- #1

pureouchies4717

- 99

- 0

it looks divergent to me

can someome please help me... even if it is convergent, i dont know how to find the sum of a trigonometric function

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter pureouchies4717
- Start date

- #1

pureouchies4717

- 99

- 0

it looks divergent to me

can someome please help me... even if it is convergent, i dont know how to find the sum of a trigonometric function

- #2

arildno

Science Advisor

Homework Helper

Gold Member

Dearly Missed

- 10,089

- 135

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

- #3

d_leet

- 1,077

- 1

nick727kcin said:

it looks divergent to me

can someome please help me... even if it is convergent, i dont 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.

- #4

pureouchies4717

- 99

- 0

thanks alot for the input guys

can someone please tell me if 1.175 is right? thanks

can someone please tell me if 1.175 is right? thanks

Last edited:

- #5

d_leet

- 1,077

- 1

nick727kcin said:thanks alot 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?

- #6

arildno

Science Advisor

Homework Helper

Gold Member

Dearly Missed

- 10,089

- 135

- #7

nrqed

Science Advisor

Homework Helper

Gold Member

- 3,765

- 295

nick727kcin said:thanks alot for the input guys

can someone please tell me if 1.175 is right? thanks

yes, that is correct.

- #8

pureouchies4717

- 99

- 0

i think the sum would be .540

- #9

nrqed

Science Advisor

Homework Helper

Gold Member

- 3,765

- 295

arildno said:

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

- #10

pureouchies4717

- 99

- 0

thanks so much for the help guys :!!)

but now im kinda confused:rofl:

but now im kinda confused:rofl:

- #11

arildno

Science Advisor

Homework Helper

Gold Member

Dearly Missed

- 10,089

- 135

- #12

pureouchies4717

- 99

- 0

arildno said:

o, its to the radians

- #13

arildno

Science Advisor

Homework Helper

Gold Member

Dearly Missed

- 10,089

- 135

Nothing in what you posted gives any suggestion of what appoximation scheme you ought to use.

- #14

pureouchies4717

- 99

- 0

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...

- #15

arildno

Science Advisor

Homework Helper

Gold Member

Dearly Missed

- 10,089

- 135

Why are you so reluctant to give your answer as the exact expression

cos(1)/(1-cos(1)) ??

- #16

pureouchies4717

- 99

- 0

o ok

hmm i just felt like rounding. thanks guys/girls

hmm i just felt like rounding. thanks guys/girls

- #17

d_leet

- 1,077

- 1

nick727kcin said:

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) + cos

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.

- #18

arildno

Science Advisor

Homework Helper

Gold Member

Dearly Missed

- 10,089

- 135

A geometric series is of the form:nick727kcin said: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...

[tex]S=a*r^{0}+ar^{2}+++ar^{k}+[/tex]

where a and r are real numbers, and |r|<1

In your case, your series is:

[tex]\cos(1)*(\cos(1))^{0}+\cos(1)*(\cos(1))^{1}+++\cos(1)*(\cos(1))^{k}+++[/tex]

Thus, a=r=cos(1)

- #19

pureouchies4717

- 99

- 0

and also,

i dont 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?

- #20

d_leet

- 1,077

- 1

nick727kcin said:thanks everyone :)

and also,

i dont 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.

- #21

arildno

Science Advisor

Homework Helper

Gold Member

Dearly Missed

- 10,089

- 135

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

2. It isn't divergent.

2. It isn't divergent.

- #22

arildno

Science Advisor

Homework Helper

Gold Member

Dearly Missed

- 10,089

- 135

Just a question:

Whatever did this thread have to do with RIEMANN sums??

Whatever did this thread have to do with RIEMANN sums??

- #23

pureouchies4717

- 99

- 0

so arctan2n doesnt have a sum right? because its divergent?

- #24

nrqed

Science Advisor

Homework Helper

Gold Member

- 3,765

- 295

nick727kcin said:ok heres 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)) )

- #25

HallsofIvy

Science Advisor

Homework Helper

- 43,021

- 970

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

The geometric sum [itex]\Sigma_{i=0}^\infty (cos(1))^n[/itex] 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 [itex]\sigma_{i= 0}{\infty}ar^n[/itex] has sum [itex]\frac{a}{1- r}[/itex], 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 [itex]\frac{cos(1)}{1- cos(1)}[/itex].

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 [itex]\sigma_{i=0}^{\infty}ar^n[/itex] would always be just a.

The geometric sum [itex]\Sigma_{i=0}^\infty (cos(1))^n[/itex] 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

Since the the geometric series [itex]\sigma_{i= 0}{\infty}ar^n[/itex] has sum [itex]\frac{a}{1- r}[/itex], 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) (

Since the common ratio is cos(1) rather than cos(1)

Last edited by a moderator:

- #26

- #27

arildno

Science Advisor

Homework Helper

Gold Member

Dearly Missed

- 10,089

- 135

No, it doesn't, at least not with radians or degrees as your angular measure.finchie_88 said:If you want the sum to infinity, is it not simply [tex] \frac{\cos1}{1 - \cos1} [/tex], which equals 1.1753426496700214107767867596561

- #28

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

Homework Helper

- 1,426

- 3

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.

[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

- #30

arildno

Science Advisor

Homework Helper

Gold Member

Dearly Missed

- 10,089

- 135

No it isn't.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..

Learn the difference between exact and approximate answers.

What do think your calculator answer is in this case?

- #31

shmoe

Science Advisor

Homework Helper

- 1,994

- 1

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

ksinclair13

- 99

- 0

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

quantumdude

Staff Emeritus

Science Advisor

Gold Member

- 5,575

- 23

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

Science Advisor

Homework Helper

Gold Member

Dearly Missed

- 10,089

- 135

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.

Share:

- Replies
- 1

- Views
- 199

- Last Post

- Replies
- 1

- Views
- 375

- Replies
- 12

- Views
- 443

- Last Post

- Replies
- 13

- Views
- 2K

- Last Post

- Replies
- 6

- Views
- 632

- Replies
- 19

- Views
- 447

- Last Post

- Replies
- 5

- Views
- 1K

- Last Post

- Replies
- 3

- Views
- 1K

- Last Post

- Replies
- 4

- Views
- 2K

- Last Post

- Replies
- 3

- Views
- 844