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We have a differential equation $$ y + y' = f(x) $$ and assume $f$ is infinitely differentiable. And we want to find particular solution. Then,I set $$ y_p = f(x)-f'(x)+f''(x)...., $$ i.e., $y_p=\sum_{i=0}^{\infty} (-1)^{n} f^{(n)}(x)$. It is easy to observe this formal sum is apparently a particular solution and it really works for all polynomial. For example, if $f(x) = x^3$ then $y_p = x^3 - 3x^2 + 6x - 6$ and it is true. But, when $f(x)=\sin x$ we have $$ y_p=(1-1+1-1....) \sin x - (1-1+1-1...) \cos x $$ since $(1-1+1-1..)=\sum_{i=0}^{\infty}(-1)^n$ is divergent we have no result. Up to here, there is nothing interesting. Euler thought that $\sum_{i=0}^{\infty}(-1)^n = 1/2$ since its result change $(1,0,1,0...)$ for finite sum (It is thought to be famous mistake of Euler. Of course, nobody blames him. In his time, convergency was not defined exactly). But, if take this sum as $1/2$ as Euler said $$ y_p = (\sin x - \cos x)/2 $$ and surprisingly, it is really the particular solution of $y+y'=\sin(x)$. Why does it work? Should we take this sum as $1/2$?

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Please consider going over your post and inserting line breaks, centered equations, etc. The combination of a single paragraph, inline equations, nested parentheses, no spaces after periods and typos make this question... hardly unreadable, but more of a chore than it needs to be. –  Jack M Feb 23 at 10:10
I am bad at typing and latex is new for me,I can try to fix it. –  user130550 Feb 23 at 10:14
That's much better, thanks. Have you heard of Césaro summation? It might be relevant here. –  Jack M Feb 23 at 10:28
"Euler thought that", it was other way around , Euler was way ahead of his time in recognising what others could not even fathom let alone comprehend. –  Arjang Feb 23 at 11:30
I think, based on the properties of linear ODEs, you could probably derive particular solutions like this using any consistent summation method for divergent series that is linear, regular, and stable. –  Tim Seguine Feb 23 at 12:49
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1 Answer

up vote 8 down vote accepted

We might justify as follows: If we replace $(-1)^n$ with $q^n$ where $q\approx -1$ but $|q|<1$, the geometric series converges to $\frac1{1-q}$ and all is good. However, if we plug $\sum q^nf^{(n)}(x)$ (where the convergence does not depend on $q$ alone, but in general also the derivatives might "explode") into the differential equation we only get an almost-solution. Or an exact solution to a slightly different equation: $$\tag1y'-qy=f(x).$$ Specifically with $f(x)=\sin(x)$ we can compute the corresponding series and obtain $$\tag2\frac{\sin x+q\cos x}{1+q^2}$$ as solution for $y'-qy=f(x)$. In this form, any special role of $q=-1$ is eliminated (and in fact this even works with $|q|>1$). In other words: Stretching the validity of limit beyond its "official" domain of convergence gives us a correct result here because the series was only a tool and the apparent singularity problems it has become invisible in the final results.

One may say that we did not answer the question: "What is $\sum(-1)^n$?" Instead we answered: "If $\sum(-1)^n$ has a value, what should it consistently be?" This leads to the investigation of summability methods such as Cesàro mean. (To quote from the linked Wikipedia page: "any summation method that possesses these properties and which assigns a finite value to the geometric series must assign this value.")

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if we can get rid of divergency problem easily,can I say that $y_p$ which I define is always a particular solution?(it always work as formal sums) –  user130550 Feb 23 at 11:15
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