derivatives of sine and consine and convergence

[sambellamy]

My question asks, for f_n(x) = sin(nx) / n²,

show that the series of derivatives diverges when x = 2nπ (n being an integer).
I am not sure how to show this, because every time I plug in 2nπ I get zero. Do I need to first show what the series converges to?

I have f_n'(x) = x·cos(nx) / n

I determined that f_n(x) converges on zero. does this have any bearing on the convergence of f_n'(x)?

Thanks for any help!

 

[pka]

My question asks, for f_n(x) = sin(nx) / n²,

show that the series of derivatives diverges when x = 2nπ (n being an integer).
I am not sure how to show this, because every time I plug in 2nπ I get zero. Do I need to first show what the series converges to?

I have f_n'(x) = x·cos(nx) / n

Your derivative is incorrect.
\(\displaystyle f_n'(x)=\dfrac{\cos(nx)}{n}\)

 

[sambellamy]

Oh man. Thanks.

I have progressed on the problem a bit. now that

f_n'(x) = n·cos(nx) / n

when I plug in different values of 2nπ I get differerent multiples of 1/2π: 1/4π, 1/6π, etc.

I tried to show that the derivative is divergent using the ratio test, but nothing was able to cancel out. I got:

| (cos(nx+n) · n )/ ((n+1) · cos(nx)) |

can I use the comparison test to show that it is divergent? I am not sure how to compare it because of the involvement of cosine, which I know to be oscillating.

 

[sambellamy]

I think I was able to show that it diverges just by replacing x with 2nπ and taking the limit as n ->∞, which I got to be ∞, thus it diverges.

The second part of the problem is to determine the values that f_n''(x) is convergent for.

I have determined that f_n''(x) = -sin(nx)

I was unable to use the ratio test, as nothing cancelled.
I don't know what I would compare it to for a comparison test.
Is this a telescoping sum? Would that be true only for certain x values? Or, since it adds infinite terms, would it be just patently true?

Any help?

 

[pka]

I think I was able to show that it diverges just by replacing x with 2nπ and taking the limit as n ->∞, which I got to be ∞, thus it diverges.

The second part of the problem is to determine the values that f_n''(x) is convergent for.

I have determined that f_n''(x) = -sin(nx)

I was unable to use the ratio test, as nothing cancelled.
I don't know what I would compare it to for a comparison test.
Is this a telescoping sum? Would that be true only for certain x values? Or, since it adds infinite terms, would it be just patently true?

Any help?

Your replies #3 & #4 are both completely wrong.

In reply #2, I gave you the correct derivative of \(\displaystyle f_n(x)\). But it appears that you are have not read it.

 

[sambellamy]

Uh, wow. I believe I had the calculations correct even though I mis-typed f'(x) as n·cos(nx) / n, I actually had it written down as cos(nx) / n, and that is the statement from which i got the fractions.

I also updated the thread with my progress, and I used cos(nx) / n to determine that f''(x) = -sin(nx), and also checked it in my calculator.

Can anyone else verify that my f''(x) is correct, and help me with a bit of direction on which method to use to determine divergence when sin/cosine is involved?