- #1
Townsend
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Ok...here is some back ground into my new found situation. I have done very well in every math class up to this point in time so I felt it was time for me to start looking at taking some more difficult classes. That being said I am technically still in my freshman year in college so I may have bitten off more than I can chew by taking a senior level class on inequalities. This is a special class in which we will be studying out the book called, "The Cauchy-Schwarz Master Class." There are all of 8 people in class who are all seniors and have all at least taken linear algebra except of course me, where the two highest math classes I have taken so far are Calculus 2 and Elementary set and logic Theory. Today was our first day of class and after discussing the syllabus and other general formalities we proceeded with a proof. I did my best to follow everything in class but I felt a little bit lost and so I am going to attempt to work through the proof here and I will ask questions as we go.
The inequality in question is
[tex]
a_1b_1+a_2b_2+\cdots+a_nb_n \leq \sqrt{a_1^2+a_2^2+\cdots+a_n^2} \sqrt{b_1^2+b_2^2+\cdots+b_n^2}
[/tex]
While it is easy to show that this is true for n=1
[tex]
\vert a_1b_1\vert\leq\sqrt{a_1^2b_1^2}
[/tex]
It is a bit harder with n=2.
Here we proceed in an algebraic manner until we get somewhere we know it is true.
[tex]
\vert a_1b_1+a_2b_2\vert\leq\sqrt{a_1^2+a_2^2}\sqrt{b_1^2+b_2^2}
[/tex]
Squaring both sides of the inequality is legal so long as both sides of the inequality are positive. So after squaring we have
[tex]
(a_1b_1)^2+2a_1b_1a_2b_2+(a_2b_2)^2 \leq\ (a_1b_1)^2+(a_1b_2)^2+(a_2b_2)^2+(a_2b_2)^2
[/tex]
So far I am good with everything that is being done. The inequality is then simplified and we end up with
[tex]
2a_1b_1a_2b_2\leq\ (a_1b_2)^2+(a_2b_1)^2
[/tex]
moving the LHS to the RHS and then reversing the entire inequality we ended up with this
[tex]
(a_1b_2)^2-2a_1b_1a_2b_2+(a_2b_1)^2\geq\ 0
[/tex]
Which factors into a perfect square which is always either positive or equal to zero and so this last inequality is always true and so we now have that Cauchy's inequality is true of n=2.
This is the basis step in our induction proof.
Our induction hypothesis is that the Cauchy inequality is true for some n so we have
[tex]
a_1b_1+a_2b_2+\cdots+a_nb_n \leq \sqrt{a_1^2+a_2^2+\cdots+a_n^2} \sqrt{b_1^2+b_2^2+\cdots+b_n^2}
[/tex]
we must show that it is true of n+1, namely we must show that
[tex]
a_1b_1+a_2b_2+\cdots+a_nb_n+a_{n+1}b_{n+1} \leq \sqrt{a_1^2+a_2^2+\cdots+a_n^2+a_{n+1}^2} \sqrt{b_1^2+b_2^2+\cdots+b_n^2+b_{n+1}^2}
[/tex]
Before we go any farther I would like it if perhaps some one could help me out a bit on induction proofs. I have no problem understanding why an induction proof works when we show it true for n=1 and then show that if it is true for some n it is also true for n+1 and then conclude that it is therefore true for all natural numbers. Where I don't really follow the logic is when our basis step is a number other than 1, like in this case where n=2.
Secondly, is it true that induction works up and down? In other words when my basis step shows something true for n=1 and I show that if it is true for n then it is true for n+1, does this mean that it is true for all integers? Or only integers greater than 1.
To keep the over all post size down I will continue this on another post
To be continued...
The inequality in question is
[tex]
a_1b_1+a_2b_2+\cdots+a_nb_n \leq \sqrt{a_1^2+a_2^2+\cdots+a_n^2} \sqrt{b_1^2+b_2^2+\cdots+b_n^2}
[/tex]
While it is easy to show that this is true for n=1
[tex]
\vert a_1b_1\vert\leq\sqrt{a_1^2b_1^2}
[/tex]
It is a bit harder with n=2.
Here we proceed in an algebraic manner until we get somewhere we know it is true.
[tex]
\vert a_1b_1+a_2b_2\vert\leq\sqrt{a_1^2+a_2^2}\sqrt{b_1^2+b_2^2}
[/tex]
Squaring both sides of the inequality is legal so long as both sides of the inequality are positive. So after squaring we have
[tex]
(a_1b_1)^2+2a_1b_1a_2b_2+(a_2b_2)^2 \leq\ (a_1b_1)^2+(a_1b_2)^2+(a_2b_2)^2+(a_2b_2)^2
[/tex]
So far I am good with everything that is being done. The inequality is then simplified and we end up with
[tex]
2a_1b_1a_2b_2\leq\ (a_1b_2)^2+(a_2b_1)^2
[/tex]
moving the LHS to the RHS and then reversing the entire inequality we ended up with this
[tex]
(a_1b_2)^2-2a_1b_1a_2b_2+(a_2b_1)^2\geq\ 0
[/tex]
Which factors into a perfect square which is always either positive or equal to zero and so this last inequality is always true and so we now have that Cauchy's inequality is true of n=2.
This is the basis step in our induction proof.
Our induction hypothesis is that the Cauchy inequality is true for some n so we have
[tex]
a_1b_1+a_2b_2+\cdots+a_nb_n \leq \sqrt{a_1^2+a_2^2+\cdots+a_n^2} \sqrt{b_1^2+b_2^2+\cdots+b_n^2}
[/tex]
we must show that it is true of n+1, namely we must show that
[tex]
a_1b_1+a_2b_2+\cdots+a_nb_n+a_{n+1}b_{n+1} \leq \sqrt{a_1^2+a_2^2+\cdots+a_n^2+a_{n+1}^2} \sqrt{b_1^2+b_2^2+\cdots+b_n^2+b_{n+1}^2}
[/tex]
Before we go any farther I would like it if perhaps some one could help me out a bit on induction proofs. I have no problem understanding why an induction proof works when we show it true for n=1 and then show that if it is true for some n it is also true for n+1 and then conclude that it is therefore true for all natural numbers. Where I don't really follow the logic is when our basis step is a number other than 1, like in this case where n=2.
Secondly, is it true that induction works up and down? In other words when my basis step shows something true for n=1 and I show that if it is true for n then it is true for n+1, does this mean that it is true for all integers? Or only integers greater than 1.
To keep the over all post size down I will continue this on another post
To be continued...
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