forked from Sasserisop/MATH201
added lec 25 and revised periodic functions
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@ -1,22 +1,27 @@
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#start of lec 19
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This lecture we will learn about periodic functions, specifically, non-sinusoidal periodic functions.
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# Periodic function
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#periodic
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Definition:
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$f$ is periodic with period $T \quad (T>0)$ if:
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$$f(t)=f(t+T), \quad \forall\ t\in \mathbb{R}$$
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![[Drawing 2023-10-20 13.06.35.excalidraw]]
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![[Drawing 2023-10-20 13.06.35.excalidraw.png]]
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We will now compute laplace transforms of these periodic functions. Computing DE's containing these periodic functions using something like #voparam would not be easy.
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Let's try taking the laplace of a periodic function $f(t)$:
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If we take the windowed version of the function (one period, where everywhere else is 0, ie:)
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$f_{T}(t)=\begin{cases}f(t)\ ,\ & 0\leq t\leq T \\0\ ,\ & \text{otherwise}\end{cases}$
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we can "glue together" many of these windows together to rebuild our $f(t)$, like this:
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$f(t)=f_{T}(t)+f_{T}(t-T)+f_{T}(t-2T)+\dots$
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We can add some unit step functions strategically such that it doesn't change the overall expression:
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$f(t)=f_{T}(t)+f_{T}(t-T)u(t-T)+f_{T}(t-2T)u(t-2T)+\dots$
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$\mathcal{L}\{f\}=\mathcal{L}\{f_{T}\}+\mathcal{L}\{f(t-T)u(t-T)\}+\dots$
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hit it with the LT!
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$\mathcal{L}\{f\}=\mathcal{L}\{f_{T}\}+\mathcal{L}\{f_{T}(t-T)u(t-T)\}+\dots$
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recall the formula from last lec: $\mathcal{L}\{u(t-a)f(t-a)\}=e^{-as}F(s)$
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then:
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$\mathcal{L}\{f\}=\mathcal{L}\{f_{T}\}(1+e^{-TS}+e^{-2TS}+e^{-3TS}+\dots)$
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$\mathcal{L}\{f\}=\mathcal{L}\{f_{T}\}(1+e^{-TS}+(e^{-TS})^{2}+(e^{-TS})^{3}+\dots)$
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$\mathcal{L}\{f\}=\mathcal{L}\{f_{T}\}(1+e^{-Ts}+e^{-2Ts}+e^{-3Ts}+\dots)$
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$\mathcal{L}\{f\}=\mathcal{L}\{f_{T}\}(1+e^{-Ts}+(e^{-Ts})^{2}+(e^{-Ts})^{3}+\dots)$
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This is a geometric series! $1+r+r^2+\dots$
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Geometric series are convergent when $|r|<1$
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and equal to: $\frac{1}{1-r}$
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@ -25,8 +30,9 @@ so:
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$$\mathcal{L}\{f\}=\mathcal{L}\{f_{T}\} \frac{1}{1-e^{-Ts}}$$
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handy formula! ^ will be used again.
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#ex
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imagine another function: (image is of a square wave with a period of 2a, oscillates between 1 and 0, starts at 1 when t=0.)
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![[Drawing 2023-10-20 13.27.58.excalidraw]]
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Imagine another function: (image is of a square wave with a period of 2a, oscillates between 1 and 0, starts at 1 when t=0.)
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![[Drawing 2023-10-20 13.27.58.excalidraw.png]]
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Let's compute its LT:
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$\mathcal{L}\{f\}=\mathcal{L}\{f_{2a}\} \frac{1}{1-e^{-2as}}$
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$f_{2a}=u(t)-u(t-a)$ (this is the first period piece)
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$\implies \mathcal{L}\{f_{2a}\}=\mathcal{L}\{u(t)\}-\mathcal{L}\{u(t-a)\}=\frac{1}{s}- \frac{e^{-as}}{s}$
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@ -34,21 +40,44 @@ plug back in:
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$\mathcal{L}\{f\}=\mathcal{L}\{f_{2a}\} \frac{1}{1-e^{-2as}}=\frac{1}{s}\cancel{ (1-e^{-as}) } \frac{1}{\cancel{ (1-e^{-as}) }(1+e^{-as})}$
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$$\mathcal{L}\{f\}=\frac{1}{s(1+e^{-as})}$$
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#ex
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$y''+3y'+2y=f(t)$ where $f(t)$ is from the previous example
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$y(0)=y'(0)=0,\ a=1$ (a is width of 1/2 period in the function f(t))
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$s^2Y+3sY+2Y=\mathcal{L}\{f\}= \frac{1}{s(1+e^{-s})}$
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#ex #IVP #periodic #second_order_nonhomogenous #LT #partial_fractions
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Solve for $y(t)$ in the following second order periodic equation:
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$$y''+3y'+2y=f(t) \qquad y(0)=y'(0)=0 \quad a=1$$where $f(t)$ is from the previous example and $a$ is the width of $\frac{1}{2}$ of a period in the function $f(t)$
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Hit it with the LT!
