User:Egm4313.s12.team14.djp

= Problem 5.3 =

Given

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

b_{1} = 2e_{1} + 7e_{2} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 1)
 * }
 * }


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b_{2} =1.5e_{1} + 3e_{2} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 2)
 * }
 * }

Find
Show that Eq. 1 and Eq. 2 are linearly independent using the Gramian.

Solution
The Grammian is defined as


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\Gamma(b_{1},b_{2}):=\det\begin{bmatrix} \langle b_{1},b_{1}\rangle & \langle b_{1},b_{2}\rangle \\ \langle b_{2},b_{1}\rangle & \langle b_{2},b_{2}\rangle \end{bmatrix} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 3)
 * }
 * }

Where the angle brackets imply the dot product of the two equations located within them. If the two equations are linearly independent, then:


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\Gamma(b_{1},b_{2})\ne 0 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 4)
 * }
 * }

Evaluating the dot products:


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\langle b_{1},b_{1} \rangle = \left \| b_{1} \right \|^2 \cos(0)=53 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 5)
 * }
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 * {| style="width:100%" border="0" align="left"

\langle b_{2},b_{2} \rangle = \left \| b_{2} \right \|^2 \cos(0)=11.25 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 6)
 * }
 * }

To find the angle between the two vectors given, we must first find the angle formed by each individual vectors with respect to the $$e_{1}$$ and $$e_{2}$$ directions.


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\theta_{1} = \arctan(7/2) = 74.05 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 7)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

\theta_{2} = \arctan(3/1.5) = 63.43 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 8)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

\phi = \theta_{1} - \theta_{2} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 9)
 * }
 * }


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

\langle b_{1},b_{2} \rangle = \langle b_{2},b_{1} \rangle = \left \| b_{1} \right \| \left \| b_{2} \right \|\cos(\phi)=24 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 10)
 * }
 * }

Plugging these values into Equation 3, we obtain:


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\Gamma(b_{1},b_{2}) = 53(11.25) - 24^2 = 596.25 - 575 = 21.25 \ne 0 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 11)
 * }
 * }

= Problem 5.8 =

Given

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\int x^n \log(1+x)dx $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 1)
 * }
 * }

Find
Solve the integral for n=0 and n=1, using integration by parts.

Solution
For n=0:

As per the suggestion in the problem, we'll solve this problem using integration by parts:


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

\int udv = uv - \int vdu $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 2)
 * }
 * }

Selecting the u and dv:


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u = \log(1+x), du=\frac{1}{1+x} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 3)
 * }
 * }


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

dv = x^n, v=\frac{x^{n+1}}{n+1} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 4)
 * }
 * }

Plugging into Eq. 2, we obtain:


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\frac{x^{n+1}}{n+1}\log(1+x) - \int\frac{x^{n+1}}{n+1} (\frac{1}{1+x}) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 5)
 * }
 * }

But n=0, so this can be simplified to:


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x\log(1+x) - \int x (\frac{1}{1+x}) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 6)
 * }
 * }

The remaining integral can be simplified using polynomial division. The result is 1 with a remainder of -1/(1+x). Thus:


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

x\log(1+x) - \int 1 -\frac{1}{1+x} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 7)
 * }
 * }

Evaluating this final integral leaves us with a final answer of:

$$\int x^n \log(1+x)dx = x\log(1+x) - x + \log(1+x)$$