User:Egm6341.s10.team3.Min/HW3

= (1)Proof the error bound for Simple Simpson' Rule = Ref: Lecture Notes [[media:Egm6341.s10.mtg17.djvu|p.17-1]]

Problem Statement
 Part 1  Replicate the proof of tighter error bound for Simple Simpson for two cases:
 * {| style="width:100%" border="0" align="left"

a) G(t) = e(t) - t^{4}e(1) $$
 * $$ \displaystyle
 * $$ \displaystyle
 * }
 * {| style="width:100%" border="0" align="left"

b) G(t) = e(t) - t^{6}e(1) $$ Point out where proof breaks down  Part 2  $$ \displaystyle G(t) = e(t) - t^{6}e(1) $$
 * $$ \displaystyle
 * $$ \displaystyle
 * }

Find $$ \displaystyle G^{(3)}(0) $$ and follow same steps in proof and see what happens

Solution
 Part 1 

a) For case 1
 * {| style="width:100%" border="0" align="left"

G(t) = e(t) - t^{4}e(1) $$
 * $$ \displaystyle
 * $$ \displaystyle
 * }

From (3) in lecture[[media:Egm6341.s10.mtg14.pdf|p.14-2]]we have
 * {| style="width:100%" border="0" align="left"

e(t) = \int_{+t}^{-t}F(t)\, dt-\frac{t}{3}[F(-t)+4f(0)+F(t)] $$
 * $$ \displaystyle
 * $$ \displaystyle
 * }


 * {| style="width:100%" border="0" align="left"

e(0)= \int_{0}^{0}f(x)\, dx -\frac{0}{3}[F(-0)+4f(0)+F(0)]=0 $$
 * $$ \displaystyle
 * $$ \displaystyle
 * }
 * {| style="width:100%" border="0" align="left"

G(0) = e(0) - 0^{4}e(1)=0 $$
 * $$ \displaystyle
 * $$ \displaystyle
 * }


 * {| style="width:100%" border="0" align="left"

G(1) = e(1) - 1^{4}e(1)= 0 $$
 * $$ \displaystyle
 * $$ \displaystyle
 * }

According to Rolle' thm,
 * {| style="width:100%" border="0" align="left"

\exists  \zeta _1 \ni ]0,1[,  G^{(1)}(\zeta _1) = 0   $$
 * $$ \displaystyle
 * $$ \displaystyle
 * }


 * {| style="width:100%" border="0" align="left"

G^{(1)}(t) = e^{(1)}(t) - 4t^{3}e(1) $$
 * $$ \displaystyle
 * $$ \displaystyle
 * }


 * {| style="width:100%" border="0" align="left"

e^{(1)}(t)& = F(-t)+F(t)- \frac{1}{3}[F(-t)+4F(0)+F(t)]-\frac{t}{3}[-F^{(1)}(-t)+F^{(1)}(t)]\\ & = \frac{2}{3}[F(-t)+F(t)] - \frac{4}{3}F(0)-\frac{t}{3}[-F^{(1)}(-t)+F^{(1)}(t)]\\ \end{align} $$
 * $$ \displaystyle \begin{align}
 * $$ \displaystyle \begin{align}
 * }
 * }


 * {| style="width:100%" border="0" align="left"

e^{(1)}(0) &= \frac{2}{3}[F(0)+F(0)] - \frac{4}{3}F(0)-\frac{0}{3}[-F^{(1)}(-0)+F^{(1)}(0)]\\ &=\frac{4}{3}-\frac{4}{3}F(0)-0\\ &=0\\ \end{align} $$
 * $$ \displaystyle \begin{align}
 * $$ \displaystyle \begin{align}
 * }
 * }


 * {| style="width:100%" border="0" align="left"

G^{(1)}(0) = e^{(1)}(0) - 4(0^{3})e(1)= 0 $$
 * $$ \displaystyle
 * $$ \displaystyle
 * }

Recall
 * {| style="width:100%" border="0" align="left"

G^{(1)}(\zeta _1) = 0  $$ Thus According to Rolle' thm,
 * $$ \displaystyle
 * $$ \displaystyle
 * }
 * {| style="width:100%" border="0" align="left"