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$s^2Y+3sY+2Y=\mathcal{L}\{f\}= \frac{1}{s(1+e^{-1s})}$
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$(s+1)(s+2)Y= \frac{1}{s(1+e^{-1s})}$
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$Y(s)=\frac{1}{s(s+1)(s+2)} \frac{1}{1+e^{-s}}$
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$=F(s) \frac{1}{1+e^{-s}}$
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What property can we use to find the inverse of that?
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> psst. We can use $\mathcal{L}\{f\}=\mathcal{L}\{f_{T}\} \frac{1}{1-e^{-Ts}}$ ;)
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$\mathcal{L}^{-1}\{F\}=\mathcal{L}^{-1}\{\frac{1}{2} \frac{1}{s}+\frac{1}{2} \frac{1}{s+2}-\frac{1}{s+1}\}$
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$=\frac{1}{2}+\frac{1}{2}e^{-2t}-e^{-t}$
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$y(t)= \dots$
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we want to use formula from earlier
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so we need to change $F(s) \frac{1}{1+e^{-s}}$
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to: $F(s) \frac{1e^{-s}}{1-e^{-2s}}$
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$y(t)=p(t)$, periodic with period of 2 ($T=2$)
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$Y(s)=F(s) \frac{1}{1+e^{-s}}$
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But that second term has a $1+e^{-Ts}$ term in the denominator, it doesn't match up in the formula. There is a fix, peep this:
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$Y(s)=F(s) \frac{1-e^{-s}}{(1+e^{-s})(1-e^{-s})}$
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$Y(s)=F(s) \frac{1-e^{-s}}{1-e^{-2s}}$ <- We are rather lucky, the 2 in the denominator matches $T$, the period of our function.
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We can tuck away the numerator into $F(s)$ :
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$Y(s)=\underbrace{ \frac{1-e^{-s}}{s(s+1)(s+2)} }_{ F(s) } \frac{1}{1-e^{-2s}}$
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Our equation is now in the correct form. We can now calculate the inverse of $F(s)$
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Split up $F(s)$ :
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$Y(s)= \underbrace{ (\frac{1}{s(s+1)(s+2)} }_{ F_{1}(s) }-\underbrace{ \frac{e^{-s}}{s(s+1)(s+2)} )}_{F_{2}(s) } \frac{1}{1-e^{-2s}}$
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We can use partial fractions for the first term and
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$\mathcal{L}\{u(t-a)f_{1}(t-a)\}=e^{-as}F_{1}(s)$ for the second term. (Where $a=1$)
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$$f_{2a}(t)=\mathcal{L}^{-1}\{F(s)-F(s)e^{-s}\}=\frac{1}{2}+\frac{1}{2}e^{-2t}-e^{-t}-\left( \frac{1}{2}+\frac{1}{2} e^{-2(t-1)}-e^{-(t-1)})u(t-1 \right)$$
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using partial fractions:
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$\frac{1}{s(s+1)(s+2)}=\frac{A}{s}+\frac{B}{s+1}+\frac{C}{s+2}$
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$A(s+1)(s+2)+Bs(s+2)+Cs(s+1)=(A+B+C)s^2+(3A+2B+C)s+2A=1$
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$2A=1\implies A=\frac{1}{2}$
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$\frac{3}{2}+2B+C=0$
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$\frac{1}{2}+B+C=0$
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subtract the two equations.
