Function proof

[Trumbone]

Would anyone have any ideas for the follow problem?

F(n) = [1-(1/2[sup:3b0b7agn]2[/sup:3b0b7agn])]*[1-(1/3[sup:3b0b7agn]2[/sup:3b0b7agn])]*...*[1-(1/n[sup:3b0b7agn]2[/sup:3b0b7agn])] where f: N ? {1}?Q

I need to use some small values of n to come up with a conjecture of F(n) and the prove it using induction. Would I start with n=2, n=3, n=4 etc?

 

[pka]

Would anyone have any ideas for the follow problem?

F(n) = [1-(1/2[supfc2svzv]2[/supfc2svzv])]*[1-(1/3[supfc2svzv]2[/supfc2svzv])]*...*[1-(1/n[supfc2svzv]2[/supfc2svzv])] where f: N ? {1}?Q?

Your function is \(\displaystyle F(n) = \prod\limits_{k = 2}^n {\left( {1 - \frac{1}{{k^2 }}} \right)}\).

That can be written as \(\displaystyle F(n) = F(n-1)\left[ {\frac{{n^2 - 1}}{{n^2 }}} \right]\)

I calculated the first ten values of the function, leaving in fraction form.
There appears to be a pattern alternating between even and odd values of n.

 

[Trumbone]

Thank you for responding.

I am a bit confused as to how the two functions are the same though as the orginal function you multiply all of the functions before it, where as the second one you gave you only use the single function preceeding the one you want.

i.e in the original function F(4) = F(2)*F(3)*F(4)

but the second function is F(4) = F(3) [(n[sup:2kk1i7bu]2[/sup:2kk1i7bu]-1)/n[sup:2kk1i7bu]2[/sup:2kk1i7bu]]

These two functions wouldn't give the same values.

 

[daon]

All pka did was take out the (n)th factor. In your example you forgot to sub in 4 for n, so it would be F(4) = F(3)[(4^2-1)/4^2].

In general, you should know there are different ways to represent the same thing. Math is often just that--finding simpler ways to express what you already know.

For a well-known example: If \(\displaystyle f(n) = 1^3+2^3+3^3+...+n^3\) Then \(\displaystyle f(n) = \frac{n^2(n+1)^2}{4}\)

That means to find the sum of the cubes of the first 300 numbers. We can just divide (300*301)^2 by 4.

 

[Trumbone]

That makes mroe sense no thank you.

Once I calculated some values I conjectured that f(n)=(n+1)/2n

So my basis is n=2, ([1-(1/2^2)] = 3/4 = (2+1)/2*2

Hypothesis is n=k, then [1-(1/2[sup:sc5h9saz]2[/sup:sc5h9saz])]*[1-(1/3[sup:sc5h9saz]2[/sup:sc5h9saz])]*...*[1-(1/k[sup:sc5h9saz]2[/sup:sc5h9saz])] = (k+1)/2k

Induction Step: n=k+1, [1-(1/2[sup:sc5h9saz]2[/sup:sc5h9saz])]*[1-(1/3[sup:sc5h9saz]2[/sup:sc5h9saz])]*...*[1-(1/k[sup:sc5h9saz]2[/sup:sc5h9saz])]*[1-(1/(k+1)[sup:sc5h9saz]2[/sup:sc5h9saz])] = [(k+1)+1]/2(k+1)

So the right side becomes k+2/(2k+2)

And the left side becomes k+1/2k*[1-(1/(k+1)[sup:sc5h9saz]2[/sup:sc5h9saz])]

I tried simplifying the left side but I am definitely not getting anything similar to the right side. Any advice?

 

[Trumbone]

Any takers?

 

[daon]

Assuming this:

\(\displaystyle \prod_{k=2}^{n} \left [1-\frac{1}{k^2} \right] = \frac{n+1}{2n}\)

Multiply both sides by \(\displaystyle [1-\frac{1}{(n+1)^2}]\):

\(\displaystyle \prod_{k=2}^{n+1} \left[1-\frac{1}{k^2} \right] = \frac{n+1}{2n} \cdot \left[1-\frac{1}{(n+1)^2} \right] = \frac{n+1}{2n} \cdot \left[\frac{n^2+2n+1-1}{(n+1)^2} \right] = \frac{n(n+2)}{2n(n+1)} = \frac{(n+1)+1}{2(n+1)}\)

 

[Trumbone]

Thank you! I guess I was struggling with cancelling things out.