forked from Sasserisop/MATH201
revised Laplace and added lec 22
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## Definition
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This is a mass-spring system with friction: (imagine the $\beta$ was $b$ instead)
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A mass-spring system with friction can be modelled by the following differential equation:
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This mass-spring system with friction can be modelled by the following differential equation:
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$$my''+by'+ky=F_{ext}(t)$$
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Where:
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$$\begin{matrix}F_{ext}(t) & \text{any external forces applied }(N) \\m & \text{mass } (kg) \\b & \text{damping constant } \left( \frac{Ns}{m} \right) \\k & \text{spring constant } \left( \frac{N}{m} \right)\end{matrix}$$
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@ -9,115 +9,150 @@ $\mathcal{L}\{0\}=0$ Look at your bank account, integrate 0 you still get 0 :D
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$\mathcal{L}\{1\}=\int_{0}^\infty e^{-st} \, dt=-\frac{1}{s}e^{-st}|_{0}^\infty=\frac{1}{s}$ if s>0
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$\mathcal{L}\{e^{at}\}$=$\int_{0}^{\infty} e^{at}e^{-st}\, dt=\int_{0}^{\infty}e^{-(s-a)t}dt=\frac{1}{s-a}$ if $s-a>0$
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$\mathcal{L}\{\sin bt\}=\int _{0}^{\infty}e^{-st}\sin(bt) \, dt=\frac{b}{s^2+b^2}$ by integration by parts
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similarly can be done for cos, but we have run out of time. (Don't worry, you will find these properties and many more on the table provided in the next lectures)
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similarly can be done for cos, but we have run out of time. (Don't worry, you will find these properties and many more on in the next lecture)
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#end of lec 14 #start of lec 15
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#ex #LT
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compute the LT of this funny function:
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$f(t)=\begin{cases}1 &\text{if } 0\leq t\leq 1 \\ 2 &\text{if } 1<t\leq 2 \\0 & \text{if } 2<t \end{cases}$
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$F(s)=\int _{0}^1 e^{-st}\, dt+2\int _{1}^2 e^{-st}\, dt+0$
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$F(s)=-\frac{1}{s}e^{-st}|_{t=0}^{t=1}-\frac{2}{s}e^{-st}|_{t=1}^{t=2}$
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$F(s)=\int _{0}^1 e^{-st}\, dt+2\int _{1}^2 e^{-st}\, dt+\int _{2} ^\infty e^{-st}0\, dt$
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$F(s)=-\frac{1}{s}e^{-st}|_{t=0}^{t=1}-\frac{2}{s}e^{-st}|_{t=1}^{t=2}+0$
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$$F(s)=-\frac{1}{s}(e^{-s}-1)-\frac{2}{s}(e^{-2s}-e^{-s})$$
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We have shown how to compute the LT of a choppy function.
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## Existence of LT
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Sometimes the LT of a function does not exist. It can be undefined over an interval, It could also be undefined everywhere (try taking the LT of $f(t)=\frac{1}{t}$ ;)). How can we know when we cant take the LT of a function? Peep these two theorems we define:
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Def: $f(t)$ is piecewise continuous on an interval $I$ if $f(t)$ is continuous on $I$, except possibly at a <u>finite</u> number of points of <u>jump</u> discontinuity.
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What is continuity? the limit exists and equals the value at that point.
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So what is jump discontinuity? Look at the picture:
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![[Drawing 2023-10-11 13.17.32.excalidraw.png]]
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Def: $f(t)$ is of exponential order $\alpha$ if there exists positive constants $\ T,\ M$ such that $f(t)\leq Me^{\alpha t}$ for all $t\geq T$
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this is important so that: $f(t)e^{-st}\underset{ \text{as } t\to \infty }{ \nrightarrow } \infty$
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$\mathcal{L}\{\alpha f(t)+\beta g(t)\}=\alpha \mathcal{L}\{f\}+\beta y\mathcal{L}\{g\}$
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the LT is a linear operator. as shown above. Proof:
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$\mathcal{L}\{\alpha f(t)+\beta g(t)\}=\alpha \int_{0}^{\infty}f(t)e^{-st} \, dt+\beta \int_{0}^{\infty}g(t)e^{-st} \, dt \quad \Box$
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### Theorem 1:
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If $f(t)$ is piecewise continuous and of an exponential order $\alpha$ on $[0, \infty)$, then $\mathcal{L}\{f\}(s)$ exists for $s>\alpha$
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### Theorem 2:
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If $F(s)$ is a LT then $\underset{ s \to \infty }{ \lim }F(s)=0$
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Def: $f(t)$ is piecewise continuous on an interval $I$ if $f(t)$ is continuous on $I$, except possibly at a <u>finite</u> number of points of <u>jump</u> discontinuity
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what is continuouity? the limit exists and equals the value at that point.
