(1-cos(1)) )

 

[HallsofIvy]

The point is that you still haven't told us precisely what the problem was!

The geometric sum \Sigma_{i=0}^\infty (cos(1))^n is a "geometric series", not a "Riemann Sum" (those are the finite sums used to define an integral). As Arildno said, you haven't stated whether that is 1 radian or 1 degree (although, I would say that, as long as we are talking about the cosine function as opposed to a trigonometry problem, radians should be assumed) and you haven't justified keeping only 4 significant figures.

Since the the geometric series \sigma_{i= 0}{\infty}ar^n has sum \frac{a}{1- r}, as long as |r|< 1, as is cos(1). In your problem, a= cos(1) (since the sum starts at k= 1, not 0) and r= cos(1) (not "(cos(1)^k)" . The sum is \frac{cos(1)}{1- cos(1)}.

Since the common ratio is cos(1) rather than cos(1)^k, you do not take the limit as k goes to infinity. Indeed, if that were correct, for |r|< 1, the sum of \sigma_{i=0}^{\infty}ar^n would always be just a.

 

[finchie_88]

If you want the sum to infinity, is it not simply \frac{\cos1}{1 - \cos1}, which equals 1.1753426496700214107767867596561. Also, 1-co1 is the same as 2(sin0.5)^2, if that helps.

 

[arildno]

If you want the sum to infinity, is it not simply \frac{\cos1}{1 - \cos1}, which equals 1.1753426496700214107767867596561

No, it doesn't, at least not with radians or degrees as your angular measure.

 

[finchie_88]

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.

 

[VietDao29]

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:
<br /> \frac{\cos 1}{1 - \cos 1} is more preferable than giving it in the form: 1.175 to 3 d.p..
No? :)
This is arildno's point, I believe.

 

[arildno]

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?

 

[shmoe]

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.

 

[ksinclair13]

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 :).

 

[quantumdude]

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)

 

[arildno]

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.