User:Egm6341.s10.team3.Min/HW1-x

= Problem 4 : Proof Taylor series by IMVT =

Given
See IMVT in [[media:egm6341.s10.p2-3.png|Lecture p.2-3]]
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$$ with $$ w(x)\geqslant 0 \quad \forall \ x \in [a,b] $$
 * $$\displaystyle \int_{a}^{b} w(x)f(x)\ dt = f(\xi)\int_{a}^{b} w(x)\ dt
 * $$\displaystyle \int_{a}^{b} w(x)f(x)\ dt = f(\xi)\int_{a}^{b} w(x)\ dt
 * }

Find
Use IMVT to show that (5) in [[media:egm6341.s10.p2-2.png|Lecture p.2-2]] is equal to (1) in [[media:egm6341.s10.p2-3.png|Lecture p.2-3]]


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= \frac{1}{n!} \frac{(x-x_0)^{n+1}}{(n+1)!}f^{(n+1)}(\xi)
 * $$\displaystyle R_n+1(x) = \frac{1}{n!} \int_{x_0}^{x} (x-t)^n f^{(n+1)}(t)\ dt

$$
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for $$\quad \xi \in [x_0,x]$$

Solution
Because $$t \in [x_0,x]$$, thus $$ (x-t)^n \geqslant 0 $$, according to IMVT we have following equations:


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= \frac{1}{n!} g(\xi)\int_{x_0}^{x} w(t)\ dt = \frac{1}{n!} f^{(n+1)}(\xi) \int_{x_0}^{x} (x-t)^n\ dt $$ $$ Int. $$ \int_{x_0}^{x} (x-t)^n\ dt $$, we have
 * $$ \displaystyle R_{n+1}(x) = \frac{1}{n!} \int_{x_0}^{x} \underbrace{(x-t)^n}_{w(t)} \underbrace{f^{(n+1)}(t)}_{g(t)}\ dt
 * $$ \displaystyle R_{n+1}(x) = \frac{1}{n!} \int_{x_0}^{x} \underbrace{(x-t)^n}_{w(t)} \underbrace{f^{(n+1)}(t)}_{g(t)}\ dt
 * $$\displaystyle (Eq. 1)
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= \cancelto{o}{-\frac{(x-x)^{n+1}}{n+1}} + \frac{(x-x_0)^{n+1}}{n+1} = \frac{(x-x_0)^{n+1}}{n+1} $$ $$
 * $$ \displaystyle \int_{x_0}^{x} (x-t)^n\ dt = \left [ -\frac{(x-t)^{n+1}}{n+1} \right ]_{t=x_0}^{t=x}
 * $$ \displaystyle \int_{x_0}^{x} (x-t)^n\ dt = \left [ -\frac{(x-t)^{n+1}}{n+1} \right ]_{t=x_0}^{t=x}
 * $$\displaystyle (Eq. 2)
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Use Eq.1 & Eq.2 we have :


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$$ \displaystyle \Rightarrow  R_{n+1}(x) = \frac{1}{n!} \int_{x_0}^{x} (x-t)^n f^{(n+1)}(t)\ dt = \frac{(x-x_0)^{n+1}}{(n+1)!}f^{(n+1)}(\xi)
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$$
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for $$\quad \xi \in [x_0,x]$$