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So what is jump discontinuity look at the picture:
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![[Drawing 2023-10-11 13.17.32.excalidraw]]
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Def: f(t) is of exponential order $\alpha$ if there $\exists\ T,\ M, T\geq0$ such that $\forall\ t>T : f(t)\leq Me^{\alpha t}$
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this is important so that $f(t)e^{-st}$ doesn't go into infinity,
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this can be proven, but not necessary
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Theorem: If $f(t)$ is piecewise continuous and of an exponential order $\alpha$ on $[0, \infty)$, then $\mathcal{L}\{f\}$ exists $s>\alpha$
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and we are guarenteed that $s>\alpha$
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$e^{\alpha t}$ is this of exponential order? Of course, $M=1,\alpha=\alpha$
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what about $\sin(t)$? yes, sin is bounded b/w -1 and 1
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What about $e^{t^2}$? No, t^2 always outgrows $\alpha t$ eventually. This function does not have a LT.
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$e^{5 t}$ is this of exponential order? Of course, $M=1,\alpha=5$
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what about $\sin(t)$? yes, sin is bounded between -1 and 1
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What about $e^{t^2}$? No, $t^2$ always outgrows $\alpha t$ eventually. This function does not have a LT.
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![[Drawing 2023-10-11 13.21.18.excalidraw]]
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Theorem if $F(s)$ is a LT then $\lim_{ s \to \infty }F(s)=0$
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![[drawings/Drawing 2023-10-11 13.21.18.excalidraw.png]]
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properties:
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## Properties
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We cover 8 key properties in this lecture, these properties are the ones you will most frequently use.
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assume LT of f(t) exists:
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first property: $\mathcal{L}\{e^{\alpha t}f(t)\}=\int _{0}^\infty e^{\alpha t}e^{-st}\, dt=\int _{0}^\infty e^{-(s-a)t}f(t)\, dt=F(s-a)$
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</br>
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Property 1:
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LT is a linear operator, meaning:
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$\mathcal{L}\{\alpha f(t)+\beta g(t)\}=\alpha \mathcal{L}\{f\}+\beta y\mathcal{L}\{g\}$
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Proof:
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$\mathcal{L}\{\alpha f(t)+\beta g(t)\}=\alpha \int_{0}^{\infty}f(t)e^{-st} \, dt+\beta \int_{0}^{\infty}g(t)e^{-st} \, dt \quad \Box$
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</br>
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Property 2: (I call this one the shifting property, it shifts the function)
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$\mathcal{L}\{e^{\alpha t}f(t)\}=\int _{0}^\infty e^{\alpha t}e^{-st}\, dt=\int _{0}^\infty e^{-(s-a)t}f(t)\, dt=F(s-a)$
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</br>
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Property 3:
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$\mathcal{L}\{e^{\alpha t}\cdot 1\}=\frac{1}{s-\alpha}$
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$\mathcal{L}\{e^{\alpha t}\sin(bt)\}=\frac{b}{(s-a)^2+b^2}$ these properties are essential for the midterm.
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</br>
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example:
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$\mathcal{L}\{e^{\alpha t}\sin(bt)\}=\frac{b}{(s-a)^2+b^2}$ These properties are essential for the midterm.
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What if we calculate the LT of $f'$ ?
