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#1 Isidia Đã gửi 31-01-2023 - 00:57

Isidia

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This series $\displaystyle\sum_{n=1}^\infty \cos(nx)=\cos(x)+\cos(2x)+\cos(3x)+...$ has attracted my curiosity because many prominent mathematicians in the XVIII century came up with contradictory results. Chiefly among these mathematicians are Euler and Lagrange. I have introduced this series in this thread, so please visit it to learn more on how Euler derived it.

Lagrange's approach

I have kept the links to many papers written by Euler, Lagrange, Daniel Bernoulli and d'Alembert on the famous problem of vibrating string. Among these papers, I am particularly interested in a paper of Lagrange entitled "Recherche sur la nature et la propagation du son" (Research on the nature of the propagation of wave) (see the entire work here). On page 110, he wrote:

"Supposons, pour simplifier le calcul, que la séries dont on veut prendre la somme soit généralement"

$$\cos(x)+\cos(2x)+\cos(3x)+\cos(4x)...$$

He rewrote the series in exponential forms, and formed a sum of two infinite terms:

$$=\dfrac{e^{ix}+e^{-ix}}{2}+\dfrac{e^{2ix}+e^{-2ix}}{2}+\dfrac{e^{3ix}+e^{-3ix}}{2}+\dfrac{e^{4ix}+e^{-4ix}}{2}...$$

$$=\dfrac{e^{ix}+e^{2ix}+e^{3ix}+e^{4ix}}{2}...+=\dfrac{e^{-ix}+e^{-2ix}+e^{-3ix}+e^{-4ix}}{2}...$$

$$=\dfrac{1}{2}\left[e^{ix}+e^{2ix}+e^{3ix}+e^{4ix}...\right]+\dfrac{1}{2}\left[e^{-ix}+e^{-2ix}+e^{-3ix}+e^{-4ix}...\right]$$

Here we have two infinite geometric series with the common ratio $e^{ix}$ and $e^{-ix}$. We can write it in the form $\dfrac{1}{1-e^{ix}}$, but we need to subtract $1$ because the geometric series has $1$ as its first term, while our series do not have this.

$$=\dfrac{1}{2}\left[\dfrac{1}{1-e^{ix}}-1\right]+\dfrac{1}{2}\left[\dfrac{1}{1-e^{-ix}}-1\right]$$

$$=\dfrac{1}{2}\left[\dfrac{1}{1-e^{ix}}-\dfrac{1-e^{ix}}{1-e^{ix}}\right]+\dfrac{1}{2}\left[\dfrac{1}{1-e^{-ix}}-\dfrac{1-e^{-ix}}{1-e^{-ix}}\right]$$

$$=\dfrac{e^{ix}}{2(1-e^{ix})}+\dfrac{e^{-ix}}{2(1-e^{-ix})}$$

$$\dfrac{e^{ix}(1-e^{-ix})+e^{-ix}(1-e^{ix})}{2(1-e^{ix})(1-e^{-ix})}$$

$$=\dfrac{e^{ix}-+e^{-ix}-2}{2(1-e^{ix}-e^{-ix}+1)}=\dfrac{e^{ix}+e^{-ix}}{2(-e^{ix}-e^{-ix}+2)}-\dfrac{2}{2(2-e^{ix}-e^{ix})}$$

$$=\dfrac{\cos(x)-1}{2-e^{ix}-e^{-ix}}==\dfrac{\cos(x)-1}{2-2\left(\dfrac{e^{ix}-e^{-ix}}{2}\right)}=\dfrac{\cos(x)-1}{2(1-\cos(x))}=-\dfrac{1}{2}$$

Thus, according to Lagrange, this series has the sum of $-\dfrac{1}{2}$.

However, we plug $x=0$ into $\cos(x)+\cos(2x)+\cos(3x)+...$, we will have a series of the form $1+1+1+1+1...=+\infty$

So this series admits two value. In our modern view, it is divergent. Yet Lagrange held on so tightly that he offered us an explanation that was very characteristic of mathematicians of his generation:

Mais, dira-t-on, comment peut-il se faire que la somme de la suite infinis $\cos(x)+\cos(2x)+\cos(3x)+...$ soit toujours égale à $-\dfrac{1}{2}$, puisque dans le cas $x=0$, elle devient nécessairement égale à une suite d'autant d'unités? Je ré ponds que cela provient des termes qui se détruisent naturellement dans tous les cas, excep té dans celui òu $x=0$. (page 111)

Lagrange then proceeds to examine the partial sum of this series.

