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I need to prove that: \begin{align} c_k =&\; \cos(k\!\cdot\!x)\\ c_k :=&\; c_{k-1} +d_{k-1}\\ d_k :=&\; 2d_0\!\cdot\!c_k +d_{k−1}\\ d_0 :=&\; −2\!\cdot\!\sin^2{(x/2)}\\ \end{align}

I've got an explicit formula for $d_k$ which should be:

\begin{align} d_k&=d_o+\sum_{i=1}^k{2\!\cdot\!d_o\!\cdot\!c_i} &&\implies& c_k &=c_{k-1}+ \sum_{i=1}^k{2\!\cdot\!d_o\!\cdot\!c_i} \end{align}

Now I want to do a proof by induction. Assuming that $c_p=\cos(p\!\cdot\!x)$ for every $p<k$. This would get me the following: $$c_k =\cos\left(\left(k-1\right)\!\cdot\!x\right)+\sum_{i=1}^k{2\!\cdot\!d_o\!\cdot\!\cos(p\!\cdot\!x)}$$ Using this formula I found:

\begin{align} \sum_{k=1}^n \cos(kx) & = \frac{\sin\left(\frac{nx}2\right)}{\sin \left(\frac{x}2\right)}\, \cos\left(\frac{(n+1)\,x}2\right) \end{align}

I tried to play around with trigonometric addition formulas but I am getting nowhere.

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    $\begingroup$ Your $c_k$ coincide with $T_k(\cos x)$ where $T_k$ is a Chebyshev polynomial of the first kind. $\endgroup$
    – Piquito
    Feb 4 '16 at 0:08
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    $\begingroup$ So the first line has to be proven, using lines 2-3-4? But then what is $c_0$? $\endgroup$ Feb 4 '16 at 13:08
  • $\begingroup$ @Justpassingby If $c_k=\cos(k*x)$, then $c_0=1$. $\endgroup$ Feb 4 '16 at 21:49
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    $\begingroup$ @simpleArt if we are required to prove the relation that $c_k=\cos(k*x)$ then how can we use this relation and say $c_0=1$ and then use this ...to again prove the same thing that is $c_k=\cos(k*x)$ ...it is logically incorrect ...to use the information that we are required to prove .... $\endgroup$
    – Freelancer
    Feb 5 '16 at 3:41
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    $\begingroup$ Very confusing. $c_k = \cos(kx)$ is the definition of $c_k$? Or you want to prove that is a consequence of the other conditions (with $c_0 = 1$)? $\endgroup$ Feb 5 '16 at 11:00
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The last equation reads $$d_0=-2\sin^2\left(\frac x2\right)=\cos(x)-1.$$

Then eliminating $d$ from the second and the third,

$$d_k=c_{k+1}-c_k,\\ c_{k+1}-2c_k+c_{k-1}=2(\cos(x)-1)c_k.$$

If we assume that $c_k=\cos(kx)$, then after some simplification,

$$\cos(kx+k)-2\cos(kx)+\cos(kx-x)=2\cos(kx)\cos(x)-2\cos(kx)\\=2(\cos(x)-1)\cos(kx).$$

Hence $c_k=\cos(kx)$ is indeed a solution, such that $c_0=1$ (not specified in the OP).

We also have

$$d_k=\cos((k+1)x)-\cos(kx).$$

Now knowing these relations, formally proving by induction isn't a problem anymore.


By the induction hypothesis, $$c_{k-1}=\cos((k-1)x),\\d_{k-1}=\cos(kx)-\cos((k-1)x).$$

Then,

$$\color{green}{c_k}=c_{k-1}+d_{k-1}=\cos((k-1)x)+\cos(kx)-\cos((k-1)x)\color{green}{=\cos(kx)}$$ and $$\color{green}{d_k}=2(\cos(x)-1)c_k+d_{k-1}\\ =2(\cos(x)-1)\cos(kx)+\cos(kx)-\cos((k-1)x)\\ =2\cos(x)\cos(kx)-2\cos(kx)+\cos(kx)-\cos(kx)\cos(x)-\sin(kx)\sin(x)\\ =\cos(kx)\cos(x)-\sin(kx)\sin(x)-\cos(kx)\\ \color{green}{=\cos((k+1)x)-\cos(kx)}.$$

And the base case is

$$\color{green}{c_0}=\cos(0\cdot x)=\color{green}{1},\\ \color{green}{d_0}=\cos(1\cdot x)-\cos(0\cdot x)=\color{green}{\cos(x)-1}.$$

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    $\begingroup$ Probably easier using $\cos((k+1)x)+\cos((k-1)x)=2\cos(kx)\cos(x)$. $\endgroup$
    – user65203
    Feb 6 '16 at 15:35

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