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The space is $C[0,2 \pi]$ with the inner product $(6)$.Show that $\|\cos k t\|^{2}=\pi$ and $\|\sin k t\|^{2}=\pi$ for $k>0$

$\pi$

Calculus 3

Chapter 6

Orthogonality and Least Square

Section 8

Applications of Inner Product Spaces

Vectors

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So let's call I m n the quantity integral from 0 to 2. Pi off co sign Empty Sign Anti DT, where m and n are positive integers. And of course, our goal is to show that I am any zero. So we do these integral by parts. So we see these as the integral from zero to buy off the derivative of Sign Mt. Divided by M And of course, he's just coast on Mt. They're multiply sign and the T d t and now by parts diseases sign MTV Alabam that multiply 17 evaluated from zero to Dubai minus the integral from 0 to 2. Pi it off sign MT Divided by AM that multiplies the derivative of their function So and co sign and d d t. Now these offers Brockett zero. They were minus. I need to go from zero to buy. Let's eyesight I n divided by am And then we have sign empty co signing team. Let's rev right in the other side. Let the other way so co sign anti sign, empty DT and we recognized as decent. A girl is nothing else but Minus and Avery by a m i n m observed that I am an is equal to something I n m So we're switching the indices, but we can do the same integral. So I am in by parts by seeing the other function as a derivative. So this isn't a girl from zero to buy off course in Mt. And then sign and TV seed as the derivative off miners co sign anti divided by N dt. Now again by parts disease consigning t that multiplies minus consigning TD Balaban have a letter from zero to buy minus So the derivative of the first function, which is minus m sign MT that multiplies Minako san and T divided van duty and again the old Rockets zero and they were minus the integral from zero to buy the two minuses constantly each are other, so we have m divided by end co sign anti sign, empty and again we recognize his expression to be just minus m a divided by N I and em. So we have two different expressions for I am n Let's they're not the 1st 1 they obtain up there in right by one and these other quality down here in green as to and we're going to use them right away. So minus and divided by a m i n m is by the first equality in red I am and which is by a quality to in green equal to minus m d valuable and I n m And it is true for every n and m great recall than one in party Are we we right? I an m as by moving the Andy Very bam on the right inside m square that divides and square I and m And this expression in blue is a number which depends on them and then in any em and end are different Then the oil quantity in blue is different than one and therefore we see that i n m must be zero if n and m are different. So our last case they we need to check Is it an NMR Rico? So we need to compute i n n is of course, the integral from zero to the pie off sign anti sign anti the teeth by now. This is easy because you can just do a simple substitution. Aunty is equal to X therefore the axes and DT and the fourth is becomes integral from zero two to buy n call, Sign of X sign of X, then DT is one divided by and the ex. Okay. And I were there because you can simply compute is a well known fact. But you can compute yourself. That didn't occur from 0 to 2 pi off course I necks Cenex the X zero and the function cause an X sign X is periodic would be to buy and therefore we see that is all integral is just equal to zero.

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