The question is: $$\lim _{n \rightarrow \infty} \prod_{i=1}^{n+1} \cos \frac{\sqrt{2 i-1}}{n} a^{2}=\mathrm{e}^{-\frac{a^{4}}{2}}$$
A limit proof problem, from a book of mathematical analysis exercises, using an idea called the "fitting method" The book prompted me to take the logarithm of the limit of the solution, convert the multiplication to addition, and then use the "fitted method" An example of the "fitted method" is as follows.
conditions: $$x \rightarrow 0$$ , $$f(x) \sim x, x_{n}=\sum_{i=1}^{n} f\left(\frac{2 i-1}{n^{2}} a\right)$$. proof：$$\lim _{n \rightarrow \infty} x_{n}=a(a>0)$$ and the given answer is: because $$1=\sum_{i=1}^{n} \frac{2 i-1}{n^{2}}$$, thereby having $$a=\sum_{i=1}^{n} \frac{2 i-1}{n^{2}} a.$$ So, $$|x_{n}-a|=\left|\sum_{i=1}^{n} f\left(\frac{2 i-1}{n^{2}} a\right)-\sum_{i=1}^{n} \frac{2 i-1}{n^{2}} a\right| \leqslant \left|\sum_{i=1}^{n}|f(\frac{2 i-1}{n^{2}} a）-\frac{2 i-1}{n^{2}} a\right| \quad (1)$$ If we can prove: $\forall \varepsilon>0, n$ is sufficiently large, $\left|f\left(\frac{2 i-1}{n^{2}} a\right)-\frac{2 i-1}{n^{2}} a\right|<\frac{2 i-1}{n^{2}} \varepsilon \quad(i=1,2, \cdots, n)$ (2)
then Eq. (1) right end $\leqslant \sum_{i=1}^{n} \frac{2 i-1}{n^{2}} \varepsilon=\varepsilon$ The problem is proved. To prove equation (2), i.e., to prove that $$\left|\frac{f\left(\frac{2 i-1}{n^{2}} a\right)}{\frac{2 i-1}{n^{2}} a}-1\right|<\frac{\varepsilon}{a} \quad (3)$$ In fact, because $f(x) \sim x(x \rightarrow 0)$, therefore $\forall \varepsilon>0, \exists \delta>0$, when $0<|x|<\delta$, there is $$\left|\frac{f(x)}{x}-1\right|<\frac{\varepsilon}{a} \quad (4)$$ So, let $N=\frac{2 a}{\delta}$, then $0<\frac{2 i-1}{n^{2}} a<\delta(i=1,2, \cdots, n)$ when $n>N$. Thus, by equation (4), equation (3) holds.
and two notes are given in the book on the "fitting method":
1.When $x \rightarrow 0$, the functions $\sin x, \tan x, \arcsin x, \arctan x, \mathrm{e}^{x}-1, \ln (1+x)$ are equivalent to $x$. Therefore, using any of these functions as $f(x)$ in this case, the conclusion holds.
2.The essence of the fitting method is to make a proper decomposition of the unit 1. This example essentially uses $1=\sum_{i=1}^{n} \frac{2 i-1}{n^{2}}$to have $a=\sum_{i=1}^{n} \frac{2 i-1}{n^{2}} a$. The essence of the idea of the $2^{\circ}$ fitting method is to make a proper decomposition of the unit 1. This example essentially uses $1=\sum_{i=1}^{n} \frac{2 i-1}{n^{2}}$ to have $a=\sum_{i=1}^{n} \frac{2 i-1}{n^{2}}\ a$. Analytical mathematics has solved a number of significant problems using this fitting method.
This is the way I just thought of implementing the "fitting method", I hope someone can see if there are any errors. $$\lim _{n \rightarrow \infty} \prod_{i=1}^{n+1} \cos \frac{\sqrt{2 i-1}}{n} a^{2}=\mathrm{e}^{-\frac{a^{4}}{2}} \quad (1)$$ The equivalent proof of this problem is： $$\lim _{n \rightarrow \infty}\ln \prod_{i=1}^{n+1} \cos \frac{\sqrt{2 i-1}}{n} a^{2}={-\frac{a^{4}}{2}}$$ i.e.$$\lim _{n \rightarrow \infty} \sum_{i=1}^{n+1} \ln \cos \frac{\sqrt{2 i-1}}{n} a^{2}=-\sum_{i=1}^{n+1} \frac{2 i-1}{n^{2}} \cdot \frac{a^{4}}{2}$$
i.e.$$\frac{1}{2} \lim _{n \rightarrow \infty} \sum_{i=1}^{n+1} \ln \cos ^{2} \frac{\sqrt{2 i-1}}{n} a^{2}=-\sum_{i=1}^{n+1} \frac{2 i-1}{n^{2}} \frac{a^{4}}{2}$$
i.e.$$\lim _{n \rightarrow \infty} \sum_{i=1}^{n+1} \ln \left(1-\sin ^{2} \frac{\sqrt{2 i-1}}{n} a^{2}\right)=-\sum_{i=1}^{n+1} \frac{2 i-1}{n^{2}} \cdot a^{4} \quad(2) $$
If we can prove$$\lim _{n \rightarrow \infty} \ln \left(1-\sin ^{2} \frac{\sqrt{2 i-1}}{n} a^{2}\right)=-\frac{2 i-1}{n^{2}} \cdot a^{4} \quad(3) $$ , we can prove (2) and thus (1)
$$\because \ln \left(1-\sin ^{2} \frac{\sqrt{2 i-1}}{n} a^{2}\right) \backsim-\sin ^{2} \frac{\sqrt{2 i-1}}{n} a^{2} \sim-\frac{2 i-1}{n^{2}} a^{4}$$ $(\frac{2 i-1}{n^{2}} a^{4} \rightarrow 0, i.e. \quad n>\frac{2 a^{4}}{\varepsilon})$
$$\therefore\left|\frac{\ln \left(1-\sin ^{2} \frac{\sqrt{2 i-1}}{n} a^{2}\right)}{-\frac{2 i-1}{n^{2}} a^{4}}-1\right|<\varepsilon$$
It can also be written as : $$\left|\ln \left(1-\sin ^{2} \frac{\sqrt{2 i-1}}{n} a^{2}\right)+\frac{2 i-1}{n^{2}} a^{4}\right|<\frac{2 i-1}{n^{2}} a^{4} \varepsilon \quad(4)$$
So equation (3) holds, and eventually equation (1) also holds.