User:EGM6341.s11.TEAM1.Yoon.HW2

1st term
Derive the first term down below from

Taking integration by parts from the integrating part of the LHS,

Back to the main equation including the factorial term,

2nd term
Derive the first term down below from

Note: Integrating by parts

Taking integration by parts from the integrating part of the LHS,

Back to the main equation including the factorial term,

3rd term
Derive the first term down below from

Taking integration by parts from the integrating part of the LHS,

Back to the main equation including the factorial term,

In conclusion,

Part.2)
Consider the remainder as a function that we derived in HW1, prob3

Applying the remainder function to below equation,

Integrating the integration part of the RHS of the equation yields,


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$$\displaystyle \begin{align} \therefore \int_{x_0}^{x} (x-t)^4 \cdot f^{(5)}(t) dx = f^{(5)}(\xi) \frac{(x-x0)^5}{5} \quad _{for} \quad w(x) \geqq 0 \end{align} $$
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 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style = | (1.17)
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Part.3)
Derive the first term down below from

With same method as used in part a), use integrating by parts again,

Back to the main equation including the factorial term,

Part.4)
Consider the remainder as a function that we derived in HW1, prob3

Applying below IMVT to equation (1.22)

Integrating the integration part of the RHS of the equation yields,


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$$\displaystyle \begin{align} \therefore \frac{1}{(n+1)!}\int_{x_0}^{x} (x-t)^{n+1} \cdot f^{(n+2)}(t) dx = \frac{(x-x_0)^{n+2}}{(n+2)!}f^{(n+2)}(\xi) \quad _{for} \quad w(x) \geqq 0 \end{align} $$
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 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style = | (1.25)
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Solution
When n = 0:

When n = 1:

When n = 2:

When n = 3:

When n = 4: