Cauchy çarpımı yardımıyla $\ sin2x$ seri açılımı

Question

$2\sin x \cos x =2 \left(\sum_{k=0}^\infty (-1)^k\frac{x^{2k+1}}{(2k+1)!}\right) \left(\sum_{k=0}^\infty (-1)^k\frac{x^{2k}}{(2k)!}\right)$

 $= 2\sum_{k=0}^\infty \sum_{i=0}^k (-1)^i \frac{x^{2i+1}}{(2i+1)!} (-1)^{k-i} \frac{x^{2(k-i)}}{(2k-2i)!}$ 

 $= \sum_{k=0}^\infty (-1)^k x^{2k+1} \sum_{i=0}^k\frac{2}{(2i+1)! (2k-2i)!}$ 

 $= \sum_{k=0}^\infty (-1)^k \frac{(2x)^{2k+1}}{(2k+1)!} = \sin(2x)$ 

 Sondaki ifade nasıl  ${(2x)^{2k+1}}$ oldu?

           

Answer 1

$(2k+1)!$ ile carpip bolelim.

$$= \sum_{i=0}^k\frac{1}{(2i+1)! (2k-2i)!}= \sum_{i=0}^k\frac{(2k+1)!}{(2k+1)!(2i+1)! (2k-2i)!}$$

$$=\frac{1}{(2k+1)!} \sum_{i=0}^k\frac{(2k+1)!}{(2i+1)! (2k-2i)!}=\frac{1}{(2k+1)!} \sum_{i=0}^k {2k+1\choose 2i+1}$$

$$=\frac{1}{(2k+1)!}\left[ \sum_{i=0}^k {2k\choose 2i}+{2k\choose 2i+1}\right]=\frac{1}{(2k+1)!}\sum_{i=0}^{2k} {2k\choose i}=\frac{2^{2k}}{(2k+1)!}$$

Comments

Teşekkürler...Euler...

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$\displaystyle\sum_{i=0}^k \frac{(2k+1)!}{(2i+1)!(2k-2i)!} = \displaystyle\sum_{i=0}^k \begin{pmatrix} 2k+1 \\ 2i+1 \end{pmatrix} =\begin{pmatrix} 2k+1 \\ 1 \end{pmatrix} + \begin{pmatrix} 2k+1 \\ 3 \end{pmatrix} +....+= 2^{2k}$

Answer 2

Şöyle oldu...

$=\dfrac{1}{(2k+1)!}\displaystyle\sum_{i=0}^k \frac{(2k+1)!}{(2i+1)!(2k-2i)!}$

$=\dfrac{1}{(2k+1)!} \displaystyle\sum_{i=0}^k \begin{pmatrix} 2k+1 \\ 2i+1 \end{pmatrix}$

$=\dfrac{1}{(2k+1)!}\displaystyle\left( \begin{pmatrix} 2k+1 \\ 1 \end{pmatrix} + \begin{pmatrix} 2k+1 \\ 3 \end{pmatrix} +\cdots+ \binom{2k+1} {2k+1} \right)$

$=\dfrac{1}{(2k+1)!}(2^{2k+1-1})$

$=\dfrac{1}{(2k+1)!}2^{2k}$

O halde ifademiz

$=2\displaystyle\sum_{k=0}^\infty (-1)^k x^{2k+1}\dfrac{1}{(2k+1)!}2^{2k}$

$=\displaystyle\sum_{k=0}^\infty (-1)^k\dfrac{(2x)^{2k+1}}{(2k+1)!}$

$=\sin2x$ olur

Comments

3. esitlikte toplam isareti olmamasi lazim..