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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

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- Thread starter pureouchies4717
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- #1

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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

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

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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.

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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

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- #5

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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?

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arildno

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nrqed

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nick727kcin said:thanks alot for the input guys

can someone please tell me if 1.175 is right? thanks

yes, that is correct.

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i think the sum would be .540

- #9

nrqed

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arildno said:

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

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thanks so much for the help guys :!!)

but now im kinda confused:rofl:

but now im kinda confused:rofl:

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arildno

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arildno said:

o, its to the radians

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arildno

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

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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

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

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

- #16

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o ok

hmm i just felt like rounding. thanks guys/girls

hmm i just felt like rounding. thanks guys/girls

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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.

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arildno

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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

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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

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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.

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arildno

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

2. It isn't divergent.

2. It isn't divergent.

- #22

arildno

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Just a question:

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

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

- #23

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

- #24

nrqed

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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

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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)

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