Prove that v is an eigenvector of operator B

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SUMMARY

The discussion centers on proving that the polynomial \( T_{4}(x) = 8x^{4} - 8x^{2} + 1 \) is an eigenvector of the linear operator \( B = (1-x^{2}) \frac{d^2}{dx^2} - x\frac{d}{dx} \). Participants detail the steps to compute \( B T_{4}(x) \) and derive the corresponding eigenvalue \( \lambda_{4} \). The final calculation leads to the conclusion that \( \lambda_{4} = -16 \), confirming that \( T_{4}(x) \) is indeed an eigenvector of operator \( B \).

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Smazmbazm
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Homework Statement



Let B be the linear operator

[itex](1-x^{2}) \frac{d^2}{dx^2}-x\frac{d}{dx}[/itex]

Show that

[itex]T_{4}(x) = 8x^{4} - 8x^{2} + 1[/itex]

is an eigenvector of B, and find the corresponding eigenvalue.

Attempt

Righto, I find these rather difficult so a step by step solution would be nice but I know that's a lot of typing and formatting so I'm grateful for any corrections.

Step 1: [itex]Bv_{4} = \lambda _{4} v_{4}[/itex]

Step 2: [itex]((1-x^{2}) \frac{d^2}{dx^2}-x\frac{d}{dx})(8x^{4} - 8x^{2} + 1) = \lambda _{4} (8x^{4} - 8x^{2} + 1)[/itex]

Step 3: This is the first point where I'm unsure of what comes next. Should I be doing something like this

[itex]\frac{d^2}{dx^2}((1-x^{2})(8x^{4} - 8x^{2} + 1)) - x\frac{d}{dx}(8x^{4} - 8x^{2} + 1) = 8\lambda _{4}x^{4} - 8\lambda _{4}x^{2} + \lambda _{4}<br /> <br /> \\ \frac{d^2}{dx^2}(-8 x^6 +16 x^4 -9 x^2 + 1) - \frac{d}{dx}(8x^{5} - 8x^{3} + x) = 8\lambda _{4}x^{4} - 8\lambda _{4}x^{2} + \lambda _{4}<br /> <br /> \\ -240x^4 + 192x^2 - 18 - 40x^4 + 18x^2 - 1 = 8\lambda _{4}x^{4} - 8\lambda _{4}x^{2} + \lambda _{4}<br /> <br /> \\ -280x^4 + 216x^2 - 19 = 8\lambda _{4}x^{4} - 8\lambda _{4}x^{2} + \lambda _{4}[/itex]

Is this all correct so far? If so, awesome. Now I don't really know how to go about solving for [itex]\lambda[/itex]

In general, is it enough to say that [itex]\lambda = \frac{-280x^4 + 216x^2 - 19}{8x^{4}- 8x^{2} + 1}[/itex]
 
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Smazmbazm said:
Step 3: This is the first point where I'm unsure of what comes next. Should I be doing something like this

[itex]= 8\lambda _{4}x^{4} - 8\lambda _{4}x^{2} + \lambda _{4}[/itex]
[itex] <br /> You can do that, but you don't need the right hand side. It's probably easier to start by calculating <br /> $$B v = \frac{d^2}{dx^2}((1-x^{2})(8x^{4} - 8x^{2} + 1)) - x\frac{d}{dx}(8x^{4} - 8x^{2} + 1)$$<br /> and after simplifying you should get a multiple of ##8x^{4} - 8x^{2} + 1## from which you can read off the value of ##\lambda##.<br /> <br /> <blockquote data-attributes="" data-quote="" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> $$ \frac{d^2}{dx^2}(-8 x^6 +16 x^4 -9 x^2 + 1) - \frac{d}{dx}(8x^{5} - 8x^{3} + x) $$ </div> </div> </blockquote><b>No!</b> You cannot move the x after the derivative operator:<br /> $$x \frac{d}{dx} x^n \neq \frac{d}{dx} x^{n+1}$$[/itex]
 
Ok great, thanks for the response. By that logic, wouldn't it imply that you can't move the [itex](1-x^2)[/itex] to the right side either? So you just take the double derivative of the eigenvector then multiply by [itex](1-x^2)[/itex] minus the first derivative of the eigenvector multiplied by x then solve for lambda. If I follow that process, I get a lambda of -16 =] so I guess that's the way to do it
 

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