This sequence of functions looks simple but

  • #1
unscientific
1,734
13

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?
 

Attachments

  • lastqn.jpg
    lastqn.jpg
    21.9 KB · Views: 383

Answers and Replies

  • #2
Sourabh N
631
0
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
unscientific
1,734
13
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
Sourabh N
631
0
Yes.
 
  • #6
Ray Vickson
Science Advisor
Homework Helper
Dearly Missed
10,706
1,722

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
 

Suggested for: This sequence of functions looks simple but

Replies
13
Views
419
Replies
9
Views
575
  • Last Post
Replies
1
Views
328
Replies
6
Views
323
Replies
5
Views
385
Replies
5
Views
730
Replies
1
Views
325
Replies
8
Views
670
  • Last Post
Replies
3
Views
304
Top