User:Egm6341.s11.team3.rakesh/hw2

 REMARK: We did the solution on our own 


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Problem
Do the integration by parts to reveal three more terms  :  $$ {f(x)=f(x_0)+\frac {x-x_0}{1!} f(x_0)+\int\limits_{x}^{x_0}(x-t)f''(t)dt } $$

Problem 2.1.2
use IMVT theorem to express the remainder $$ f^5(\S)for (x,x_0) $$

Problem
use IMVT theorem to express the remainder $$ f^5(\S)for (x,x_0) $$

Solution
Given That 
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$$ {f(x)=f(x_0)+\frac {x-x_0}{1!} f(x_0)+\int\limits_{x}^{x_0}(x-t)f''(t)dt } $$|| $$ \displaystyle $$
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$$ \int\limits_{x}^{x_0}(x-t)f(t)dt = -\frac {(x-t)^2}{2} f(t)-\int\limits_{x}^{x_0}\frac{-(x-t)^2}{2}f'''(t)dt $$| $$ \displaystyle $$
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$$ \int\limits_{x}^{x_0}(x-t)f(t)dt = \frac {(x-x_0)^2}{2} f(x_0)+\int\limits_{x}^{x_0}\frac {(x-t)^2}{2}f'''(t)dt $$| $$ \displaystyle $$
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$$ \int\limits_{x}^{x_0}\frac {(x-t)^2}{2}f(t)dt = -\frac {(x-t)^3}{2*3} f(t)-\int\limits_{x}^{x_0}\frac {-(x-t)^3}{2*3}f(t)dt $$| $$ \displaystyle $$
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$$ \int\limits_{x}^{x_0}\frac {(x-t)^2}{2}f(t)dt = \frac {(x-x_0)^3}{3!} f(t)+\int\limits_{x}^{x_0}\frac {(x-t)^3}{3!}f(t)dt $$| $$ \displaystyle $$
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$$ \int\limits_{x}^{x_0}\frac {(x-t)^3}{3}f'(t)dt = -\frac {(x-t)^4}{4*3!} f'(t)-\int\limits_{x}^{x_0}-\frac {(x-t)^4}{4!}f'(t)dt $$| $$ \displaystyle $$
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$$ \int\limits_{x}^{x_0}\frac {(x-t)^3}{3}f'(t)dt = \frac {(x-x_0)^4}{4!} f'(t)+\int\limits_{x}^{x_0}\frac {(x-t)^4}{4!}f(t)dt $$| $$ \displaystyle $$
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$$ \int\limits_{x}^{x_0}\frac {(x-t)^4}{4}f(t)dt $$| $$ \displaystyle $$
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<span id="(1)">
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$$ {f(x)=f(x_0)+\frac {x-x_0}{1!} f(x_0)+\frac {(x-x_0)^2}{2} f(x_0)+ \frac {(x-x_0)^3}{3!} f'(t)+\frac {(x-x_0)^4}{4!} f'(t)+\int\limits_{x}^{x_0}\frac {(x-t)^4}{4!}f'(t)dt} $$||<p style="text-align:right"> $$ \displaystyle $$
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APPLYING IMVT THEOREM TO ABOVE EQUATION <span id="(1)">
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$$ \int\limits_{x}^{x_0}\frac {(x-t)^4}{4}f^(4)(t)dt =  f^5(\xi)\frac {(x-x_0)^5}{5!} $$, $$   \xi  \in (x,x_0) $$
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APPLYING IMVT TO EQUATION

<span id="(1)">
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$$\int\limits_{x}^{x_0}\frac{(x-t)^n}{n!}f^{n+1}(t)dt$$ $$
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<span id="(1)">
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$$f^{n+1}(\xi)\int\limits_{x}^{x_0}\frac {(x-t)^n}{n!}dt $$ $$ \xi \in (x,x_0)$$ $$
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