"Pour rendre la chose plus sensible, cherchons la somme de la suite:"

$$\cos(x)+\cos(2x)+\cos(3x)+\cos(4x)+...\cos(mx)$$

He used the formula for summing finite geometric series $\dfrac{1-r^{n+1}}{1-r}$. Rewrite the finite given series above as:

$$=\dfrac{e^{ix}+e^{2ix}+e^{3ix}+e^{4ix}+...e^{mix}}{2}+\dfrac{e^{-ix}+e^{-2ix}+e^{-3ix}+e^{-4ix}+...e^{-mix}}{2}$$

Apply the formula, substracting $1$ from the series as above, we have:

$$\dfrac{1-e^{(m+1)ix}}{1-e^{ix}}-\dfrac{1-e^{ix}}{1-e^{ix}}+\dfrac{1-e^{-(m+1)ix}}{1-e^{-ix}}-\dfrac{1-e^{-ix}}{1-e^{-ix}}=\dfrac{e^{ix}-e^{(m+1)ix}}{2(1-e^{ix})}+\dfrac{e^{-ix}-e^{-(m+1)ix}}{2(1-e^{ix})}$$

$$=\dfrac{(e^{ix}-e^{(m+1)ix})(2-2e^{-ix})+(2-2e^{-ix})(e^{-ix}-e^{-(m+1)ix})}{4(2-e^{ix}-e^{-ix})}$$

$$=\dfrac{\color{red}{2}(e^{ix}-1-e^{(m+1)ix}+e^{mix}+e^{-ix}-e^{-(m+1)ix}-1+e^{-mix})}{\color{red}{4}(2-e^{ix}-e^{-ix})}$$

$$=\dfrac{(e^{ix}-1-e^{(m+1)ix}+e^{mix}+e^{-ix}-e^{-(m+1)ix}-1+e^{-mix})}{2(2-e^{ix}-e^{-ix})}$$

$$=\dfrac{(e^{ix}+e^{-ix}-e^{mix}+e^{-mix}-(e^{(m+1)ix}+e^{-(m+1)ix})-2)}{2(2-e^{ix}-e^{-ix})}$$

$$=\dfrac{e^{ix}+e^{-ix}}{2(2-e^{ix}-e^{-ix})}+\dfrac{e^{mix}+e^{-mix}}{2(2-e^{ix}-e^{-ix})}-\dfrac{e^{(m+1)ix}+e^{-(m+1)ix}}{2(2-e^{ix}-e^{-ix})}-\dfrac{\color{red}{2}}{\color{red}{2}(2-e^{ix}-e^{-ix})}$$

$$=\dfrac{\cos(x)+\cos(mx)-\cos[(m+1)x]-1}{2-2\left(\dfrac{e^{ix}+e^{-ix}}{2}\right)}=\dfrac{\color{red}{\cos(x)}+\cos(mx)-\cos(m+1)\color{red}{-1}}{2\color{red}{(1-\left(\cos(x)\right)})}$$

$$=\dfrac{\cos(mx)-\cos[(m+1)x]}{2{(1-\left(\cos(x)\right)})}-\dfrac{1}{2}$$

Proof that the concerned series is divergent for all $x$

In order to prove this, we need two theorems that are encountered in Calculus and Analysis:

Theorem 1: If a series is convergent, this implies that the limit of the sequence that makes up that series must tend to 0.

Theorem 2 (this theorem is pointed out to me by Nguyễn Mạnh Linh, but the blog that contains this theorem along with its proof has been deleted):

Let $(x_{n})_{n=1}^{+\infty}$ be a sequence of real numbers, with $x$ being given. If all subsequences of $(x_{n})_{n=1}^{+\infty}$ contain a subsequence that converges to a limit, then $(x_{n})_{n=1}^{+\infty}$ also converges to that limit.

Now, using theorem 1, let assume that the series $\displaystyle\sum_{n=1}^\infty cos(nx)$ is convergent. This implies that the sequence $(cos(nx))_{n=1}^{+\infty}$ converges to zero as n tends to infinity. Applying theorem 2, this imples that all subsequences of $(cos(nx))_{n=1}^{+\infty}$ contain a subsequence that converges to 0. Now, for the sequence $(cos(nx))_{n=1}^{+\infty}=\cos(x), \cos(2x), \cos(3x), \cos(4x),...$, we can extract a subsequence $(cos(2nx))_{n=1}^{+\infty}=\cos(2x), \cos(4x),...$ that tends to zero as n approaches infinity. This means $\displaystyle\lim_{n\to\infty} \cos(2nx)=0$.

From trigonometry, we know that $\cos(2nx)=2\cos^2(nx)-1$. If $\displaystyle\lim_{n\to\infty} \cos(nx)=0$, this means $\displaystyle\lim_{n\to\infty} 2\cos^2(nx)-1=2\cdot 0-1=-1$. This proves that  $\displaystyle\lim_{n\to\infty} \cos(2nx)=-1$, hence  $\displaystyle\lim_{n\to\infty} \cos(nx)=0$ is impossible. Thus, the series is divergent.

