node and inverse mathematical functions in conjunction with a combination of electronic switches analog and digitally

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Inverse Functions

Two functions are inverses of one another if they "undo each other" in the following sense: if the output of one is used as input to the other, they leave the original input unchanged.

To be precise, two functions f $f$ and g $g$ are inverses of each other if and only if f(g(x))=x $f(g(x))=x$ for every value of x $x$ in the domain of g $g$ and g(f(x))=x $g(f(x))=x$ for every value of x $x$ in the domain of f $f$ .

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Example
Given f(x)=x 3  $f(x)=x^3$ and g(x)=x − −  √ 3  $g(x)=\sqrt[3]{x}$ , we know f $f$ and g $g$ are inverses since

f(g(x))g(f(x)) == f(x − −  √ 3 )g(x 3 ) == (x − −  √ 3 ) 3 x 3  − −  √ 3  == xandx  $$\begin{array}{rcccccl}f(g(x))& =& f(\sqrt[3]{x})& =& (\sqrt[3]{x}{)}^3& =& x\text{and}\\ g(f(x))& =& g(x^3)& =& \sqrt[3]{x^3}& =& x\end{array}$$

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We also say a function f $f$ is invertible if there exists a function g $g$ such that f $f$ and g $g$ are inverses of one another.

Notation for Inverse Functions

Historically, we let f −1  $f^{-1}$ denote the inverse of an invertible function f $f$ .
A common source of confusion for students stems from how similar this notation looks to exponentiation.
However, if we wish to apply a power of −1 $-1$ to something, that something must be a value. The letter f $f$ represents a function, not a value!
On the other hand, f(x) $f(x)$ is a value -- it's the output of function f $f$ given an input of x $x$ . So f(x) $f(x)$ actually can be raised to the −1 $-1$ power...
Putting these thoughts together into an example, suppose f(x)=x 3  $f(x)=x^3$ . Then,

f −1 (x)=x − −  √ 3 while,[f(x)] −1 =1x 3   $$f^{-1}(x)=\sqrt[3]{x}\text{while,}[f(x){]}^{-1}=\frac{1}{x^3}$$

Graphs of Inverse Functions

We have already established that f(x)=x 3  $f(x)=x^3$ and g(x)=x − −  √ 3  $g(x)=\sqrt[3]{x}$ are inverses. Let us consider the graphs of these two functions, as shown below:

There is a striking symmetry between these two graphs -- one that is shared between any pair of inverse functions we might choose to draw.
In every case, two functions that are inverses of one another will be symmetric about the line y=x $y=x$ .
Equivalently, if (x,y) $(x,y)$ will be a point on the graph of some invertible function f $f$ if and only if (y,x) $(y,x)$ is a point on the graph of y=f −1 (x) $y=f^{-1}(x)$ .
Just consider the graphs of f(x)=x 3  $f(x)=x^3$ and g(x)=x − −  √ 3  $g(x)=\sqrt[3]{x}$ shown above. The points (−2,−8) $(-2,-8)$ , (−1,−1) $(-1,-1)$ , (0,0) $(0,0)$ , (1,1) $(1,1)$ , and (2,8) $(2,8)$ are all points on the graph of the former. While (−8,−2) $(-8,-2)$ , (−1,−1) $(-1,-1)$ , (0,0) $(0,0)$ , (1,1) $(1,1)$ , and (8,2) $(8,2)$ are all on the graph of the latter.
To see why this is so, suppose (x 0 ,y 0 ) $(x_0,y_0)$ is a point on the graph of y=f(x) $y=f(x)$ and f $f$ is invertible.
Consequently, f(x 0 )=y 0  $f(x_0)=y_0$ , and thus, f −1 (f(x 0 ))=f −1 (y 0 ) $f^{-1}(f(x_0))=f^{-1}(y_0)$ .
However, since f −1  $f^{-1}$ is the inverse of f $f$ , we also know f −1 (f(x 0 ))=x 0  $f^{-1}(f(x_0))=x_0$ .
Hence, f −1 (y 0 )=x 0  $f^{-1}(y_0)=x_0$ , which tells us that (y 0 ,x 0 ) $(y_0,x_0)$ is a point on the graph of y=f −1 (x) $y=f^{-1}(x)$ .
Since every point (x,y) $(x,y)$ on the graph of an invertible function yields a point (y,x) $(y,x)$ on the graph of its inverse and vice-versa, and given that the domain of a function is the set of allowable inputs x $x$ , and the range is the set of obtainable outputs y $y$ -- we have as an immediate consequence that the domain of an invertible function is the range of its inverse, and the range of such a function is the domain of its inverse.

