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I want to find the minimum value of the series cos(2x)+cos(4x)+cos(6x)+cos(8x)+...+cos(2nx). x could be 2pit. Anyone can share a method of how to determine the minimum value of the series?

Dave Du
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    Welcome to MSE. Please use MathJax to format your posts. To begin with, surround math expressions (including numbers) with $ signs and use _ for subscripts. $x_1$ comes out as $x_1$. – saulspatz Feb 02 '21 at 16:46
  • Just take the derivative and find the critical points would be okay? – Yuki.F Feb 02 '21 at 16:48
  • You can isolate sin(x) and find the critical points associated with sin(x), but not sure whether those critical points for sin(x) could lead to the minimum value. – Dave Du Feb 02 '21 at 16:55
  • @DaveDu - differentiate the function again (ie, 2nd derivative) and check the values at critical points. If the second derivative is positive, the point would give minima. (I'd expect summation of cosine to n terms to form continuous function). Alternatively, you can first shorten your summation using the cosine summation identity when terms are in A.P – SteelCubes Feb 02 '21 at 17:15
  • Last term is $\cos(20x)$ or $\cos(2nx)$? – Raffaele Feb 02 '21 at 17:28
  • let's say 2nx in order to be generic. Want to know the method of determining the minimum value. – Dave Du Feb 02 '21 at 18:21
  • @SteelCubes - It's not so easy to get the critical points. Please show me how to get the critical points step by step if you have worked it out. – Dave Du Feb 02 '21 at 18:23

3 Answers3

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You can use $\displaystyle \sum_{k=1} \cos(kx)=R\left(\sum_{k=1} e^{2ikx}\right)$

Luis Alexandher
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  • You can find the sum by doing this way, but the question is how to determine the minimum value. – Dave Du Feb 02 '21 at 17:01
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Comment:

Let $t=2x$ , you get:

$s=\cos (t)+\cos(2t)+\cos(3t)+\cdot\cdot\cdot +\cos(10t)$

Now use this formula:

$$s=\cos (t)+\cos(2t)+\cos(3t)+\cdot\cdot\cdot +\cos(nt)=\frac{\sin \frac n2 t\cos\frac {n+1}2 t}{\sin (\frac t 2)}$$

So you have to find minimum of this function:

$$y=\frac{\sin(10x)\cos(11x)}{\sin(x)}$$

By plotting in Wolfram minimum of y is about -2.8.

sirous
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  • Finding the sum is easy. The difficult part is to take the derivative and equate it to zero in order to find the critical points. Could you please show me how to do that? – Dave Du Feb 02 '21 at 19:48
  • No doubt that the maximum value is 10, but the minimum value is -2.78 by plotting the graph. The reasoning for minimum value is not correct. – Dave Du Feb 03 '21 at 16:27
  • I plotted in wolfram and it can be seen that max and min are equal in absolute value although not for a certain value of x. minimum , due to wolfram plot is certainly less than -2.78.it looks less than 3 for sure. – sirous Feb 03 '21 at 16:46
  • Sorry I am looking for the minimum value for $x$ =\pi $t$, then it will be periodic for $t$ in [0,1] and the minimum value is around -2.78. But really this should not be matter and the minimum value should not change. Your sum expression is not correct. The cos(21$x$) should be cos(11$x$) for $n$ = 10. – Dave Du Feb 03 '21 at 17:25
  • -10 can't be the minimum value as there is no such x that can let all the individual item equate -1 simultaneously as their frequency is different. – Dave Du Feb 03 '21 at 17:40
  • @DaveDu, thanks for pointing out my error. I corrected it. I will submit the plot by Wolfram.You can also plot it in Wolfram and see the curve is symmetric about x axis. – sirous Feb 03 '21 at 17:45
  • @DaveDu, You are write . I plotted the corrected function and minimum is about -2.78. It is interest to know how this was found analytically. – sirous Feb 03 '21 at 17:58
  • it is not easy to take the derivative and let it equate $0$ in order to find the critical points. – Dave Du Feb 03 '21 at 18:07
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Welcome to MSE. As a hint, you might want to look into Lagrange's trignometric identities. (They can be derived with complex numbers like Luis is suggesting.

Sam Freedman
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