\exists  \zeta _2 \ni ]0,\zeta _1[,  G^{(2)}(\zeta _2) = 0   $$
 * $$ \displaystyle
 * $$ \displaystyle
 * }

Again
 * {| style="width:100%" border="0" align="left"

G^{(2)}(t) = e^{(2)}(t) - 12t^{2}e(1) $$
 * $$ \displaystyle
 * $$ \displaystyle
 * }


 * {| style="width:100%" border="0" align="left"

G^{(2)}(0) = e^{(2)}(0) - 12(0^{2})e(1) = e^{(2)}(0)  $$
 * $$ \displaystyle
 * $$ \displaystyle
 * }

Where
 * {| style="width:100%" border="0" align="left"

e^{(2)}(t) &= \frac{2}{3}[-F^{(1)}(-t)+F^{(1)}(t)] - \frac{1}{3}[-F^{(1)}(-t)+F^{(1)}(t)]-\frac{1}{3}[F^{(2)}(-t)+F^{(2)}(t)]\\ &=\frac{1}{3}[-F^{(1)}(-t)+F^{(1)}(t)]-\frac{t}{3}[F^{(2)}(-t)+F^{(2)}(t)]\\ \end{align} $$
 * $$ \displaystyle \begin{align}
 * $$ \displaystyle \begin{align}
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

e^{(2)}(0) = \frac{1}{3}[-F^{(1)}(-0)+F^{(1)}(0)]-\frac{0}{3}[F^{(2)}(-0)+F^{(2)}(0)]=0 $$ Thus
 * $$ \displaystyle
 * $$ \displaystyle
 * }
 * }
 * {| style="width:100%" border="0" align="left"

G^{(2)}(0) = e^{(2)}(0)=0   $$ Recall
 * $$ \displaystyle
 * $$ \displaystyle
 * }
 * {| style="width:100%" border="0" align="left"

G^{(2)}(\zeta _2) = 0  $$ Thus According to Rolle' thm,
 * $$ \displaystyle
 * $$ \displaystyle
 * }
 * {| style="width:100%" border="0" align="left"

\exists  \zeta _3 \ni ]0,\zeta _2[,  G^{(3)}(\zeta _3) = 0   $$
 * $$ \displaystyle
 * $$ \displaystyle
 * }

Again
 * {| style="width:100%" border="0" align="left"

G^{(3)}(t) = e^{(3)}(t) - 24te(1) $$ Where
 * $$ \displaystyle
 * $$ \displaystyle
 * }
 * {| style="width:100%" border="0" align="left"

e^{(3)}(t) = -\frac{t}{3}[F^{(3)}(t)+F^{(3)}(-t)] $$
 * $$ \displaystyle
 * $$ \displaystyle
 * }
 * }


 * {| style="width:100%" border="0" align="left"

G^{(3)}(\zeta_3) & = -\frac{\zeta_3}{3}\underbrace{[F^{(3)}(\zeta_3)+F^{(3)}(-zeta_3)]}_{use DMVT}- 24\zeta_3e(1)\\ &=-\frac{\zeta_3}{3}[2\zeta_3F^{(4)}(\zeta_4)]- 24\zeta_3e(1)\\ &=0 \end{align} $$
 * $$ \displaystyle \begin{align}
 * $$ \displaystyle \begin{align}

$$
 * $$\displaystyle (Eq. 1)
 * }
 * }

Where $$\zeta _{4} \in [-\zeta _{3}, \zeta _{3}]$$ Since $$\zeta _{3} \neq 0, $$

Solve for


 * {| style="width:100%" border="0" align="left"

$$\displaystyle e(1)= -\frac{\zeta_3}{36}F^{(4)}(\zeta_4) $$
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style = |
 * }
 * }

From the relation ship between $$\zeta_4 and \xi $$ See HW2, No 15 for details

$$ \displaystyle F(t)=f(x(t)) $$

$$ \displaystyle x(t)=x_{1}+ht $$

$$ \displaystyle F^{(4)}(t) = \frac{d\left(F(t)^{(3)}\right)}{dt} = \frac{d\left(h^{3}f(x)^{(3)}\right)}{dx}\frac{dx}{dt} = h^{3}f^{(4)}(x) \times h = h^{4}f^{(4)}(x) $$