The Horizontal Line Test

Not every function is invertible. Fortunately, we can tell which functions are invertible from those that are not by a quick examination of their graphs.
As an example, suppose you are interested in determining whether or not f(x)=x 2  $f(x)=x^2$ is invertible.
Keeping in mind the aforementioned symmetry between inverse functions about the line y=x $y=x$ , suppose we graph both y=x 2  $y=x^2$ (shown in blue below) and its reflection about the line y=x $y=x$ (shown in red below).

Notice, the red curve doesn't pass the vertical line test! That is to say, there is a vertical line that crosses the red curve more than once [e.g., the line through (4,2) $(4,2)$ and (4,−2) $(4,-2)$ ].
Equivalently, there is at least one x $x$ (e.g., x=4 $x=4$ ) that is associated with more than one y $y$ -value (e.g., both 2 $2$ and −2 $-2$ ). Consequently, the red curve does not represent a function.
As the inverse to a function must by definition be a function itself, it must be the case that f(x)=x 2  $f(x)=x^2$ is not invertible.
Note, we could have been more efficient by realizing that any intersections along a vertical line in the graph of the reflection across the line y=x $y=x$ must correspond to intersections along a horizontal line in the graph of y=f(x) $y=f(x)$ .
Thus, if any horizontal line crosses the graph of y=f(x) $y=f(x)$ more than once, then f $f$ will not have an inverse function associated with it, and is consequently not invertible.
Consequently, if a function is invertible it must be the case that for every x $x$ value in the domain, there is one and only one y $y$ value associated with that x $x$ value, and for every y $y$ value in the range, there is one and only one x $x$ value associated with that y $y$ value. As such, we also describe invertible functions as functions that are one-to-one.

Solving for an Inverse Function

In certain circumstances, given a formula for some invertible function f $f$ , we can determine a formula for its inverse f −1  $f^{-1}$ .
To do this, we take advantage of the previously discovered symmetry between the coordinates of points on the graphs of a function and its inverse, as the following suggests:
Suppose we wished to find a formula for the inverse of a function f $f$ defined by f(x)=1+x1−2x  ${\displaystyle f(x)=\frac{1+x}{1-2x}}$ .
First, we note that if (x,y) $(x,y)$ is a point on the graph of f −1  $f^{-1}$ , then (y,x) $(y,x)$ must be a point on the graph of f $f$ .
Then, it must be the case that

x = 1+y1−2y   $${\displaystyle \begin{array}{rcl}x& =& {\displaystyle \frac{1+y}{1-2y}}\end{array}}$$

From here, we can simply solve for y $y$ , which then gives us a formula for f −1 (x) $f^{-1}(x)$ :

Inverse Function of ln(x)

What is the inverse function of the natural logarithm of x?
The natural logarithm function ln(x) is the inverse function of the exponential function e^x.
When the natural logarithm function is:

f (x) = ln(x),  x>0

Then the inverse function of the natural logarithm function is the exponential function:

f ^-1(x) = e^x

So the natural logarithm of the exponent of x is x:

f (f ^-1(x)) = ln(e^x) = x

Or

f ^-1(f (x)) = e^ln(x) = x

  

What is the natural logarithm of 1?

What is the natural logarithm of one.

ln(1) = ?

The natural logarithm of a number x is defined as the base e logarithm of x:

ln(x) = log_e(x)

So

ln(1) = log_e(1)

Which is the number we should raise e to get 1.

e⁰ = 1

So the natural logarithm of one is zero:

ln(1) = log_e(1) = 0

 

What is the natural logarithm of e?

What the natural logarithm of the e constant (Euler's constant)?

ln(e) = ?

The natural logarithm of a number x is defined as the base e logarithm of x:

ln(x) = log_e(x)

So the natural logarithm of e is the base e logarithm of e:

ln(e) = log_e(e)

ln(e) is the number we should raise e to get e.

e¹ = e

So the natural logarithm of e is equal to one.

ln(e) = log_e(e) = 1

 

 

Natural Logarithm of Infinity

What is the natural logarithm of infinity?

ln(∞) = ?

Since infinity is not a number, we should use limits:

x approaches infinity

The limit of the natural logarithm of x when x approaches infinity is infinity:

lim ln(x) = ∞

  x→∞

x approaches minus infinity

The opposite case, the natural logarithm of minus infinity is undefined for real numbers, since the natural logarithm function is undefined for negative numbers:

lim ln(x) is undefined

  x → -∞

So we can summarize

ln(∞) = ∞

 

ln(-∞) is undefined

 

 

 

 

 



 

 

   

               

   

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Inverse Operations and Functions

An operation we might do with a glove is put on. Another operation that could be done is take off. If we start with a bare hand:

If we start with a glove on:

The operations put on and take off undo each other. If we do one operation then the other, we end up where we started. Put on is the inverse operation to take off.  Take off is the inverse operation of put on.  Such operations form an operation-inverse operation pair.
The same is true in mathematics.  Most operations have an inverse operation. Starting with the simplest operations:

and

Add and Subtract are inverse operations. Similarly multiply and divide are inverse operations, except division by zero is not allowed.