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using integration by parts:
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$\mathcal{L}\{f'(t)\}(s)=\int _{0}^\infty e^{-st}f'(t)\, dt=e^{-st}f(t)|_{t=0}^{t\to \infty}+ \underbrace{s \int e^{-st}f(t) \, dt }_{ sF(s) }$$=sF(s)-f(0)$
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$\mathcal{L}\{f''\}=s^2F(s)-sf(0)-f'(0)$
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in general, we can use proof by induction to show the following (try at home!):
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And in general, we can use proof by induction to show property 4 (try at home!):
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Property 4:
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$\mathcal{L}\{f^{(m)}\}=s^mF(s)-s^{m-1}f(0)-\dots-f^{m-1}(0)$
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trig stuff:
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Property 5:
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$\mathcal{L}\{\sin bt\}=\frac{b}{s^2+b^2}$
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Property 6:
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$\mathcal{L}\{\cos bt\}=\frac{s}{s^2+b^2}$
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$\cos(bt)=\frac{1}{b}(\sin(bt))'$
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$F'(s)=\frac{dF}{ds}(s)=\frac{d}{ds}\int _{0 } ^\infty e^{-st}f(t)\, dt=\int _{0} ^\infty f(t) \frac{d}{ds}(e^{-st}) \, dt$
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$=-\int _{0} ^\infty e^{-st}tf(t)\, dt$
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so that means:
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$-\frac{dF}{ds}=\mathcal{L}\{tf(t)\}$; again we can use induction to prove:
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$\mathcal{L}\{t''f(t)\}=(-1)^{n} \frac{d^nF}{ds^n}$
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$-\frac{dF}{ds}=\mathcal{L}\{tf(t)\}$; again we can use induction to prove the general formula:
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Property 7:
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$\mathcal{L}\{t^nf(t)\}=(-1)^{n} \frac{d^nF}{ds^n}$
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$\mathcal{L}\{t 1\}=-(s^{-1})^1=s^{-2}$
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$\mathcal{L}\{t^n\}= \frac{n!}{s^{n+1}}$ (can be proven by induction)
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today covers all midterm material. Yay!
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#end of lec 15
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#start of lec 16
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Again, using induction:
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Property 8:
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$\mathcal{L}\{t^n\}= \frac{n!}{s^{n+1}}$
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Today covers all midterm material. Yay!
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#end of lec 15 #start of lec 16
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He advises us to learn the table of common LT's, however a sheet will be provided for the exam.
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You can find the table of common LT's in the course textbook. There's a big table and a small table, the small table is the one he recommends we learn.
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## Examples
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#ex #LT
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lets try to compute the LT of:
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Lets try to compute the LT of:
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$\mathcal{L}\{t\cos (t)e^t\}= ?$
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Look at the table, which one would be useful? (the bottom one?)
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Look at the small table, which one would be useful? (the bottom one?)
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We know $\mathcal{L}\{t^nf(t)\}=(-1)^n \frac{d^nF}{ds^n}$ this property was shown last lecture, can be proven by induction. It's good to learn this by heart.
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$\mathcal{L}\{t\cdot\underbrace{ \cos (t)e^t }_{ f(t) }\}=-\frac{d}{ds}\mathcal{L}\{e^t\cos t\}$
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how do we compute that $\mathcal{L}$? Well now we can use that last row in the table!
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How do we compute that $\mathcal{L}$? Well now we can use the third row!
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$\mathcal{L}\{e^{\alpha t}f(t)\}=F(s-a)$ <-The shifting property
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$\mathcal{L}\{e^t\cos t\}=\dots$
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Recall property 6: $\mathcal{L}\{\cos(bt)\}=\frac{s}{s^2+b^2}$
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$=-\frac{d}{ds}\left( \frac{{s-1}}{(s-1)^2+1^2} \right)$
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$=-\frac{(s-1)^2+1-(s-1)2(s-1)}{((s-1)^2+1)^2}$
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$$=\frac{{(s-1)^2-1}}{((s-1)^2+1)^2}$$
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Doing this normally would be hard, with the table its a piece of cake!
