User:Egm6341.s11.team4.BM/HW6

=Problem 6.1 Find $$ P_{6} $$ and $$ P_{7} $$ in the HOTRE=

Problem Statement
Given on Lecture Slide 31-1

Given
$$ E=[P_{2}(t)g^{(1)}(t)+P_{4}(t)g^{(3)}(t)]_{-1}^{+1}-\int\limits_{-1}^{+1} P_{5}(t)g^{(5)}(t)dt $$

Objective
1. Perform steps 4ab to find $$(P_{6}, P_{7}) $$ and E

2.1 Find $$ t_{k}(x) $$

2.2 Find $$ d_{1}, d_{2} $$ and $$ d_{3} $$

Part 1.1:$$(P_{6},P_{7}) $$
We will start with the values we obtained for polynomials 4 and 5 in class

$$ P_{4}(t)=c_{1}(\frac{t^{4}}{4!})+c_{3}(\frac{t^{2}}{2!})+c_{5}$$

$$ P_{5}(t)=c_{1}(\frac{t^{5}}{5!})+c_{3}(\frac{t^{3}}{3!})+c_{5}$$


 * $$ c_{1}=-1 $$
 * $$ c_{3}=\frac{1}{6} $$
 * $$ c_{5}=\frac{-7}{360} $$

We will now perform step 4a as was done with 3a in class using integration by parts. This was performed in class on Lecture slides 30-5 and 30-6

$$ E=[P_2(t)g^{(1)}(t)+P_4(t)g^{(3)}(t)]_{-1}^{1}+ [P_6(t)g^{(5)}(t)]_{-1}^{1}-\int_{-1}^{1}P_6(t)g^(4)(t)dt$$

$$ P_6(t)=\int P_5(t)=c_1\frac{t^6}{6!}+c_3\frac{t^4}{4!}+c_5\frac{t^2}{2!}+c_7 $$

Now we must perform step 4b, once again using integration by parts.

$$ E=[P_2(t)g^{(1)}(t)+P_4(t)g^{(3)}(t)+P_{6}(t)g^{(5)}(t)]_{-1}^{1} +[P_7g^{(6)}(t)]_{-1}^{1} -int_{-1}^{1} P_6(t)g^{(5)}(t)dt$$

$$ P_7(t)=\int P_6(t)dt=c_{1}\frac{t^7}{7!}+c_3\frac{t^5}{5!}+c_5\frac{t^{3}}{3!}+c_{7}t+c_8 $$ If we now proceed such that

$$ P_5(0)=0 $$ \qquad $$ P_5(\pm 1)=0 $$

we can determine the higher powered coefficents.

The equation at t=0 becomes

$$ P_7(0)=c_{1}\frac{0^7}{7!}+c_3\frac{0^5}{5!}+c_5\frac{0^{3}}{3!}+c_{7}0+c_8 $$ $$ c_8=0 $$

and at $$ t=\pm 1 $$

$$ P_7(\pm 1)=c_{1}\frac{1^7}{7!}+c_3\frac{1^5}{5!}+c_5\frac{1^{3}}{3!}+c_{7}1+c_8 $$

and now through substitution of $$ c_8=0 $$ we obtain

$$ P_7(\pm 1)=0=c_{1}\frac{1^7}{7!}+c_3\frac{1^5}{5!}+c_5\frac{1^{3}}{3!}+c_{7}1 $$

Through substituting in the previously determined values for our other coefficients, we obtain

$$ c_7=\frac{31}{15120} $$

Now if we re-express our polynomials with our obtained coefficients.


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$$ P_6(t)=\frac{-t^6}{6!}+\frac{t^4}{144}-\frac{7t^2}{720}+\frac{31}{151210} $$


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$$ P_7(t)=\frac{-t^7}{7!}+\frac{t^5}{720}-\frac{7t^3}{2160}+\frac{31}{151210}t $$


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And our E value is


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$$E=[P_2(t)g^{(1)}(t)+P_4(t)g^{(3)}(t)+P_6(t)g^{(5)}(t)]_{-1}^{1}-\int_{-1}^{1}P_7(t)g^{(7)}(t)dt $$


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Part 2.1: Find $$ t_k(x) $$
Starting from the given equation.

$$ x(t) := \frac{x_k+x_{k+1}}{2}+t\frac{h}{2} $$

$$ t \in [-1, 1] $$

$$ t_k $$ can be found from algebraic manipulation.

$$ x(t_k) := \frac{x_k+x_{k+1}}{2}+t_k\frac{h}{2} $$

$$ t_k=\frac{2}{h}(x(t_k)-\frac{x_k+x_{k+1}}{2}) $$

$$ t_k=\frac{2}{h}(x-\frac{x_k+x_{k+1}}{2}) $$

As expected, this relation yields a value of -1 when evaluated at $$ x_k $$ and a value of 1 when evaluated at $$ x_{k+1} $$. This is due to the definition of h ($$ h:=x_{k+1}-x_k $$).