You may have thought multiply and multiply by the reciprocal are the inverse pair. Since divide and multiply by the reciprocal are equivalent operations this is quite true.
Let's think about exponents.  We can get from a number to that number to the power of 2 by squaring the number.  To get back to the original number we need to take the square root.

And in general, raising to a power and taking the root are inverse operations. Another common pair is cube-cube root.

Raising the base to a power and getting the logarithm (to that base) are also inverse operations.Recall that the expression y = 10^x means y is equal to 10 raised to the power of x.  x is the exponent and 10 is the base.  This can also be written as x = log₁₀ y.

A pair that are very common from the various logarithms is the natural logarithm, ln, and the exponent, e.

and

The inverse trigonometric pairs are sin and sin^-1, cos and cos^-1 and tan with tan^-1.These are dealt with in detail in Inverse Trigonometric Functions.
Sometimes an operation is its own inverse. Take a bus is an example.

A mathematical example is the reciprocal.

 

 

 

 

  

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Laplace Transform

 

The Laplace transform is an integral transform perhaps second only to the Fourier transform in its utility in solving physical problems. The Laplace transform is particularly useful in solving linear ordinary differential equations such as those arising in the analysis of electronic circuits.

The (unilateral) Laplace transform (not to be confused with the Lie derivative, also commonly denoted ) is defined by

(1)

where is defined for (Abramowitz and Stegun 1972). The unilateral Laplace transform is almost always what is meant by "the" Laplace transform, although a bilateral Laplace transform is sometimes also defined as

(2)

(Oppenheim et al. 1997). The unilateral Laplace transform is implemented in the Wolfram Language as LaplaceTransform[f[t], t, s] and the inverse Laplace transform as InverseRadonTransform.
The inverse Laplace transform is known as the Bromwich integral, sometimes known as the Fourier-Mellin integral (see also the related Duhamel's convolution principle).
A table of several important one-sided Laplace transforms is given below.

		conditions
1

In the above table, is the zeroth-order Bessel function of the first kind, is the delta function, and is the Heaviside step function.
The Laplace transform has many important properties. The Laplace transform existence theorem states that, if is piecewise continuous on every finite interval in satisfying

(3)

for all , then exists for all . The Laplace transform is also unique, in the sense that, given two functions and with the same transform so that

(4)

then Lerch's theorem guarantees that the integral

(5)

vanishes for all for a null function defined by

(6)

The Laplace transform is linear since

			(7)
			(8)
			(9)

The Laplace transform of a convolution is given by

(10)

Now consider differentiation. Let be continuously differentiable times in . If , then

(11)

This can be proved by integration by parts,

			(12)
			(13)
			(14)
			(15)

Continuing for higher-order derivatives then gives

(16)

This property can be used to transform differential equations into algebraic equations, a procedure known as the Heaviside calculus, which can then be inverse transformed to obtain the solution. For example, applying the Laplace transform to the equation

(17)

gives

(18)

(19)

which can be rearranged to

(20)

If this equation can be inverse Laplace transformed, then the original differential equation is solved.
The Laplace transform satisfied a number of useful properties. Consider exponentiation. If for (i.e., is the Laplace transform of ), then for . This follows from

			(21)
			(22)
			(23)

The Laplace transform also has nice properties when applied to integrals of functions. If is piecewise continuous and , then

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

How to Invert a Function to Find Its Inverse

 

If you’re given a function and must find its inverse, first remind yourself that domain and range swap places in the functions. Literally, you exchange f(x) and x in the original equation. When you make that change, you call the new f(x) by its true name — f^–1(x) — and solve for this function.

For example, follow the steps to find the inverse of this function:

1. Switch f(x) and x.
   When you switch f(x) and x, you get
   
   
   (Note: To make the notation less clumsy, you can rewrite f(x) as y and then switch x and y.)
2. Change the new f(x) to its proper name — f^–1(x).
   The equation then becomes
   
3. Solve for the inverse.
   This step has three parts:
   1. Multiply both sides by 3 to get 3x = 2f^–1(x) –1.
   2. Add 1 to both sides to get 3x + 1 = 2f^–1(x).
   3. Lastly, divide both sides by 2 to get your inverse:
         



How to inverse behaviour of a switch?