[Twelve] Laplace transform --1

Author: LeeKunHwee
link: https: //www.jianshu.com/p/400d93229ff5
Source: Jane books

A more interesting explanation of the Laplace transform:

A good way of thinking of where the Laplace Transform comes from, and a way which dispels some of its mystery is by thinking of power series .-- Arthur Mattuck (MIT Department of Mathematics Fanpin professor, former MIT math department)

A better explanation about the Laplace transform method is to start from the Power Series (Power Series).

Pierre-Simon marquis de Laplace (1749-1827) (Source: Wikipedia)

Pierre - Simon Laplace Marquis (French: Pierre-Simon marquis de Laplace), the famous French astronomer and mathematician, his work has a pivotal development of celestial mechanics and statistics. He is also the discoverer of the Laplace transform and Laplace equation, outstanding contribution to the development of mathematics and physics.

Learned to control all know Laplace transforms (Laplace Transform), but you are also not in doubt, Laplace transforms S in the end is what the hell? Pierre - Simon Laplace Marquis then why could come up with such mathematical transformation formula?

Since I have been in contact with the Laplace transform have such a question, feel this kind of thing very strong line, you do not understand but can not refuse. Until one day, I looked at the differential equation Arthur Mattuck came before, and worship the great God!

We know that a power series can be written as follows:

$\large \dpi{80} \mathbf{A(x)=\sum_{n=0}^{\infty }a_{n}x^{n}}$

If the A n- as a discrete series of functions, then the above equation may be written as:

$\large \mathbf{A(x)=\sum_{n=0}^{\infty }a(n)x^{n}}$

To $a(n)$ as a set of discrete functions as a power series coefficients, the equation can be seen as a function $A(x)$ configuration process, i.e., a long input $a(0),a(1),a(2),...$ sequences, one can be output $A(x)$ , wherein $x$ the output function $A(x)$ arguments.

I understand: any of these functions can be written in the form of power series.

This transformation can be seen as a map, $a(n)$ to $A(x)$ , accurate to say that $a(n)$ to $A(x^n)$ .

$a(n)$ The group is $x$ , $A(x^n)$ the group is $x^n$ .

Now, give an example, if you take $\large a(n)=1$ that $\large a(0)=1,a(1)=1,a(2)=1,...$ we would get:

$\large A(x)=1+x^{1}+x^{2}+x^{3}+...$

Some people say that the style is equal to the last $\large \frac{1}{1-x}$ , but say that is not really accurate, because not all of $\large x$ the above formulas are true only if it is established a convergent series! The above formula $x$ convergence domain $\large (-1,1)$ , so that the formula can be rewritten as:

$\large A(x)=1+x^{1}+x^{2}+x^{3}+...\frac{1}{1-x},\left | x \right |<1$

As another example, if $\large a(n)=\frac{1}{n!}$ there are:

$\large A(x)=1+\frac{1}{1!}x^{1}+\frac{1}{2!}x^{2}+\frac{1}{3!}x^{3}+...=e^x$

I understand: is not to be understood as the $\large a(n)=\frac{1}{n!}$ coordinate transformation obtained $\large A(x)$ . Is transformed coordinates $\large (1,x^{1},x^{2},x^{3},...,x^n)$ , and the coordinate value is $\large (1,\frac{1}{1!},\frac{1}{2!},\frac{1}{3!}x^{3},...,\frac{1}{n!})$ .

In this case, $\large x$ for any real numbers are true, in fact, the formula is $\large e^x$ in $\large x=0$ at the Taylor expansion.

As can be seen from the above example, taking a discrete function defined on negative integer or a positive integer, followed by adding and operation, the result is a continuous function can be generated. Note that one of the discrete function $\large a_{n}$ variable is $\large n$ plus and the result is about the variables $x$ . In short, this is a property of power series, also belong to the situation Discrete sum.

This assumption allows continuous rather than discrete summing becomes, i.e., the variable is not so $\large n=0,1,2,3...$ , another definition of a variable $\large t$ , and $\large 0\leqslant t< \infty$ that t can be $\large [0,\infty)$ arbitrary real numbers in.

If you want $\large t$ to take an alternative $\large n$ , obviously can not then the above approach to discrete sequence summed over all real numbers, but through integration. which is:

$\large \mathbf{A(x)=\int_{0}^{\infty }a(t)x^{t}}dt$

We can maintain this form, but did not want to do that mathematicians, and engineers rarely do so, because when performing integral and differential operation, no one wants that contains the end of an exponential function is x integration or differentiation and the like items, which people seem a headache. The only convenience is the natural base- E .

Only e is for people like integral or differential, because of the exponential function is the natural base-end of Y = e ^ A X be the result of integration or differentiation or whether it itself or simply multiplied by a coefficient, satisfying only this one, no branch function of the nature of the world! ! ! Want to know the source of e, see " natural base e how it is" natural "a ."

So we will be x raised to the exponential transformation into a e exponent in base of:

$\large \mathbf{A(x)=\int_{0}^{\infty }a(t)(e^{\ln x})^{t}}dt$ 。

Now, we look at the points, obviously, we write this integration certainly hope it is solvable, or convergence. After all, this is a zero to infinity from generalized integral, we need special treatment, only if $\large x$ the possible points of convergence when a number is less than 1, the only way, when power is growing, the number will get more smaller and smaller, it is required here $\large x< 1$ . Then, we also want to $x$ be positive, otherwise it will encounter negative powers of trouble, such as when $\large x=-1$ , $\large t=1/2$ when, will be imaginary, this is what we want to see, so they requested $\large 0<x<1$ that we do so in order to make the integral convergence. So in this case, $\large lnx$ and it looks like it? Obviously, $\large 0<x<1$ when $\large lnx<0$ .

$\large lnx$ This variable seems, a bit complicated, we do not have a variable to replace what it does?

Then use s it!

Now let $\large s= -lnx$ or $\large -s= lnx$ because $\large lnx<0$ , take $\large -s= lnx$ it, $\large s$ it is always positive, and positive treatment course more in line with people's habits. In addition, we use $\large f(x)$ instead of $\large a(x)$ , it looks like we are familiar with functional form. Various alternatives are just above us in order to modify, we will replace these into the equation, yielding:

$\large F(s)=\int_{0}^{\infty }f(t)e^{-st}dt$ 。

We actually got the Laplace Transform this way! ! !

If you use symbols instead, the formula can be written as: $\large F(s)=\pounds [f(t)]$

This is the Laplace transform, when one $t$ of the input function, the obtained about $s$ the function.

Also to mention here that the " transformation ", in fact, in mathematics there is a concept called " operator ", and transform and calculate the most essential difference is that the child, after "operator" operation, the variable has not changed, such as differentiation is a typical operator, and after "transformed" as variables will change the operation.

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[Twelve] Laplace transform --1

A more interesting explanation of the Laplace transform:

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