Question

Help, what does this even mean?

Answer 1

\[f(x)=\sum_{N}^{n=1}a_n \cos(nx)\], find \[\int\limits_{0}^{\pi}f(x)\cos(mx)dx\] where m,n are integers.

Answer 2

The bounds on your summation are upside down :)
No big deal though, thinking

Answer 3

whoops hehe. thanks.

Answer 4

\[\large\rm \int\limits_0^{\pi}\color{orangered}{f(x)}\cos(mx)dx\quad=\quad\int\limits_0^{\pi}\color{orangered}{\sum_{n=1}^{N}\cos(nx)}\cos(mx)dx\]And uhhhh,
I guess we're allowed to do this, ya?\[\large\rm \sum_{n=1}^{N}~\int\limits_0^{\pi}\cos(nx)\cos(mx)dx\]And then you can apply this weird trig identity,\[\large\rm \sum_{n=1}^{N}~\int\limits\limits_0^{\pi}\frac{1}{2}\left(\cos\left[(m-n)x\right]+\cos\left[(m+n)x\right]\right)dx\]Hmm that's too many brackets..\[\large\rm \frac{1}{2}\sum_{n=1}^{N}~\int\limits\limits\limits_0^{\pi}\cos\left[(m-n)x\right]+\cos\left[(m+n)x\right]~dx\]

Answer 5

And then ummmm,
is this one of those things where all they all work out to zero except the case where m=n or something?
mmm thinking...

Answer 6

yes :P

Answer 7

\(\int \limits_0^\pi \cos mx \cos nx dx = 0 \\ m \ne n\)

Answer 8

so that bound is a constant?

Answer 9

and what happened to the "a_n"?

Answer 10

\(\int_0^\pi \cos^2 x = \pi/2 +c\)

Answer 11

Oh I dropped the A_n's woops :[

Answer 12

note: when n = m, the co-efficient is a_m which is anyways a constant
\(\int_0^\pi a_m \cos^2 x =a_m \pi/2 +c\)

Answer 13

and all the other integral are anyways 0 :)

Answer 14

a_n not a_m o.0

Answer 15

n goes from 1 to N

when n = 1, its a1
when n = 2, its a2
...
when n = m, its a_m

Answer 16

but..at the same time we don't when n=m ? because we want it to =0 ?

Answer 17

\[\sum_{N}^{n=1}\int\limits_{0}^{\pi}a_ncos(nx)\cos(mx)dx\] is that where the a_n would go? haha I got a bit lost after this point xD

Answer 18

yes, thats right.

now we separate that out into 2 terms,

one with n =m and the other with n NOT = m

and we know, all the terms with n NOT = m are 0

Answer 19

\(\sum_{n=1}^{N, n \ne m}\int\limits_{0}^{\pi}a_ncos(nx)\cos(mx)dx + \int\limits_{0}^{\pi}a_m\cos^2(mx)dx \\ = 0+\int\limits_{0}^{\pi}a_m\cos^2(mx)dx \\= a_m \pi/2 + c \)

Answer 20

hm probably a silly question, where did you get the second term from? the a_mcos^2(mx) ?
and also, then we have to do this whole process twice? (with m=n and n=/=m) why?

Answer 21

when n=m,
\(a_n = a_m\)

\(\cos mx \times \cos mx = \cos^2 mx\)

we need to separate it out because the result of 2 integrals are different:

\(\int \limits_0^\pi \cos mx \cos nx dx = 0 \\ m \ne n \\ \int_0^\pi a_m \cos^2 x =a_m \pi/2 +c\)

Answer 22

and we don't need that trig identity that zep brought up?

Answer 23

ignore that "+c", its definite integral!

We would need that to prove that
\(\int_0^\pi a_n \cos mx \cos nx =0\)

Answer 24

if we can directly not use it

Answer 25

right. o.o we should probably prove that haha

Answer 26

Sorry, my interwebs got ruined for a bit

Answer 27

Also when you read the
\[a_m(\pi/2)\] did you forget to distribute the bound thingy? or is that just a bound?

Answer 28

its after plugging in upper and lower bounds.

Answer 29

hm ok

Answer 30

so we want to prove it = 0 ?

\(\large\rm ~\int\limits\limits\limits_0^{\pi}\cos\left[(m-n)x\right]+\cos\left[(m+n)x\right]~dx\)

whats the integral of cos?

Answer 31

Right.
and in that we already have the a_n distributed?
and isn't there a half in front of that?

Answer 32

sure sure...but since its anyways going to get 0, it won't matter :P

Answer 33

\[\sum_{N}^{n=1}\int\limits_{0}^{\pi}a_ncos(nx)\cos(mx)dx\]\[\sum_{N}^{n=1}\int\limits_{0}^{\pi}\frac{ 1 }{ 2 }[\cos[(m-n)x]+\cos[(m+n)x]dx\]
so the..a_n just sort of disappeared lol

Answer 34

Sorry, I like to make sure all the small things are taken care of xD #perfectionist o_o

Answer 35

its there
\(\sum_{N}^{n=1}\int\limits_{0}^{\pi}a_n \frac{ 1 }{ 2 }[\cos[(m-n)x]+\cos[(m+n)x]dx\)

and thats a good habit actually

Answer 36

haha thanks. It annoys some people :>
ok so theen..we would pull out the 1/2 to the front as a constant

Answer 37

you can pull out a_n/2 altogether,

but don't put it before the sum operator

Answer 38

\[\sum_{N}^{n=1}\frac{ 1 }{ 2 }a_n \int\limits_{0}^{\pi}\cos[(m-n)x]+\cos[(m+n)x]dx\]

Answer 39

yes,
now whats the integral of cos ax ?