I wish to thank Vũ Tuấn Hiền for providing me guidance to this problem. His direction is important for me to decisively prove that this series is divergent.

Historical commentary:

This series and the method of using to sum it is demonstrated by the famed French mathematician Henri Lebesgue in his small lecture note on trigonometric series entitled "Leçons sur les séries trigonométriques professées au Collège de France" (page 35, section 21). He wrote: "Que l'on donne à la methode l'une ou l'autre forme, elle n'est rigoureuse que si l'on a étudié pour $|z|=1$, la série qui joue le rôle de (Z). Dans ce cas particulier, elle conduit à un résulta exact, mais, dans d'autres cas, elle peut conduire à des résultas incorrects, c'est ainsi que Lagrange écrivait l'égailité

$$0=\dfrac{1}{2}+\cos(x)+\cos(2x)+\cos(3x)+...$$

alors que la série du second membre est divergent, comme on le voit en calculant la somme de ses n premiers termes."

Bài viết đã được chỉnh sửa nội dung bởi Isidia: 31-01-2023 - 07:11

There is no mathematical model that can predict your future or tell you how your life will unfold. All strength and power lies within your soul, and that's all what you need.

#2 Nesbit Đã gửi 31-01-2023 - 06:48

Nesbit

...let it be...

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Không biết "theorem 2" mà bạn nhắc đến là định lý nào? Ngoài ra những đường link trên blogger mà bạn đăng đều là private, người khác không xem được.

Không đọc tin nhắn nhờ giải toán.

Góp ý về cách điều hành của mod

#3 Isidia Đã gửi 31-01-2023 - 07:16

Isidia

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Không biết "theorem 2" mà bạn nhắc đến là định lý nào? Ngoài ra những đường link trên blogger mà bạn đăng đều là private, người khác không xem được.

Mình đã sửa chữa hết các link và xóa đoạn này khỏi bài chính vì mình không hiểu Lagrange đang làm gì trong bài viết gốc:

He stated that in the case when $m$ is an infinite number, $1$ vanishes along with $m$, where the term $\cos(m+1)$ equals to $\cos(mx)$, the formula is equal to:

$$\dfrac{\cos(x)+\cos(mx)-\cos[(m+1)x]}{2{(1-\left(\cos(x)\right)})}=0$$

Differentiating the expression, we have:

$$\dfrac{(m+1)\sin[(m+1)x]-m\sin(mx)}{2\sin(x)}$$

Let $x=0$, the expression has the form $\frac{0}{0}$

Differentiating twice, we have:

$$\dfrac{(m+1)^2\sin[(m+1)x]-m^2\sin(mx)}{2\cos(x)}$$

And by making $x=0$

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

So the value is:

$$m+\dfrac{1}{2}-\dfrac{1}{2}=m$$

Không biết "theorem 2" mà bạn nhắc đến là định lý nào? Ngoài ra những đường link trên blogger mà bạn đăng đều là private, người khác không xem được.

Theorem 2 được phát biểu như sau:

Theorem 2 (this theorem is pointed out to me by Nguyễn Mạnh Linh, but the blog that contains this theorem along with its proof has been deleted):

Let $(x_{n})_{n=1}^{+\infty}$ be a sequence of real numbers, with $x$ being given. If all subsequences of $(x_{n})_{n=1}^{+\infty}$ contain a subsequence that converges to a limit, then $(x_{n})_{n=1}^{+\infty}$ also converges to that limit.

Theorem này được mình tìm thấy trong một blog cũ của bạn Nguyễn Mạnh Linh, blog ấy nay đã bị xóa vì...không còn xứng tầm bạn ấy nữa =))

Mình không biết chứng minh cả 2 theorem nên chỉ dùng nó để chứng minh mà thôi.

Lí do mình xóa theorem 2 này là vì sau khi trao đổi với bạn Linh, bạn ấy nói mình không cần đến nó. Nhưng lúc xóa chưa edit toàn bài nên còn lại dấu vết ấy.

There is no mathematical model that can predict your future or tell you how your life will unfold. All strength and power lies within your soul, and that's all what you need.

#4 Isidia Đã gửi 31-01-2023 - 07:17

Isidia

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Đây là blog toán của mình: https://analysisbeauty.blogspot.com/

Còn đây là bài gốc: https://analysisbeau...rd-of-name.html

There is no mathematical model that can predict your future or tell you how your life will unfold. All strength and power lies within your soul, and that's all what you need.

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