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$$\mathcal{L}\{t\cos (t)e^t\}=\frac{{(s-1)^2-1}}{((s-1)^2+1)^2}$$
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Doing this normally would be hard, with the table it's a piece of cake!
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# Inverse Laplace Transform
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we know LT is a operator, we can also define an inverse transform! The inverse of LT. #inv_LT
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We know LT is a operator, we can also define an inverse transform! The inverse of LT. #inv_LT
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Here's the definition:
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$$\mathcal{L}^{-1}\{F(s)\}=f(t)$$
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Recall the forward LT was: $\mathcal{L}\{f(t)\}=F(s)$
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Next time hopefully, we will see how this will all be useful for solving DE!
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Next time hopefully, we will see how all this will be useful for solving DE's!
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We know the forward transform is linear, the inverse is also linear.
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$\mathcal{L}^{-1}\{\alpha F(s)+\beta G(s)\}=\alpha \mathcal{L}^{-1}\{F\}+\beta \mathcal{L}^{-1}\{G\}$
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This can be proven rather easily due to the linearity of the forward transform. (wasn't done in class unfortunately.)
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This can be proven rather easily due to the linearity of the forward transform (wasn't done in class unfortunately).
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## Examples
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#ex
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$\mathcal{L}^{-1}\left\{ \frac{1}{s^5}+\frac{3}{(2s+5)^2}+\frac{1}{s^2+4s+8}+ \frac{{s-1}}{s^2+2s+10} \right\}$ notice that all these terms approach 0 as s approaches inf.
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using linearity:
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second term doesn't look like its in the table, 3rd term looks like second last row
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$\mathcal{L}^{-1}\left\{ \frac{1}{s^5} \right\}+\frac{3}{4}\mathcal{L}^{-1}\left\{ \frac{1}{\left( s+\frac{5}{2} \right)^2} \right\}+\frac{1}{2}\mathcal{L}^{-1}{\frac{2}{(s+2)^2+2^2}}+\mathcal{L}^{-1}\left\{ \frac{{s+1}}{(s+1)^2+3^2} \right\}$
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multiply the first term by 4!, divide outside by 4!
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=$\frac{1}{4!}t^4+\frac{3}{4}te^{-5t/2}+\frac{1}{2}\sin(2t)e^{-2t}+e^{-t}\cos(3t)$
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it's kinda fun, try to match each term with something in the table, like a puzzle.
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We are done.
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Compute this inverse LT:
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$$\mathcal{L}^{-1}\left\{ \frac{1}{s^5}+\frac{3}{(2s+5)^2}+\frac{1}{s^2+4s+8}+ \frac{{s+1}}{s^2+2s+10} \right\}$$
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Notice that all these terms approach 0 as s approaches inf.
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Using linearity:
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$\mathcal{L}^{-1}\left\{ \frac{1}{s^5} \right\}+\mathcal{L}^{-1}\left\{ \frac{3}{\left( 2s+5 \right)^2} \right\}+\mathcal{L}^{-1}\left\{ \frac{1}{s^2+4s+8} \right\}+\mathcal{L}^{-1}\left\{ \frac{{s+1}}{s^2+2s+10} \right\}$
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Now we use the table! Um, but a lot of these terms don't really fit the table.
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You'll see how computing an inverse LT is kinda like solving a puzzle, or cooking imo. You look at the terms you are given and you try to manipulate it to fit something in the table. Kinda like how you stare at your pantry and think of what food you can cook with it. It's kinda fun!
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First term: take out an $\frac{1}{4!}$. This is possible due to linearity:
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$\frac{1}{4!}\mathcal{L}^{-1}\left\{ \frac{4!}{s^5} \right\}$ <- Now we can use $\mathcal{L}\{t^n\}=\frac{n!}{s^{n+1}}$
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Second term: get rid of that 2 beside the $s$ by dividing the top and bottom by 4. Move the $\frac{3}{4}$ outside.
