User:EGM6321.f12.team7.Marjanovic/report2

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Problem 4: Show linear independence
Based on lecture notes section 9

Given: Homogeneous solutions to a Legendre differential operator
The homogeneous solutions $$\displaystyle y^1_H(x) $$ in (3) p.7-1 and $$\displaystyle y^2_H(x) $$ in (4) p.7-1 are linearly independent.

Find: Linear independence
Show that $$\displaystyle \forall \alpha \in \mathbb{R}, y^1_H(\cdot) \ne \alpha y^2_H(\cdot) $$

i.e., for any given $$\displaystyle \alpha, $$ show that $$\displaystyle \exists \hat x $$ such that $$\displaystyle y^1_H(\hat x) \ne \alpha y^2_H(\hat x) $$

Plot $$\displaystyle y^1_H(x) $$ and $$\displaystyle y^2_H(x) $$

Solution: The solutions are linearly indepenent

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This solution was prepared without referring to previous solutions. The Legendre Differential Operator is:
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When n=1, the Legendre Differential Operator becomes:

The two linearly-independent homogeneous solutions are:

Suppose that there exists an $$\displaystyle \alpha_1 $$ such that $$\displaystyle y^1_H(\hat x_1)=\alpha_1 y^2_H(\hat x_1) $$ and that there exists an $$\displaystyle \alpha_2 $$ such that $$\displaystyle y^1_H(\hat x_2)=\alpha_2 y^2_H(\hat x_2) $$ The two homogeneous solutions are linearly independent if $$\displaystyle \alpha_1 \ne \alpha_2 $$ Let $$\displaystyle \hat x_1 = 0.1 $$ Then we have: $$\displaystyle 0.1 = \alpha_1 [\frac{0.1}{2} \log \left( \frac{1+0.1}{1-0.1} \right) - 1] \rightarrow \alpha_1 = -0.1004 $$ And let $$\displaystyle \hat x_2 = 0.2 $$ Then we have: $$\displaystyle 0.2 = \alpha_2 [\frac{0.2}{2} \log \left( \frac{1+0.2}{1-0.2} \right) - 1] \rightarrow \alpha_2 = -0.2036 $$ Thus $$\displaystyle \alpha_1 \ne \alpha_2 $$ and therefore the 2 homogeneous solutions are linearly independent.