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This sequence of functions looks simple but

  • #1
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Homework Statement




The question is attached in the picture.

The Attempt at a Solution



Since V0 = 1,

Thus V1 = [itex]\stackrel{1}{2}[/itex]∏R2 which is a constant.

Then shouldn't Vn be as in the picture?
 

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Answers and Replies

  • #2
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I think there is a typo in the question.

[itex](\sqrt{R^2 - x^2})[/itex] should be [itex](\sqrt{R^2 - x^2})^n[/itex]
 
  • #3
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I think there is a typo in the question.

[itex](\sqrt{R^2 - x^2})[/itex] should be [itex](\sqrt{R^2 - x^2})^n[/itex]
hmm are u sure about that?
 
  • #4
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Yes.
 
  • #5
Bacle2
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  • #6
Ray Vickson
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Homework Statement




The question is attached in the picture.

The Attempt at a Solution



Since V0 = 1,

Thus V1 = [itex]\stackrel{1}{2}[/itex]∏R2 which is a constant.

Then shouldn't Vn be as in the picture?
[tex] V_1(R) =
\int_{-R}^R V_0 \left(\sqrt{R^2-x^2} \right) \, dx = \int_{-R}^R \; 1 \, dx = 2R, [/tex]
etc.

RGV
 

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