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$\frac{3}{4}\mathcal{L}^{-1}\left\{ \frac{1}{\left( s+\frac{5}{2} \right)^2} \right\}$ <- Oh we are cooking! We can use the shifted property: $\mathcal{L}\{e^{\alpha t}t^n\}=\frac{n!}{(s-\alpha)^{n+1}}$
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Third term: complete the square, and move a $\frac{1}{2}$ term outside:
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$+\frac{1}{2}\mathcal{L}^{-1}\left\{ \frac{2}{(s+2)^2+2^2} \right\}$ <- We can use $\mathcal{L}\{e^{\alpha t}\sin bt\}=\frac{b}{(s-\alpha)^2+b^2}$
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Fourth term: Complete the square
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$+\mathcal{L}^{-1}\left\{ \frac{{s+1}}{(s+1)^2+3^2} \right\}$ <- We can use $\mathcal{L}\{e^{\alpha t}\cos bt\}=\frac{s-\alpha}{(s-\alpha)^2+b^2}$
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All in all we get:
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$$\mathcal{L}^{-1}\left\{ \frac{1}{s^5}+\frac{3}{(2s+5)^2}+\frac{1}{s^2+4s+8}+ \frac{{s+1}}{s^2+2s+10} \right\}=\frac{1}{4!}t^4+\frac{3}{4}te^{-5t/2}+\frac{1}{2}\sin(2t)e^{-2t}+e^{-t}\cos(3t)$$
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Isn't that fun!
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Well, every good thing must come to an end, we are done.
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#ex #inv_LT
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$\mathcal{L}^{-1}\{\frac{1}{(s-3)(s^2+2s+2)}\}$ notice the numerator is at least one degree lower than the denominator. The limit of the overall term is zero as s->inf
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it doesn't look like anything in the table, can we factor the denominator? not really, they have complex solutions. So maybe split the terms using partial fractions!
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$\mathcal{L}^{-1}\{\frac{1}{(s-3)(s^2+2s+2)}\}$ notice the numerator is at least one degree lower than the denominator. The limit of the overall term is zero as $s\to \infty$
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It doesn't look like anything in the table, can we factor the denominator? not really, they have complex solutions. So maybe split the terms using partial fractions!
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$=\mathcal{L}^{-1}\left\{ \frac{A}{{s-3}}+\frac{{Bs+C}}{s^2+2s+2} \right\}$
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$=\mathcal{L}^{-1}\left\{ \frac{{As^2+2As+2A+Bs^3+3Bs+Cs-3C}}{(s-3)(s^2+2s+2)} \right\}$
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$=\mathcal{L}^{-1}\left\{ \frac{{As^2+2As+2A+Bs^2-3Bs+Cs-3C}}{(s-3)(s^2+2s+2)} \right\}$
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we get a linear system of equations:
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$\begin{bmatrix}A+B&=0 \\2A-3B+C&=0 \\ 2A-3C&=1\end{bmatrix}$
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skipping some computations:
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$A=\frac{1}{17},\ B=-\frac{1}{17},\ C=-\frac{5}{17}$
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$=\frac{1}{17}\mathcal{L}^{-1}\left\{ \frac{1}{(s-3)} \right\}-\frac{1}{17}\mathcal{L}^{-1}\left\{ \frac{s}{(s+1)^2+1^2} \right\}-\frac{5}{17}\mathcal{L}^{-1}\left\{ \frac{1}{(s+1)^2+1^2} \right\}$
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$=\frac{1}{17}e^{3t}-\frac{1}{17}$
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for second term we use an important identity: $s=s+1-1$:
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$-\frac{1}{17}\mathcal{L}^{-1}\left\{ \frac{{s+1-1}}{(s+1)^2+1^2} \right\}$
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$=\frac{1}{17}\mathcal{L}^{-1}\left\{ \frac{1}{(s-3)} \right\}-\frac{1}{17}\mathcal{L}^{-1}\left\{ \frac{s+1-1}{(s+1)^2+1^2} \right\}-\frac{5}{17}\mathcal{L}^{-1}\left\{ \frac{1}{(s+1)^2+1^2} \right\}$
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$=\frac{1}{17}\mathcal{L}^{-1}\left\{ \frac{1}{(s-3)} \right\}-\frac{1}{17}\mathcal{L}^{-1}\left\{ \frac{s+1}{(s+1)^2+1^2} \right\}-\frac{5-1}{17}\mathcal{L}^{-1}\left\{ \frac{1}{(s+1)^2+1^2} \right\}$
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final answer:
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Final answer:
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$$\frac{1}{17}e^{3t}-\frac{1}{17}e^{-t}\cos(t)-\frac{4}{17}\sin(t)e^{-t}$$
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partial fractions, hopefully you remember from math 101:
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@ -0,0 +1,73 @@
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#start of lec 22
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Finished chapter 7 of the course textbook, Let's begin chapter 8!
