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$$\cos{x} + \cos{2x} + \cos{3x} \ldots = y$$ $$ 2\cos{x} + 2\cos{2x} + 2\cos{3x} \ldots = 2y $$ By grouping every alternate term by $\cos{A} + \cos{B} = 2\cos(\frac{A+B}{2})\cos(\frac{A-B}{2})$ $$ \cos{x} + \cos{2x} + 2\cos{x}\left(\cos{2x} + \cos{3x} + \cos{4x} \ldots\right) = 2y$$ $$\cos{x} - 1 + 2\cos{x}\left(\cos{x} + \cos{2x} + \cos{3x} \ldots\right) = 2y$$
$$ \cos{x} - 1 + 2\cos{x}\cdot y = 2y$$ $$y = -\frac{1}{2}$$

Me and my friend are confused right now. We don't see anything mathematically wrong but still by logic, it doesn't seem right. Please explain us our errors . We are high school students with basic knowledge of limits and calculus. If possible, please elaborate on the solution.

J.D
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dhruvk
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  • Well , as the problem stands ,this is not correct: $cos(0)+cos(0)+....\neq \frac{-1}{2}$ – J.D May 19 '24 at 09:38
  • @J.Dmaths but what about the algebra we did here?? what is wrong with this mathematially? – dhruvk May 19 '24 at 09:41
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    I am unsure what "clubbing each alternate term" means . Also, this may be a minor formatting issue, but do you mean $cos^3(x)$ or $cos(3x)$ to be clear? – J.D May 19 '24 at 09:44
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    The identity $\sum_{j=0}^\infty a_j=\sum_{j=0}^\infty a_{2j}+\sum_{j=0}^\infty a_{2j+1}$ is true when both $\sum_{j=0}^\infty a_j$ and $\sum_{j=0}^{\infty}a_{2j+1}$ exist in $\Bbb R$. You may take that thing you've written as a portion of a proof that for all the $x$ such that $\cos x\ne 1$ the condition I've written after "is true when" is false. – Sassatelli Giulio May 19 '24 at 09:46
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    @J.Dmaths I think "grouping" but not quite sure myself – Antony Theo. May 19 '24 at 09:46
  • @J.Dmaths we have 2 terms of each cosnx. now i have clubbed or used the factorization formula for cos A + cos B ie 2cos((A+B)/2)cos((A-B)/2). – dhruvk May 19 '24 at 09:48
  • @SassatelliGiulio i am sorry but i didnt quite get your soln.. could you please put it in more simpler language – dhruvk May 19 '24 at 09:49
  • @dhruvk Simplified. And it's not a solution, it's the statement of a fact. – Sassatelli Giulio May 19 '24 at 09:50
  • @dhruvk are you aware of complex numbers? There is a more elegant solution in that manner. – J.D May 19 '24 at 09:51
  • @J.Dmaths yeah i am. tho i havent revised it much, i learnt it last year. but still i could catch up if its not that complicated lol – dhruvk May 19 '24 at 09:52
  • Right, I'll write the answer in terms of complex numbers. – J.D May 19 '24 at 09:52
  • @SassatelliGiulio then i am not able to understand that statement – dhruvk May 19 '24 at 09:53
  • This is a duplicate of a recent question from a separate account. I'm happy to see that this one has transcribed the equations from an image into MathJax. (We lost the indication of the grouping strategy.) Anyway ... Instead of posting a new question, you(? your friend?) should simply have edited the previous version, because we now have two copies floating around on the site. Since this version has an answer, I recommend deleting the previous one, as it's no longer needed. (No harm done, but keep this in mind for the future.) ... Good luck! – Blue May 19 '24 at 10:25
  • Refer to https://math.stackexchange.com/q/800489/505767 – user May 19 '24 at 10:26
  • @ράτ Yes, it is a popular expression in India. It means to gather together items. I don't know the origin. I have wondered whether it might be related to "club sandwich" but that's just my guess. – badjohn May 19 '24 at 10:58
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    @Blue yeah we are sorry about that.. this was originally my question and my friend suggested me to ask it in stack exchange... he posted it himself and i didnt notice that lol. will ask him to delete that.. – dhruvk May 19 '24 at 13:27

2 Answers2

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Let :

$C_n=\cos(x)+\cos(2x)+\cos(3x)+...\cos(nx)$

$S_n=\sin(x)+\sin(2x)+\sin(3x)+...\sin(nx)$

Therefore, $$C_n+iS_n=\cos(x)+i\sin(x)+\cos(2x)+i\sin(2x)+....$$ $$=e^{ix}+e^{2ix}+e^{3ix}+...e^{nix}$$

$$=\frac{e^{ix}(1-e^{nix})}{1-e^{ix}}= \frac{e^{ix}(1-e^{nix})(1-e^{-ix})}{(1-e^{ix})(1-e^{-ix})}=\frac{e^{ix}-e^{(n+1)ix}-1+e^{nix}}{2-2\cos(x)}$$ Then , $C_n=Re(C_n+iS_n)= \frac{\cos(x)-\cos((n+1)x)-1+\cos(nx)}{2-2\cos(x)}$. $lim_{n\rightarrow \infty}C_n $ does not exist. So the infinite sum diverges.

J.D
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  • From here you can also prove that the Cesaro summation of $\cos(n x)$ is $-1/2$, which means that there is something to the OP's logic, but of course the issue is in the presumption of convergence. – Dark Malthorp Jun 05 '24 at 10:19
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You don't see what is mathematically wrong, but the first line is already mathematically wrong because when you write \begin{equation} \cos x + \cos 2x + \cos 3x+ \ldots \end{equation} we don't know what the $\ldots$ means: it has not been defined yet.

Normally, one tries to define this by taking the limit \begin{equation} \lim_{n\to\infty}\left(\cos x + \cos 2 x + \ldots + \cos nx\right) \end{equation} unfortunately, this limit does not exist in this case. Your reasoning is that if this object exists and satisfies certain rules of calculus then we can prove that $y=-\frac{1}{2}$, but both assertions are false.

When you work with undefined objects, anything can happen.

Gribouillis
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