Question

i really need some help please?

Answer 1

how do i do a fourier series?

Answer 2

f(x) = 1 for -5 to 0
fx = 1+x for 0 to 5

Answer 3

if someone has a link that'd be great

Answer 4

1 worked out a0 = 9/2
can't get what the teacher got for an and bn, so cant combine properly

Answer 5

wikipedia has a decent explanation. you basically just need to findt he constants an and bn

Answer 6

\[a_n = \frac{ 1 }{ 5 }(\int\limits_{-5}^{0}\cos(\frac{ n \pi x }{ 5 })dx + \int\limits_{0}^{5}(1 + x)\cos(\frac{ n \pi x }{ 5 })dx)\]

Answer 7

i got that to, but then he's reduced the answer to
= 5/(n^2*pi) ((-1)^n)-1)

Answer 8

Man don't start a calculus question saying just "i really need some help please". I was sure it was an 8yo to lazy to even type the question.

Answer 9

oh. what you want to know is how he got that answer. what is your final answer? ill show u how to convert it

Answer 10

i did the integral within those limits and got values, but mine does'nt reduce down the same way his does

Answer 11

final answer is:

Answer 12

top line:
5x cos(n*pi) + 7n * sin (n*pi) *pi-5

over bottom line
n^2 pi^2

Answer 13

so :(5x cos(n*pi) + 7n * sin (n*pi) *pi-5) / (n^2 pi^2)

Answer 14

sorry, that's for a n

Answer 15

mine reduces, since the sins are all sin(npi) = 0 to:

\[\frac{ 5 }{ n^2 \pi^2 }\cos(\frac{ n \pi x }{ 5}) \] for x from 0 to 5

Answer 16

sin n pi = 0...?
how do we assume that sorry?

Answer 17

which gives:

\[\frac{ 5 }{ n^2 \pi^2 }(\cos(n \pi) - \cos(0)) = \frac{ 5 }{ n^2 \pi^2 }(\cos(n \pi) - 1)\]

Answer 18

no matter what n is sin(n pi) = 0. like sin (2pi) sin (3 pi) ect

Answer 19

if it were sin(npi/2), then n would alternate between 1 and -1

Answer 20

ok, so we only assume whole intergers for n

Answer 21

ya because were using a series which goes through integers

Answer 22

from 1 to inf., (since a0 is already in front of the series)

Answer 23

so cos(n pi) takes values of 1 and - 1. so we can write it as (-1)^n

Answer 24

ur a genius!
cheers bro!

Answer 25

that explains heaps, think i can duplicate from here

Answer 26

im glad i could help and that you understood :)

im actually applying it in engineering atm so i had to brush up on it. i was happy to see a question about it :P

Answer 27

thanks again hey, happy to help

Answer 28

@Euler271
dude, quick query, his answer's :
\[\frac{ 5 }{ n ^{2} \pi ^{2} } \times (-1^{n}-1)\]

Answer 29

where'd the last "-1"come from?

Answer 30

ah, not pi sqrd, my bad

Answer 31

mine has it too.

it was fromt he limits of the integral (x from 0 to 5)

you get cos ( n pi (5) / 5 ) - cos (n pi (0) / 5 ) = ( cos(npi) - 1) for all n

Answer 32

ah, from limits, cool cool
cheers