Let

$$ \displaystyle \xi = x_{1}+h \zeta_{4} $$


 * {| style="width:100%" border="0" align="left"

$$\displaystyle e(1)= -\frac{\zeta_3}{36}h^{4}f^{(4)}(\xi )=-\frac{\zeta_3(b-a)^{4}}{576}f^{(4)}(\xi )
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style="width:10%; padding:10px; border:2px solid #8888aa" |

$$ We know the Error for Simple Simpson's rule from lecture[[media:Egm6341.s10.mtg14.pdf|p.14-2]] $$ \displaystyle
 * style = |
 * }
 * }

E_{2} = I - I_2 = \frac{b-a}{2}e(1) = h e(1) $$

Thus the error is
 * {| style="width:100%" border="0" align="left"

$$\displaystyle E_{2} = h e(1)=-\frac{\zeta_3(b-a)^{5}}{1152}f^{(4)}(\xi )
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style="width:10%; padding:10px; border:2px solid #8888aa" |

$$
 * style = |
 * }
 * }

Sum: the proof breaks though at Eq 1, where we can not cancel $${\zeta_3}^2$$, because the third derivative of G(t) at $$\zeta_3$$ gives $$\zeta_3 e(1)$$ instead of $${\zeta_3}^2 e(1)$$

b) For case 2
 * {| style="width:100%" border="0" align="left"

G(t) = e(t) - t^{6}e(1) $$ The proof procedure is similar to case 1 <\br> From (3) in lecture[[media:Egm6341.s10.mtg14.pdf|p.14-2]]we have
 * $$ \displaystyle
 * $$ \displaystyle
 * }
 * {| style="width:100%" border="0" align="left"

e(t) = \int_{+t}^{-t}F(t)\, dt-\frac{t}{3}[F(-t)+4f(0)+F(t)] $$
 * $$ \displaystyle
 * $$ \displaystyle
 * }


 * {| style="width:100%" border="0" align="left"

e(0)= \int_{0}^{0}f(x)\, dx -\frac{0}{3}[F(-0)+4f(0)+F(0)]=0 $$
 * $$ \displaystyle
 * $$ \displaystyle
 * }
 * {| style="width:100%" border="0" align="left"

G(0) = e(0) - 0^{6}e(1)=0 $$
 * $$ \displaystyle
 * $$ \displaystyle
 * }


 * {| style="width:100%" border="0" align="left"

G(1) = e(1) - 1^{6}e(1)= 0 $$
 * $$ \displaystyle
 * $$ \displaystyle
 * }

According to Rolle' thm,
 * {| style="width:100%" border="0" align="left"

\exists  \zeta _1 \ni ]0,1[,  G^{(1)}(\zeta _1) = 0   $$
 * $$ \displaystyle
 * $$ \displaystyle
 * }


 * {| style="width:100%" border="0" align="left"

G^{(1)}(t) = e^{(1)}(t) - 6t^{5}e(1) $$
 * $$ \displaystyle
 * $$ \displaystyle
 * }


 * {| style="width:100%" border="0" align="left"

e^{(1)}(t)& = F(-t)+F(t)- \frac{1}{3}[F(-t)+4F(0)+F(t)]-\frac{t}{3}[-F^{(1)}(-t)+F^{(1)}(t)]\\ & = \frac{2}{3}[F(-t)+F(t)] - \frac{4}{3}F(0)-\frac{t}{3}[-F^{(1)}(-t)+F^{(1)}(t)]\\ \end{align} $$
 * $$ \displaystyle \begin{align}
 * $$ \displaystyle \begin{align}
 * }
 * }


 * {| style="width:100%" border="0" align="left"

e^{(1)}(0) &= \frac{2}{3}[F(0)+F(0)] - \frac{4}{3}F(0)-\frac{0}{3}[-F^{(1)}(-0)+F^{(1)}(0)]\\ &=\frac{4}{3}-\frac{4}{3}F(0)-0\\ &=0\\ \end{align} $$
 * $$ \displaystyle \begin{align}
 * $$ \displaystyle \begin{align}
 * }
 * }