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$1+B=0\implies B=-1$
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$\implies C=\frac{1}{2}$
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$\mathcal{L}^{-1}\{F_{1}\}=\mathcal{L}^{-1}\{\frac{1}{2} \frac{1}{s}-\frac{1}{s+1}+\frac{1}{2} \frac{1}{s+2}\}$
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$\mathcal{L}^{-1}\{F_{1}\}=f_{1}=\frac{1}{2}-e^{-t}+\frac{1}{2}e^{-2t}$
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Second term ($F_{2}(s)$): use $\mathcal{L}\{u(t-a)f_{1}(t-a)\}=e^{-as}F_{1}(s)=F_{2}(s)$
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$f_{2}(t)=u(t-1)(\frac{1}{2}-e^{-(t-1)}+\frac{1}{2}e^{-2(t-1)})$
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$y(t)=p(t)=f_{2a}(t)$ meaning $y(t)$ is a periodic function, with period of $T=2$
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$$f_{2a}(t)=\mathcal{L}^{-1}\{F_{1}(s)-F_{2}(s)\}=\frac{1}{2}+\frac{1}{2}e^{-2t}-e^{-t}-u(t-1)(\frac{1}{2}-e^{-(t-1)}+\frac{1}{2}e^{-2(t-1)})$$
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@ -197,3 +197,79 @@ To calculate distance: $\alpha_{1}+\beta_{1}i, \qquad \alpha_{2}+\beta_{2}i$
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$\sqrt{ (\alpha_{1}-\alpha_{2})^2+(\beta_{1}-\beta_{2})^2 }$
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$\rho\geq \sqrt{ 1^2+1^2 }=\sqrt{ 2 }$
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#end of lec 24
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#start of lec 25
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find the zeros of
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$xy''-y'+y=0 \qquad x_{0}=2$
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is this function analytic about x_0=2?
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DONT FORGET! put it in standard form:
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$y''-\frac{1}{x}y'+\frac{y}{x}=0$ <- now we can see that there are singular points at x=0
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so we have a radius convergence of $\rho=2$
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$y(x)=\sum_{n=0}^\infty a_{n}(x-2)^n \quad x\in(0,4)$
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$x-2=t \qquad t\in(-2,2)$
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$y(t)=\sum_{n=0}^\infty a_{n}t^n$
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$y'(t)=\sum_{n=1}^\infty a_{n}nt^{n-1}$
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$y''(t)=\sum_{n=2}^\infty a_{n}n(n-1)t^{n-2}$
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$\sum_{n=2}^\infty a_{n}n(n-1)t^{n-1}+2\sum_{n=2}^\infty a_{n} n(n-1)t^{n-2}-\sum_{n=1}^\infty a_{n}nt^{n-1}+\sum_{n=0}^\infty a_{n}t^n=0$
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n-1=k n-2=k n-1=k
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first 5 non-zeros:
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$\sum_{k=1}^\infty a_{k+1}(k+1)kt^k+\sum_{k=0}^\infty 2(k+2)(k+1)a_{k+2}t^k-\sum_{k=0}^\infty a_{k+1}(k+1)t^k+\sum_{k=0}^\infty a_{k}t^k$
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$\underbrace{ 4a_{2}-a_{1}+a_{0} }_{ =0 }+\sum_{k=1}^\infty \underbrace{ (a_{k+1}(k+1)k+4a_{k+2}-a_{k+1}(k+1)+a_{k}) }_{ =0 }t^k=0$
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$a_{2}=\frac{a_{1}-a_{0}}{4}$
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$12a_{3}+2a_{2}-2a_{2}+a_{1}=0$
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$a_{3}=-\frac{a_{1}}{12}$
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$a_{4}=-\frac{1}{24}(3a_{3}+a_{2})=\frac{1}{96}a_{1}-\frac{a_{1}}{96}+\frac{a_{0}}{96}=\frac{a_{0}}{96}$
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$y(x)=a_{0}+a_{1}(x-2)+\frac{a_{1}-a_{0}}{4}(x-2)^2-\frac{a_{1}}{12}(x-2)^3+\frac{a_{0}}{96}(x-2)^4+\dots$
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in this case we cant go much further, cant explicitly find the coefficients for each term. in the last lecture's example we were lucky.
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So we are done.
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#ex
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Find first four non-zero terms of the power series for $y(x)$ about $x_{0}=\pi$
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of the IVP:
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$$y''-\sin (x)y=0 \qquad y(\pi)=1 \qquad y'(\pi)=0$$
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This is already in standard form.
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clearly this is analytical over the entire real axis, sin(x) and 1 are both infinitely differentiable. no singular points real or complex.
