Properties of Integrals and differentials

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I had a couple of questions.

1. Why does the integral ∫exf(t) dt transform to exf(t) dt? Shouldn't ex be a part of the integrand too?

2. Why is the difference dy - dy1 = d(y - y1)?
 
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  • #2
lavinia
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I had a couple of questions.

1. Why does the integral ∫exf(t) dt transform to exf(t) dt? Shouldn't ex be a part of the integrand too?

2. Why is the difference dy - dy1 = d(y - y1)?
1. The integral is with respect to t so ex is a constant.

2. Differentials are linear maps.
 
  • #3
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2. Differentials are linear maps.
What does that mean?
 
  • #4
MarneMath
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A linear mapping is one such that f(x*y) = f(x)*f(y) and f(ax) = af(x), where * is some operation and a is a constant.
 
  • #5
Svein
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What does that mean?
Try it yourself: Add a small value and subtract the original. Thus: d(y - y1) = ((y + h) - (y1 + h1)) - (y - y1) . Rearrange and get ((y +h) -y) - ((y1 +h1) - y1) = dy - dy1
 
  • #6
Stephen Tashi
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What does that mean?
A function (also called a "map" or "mapping") has a domain and codomain that are sets. These sets need not be sets of real numbers. They need not be sets of any kind of numbers. They can be sets whose elements are very complicated things.

You can consider "taking the derivative" as function who domain and codomain are each sets of real valued functions of one real variable.

If we write "take the derivative with respect to [itex] t [/itex]" in function notation we could write the result of doing this as [itex] D_t(f(t)) [/itex] or just [itex] D() [/itex] when we are talking about real valued functions of one real variable.

The function [itex] D() [/itex] satisfies:
1) [itex] D(f + g) = D(f) + D(g) [/itex]
2) [itex] D(kf) = k D(f) [/itex] when [itex] k [/itex] is a constant real number.

Those are the properties that define a "linear mapping" or "linear operator".

As an example of a function from a set of functions to a set of functions that is not linear, consider the function [itex] H(f) = f^2 [/itex].

It's easy to get the impression that the word "linear" is always denotes something that has to do with a line (such as a "linear equation"). Often the word "linear" is used as an adjective without any implication that a straight line is involved. For example there are "linear operators", "linear differential equations" and the subject of "linear algebra". Sometimes a "linear" thing has indirect associations with a line. For example, in the case of the derivative operator, it produces a function that is interpreted as slopes of tangent lines.

As another example of an operator that is not linear consider [itex] H(f) = af + b [/itex] where [itex] a [/itex] and [itex] b [/itex] are constants and [itex] b\ne 0 [/itex]. It reminds us of the equation of a line, but it does not satisfy property 1) above.
 
  • #7
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Try it yourself: Add a small value and subtract the original. Thus: d(y - y1) = ((y + h) - (y1 + h1)) - (y - y1) . Rearrange and get ((y +h) -y) - ((y1 +h1) - y1) = dy - dy1
Yeah! That was a quick and simple explanation..Thanks! :)
 
  • #8
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Try it yourself: Add a small value and subtract the original. Thus: d(y - y1) = ((y + h) - (y1 + h1)) - (y - y1) . Rearrange and get ((y +h) -y) - ((y1 +h1) - y1) = dy - dy1
But wait. On second thought, d(y - y1) ≠ ((y + h) - (y1 + h1)) - (y - y1) because d(y - y1) = ((y - y1) + h) - (y - y1). How would you rearrange that to make it equal to dy - dy1?
 
  • #9
Svein
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But wait. On second thought, d(y - y1) ≠ ((y + h) - (y1 + h1)) - (y - y1) because d(y - y1) = ((y - y1) + h) - (y - y1). How would you rearrange that to make it equal to dy - dy1?
It depends on how you view y1. If y1 is a constant, then your reasoning is correct. But if y and y1 are two different functions of x, say y(x) and y1(x), you must treat them as I did.
 
  • #10
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It depends on how you view y1. If y1 is a constant, then your reasoning is correct. But if y and y1 are two different functions of x, say y(x) and y1(x), you must treat them as I did.
They are two different functions. But why should that mean that we do it your way and not mine?
 
  • #11
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Is it because we look at dy as corresponding change in y when we change x? That is, dy by itself makes no sense, it is always accompanied by a change in x which further creates a change in y?
 
  • #12
Svein
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That is, dy by itself makes no sense, it is always accompanied by a change in x which further creates a change in y?
Yes. I apologize for being a trifle sloppy with my notation. The correct (and long-winded) way is: [itex] \Delta y(x) = y(x+\Delta x) -y(x)[/itex] and the same for y1(x). Then [itex]lim_{\Delta x\rightarrow 0}\frac{\Delta y(x)}{\Delta x}=y'(x) = \frac{dy}{dx} [/itex] (and the same for y1). So, when we say y + h, we are in reality saying y(x + Δx), so h is a function of x.
 

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