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# Induction Formula watch

1. I(n) = integral of (sin(x))^n between 0 and pie/2

I get:

I(n) = (n-1)(I(n-2)-I(n))

But, the correct answer is supposed to be:

(n+2)I(n+2)=(n+1)I(n)

What has gone wrong?
2. (Original post by natashabedford)
I(n) = integral of (sin(x))^n between 0 and pie/2

I get:

I(n) = (n-1)(I(n-2)-I(n))

But, the correct answer is supposed to be:

(n+2)I(n+2)=(n+1)I(n)

What has gone wrong?
You're not wrong as such. If you replace "n" with "n+2" in your working and re-arrange, you get the answer they give.

What does the question ask exactly?
3. The question says; 'Given that I(n) = integral of (sin(x))^n between 0 and pie/2, where n is a non-negative integer, show that;

(n+2)I(n+2)=(n+1)I(n)

Hence find the exact values of I(4) and I(5)
But, why should I replace n with n+2? What was wrong with my original attempt? Thanks
4. (Original post by ghostwalker)
You're not wrong as such. If you replace "n" with "n+2" in your working and re-arrange, you get the answer they give.

What does the question ask exactly?
Actually, if you replace n with n+2 you don't quite get the answer they want, i don't think

you get (n+2)I(n) = (n+1)I(n-2)
5. (Original post by natashabedford)
I(n) = integral of (sin(x))^n between 0 and pie/2

I get:

I(n) = (n-1)(I(n-2)-I(n))
So,

I(n) = (n-1)I(n-2)- (n-1)I(n)

Re-arranging.

nI(n)=(n-1)I(n-2)

And replacing n with n+2, we get:

(n+2) I(n+2) = (n+1)I(n)

There was nothing wrong with your original attempt as far as it went. But since they have asked you to show something specific, anything else won't do. In addition it's not true, as you had it, for all non-negative integers.
6. (Original post by ghostwalker)
So,

I(n) = (n-1)I(n-2)- (n-1)I(n)

Re-arranging.

nI(n)=(n-1)I(n-2)

And replacing n with n+2, we get:

(n+2) I(n+2) = (n+1)I(n)

There was nothing wrong with your original attempt as far as it went. But since they have asked you to show something specific, anything else won't do. In addition it's not true, as you had it, for all non-negative integers.
If you replace n with n+2, you dont get that...

nI(n)=(n-1)I(n-2)

, where n = n+2

(n+2)I(n) = (n+1)I(n-2)

7. (Original post by natashabedford)
If you replace n with n+2, you dont get that...

nI(n)=(n-1)I(n-2)

Try replacing all occurrences of "n" with "M+2" and see what you get. There should be no "n" left in the equation when you've done - no, not one.
8. (Original post by ghostwalker)

Try replacing all occurrences of "n" with "M+2" and see what you get. There should be no "n" left in the equation when you've done - no, not one.
ah i see, sorry!

What is the justification for replacing n with n+2 though?
9. (Original post by natashabedford)
ah i see, sorry!

What is the justification for replacing n with n+2 though?
Your formula was valid for all integer n>=2. n is just a representation for the value of the integer, so you can replace it with any other representation.

There is a subtle difference between the two representations. Your original representation was only valid for n >= 2 (I think, not having worked through the calculation), whereas the revised formulation is valid for n >= 0. If this was a graph, we have effectedly shifted the y-axis 2 units to the left (I think that's the correct direction).

Edit: Correct inequalities.

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