# Wallis's Formula

1. Aug 18, 2013

### Miike012

The portion highlighted in the document seems wrong.

(n-1)/n < (n-2)/(n-1)
n2 - 2n + 1 < n2 - 2n

1<0 (WRONG)

So how can I believe what the book is saying?

Now if I let n be n - 2 then I still have 1<0. Therefore the element of (1) is greater than the corresponding element of (2), and not the other way around as the book says

#### Attached Files:

• ###### Integrall.jpg
File size:
23.8 KB
Views:
96
2. Aug 18, 2013

### Ray Vickson

It might not be saying what you claim (although it is badly worded); maybe it asserts that $\sin^{n-1}(x) < \sin^{n}(x)$ for every $x \in (0, \pi/2),$ and that is certainly true.

3. Aug 18, 2013

### Miike012

But it never said that. It says sin^{n-1}(x) > \sin^{n}(x)

4. Aug 18, 2013

### vanhees71

Yes and that's correct as long as $\sin x$ is not negative. Proof: Let $0<q \leq 1$. Then you can multiply the inequality $q \leq 1$ by $q^{n-1}>0$, leading to $q^n \leq q^{n-1}$.

5. Aug 18, 2013

### Miike012

Well why doesn't it work in the math in post #1?

and is q equal to sin(theta)?

Last edited: Aug 18, 2013
6. Aug 18, 2013

### Miike012

Why would this argument not work..
If n>0 then we have n>(n-1) and sin(x)n ≥ sin(x)n-1 for sin(x)≥0

is it because for 0≤x≤pi/2, 0≤sin(x)≤1?

Last edited: Aug 18, 2013
7. Aug 18, 2013

### Dick

Well, yes. If 0<=q<=1 then q^n=q*q^(n-1)<=q^(n-1). As vanhees71 already said. E.g. (1/2)^4<=(1/2)^3.

Last edited: Aug 18, 2013