UnkleRhaukus
  • UnkleRhaukus
\(F\)ourier Series
Differential Equations
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
UnkleRhaukus
  • UnkleRhaukus
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UnkleRhaukus
  • UnkleRhaukus
please check my work for; •mathematical •typographical •formatting errors
UnkleRhaukus
  • UnkleRhaukus

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richyw
  • richyw
man I can't wait to learn this stuff
anonymous
  • anonymous
i would but i'm studying for my physics final =/
anonymous
  • anonymous
that was Fourier series. And also where is your works? That was you typed?
UnkleRhaukus
  • UnkleRhaukus
my work is in 1.pdf,
anonymous
  • anonymous
Can't wait till I learn tall this stuff.
anonymous
  • anonymous
Why wait? Basic Fourier series are rather intuitive and simple to figure out!
anonymous
  • anonymous
I didn't see any mistakes, though I didn't have time to check all of the integrations. I think it's good.
UnkleRhaukus
  • UnkleRhaukus
thanks !
anonymous
  • anonymous
1 a) ii bounds on bn remain -pi to pi after you pull out f(x)
anonymous
  • anonymous
didn't see anything else wrong with the integrals. this is a great primer for fourier. i like the complex form once i saw it
UnkleRhaukus
  • UnkleRhaukus
Your ,... Right , thank-you so much @Mathmuse \[\begin{align*} b_n&=\frac1\pi\int\limits_{-\pi}^\pi f(x)\sin(nx)\text dx\\ &=\frac1\pi\int\limits_{\color{red}{\cancel{-\pi}}0}^\pi\sin(nx)\text dx\\ &=\dots \end{align*}\]
UnkleRhaukus
  • UnkleRhaukus
\[\textbf 2 (b) \]\begin{align*} f(x)&=\cos(x),\qquad0
UnkleRhaukus
  • UnkleRhaukus
im going wrong somewhere here, because the correct Series should be \[S(x)=\frac{\sin(3)}3+6\sin(3)\sum\limits_{n=1}^\infty\frac{\cos(\tfrac23n\pi x)}{9-16n^2\pi^2}+4\pi(1-\cos(3))\sum\limits_{n=1}^\infty\frac{n\sin(\tfrac23n\pi x)}{4\pi^2n^2-9}\]
anonymous
  • anonymous
Let me look over it.
anonymous
  • anonymous
Okay \[=-\frac13\left(\frac{\cos\left((\tfrac23\pi n+1)x\right)}{\tfrac23\pi n+1}-\frac{\cos\left((\tfrac23\pi n-1)x\right)}{\tfrac23\pi n-1}\right)\Big|_0^3\] Since you integrate both sines they both acquire a negative so when you pull out the negative shouldn't that middle one be positive?
anonymous
  • anonymous
because when you redistribute it the signs are different.
UnkleRhaukus
  • UnkleRhaukus
it turns out i didn't use the trig formula correctly for a_n this somehow didnt matter , but for b_n it did should be like this \[\begin{align*} % a_n a_n&=\frac23\int\limits_{0}^{3}\cos(x)\cos\left(\frac{\pi n x}{3/2}\right)\text dx\\ &=\frac13\int\limits_{0}^{3}\cos\left((1-\tfrac23\pi n )x\right)+\cos\left((1+\tfrac23\pi n)x\right)\text dx\\ &=-\frac13\left.\left(\frac{\sin\left((1-\tfrac23\pi n )x\right)}{1-\tfrac23\pi n }-\frac{\sin\left((1+\tfrac23\pi n )x\right)}{1+\tfrac23\pi n}\right)\right|_0^3\\ &=\frac{\sin\left(3-2\pi n \right)}{3-2\pi n }+\frac{\sin\left(3+2\pi n \right)}{3+2\pi n }\\ &=\frac{\sin\left(3 \right)}{3-2\pi n }+\frac{\sin\left(3\right)}{2\pi n +3}\\ &=\sin\left(3 \right)\left(\frac{1}{3-2\pi n }+\frac{1}{3+2\pi n}\right)\\ &=\sin\left(3 \right)\left(\frac{3+2\pi n }{(3-2\pi n)(3+2\pi n) }+\frac{3-2\pi n}{(3+2\pi n)(3-2\pi n)}\right)\\ &=\left(\frac{6\sin\left(3 \right)}{9-4\pi^2n^2}\right)\\ &=\left(\frac{6\sin\left(3 \right)}{9-4\pi^2n^2}\right) \end{align*}\] \[\begin{align*} % b_n b_n&=\frac23\int\limits_{0}^{3}\cos(x)\sin\left(\frac{\pi n x}{3/2}\right)\text dx\\ &=\frac13\int\limits_{0}^{3}\sin\left((1+\tfrac23\pi n)x\right)-\sin\left((1-\tfrac23\pi n)x\right)\text dx\\\ &=-\frac13\left.\left(\frac{\cos\left((1+\tfrac23\pi n)x\right)}{1+\tfrac23\pi n}-\frac{\cos\left((1-\tfrac23\pi n)x\right)}{1-\tfrac23\pi n}\right)\right|_0^3\\ &=\frac{1-\cos\left(3+2\pi n\right)}{2\pi n+3}-\frac{1-\cos\left(3-2\pi n\right)}{2\pi n-3}\\ &=\frac{1-\cos\left(3\right)}{3+2\pi n}-\frac{1-\cos\left(3\right)}{3-2\pi n}\\ &=\left(1-\cos\left(3\right)\right)\left(\frac{1}{3+2\pi n}-\frac{1}{3-2\pi n}\right)\\ &=\left(1-\cos\left(3\right)\right)\left(\frac{3-2\pi n}{(3+2\pi n)(3-2\pi n)}-\frac{3+2\pi n}{(3-2\pi n)(3+2\pi n)}\right)\\ &=\left(1-\cos\left(3\right)\right)\left(\frac{-4\pi n}{9-4\pi^2n^2}\right)\\ &=\left(\frac{4\pi\left(1-\cos\left(3\right)\right)n}{4\pi^2n^2-9}\right)\\ \end{align*}\]
UnkleRhaukus
  • UnkleRhaukus
i think
UnkleRhaukus
  • UnkleRhaukus
how did i get a minus for a_0? lolsz
UnkleRhaukus
  • UnkleRhaukus
so\[S(x)=\frac{\sin(3)}3+6\sin(3)\sum\limits_{n=1}^\infty\frac{\cos(\tfrac23n\pi x)}{9-4n^2\pi^2}+4\pi(1-\cos(3))\sum\limits_{n=1}^\infty\frac{n\sin(\tfrac23n\pi x)}{4\pi^2n^2-9}\checkmark\]
UnkleRhaukus
  • UnkleRhaukus
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