Answer 40

\[\frac{ 1 }{ a }\sin(ax)+C\]

Answer 41

good so you you'll get the terms like
\(\sin (m-n)\pi , \sin (m+n)\pi, \sin (m-n)0 , \sin (m+n)0 \)

which are all 0!
wanna give it a try yourself?

Answer 42

and the "1/a" part? o.0

Answer 43

1/a(0) = 0 :P

Answer 44

ou xD

Answer 45

let me seee

Answer 46

you'll get a m-n in the denominator for one terms.
thats why n cannot be = m, otherwise denominator = 0

thats the reason we separate it out!

Answer 47

ok so.. Let me see if I understand correctly lol

Answer 48

\[\int\limits_{0}^{\pi}\cos[(m-n)x]dx\]\[u=mx-nx\]\[du=mdx-ndx => (m-n)dx\]\[dx=\frac{ du }{ (m-n) }\]\[\frac{ 1 }{ (m-n)} \int\limits_{0}^{\pi}\cos(u)du\]\[\frac{ 1 }{ (m-n) }[\sin[(m-n)x]]|^\pi_0\]

Answer 49

good going!
continue

Answer 50

\[\frac{ 1 }{ (m-n) }[\sin[(m-n)\pi]-\sin[(m-n)0]]\]now sure how to evaluate those?

Answer 51

exactly at this step:

1. we say that \(m\ne n\), as denominator can't be 0

2. use the fact that \(\sin a\pi = 0\)
for a = 0, \pi, 2pi, ...

Answer 52

hmhm

Answer 53

so
\(\sin (m-n)\pi = 0 \\ \sin 0 = 0 \)

Answer 54

and the other one is just
sin[(m-n)0] = 0 also :P yayay

Answer 55

Then the
1/(m-n) * 0 = 0

Answer 56

same thing with m+n :)

Answer 57

the m+n is the entire same process I just have to put a little + instead of - everywhere? xD

Answer 58

exactly

Answer 59

\[\sum_{N}^{n=1}\frac{ 1 }{ 2 }a_n[0-0]\]then...in the end we would have something like this?

Answer 60

oops that would be a 0+0 but whatever haha

Answer 61

right doesn't make a difference

Answer 62

and then that..just = 0?

Answer 63

but remember we said
\(m \ne n\)

so we have to consider the case when m= n

Answer 64

\[\int\limits_{0}^{\pi}\sum_{N}^{n=1}a_ncos(nx)\cos(mx)dx\]\[\sum_{N}^{n=1}a_n \int\limits_{0}^{\pi}\cos(nx)\cos(mx)dx\]\[\sum_{n=1}^{N, n \neq m}[\int\limits_{0}^{\pi}a_ncos(nx)\cos(mx)dx]+\sum_{n=1}^{N,n=m} \int\limits_{0}^{\pi}a_mcos^2m(x)dx\] see proof
\[0+\sum_{n=1}^{N}a_m \int\limits_{0}^{\pi}\cos^2(mx)dx\]

Answer 65

Does writing it like that make sense? xD lol aah so many parts.

Answer 66

when n=m we have
\(a_m \int\limits_{0}^{\pi}\cos^2(mx)dx\)

Answer 67

so your last step should not have the sum sign

Answer 68

to the right of the
0+ ?

Answer 69

yeah

Answer 70

its just one term, when
n=m, so need of addition sign

Answer 71

I see :)

Answer 72

so now..we just need to calculate
\[0+a_m \int\limits_{0}^{\pi}\cos^2(mx)dx\]

Answer 73

which is an identity! yay ehhe

Answer 74

thank god lol

Answer 75

any more doubts? :)

Answer 76

Ikr.
ok so because the mx is there would it be?
\[\int\limits_{0}^{\pi}\frac{ 1+\cos(2mx) }{ 2 }dx\]

Answer 77

correct

Answer 78

also there is an a_m

Answer 79

but its outside the integral,
so nevermind

Answer 80

\[a_m \frac{ 1 }{ 2 }[x-\frac{ \sin(2mx) }{ 2m }]|^\pi_0\]...is that right? lol

Answer 81

why negative in between?

Answer 82

...I..felt like it xP
yeah that should be a +

Answer 83

then good, continue with + in between

Answer 84

\[a_m[\frac{ x }{ 2 }+\frac{ \sin(2mx) }{ 4m }]|^\pi_0\]\[a_m[(\frac{ \pi }{ 2 }+\frac{ \sin(2m \pi) }{ 4m })-(\frac{ 0 }{ 2 }+\frac{ \sin(2m0) }{ 4m })]\]

Answer 85

you know whats sin 2mpi = ... ?

Answer 86

er..no o.0

Answer 87

0?

Answer 88

m is an integer, so 2m is also an integer

sin 0 = sin 0pi = 0
sin 1pi = 0
sin 2 pi = 0
sin 3pi = 0
...

sin (integer)pi = 0

sin (2m pi) = 0
makes sense?

Answer 89

oh yes, because sin(pi) always equals 0
I see i se

Answer 90

that makes so much sense haha

Answer 91

\[a_m[\frac{ \pi }{ 2 }+0-0-0]\]\[0+a_m(\pi/2)\]\[=a_m(\frac{ \pi }{ 2 })\]wooo!!

Answer 92

yay! :D

Answer 93

You, sir, are a genius.

Answer 94

I just have a practice of these type of problems,
but thanks anyways ^_^

Answer 95

:) it's a bunch of tiny steps o.o

Answer 96

What exactly is that trig identity that we used to get the 1/2 with the cos thing?

Answer 97

\(\Large \cos(A+B) + \cos(A-B) = 2 \cos A \cos B\)

Answer 98

aah glorious. ok:)
So I think this is all done!! thanks!