Part 2.2: Find $$ d_1,d_2, $$ and $$ d_3 $$
These can all be obtained from their definitions (which are all composed entirely of terms we've already developed.

$$ d_1=\overline{d}_{2.1}=\overline{d}_{2}=\frac{P_2(1)}{2^2}$$

$$ d_2=\overline{d}_{4}=\frac{P_4(1)}{2^4} $$

$$ d_3=\overline{d}_{6}=\frac{P_6(1)}{2^6} $$

Since we know our values of $$ P_2,P_4 $$ and $$ P_6 $$ we can evaluate these expressions

$$ P_2(1)=-\frac{1^2}{2!}+\frac{1}{6}=-\frac{1}{3} $$

$$ P_4(1)=-\frac{1^4}{4!}+\frac{1^2}{12}+\frac{-7}{360}=\frac{1}{45} $$

$$ P_6(1)=\frac{-1^6}{6!}+\frac{1^4}{144}-\frac{7*1^2}{720}+\frac{31}{151210}=-\frac{1}{945} $$

Now substituting these values into our expressions for our d values we obtain


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$$ d_1=-\frac{1}{12} $$


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$$ d_2=\frac{1}{720} $$


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$$ d_3=-\frac{1}{30240} $$


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This problem was solved by Brendan Mahon

Problem Statement
Given on Lecture slide 33-2

Given
Triangle OAB

Objective
Prove


 * $$ \overline{PQ}= \int_{\theta_{P}}^{\theta_{Q}} \left[ r^2+ \left( \frac{dr}{d\theta}\right)^2\right]^{\frac{1}{2}} d\theta$$

Solution
Using the Law of Cosines We can express the length as.

$$ \overline{AB}^{2}=\overline{OA}^{2}+\overline{OB}^{2}-2(\overline{OA})( \overline{OB})cos(d\theta) $$

Squaring both sides

$$ \overline{AB}=\sqrt{\overline{OA}^{2}+\overline{OB}^{2}-2(\overline{OA} )(\overline{OB})cos(d\theta)} $$

If we now make the following substitutions

$$ \overline{AB}=dl $$

$$ \overline{OA}=r(\theta) $$

$$ \overline{OB}=r(\theta+d\theta) $$

$$ \overline{AB}=dl $$

We obtain

$$ dl=\sqrt{(r(\theta))^{2}+(r(\theta+d\theta))^{2}-2(r(\theta))(r(\theta+d\theta))cos(d\theta)} $$

This equation can be better expressed if we regroup the first term into a quadratic. This will allow us to use some simplifying assumptions later on.

$$ dl=\sqrt{(r(\theta)+r(\theta+d\theta))^{2}+2(r(\theta))(r(\theta+d\theta))(1-cos(d\theta))} $$

We can now use the following given information to simplify the first component of this equation.

$$ r(\theta+d\theta)-r(\theta) \approx dr $$

$$ dl=\sqrt{(dr)^{2}+2(r(\theta))(r(\theta+d\theta))(1-cos(d\theta))} $$

We may now use the cosine double angle formula to simplify the second component.


 * $$ cos(2\theta)=1-2sin^{2}(\theta) $$

When applied to the second component we get

$$ (1-cos(d\theta))=1-(1-2sin^{2}(\frac{\theta}{2})=2sin^{2}(\frac{\theta}{2}) $$

If we continue to assume that the angle is small, this can be simplified to

$$ (1-cos(d\theta))\approx2(\frac{d\theta}{2})^{2} =\frac{d\theta^{2}}{2} $$

Through substitution, we now have

$$ dl \approx \sqrt{(dr)^{2}+2(r(\theta))(r(\theta+d\theta))\frac{d\theta^{2}}{2}} $$

$$ dl \approx \sqrt{dr^{2}+r(\theta)r(\theta+d\theta)d\theta^{2}} $$

$$ dl \approx d\theta \sqrt{(\frac{dr}{d\theta})^{2}+r(\theta)r(\theta+d\theta)} $$

In keeping with our small angle assumption we can say that the term $$ r(\theta)r(\theta+d\theta) $$ will be dominated by the $$r(\theta)$$ components, leading us to the following small angle approximation.

$$ r(\theta)r(\theta+d\theta) $$

Substituting this back into our equation, we obtain

$$ dl \approx d\theta \sqrt{(\frac{dr}{d\theta})^{2}+r(\theta)^{2}} $$

If we now integrate this to find the arc length $$\overline{PQ} $$ we come up with the following definite integral.


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$$ \overline{PQ} = \int_{\theta_{P}}^{\theta_{Q}} \sqrt{r^{2}+(\frac{dr}{d\theta})^{2}} d\theta$$


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Solved by Brendan Mahon