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# Power series
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A power series is defined by:
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$$\sum_{n=0}^\infty a_{n}(X-X_{0})^n=a_{0}+a_{1}(x-x_{0})+a_{2}(x-x_{0})^2+\dots$$
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It is convergent if:
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$$\sum_{n=0} ^ \infty a_{n}(x-x_{0})^n<\infty \text{ at a given x}$$
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Otherwise, it is divergent.
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If $\sum_{n=0}^\infty \mid a_{n}(x-x_{0})^n\mid$ is convergent
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$\implies\sum_{n=0}^\infty a_{n}(x-x_{0})^n$ is absolutely convergent
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Just because something is absolutely convergent doesn't mean it is conditionally convergent. think of the harmonic series. It is absolutely convergent but also divergent (conditionally divergent).
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Theorem: With each $\sum_{n=0}^{\infty}a_{n}(x-x_{0})^n$ we can associate $0\leq \rho\leq \infty$ such that
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$\sum_{n=0} ^\infty a_{n}(x-x_{0})^n$ is absolutely convergent
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for all x such that $\mid x-x_{0}\mid<\rho$, divergent for all x where $\mid x-x_{0}\mid>\rho$
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"Who keeps stealing the whiteboard erases? (jokingly) It's a useless object, anyways"
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![[Drawing 2023-10-30 13.12.57.excalidraw.png]]
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how can we find $\rho$?
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Ratio test: If $\lim_{ n \to \infty }\mid \frac{a_{n+1}}{a_{n}}\mid=L$
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then $\rho=\frac{1}{L}$
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## Examples:
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#ex
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is this convergent? Divergent? and where so?
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$\sum_{n=0}^\infty \frac{2^{-n}}{n+1}(x-1)^n$
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determine the convergent set.
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Use ratio test:
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$\lim_{ n \to \infty }\mid \frac{a_{n+1}}{a_{n}}\mid=\lim_{ n \to \infty } \frac{2^{-(n+1)}}{n+2} \frac{n+1}{2^{-n}}=\frac{1}{2}\implies \rho=2$
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so it's convergent on $-1<x<3$, divergent on $\mid x-1\mid>2$
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But what about on the points $-1$ and $3$?
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plug in $x_{0}=-1$
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$\sum_{n=0}^{\infty} \frac{2^{-n}}{n+1}(-2)^n=\sum_{n=0}^\infty \frac{(-1)^n}{n+1}<\infty$ <- That is the alternating harmonic series, it is convergent.
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plug in $x_{0}=3$:
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$\sum_{n=0}^\infty \frac{2^{-n}}{n+1}2^n=\sum_{n=0}^\infty \frac{1}{n+1}>\infty$ <- harmonic series, this diverges.
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so the power series is convergent on $[-1,3)$ divergent otherwise.
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$$\text{ converges only on: } [-1,3)$$
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Assume that $\sum_{n=0}^\infty a_{n}(x-x_{0})^n$ and $\sum_{n=0}^\infty b_{n}(x-x_{0})^n$ are converget with $\rho>0$
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Then:
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1.) $\sum_{n=0}^\infty a_{n}(x-x_{0})^n+\sum_{n=0}^{\infty}b_{n}(x-x_{0})^n=\sum_{n=0}^\infty(a_{n}+b_{n})(x-x_{0})^n$
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That has a radius of convergence of at least $\rho$.