 * {| style="width:100%" border="0" align="left"

G^{(1)}(0) = e^{(1)}(0) - 6(0^{5})e(1)= 0 $$
 * $$ \displaystyle
 * $$ \displaystyle
 * }

Recall
 * {| style="width:100%" border="0" align="left"

G^{(1)}(\zeta _1) = 0  $$ Thus According to Rolle' thm,
 * $$ \displaystyle
 * $$ \displaystyle
 * }
 * {| style="width:100%" border="0" align="left"

\exists  \zeta _2 \ni ]0,\zeta _1[,  G^{(2)}(\zeta _2) = 0   $$
 * $$ \displaystyle
 * $$ \displaystyle
 * }

Again
 * {| style="width:100%" border="0" align="left"

G^{(2)}(t) = e^{(2)}(t) - 30t^{3}e(1) $$
 * $$ \displaystyle
 * $$ \displaystyle
 * }


 * {| style="width:100%" border="0" align="left"

G^{(2)}(0) = e^{(2)}(0) - 30(0^{4})e(1) = e^{(2)}(0)  $$
 * $$ \displaystyle
 * $$ \displaystyle
 * }

Where
 * {| style="width:100%" border="0" align="left"

e^{(2)}(t) &= \frac{2}{3}[-F^{(1)}(-t)+F^{(1)}(t)] - \frac{1}{3}[-F^{(1)}(-t)+F^{(1)}(t)]-\frac{1}{3}[F^{(2)}(-t)+F^{(2)}(t)]\\ &=\frac{1}{3}[-F^{(1)}(-t)+F^{(1)}(t)]-\frac{t}{3}[F^{(2)}(-t)+F^{(2)}(t)]\\ \end{align} $$
 * $$ \displaystyle \begin{align}
 * $$ \displaystyle \begin{align}
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

e^{(2)}(0) = \frac{1}{3}[-F^{(1)}(-0)+F^{(1)}(0)]-\frac{0}{3}[F^{(2)}(-0)+F^{(2)}(0)]=0 $$ Thus
 * $$ \displaystyle
 * $$ \displaystyle
 * }
 * }
 * {| style="width:100%" border="0" align="left"

G^{(2)}(0) = e^{(2)}(0)=0   $$ Recall
 * $$ \displaystyle
 * $$ \displaystyle
 * }
 * {| style="width:100%" border="0" align="left"

G^{(2)}(\zeta _2) = 0  $$ Thus According to Rolle' thm,
 * $$ \displaystyle
 * $$ \displaystyle
 * }
 * {| style="width:100%" border="0" align="left"

\exists  \zeta _3 \ni ]0,\zeta _2[,  G^{(3)}(\zeta _3) = 0   $$
 * $$ \displaystyle
 * $$ \displaystyle
 * }

Again
 * {| style="width:100%" border="0" align="left"

G^{(3)}(t) = e^{(3)}(t) - 120t^{3}e(1) $$ Where
 * $$ \displaystyle
 * $$ \displaystyle
 * }
 * {| style="width:100%" border="0" align="left"

e^{(3)}(t) = -\frac{t}{3}[F^{(3)}(t)+F^{(3)}(-t)] $$
 * $$ \displaystyle
 * $$ \displaystyle
 * }
 * }


 * {| style="width:100%" border="0" align="left"

G^{(3)}(\zeta_3) & = -\frac{\zeta_3}{3}\underbrace{[F^{(3)}(\zeta_3)+F^{(3)}(-\zeta_3)]}_{use DMVT}- 120{\zeta_3}^{3}e(1)\\ &=-\frac{\zeta_3}{3}[2\zeta_3F^{(4)}(\zeta_4)]- 120{\zeta_3}^{3}e(1)\\ &=0 \end{align} $$
 * $$ \displaystyle \begin{align}
 * $$ \displaystyle \begin{align}

$$
 * $$\displaystyle (Eq. 2)
 * }
 * }

Where $$\zeta _{4} \in [-\zeta _{3}, \zeta _{3}]$$ Since $$\zeta _{3} \neq 0, $$

Solve for


 * {| style="width:100%" border="0" align="left"

$$\displaystyle e(1)= -\frac{1}{180\zeta_3}F^{(4)}(\zeta_4) $$
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style = |
 * }
 * }

