Can somebody explain how they got equation (4)
http://ocw.mit.edu/courses/mathematics/18-03sc-differential-equations-fall-2011/unit-ii-second-order-constant-coefficient-linear-equations/linear-operators-linear-time-invariance/MIT18_03SCF11_s17_6text.pdf

Question

Can somebody explain how they got equation (4)
http://ocw.mit.edu/courses/mathematics/18-03sc-differential-equations-fall-2011/unit-ii-second-order-constant-coefficient-linear-equations/linear-operators-linear-time-invariance/MIT18_03SCF11_s17_6text.pdf

ganeshie8 · Answer

this particular underlined line

|dw:1445706431668:dw|

ganeshie8 · Answer

@baru and @Nishant_Garg  also want to know this

freckles · Answer

http://openstudy.com/study#/updates/501dcdb1e4b0be43870e7260 I don't understand this explanation. Do you?

ikram002p · Answer

Okay

anonymous · Answer

The way I see it, it's an application of undetermined coefficients.

Let's consider $P(x)=x^2+x+1$ as an example. Clearly this is a function of $x$, but we can write it as a function of any shift in $x$ while still having the polynomial equivalent to $P(x)$. Suppose we want to express $P(x)$ in terms of $x-1$. Then
$$\begin{align*}
x^2+x^2+1&=a(x-1)^2+b(x-1)+c\
&=ax^2+(-2a+b)x+(a-b+c)
\end{align*}$$giving
$$\begin{cases}a=1\-2a+b=1\a-b+c=1\end{cases}\implies\begin{cases}a=1\b=3\c=3\end{cases}$$
So to make the connection with the PDF: you're basically writing $q(D)$ not as a polynomial in $D$, but rather in $D-a$, where the $c_i$, $i=1,\ldots,k$, are the same as the coefficients $a,b,c$ in my example.

anonymous · Answer

Notice that in my example, you can equate $q(D)=D^2+D+1$, then if $a=1$ you get $\color{red}{q(1)}=1^2+1+1=3$, and
$$q(D)=D^2+D+1=(D-1)^2+3(D-1)+\color{red}3$$

baru · Answer

yes that makes sense....
we re-write q(D) shifted by a as a polynomial with unkown co-eff.
$$q(D)= c_0 +c_1(d-a) + c_2(d-a)^2....$$
if we set D=a , all the terms on the RHS other than c_0 vanishes..
thus
$$c_0=q(a)$$

baru · Answer

also i just googled this in general terms: f(x)=f(a) -c1(x-a)....
it showed power series...any connection with that ?? ( i dont know power series)

anonymous · Answer

There is! Recall that any function can be approximated to a certain degree with an appropriate finite-term polynomial. For example, linear approximations are often used if you can find a reliable one, and also they're easy to compute: https://en.wikipedia.org/wiki/Linear_approximation A power series extends this idea to polynomials with an infinite number of terms. In fact, you can represent some functions *exactly* with power series representations. For instance, $$e^x=1+x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24}+\underbrace{\cdots}_{\text{goes on forever}}=\sum_{k=0}^\infty \frac{x^k}{k!}$$Here are some neat animations that demonstrate this: http://www.physics.miami.edu/~nearing/mathmethods/power-animations.html Basically, as you add more terms, the polynomial follows a general trend of behaving more and more like the function they approximate. Not every function has a "perfect" power series representation. Discontinuities can easily disrupt approximations, while even a function that's continuous might only have a power series representation over part of its domain. There are quite a few contextual advantages to using polynomial/power series representations in place of functions. For example, computing $$\quad\quad\quad\quad I=\int_0^1e^{-x^2}\,dx$$would be really difficult without prior knowledge of the non-elementary antiderivative of $e^{-x^2}$ or familiarity with probability distributions. However, you can use the power series for $e^x$ and integrate to find at least $$\begin{align*}I&=\int_0^1\sum_{k=0}^\infty \frac{(-x^2)^k}{k!}\,dx\[1ex] &=\sum_{k=0}^\infty \frac{(-1)^k}{k!}\int_0^1x^{2k}\,dx\[1ex] &=\sum_{k=0}^\infty \frac{(-1)^k}{k!}\left[\frac{x^{2k+1}}{2k+1}\right]_0^1\[1ex] &=\sum_{k=0}^\infty \frac{(-1)^k}{k!(2k+1)}\end{align*}$$ Of course, this sum isn't much use in its current form, but in practice we can make do with just the first few terms to approximate the value of the integral. For example, $$I\approx\sum_{k=0}^3 \frac{(-1)^k}{k!(2k+1)}=1-\frac{1}{3}+\frac{1}{10}-\frac{1}{42}=0.742857$$Compare to the actual value: $$I=\frac{\sqrt{\pi}}{2}\text{erf}(1)\approx 0.746824$$

anonymous · Answer

@freckles That answer isn't really an explanation, but rather the general procedure for determining the values of the $c_i$.