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$y(x)=\sum_{n=0}^\infty a_{n}(x-\pi)^n \qquad x-\pi=t$
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$y(t)=\sum_{n=0}^\infty a_{n}t^n$ <- we are abusing notation, the y here isn't the same as above. But it's all good.
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$y''-\sin(t+\pi)\sum_{n=0}^\infty a_{n}t^n=0$
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$y''+\sin(t)\sum_{n=0}^\infty a_{n}t^n=0$
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$\sum_{n=2}^\infty a_{n}n(n-1)t^{n-2}+\left( \sum_{n=0}^\infty(-1)^n \frac{t^{2n+1}}{(2n+1)!} \right)\left( \sum_{n=0}^\infty a_{n}t^n \right)$ remember, sin is odd so its infinite series has odd powers.
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now from:
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$y''+\sin(t)\sum_{n=0}^\infty a_{n}t^n=0$
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$y(t)=\sum_{n=0}^\infty a_{n}t^n$
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this implies $y(0)=1=a_{0} \quad y'(0)=0=a_{1}$
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from the big summ-y equation:
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$(2a_{2}+6a_{3}t+12a_{4}t^2+20a_{5}t^3+\dots)+\left( t-\frac{t^3}{6}+\frac{t^5}{120}-\dots \right)(a_{0}+a_{1}t+a_{2}t^2+a_{3}t^3+\dots)=0$
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the only constant factor is $a_{2}$
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this implies: $2a_{2}=0 \implies a_{2}=0$
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what about the factors of $t$?
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$(6a_{3}+a_{0})t=0$
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$a_{3}=-\frac{a_{0}}{6}=-\frac{1}{6}$
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$t^2$:
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$(12a_{4}+a_{1})t^2=0$
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$a_{2}=-\frac{a_{1}}{12}$
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$t^3$:
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$\left( 20a_{5}+a_{2}-\frac{a_{0}}{6} \right)=0 \implies a_{5}=\frac{1}{120}$
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$t^4$:
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$\left( 30a_{6}+a_{3}-\frac{a_{1}}{6} \right)t^4=0 \implies a_{6}=\frac{1}{180}$
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$y(x)=1-\frac{1}{6}(x-\pi)^3+\frac{1}{120}(x-\pi)^5+\frac{1}{180}(x-\pi)^6+\dots$
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theres no general formula here for the constants? (or maybe no formula for y(x)?), but we can write the solution in the following form^.
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#ex
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$$y'-xy=e^x \qquad x_{0}=0$$
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$y(x)=\sum_{n=0}^\infty a_{n}x^n$
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$y'(x)=\sum_{n=1}^\infty a_{n}nx^{n-1}$
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$\sum_{k=0}^\infty a_{k+1}(k+1)x^k-\sum_{k=1}^\infty a_{k-1}x^k-\sum_{k=0}^\infty \frac{x^k}{k!}=0$
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$a_{1}-1=0 \implies a_{1}=1$
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$a_{k+1}=\frac{a_{k-1}+\frac{1}{k!}}{k+1}$
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$k=1\implies a_{2}=\frac{a_{0}}{2}+\frac{1}{2}$
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$k=2\implies a_{3}=\frac{1}{2}$
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$k=3\implies a_{4}=\frac{ \left( \frac{a_{0}}{2}+\frac{1}{2} \right)+\frac{1}{6}}{4}$
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We are lucky, in this course fubini's method is not needed. (what?)
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and with that, we are finished this chapter on power series.
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#end of lec 25
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@ -22,7 +22,7 @@ Good luck on midterms! <3 -Oct 18 2023
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[Periodic functions (lec 19)](periodic-functions-lec-19.html) (raw notes, not reviewed or revised yet.)
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[Convolution (lec 19-20)](convolution-lec-19-20.html) (raw notes, not reviewed or revised yet.)
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[Dirak δ-function (lec 21)](dirak-δ-function-lec-21.html) (raw notes, not reviewed or revised yet.)
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[Power series (lec 22-24)](power-series-lec-22-24.html) (raw notes, not reviewed or revised yet.)
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[Power series (lec 22-25)](power-series-lec-22-25.html) (raw notes, not reviewed or revised yet.)
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</br>
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[How to solve any DE, a flow chart](Solve-any-DE.png) (Last updated Oct 1st, needs revision. But it gives a nice overview.)
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[Big LT table (.png)](drawings/bigLTtable.png)
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