From the relation ship between $$\zeta_4 and \xi $$ See HW2, No 15 for details

$$ \displaystyle F(t)=f(x(t)) $$

$$ \displaystyle x(t)=x_{1}+ht $$

$$ \displaystyle F^{(4)}(t) = \frac{d\left(F(t)^{(3)}\right)}{dt} = \frac{d\left(h^{3}f(x)^{(3)}\right)}{dx}\frac{dx}{dt} = h^{3}f^{(4)}(x) \times h = h^{4}f^{(4)}(x) $$

Let

$$ \displaystyle \xi = x_{1}+h \zeta_{4} $$


 * {| style="width:100%" border="0" align="left"

$$\displaystyle e(1)= -\frac{1}{180\zeta_3}h^4F^{(4)}(\xi)= -\frac{(b-a)^4}{2880\zeta_3}F^{(4)}(\xi)
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style="width:10%; padding:10px; border:2px solid #8888aa" |

$$
 * style = |
 * }
 * }

We know the Error for Simple Simpson's rule from lecture[[media:Egm6341.s10.mtg14.pdf|p.14-2]] $$ \displaystyle

E_{2} = I - I_2 = \frac{b-a}{2}e(1) = h e(1) $$

Thus the error is
 * {| style="width:100%" border="0" align="left"

$$\displaystyle E_{2} = h e(1)= -\frac{(b-a)^4}{2880\zeta_3}hF^{(4)}(\xi)=-\frac{(b-a)^5}{5760\zeta_3}F^{(4)}(\xi)
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style="width:10%; padding:10px; border:2px solid #8888aa" |

$$
 * style = |
 * }
 * }

Sum: the proof breaks though at Eq 2, where we can not cancel $${\zeta_3}^2$$, but the third derivative of G(t) at $$\zeta_3$$ gives $${\zeta_3}^2 e(1)$$ instead of $${\zeta_3}^2 e(1)$$

 Part 2  Continue the proof from lecture [[media:Egm6341.s10.mtg15.djvu|Lecture p.15-2]] where we have
 * {| style="width:100%" border="0" align="left"

G^{(3)}(t) = e^{(3)}(t) - 60t^{2}e(1) $$
 * $$ \displaystyle
 * $$ \displaystyle

$$
 * $$\displaystyle (Eq. 3)
 * }
 * }


 * {| style="width:100%" border="0" align="left"

G^{(3)}(\zeta _3)=0 $$
 * $$ \displaystyle
 * $$ \displaystyle
 * }
 * {| style="width:100%" border="0" align="left"

e^{(3)}= \frac{t}{3}[F^{(3)}(t)-F^{(3)}(-t)] $$ $$
 * $$ \displaystyle
 * $$ \displaystyle
 * $$\displaystyle (Eq. 4)
 * }
 * }

From Eq 3. and Eq 4, we have


 * {| style="width:100%" border="0" align="left"

G^{(3)}(t) = \frac{t}{3}[F^{(3)}(t)-F^{(3)}(-t)] - 60t^{2}e(1) $$ Thus
 * $$ \displaystyle
 * $$ \displaystyle
 * }
 * {| style="width:100%" border="0" align="left"

$$\displaystyle G^{(3)}(0) = \frac{0}{3}[F^{(3)}(0)-F^{(3)}(-0)] - 60(0)^{2}e(1) = 0 $$
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style = |
 * }
 * }

Again, according to Rolle's thm,
 * {| style="width:100%" border="0" align="left"

\exists  \zeta _4 \ni ]0,\zeta _3[,  G^{(4)}(\zeta _4) = 0   $$
 * $$ \displaystyle
 * $$ \displaystyle
 * }

