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Prove that ( Cauchy Method )

$\quad \qquad { \bigstar { \underline { \boxed { \red { \bf { \int_{0}^{x} \int_{0}^{{x}_{n}} \int_{0}^{{x}_{n-1}} ... \int_{0}^{x_2} \phi(t) (dt)^n = \dfrac{1}{(n-1)!} \int_{0}^{x} ( x - \xi) \phi(\xi) d \xi }}}}}}{\bigstar}$

Subject :- Maths

Standard :- Msc 3rd semester ​

Answer 1

Answer:

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PROOF:-

The limit of integrals is not equal to the integral of the limit.

Consider the sequence {f}, where

f(x) = n x(1 − x 2) n , 0 ≤ x ≤ 1, n = 1, 2, 3,…

For 0 < x ≤ 1, n→∞

limit f(x) = 0

At x = 0, each f(0) = 0, so that n→∞

limit f(0) = 0

Thus the limit function f(x) = n→∞

limit f(x) = 0, for 0 ≤ x ≤ 1

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Answer 2

To prove the Cauchy method, we can use integration by parts repeatedly. Let's consider the integral:

$\begin{aligned} I(x) &= \int_0^x \int_0^{x_n} \cdots \int_0^{x_2} \phi(t) (dt)^n \\ &= \int_0^x \int_0^{x_n} \cdots \int_0^{x_3} (x_2 - t) \phi(t) (dt)^{n-1} dx_2 \\ &= \int_0^x \int_0^{x_n} \cdots \int_0^{x_4} \frac{(x_2-t)^2}{2} \phi'(t) (dt)^{n-2} dx_2 \\ &= \cdots \\ &= \frac{1}{(n-1)!} \int_0^x (x-\xi)^{n-1} \phi(\xi) d\xi \end{aligned}$

where we have used integration by parts repeatedly, and the fact that ${\bf \int_0^a (a-t)^k dt = \frac{a^{k+1}}{k+1}}$ for any non-negative integer ${\bf k}$. Therefore, we have:

$I(x) = \frac{1}{(n-1)!} \int_0^x (x-\xi)^{n-1} \phi(\xi) d\xi$

which is the desired result.

Note that this formula is sometimes called the Cauchy method, or the Cauchy formula, and is a useful tool in the study of differential equations and other areas of mathematical analysis.