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2.) $\left( \sum_{n=0}^\infty a_{n}(x-x_{0})^n \right)\left( \sum_{n=0}^\infty b_{n}(x-x_{0})^n \right) \qquad c_n=\sum_{k=0}^n a_{k}b_{n-k}$(Cauchy)
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$=(a_{0}+a_{1}(x-x_{0})+a_{2}(x-x_{0})^2+\dots)(b_{0}+b_{1}(x-x_{0})+b_{2}(x-x_{0})^2+\dots)$
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$=a_{0}b_{0}+(a_{0}b_{1}+a_{1}b_{0})(x-x_{0})+(a_{0}b_{2}+a_{1}b_{1}+a_{2}b_{1})(x-x_{0})^2+\dots$ (Cauchy multiplication)
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more Definitions of power series:
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||||
If $\sum_{n=0}^{\infty}a_{n}(x-x_{0})^n$ is convergent with $\rho>0$
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$\mid x-x_{0}\mid<\rho$
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we can differentiate this infinite sum and get:
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$\implies y'(x)=\sum_{n=1}^\infty a_{n}n(x-x_{0})^{n-1}$
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$y''(x)=\sum_{n=2}^\infty a_{n}n(n-1)(x-x_{0})^{n-2}$
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Theorem: If $y(x)$ is infinitely many times differentiable on some interval: $\mid x-x_{0}\mid<\rho$
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then: $\sum_{n=0}^\infty \frac{y^{(n)}(x_{0})}{n!}(x-x_{0})^n$ (Taylor series)
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"believe me, taylor series is the most important theorem in engineering."
|
||||
"I mean engineering is all about approximations, do you know how your calculator computes ...? Taylor series!"
|
||||
"Applied mathematics is all about approximating and then measuring how good your approximation is, it's what engineering is all about." -Prof (loosy quotes, can't keep up with how enthusiastic he is!)
|
||||
|
||||
Definition: If $y(x)$ can be represented with a power series on $\mid x-x_{0}\mid$ then $y(x)$ is an analytic function on $(x_{0}-\rho,x_{0}+\rho)$
|
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btw analytic functions are very important in complex calculus MATH301. (i don't have that next term)
|
||||
|
||||
$f(x)=\sum_{n=0}^\infty a_{n}(x-x_{0})^n<\infty$
|
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$f'(x)=\sum_{n=1}^\infty a_{n}n(x-x_{0})^{n-1}$
|
||||
$f(x)+f'(x)=\sum_{n=0}^\infty a_{n}(x-x_{0})^n+\sum_{n=1}^\infty a_{n}n(x-x_{0})^{n-1}$
|
||||
let $n-1=k$
|
||||
$=\sum_{n=0}^\infty a_{n}(x-x_{0})^n+\sum_{k=0}^\infty a_{n}(k+1)(x-x_{0})^{k}$
|
||||
$=\sum_{n=0}^\infty(a_{n}+a_{n}(n+1))(x-x_{0})^n$
|
||||
|
||||
Last theorem fo' da day:
|
||||
If $\sum_{n=0}^\infty a_{n}(x-x_{0})^n=0$ for all x$\in(x_{0}-\rho,x_{0}+\rho)$ where $\rho>0$
|
||||
$\implies a_{n}=0$, $n=0,1,2,\dots$
|
||||
#end of lec 22
|
||||
|
|
@ -22,6 +22,7 @@ Good luck on midterms! <3 -Oct 18 2023
|
|||
[Periodic functions (lec 19)](periodic-functions-lec-19.html) (raw notes, not reviewed or revised yet.)
|
||||
[Convolution (lec 19-20)](convolution-lec-19-20.html) (raw notes, not reviewed or revised yet.)
|
||||
[Dirak δ-function (lec 21)](dirak-δ-function-lec-21.html) (raw notes, not reviewed or revised yet.)
|
||||
[Power series (lec 22)](power-series-lec-22.html) (raw notes, not reviewed or revised yet.)
|
||||
</br>
|
||||
[How to solve any DE, a flow chart](Solve-any-DE.png) (Last updated Oct 1st, needs revision. But it gives a nice overview.)
|
||||
</br>
|
||||
Loading…
Reference in New Issue