Next is to find the 4th derivative of G(t), we calculate the 4th derivative of e(t) first
 * {| style="width:100%" border="0" align="left"

e^{(4)}= -\frac{1}{3}[F^{(3)}(t)-F^{(3)}(-t)]-\frac{t}{3}[F^{(4)}(t)+F^{(4)}(-t)] $$
 * $$ \displaystyle
 * $$ \displaystyle
 * }

The 4th derivative of G(t) is given as:
 * {| style="width:100%" border="0" align="left"

G^{(4)}(t) & = e^{(4)}(t) - 120te(1) \\ & = -\frac{1}{3}[F^{(3)}(t)-F^{(3)}(-t)]-\frac{t}{3}[F^{(4)}(t)+F^{(4)}(-t)] - 120te(1)\\ \end{align} $$
 * $$ \displaystyle \begin{align}
 * $$ \displaystyle \begin{align}
 * }
 * }


 * {| style="width:100%" border="0" align="left"

G^{(4)}(\zeta _{4}) &= -\frac{1}{3}\underbrace{[F^{(3)}(\zeta _{4})-F^{(3)}(-\zeta _{4})]}_{use DMVT}-\frac{\zeta _{4}}{3}[F^{(4)}(\xi_4)+F^{(4)}(-\xi_4)] - 120\xi_4e(1)\\ &=-\frac{1}{3}[2\zeta _{4}F^{(4)}(\xi_5)]-\frac{\zeta _{4}}{3}[F^{(4)}(\xi_4)+F^{(4)}(-\xi_4)]-120\zeta _{4}e(1)\\ &=0 \end{align}$$ Where $$\zeta _{5} \in [-\zeta _{4}, \zeta _{4}]$$ Since $$\zeta _{4} \neq 0, $$
 * $$ \displaystyle \begin{align}
 * $$ \displaystyle \begin{align}
 * }
 * }

Solve for


 * {| style="width:100%" border="0" align="left"

$$\displaystyle e(1)= -\frac{1}{360}[2F^{(4)}(\zeta _{5})+F^{(4)}(\zeta _{4})+F^{(4)}(-\zeta _{4})] $$
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style = |
 * }
 * }

Since

$$ \displaystyle F(t)=f(x(t)) $$

$$ \displaystyle x(t)=x_{1}+ht $$

$$ \displaystyle F^{(4)}(t) = \frac{d\left(F(t)^{(3)}\right)}{dt} = \frac{d\left(h^{3}f(x)^{(3)}\right)}{dx}\frac{dx}{dt} = h^{3}f^{(4)}(x) \times h = h^{4}f^{(4)}(x) $$

See HW2, No 15 for details 

Let

$$ \displaystyle \xi_1 = x_{1}+h \zeta_{4} $$

$$ \displaystyle \xi_2 = x_{1}+h \zeta_{5} $$


 * {| style="width:100%" border="0" align="left"

$$\displaystyle e(1)= -\frac{1}{360}[2h^{4}f^{(4)}(\xi_2 )+h^{4}f^{(4)}(\xi_1 )+h^{4}f^{(4)}(-\xi_1 )] =-\frac{(b-a)^4}{5760}[2f^{(4)}(\xi_2 )+f^{(4)}(\xi_1 )+f^{(4)}(-\xi_1 )] $$
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style = |
 * }
 * }

We know the Error for Simple Simpson's rule from lecture[[media:Egm6341.s10.mtg14.pdf|p.14-2]] $$ \displaystyle

E_{2} = I - I_2 = \frac{b-a}{2}e(1) = h e(1) $$

Thus the error is
 * {| style="width:100%" border="0" align="left"

$$\displaystyle E_{2} = h e(1)= -\frac{(b-a)^5}{11520}[2f^{(4)}(\xi_2 )+f^{(4)}(\xi_1 )+f^{(4)}(-\xi_1 )]
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style="width:10%; padding:10px; border:2px solid #8888aa" |

$$
 * style = |
 * }
 * }

Sum: Simple simpson is exact for poly deg less than 3. Because $$f ^4(t) = 0$$, thus $$ E_{2} = 0 $$

-- By Min Zhong 12:00, 17 February 2010 (UTC)