Understanding Wallis's Formula: Debunking Misconceptions About Inequalities"

[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

 

[Ray Vickson]

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

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.

 

[Miike012]

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.

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

 

[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 $q^n \leq q^{n-1}$.

 

[Miike012]

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 $q^n \leq q^{n-1}$.

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

and is q equal to sin(theta)?

 

[Miike012]

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 $q^n \leq q^{n-1}$.

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?

 

[Dick]

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?

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.