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Friday, 17 July 2015

Fundamentals of calculus Chapter 2: Integration

Fundamentals of calculus
Chapter 2: Integration

Integration is the converse of differentiation. To understand this better, it must be known that during differentiation an important information is lost. Let $\frac{dz(x)}{dx} = z’(x)$. Then, we cannot get back $z(x)$ from $z’(x)$ because, while taking the difference $z(x+\Delta x) – z(x)$ only the information regarding the difference between a point in Z axis and a neighbourhood of that point is retained. But the actual orientation is lost. Throughout this article, unless stated $\Delta x$ is limited to zero. To see how this could cause the loss of a piece of information, let me elaborate this with a simple algebraic example. Consider $a-b = c$ to be true. Then we may write $a – (a-c) = c$, here $b=a-c$.

It’s obvious that just from the value of $c$ it isn’t possible to incur the values of $a$ and $b$. But it is possible to incur that $c$ is some value obtained by subtracting $a$ and $b$, though the two values may be anything they are interrelated. So there is a possibility that after performing the differentiation operation, it is not possible to say that we can get back the original function. After computing the differentiation, we have $\frac{z(x+\Delta x) – z(x)}{\Delta x} = z’(x) = \frac{z(x) }{\Delta x} + z’(x) -\frac{z(x) }{\Delta x} = z’(x) $. Note that the expression $\frac{z(x) }{\Delta x}$ can be anything and is independent of $z’(x)$, if we treat them as algebraic constants. This is not necessarily same as $a – (a-c) = c , b=a-c$, which clearly shows that $a$ is totally independent of $c$. But in this case, both are a function of $x$.

On rearranging we may write, $\frac{z(x + \Delta x) }{\Delta x} –z’(x) = \frac{z(x )}{\Delta x}$. We need a function $z’(x)$ such that $z(x+\Delta x) – z’(x)\Delta x = z(x)$. Now let’s explore and find how we may obtain $z(x)$ from $z’(x)$. We know that $z’(b) \Delta x + z(b) = z(b+\Delta x)$ and $z’(b+\Delta x)\Delta x + (z’(b) \Delta x + z(b)) = z’(b+\Delta x) + z(b + \Delta x) = z(b+ 2\Delta x) $. So generalising this to obtain $z(x)$ from $z(b), x > b$ by implementing the above property recursively, we get $z(x) = z(b) + \Delta x (z’(b+\Delta x) + z’(b + 2\Delta x) + z’(b + 3\Delta x) +… +z’(b+ \frac{x – b}{\Delta x} \Delta x) = z(x)$. We may write this using the summation notation in the following way,
                        $z(x) = z(b) + \Delta x \times \sum \limits_{i=0}^{\frac{x-b}{\Delta x}} {z’(b + i\times \Delta x)}$.
The above holds good even if $x < b$, because for this case, instead of $z’(b + |i|\times \Delta x)$ we will be needing $z’(b - |i|\times \Delta x)$; this will automatically be true if $i$ ranges from $0\to \frac{x-b}{\Delta x}$. So now let’s use this definition to find the differential of $z(x)$. 
$\frac{z(x+\Delta x) – z(x)} {\Delta x} = \frac{1}{\Delta x}\times((z(b) + \Delta x \times\sum \limits_{i=0}^{\frac{x+\Delta x-b}{\Delta x}} {z’(b + i\times \Delta x)})– (z(b) + \Delta x \times \sum \limits_{i=0}^{\frac{x-b}{\Delta x}} {z’(b + i\times \Delta x)}))$
Which can be written as,
$z’(x+\Delta x) = z’(x)$.

So what is the information that is being freely lost? It’s obvious that the constant $z(b)$ is the term present in both $z(x)$ expression and $z(x+\Delta x)$ expression. So the information that is lost is the starting point! From the article ‘fundamentals of differentiation, chapter 1’ know that $z’(x)$ is the rate of change of $z(x)$ for each change in $x$. So if we know the rate of change, we also need to know the point from where the function is to be generated on the graph; without which we will not be able to plot the right graph. Note that from $z’(x)$ we can construct $z(x)$ if we know $z(b)$, where $b$ being a constant that lies on the X axis and $z(b)$ being a constant that lies on the Z axis. So the term that gets lost is $z(b)$, so while integrating (or obtaining the function $z(x)$ from $z’(x)$), we will not be able to recover $z(b)$ and hence we just leave it as a constant. This unknown constant shifts the graph generated by $z(x)$ along the Z axis.

So if we write $g(x) = z(x)+c$ then the graph $g(x)$ is shifted upwards by $c$ units. So from the above explanation, it is obvious that we just lose the information that denotes the orientation of the graph along the Z axis.  

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