My Math Forum Is dx the limit of Δx?

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 March 18th, 2019, 05:35 PM #1 Senior Member   Joined: Oct 2015 From: Greece Posts: 115 Thanks: 8 Is dx the limit of Δx? So, I'm trying to deeply understand some things that I take for granted in mathematics. Let's start with some displacements. Given two points A(x, f(x)) and B(x0, f(x0)) $\displaystyle \frac{\Delta{f}}{\Delta{x}} = \frac{f(x_0) - f(x)}{x_0 - x}$ Lets name $\displaystyle h = x_0 - x$ Solving for x0 $\displaystyle x_0 = x + h$ Substituting to the original displacement ration $\displaystyle \frac{\Delta{f}}{\Delta{x}} = \frac{f(x + h) - f(x)}{h}$ Taking the limit on both sides such us h---->0 $\displaystyle \lim_{h\to0}\frac{\Delta{f}}{\Delta{x}} = \lim_{h\to0}\frac{f(x + h) - f(x)}{h} \Leftrightarrow \frac{\lim_{h\to0}\Delta{f}}{\lim_{h\to0}\Delta{x} } = \lim_{h\to0}\frac{f(x + h) - f(x)}{h} \Leftrightarrow \frac{df}{dx} = \lim_{h\to0}\frac{f(x + h) - f(x)}{h}$ Question 1 So in other words dx is still a length like Δx but dx <<<<<< Δx in such a way that dx is as small as possible. So is the below an accepted writing for dx? $\displaystyle \lim_{x_0-x\to0}\Delta{x} = dx$ And $\displaystyle \lim_{x_0-x\to0}\Delta{f} = df$ Question 2 Is this why we can rearrange equations like this? (We treat dx, df as limit values) $\displaystyle v = \frac{dx}{dt} \Leftrightarrow dx = v\cdot dt$ Question 3. If Question 2 is true then it means I can cancel any df dx? $\displaystyle v = \frac{dx}{dt} \Leftrightarrow dt \cdot v = \frac{dx}{dt} \cdot dt$ Question 4: Can I treat the symbol of integration exactly like I do with numbers when I multiply or divide both sides of an equation? $\displaystyle a = \frac{dv}{dt} \Leftrightarrow dv = a \cdot dt$ Multiplying both sides with integration??? lol but I know this can happen $\displaystyle \int dv = \int a \cdot dt$ Question 5: When I have a differential equation I can multiply both sides with a dx and an integration to complete the form of integration? $\displaystyle \frac{dy}{dx} = yx \Leftrightarrow \frac{dy}{dx} \cdot \frac{1}{y} = x$ Multiplying both sides with dx (dx's on the left cancel out) $\displaystyle \frac{dy}{dx} \cdot \frac{1}{y} \cdot dx = x \cdot dx \Leftrightarrow \frac{1}{y} \cdot dy = x \cdot dx$ Integrating both sides: $\displaystyle \int \frac{1}{y} \cdot dy = \int x \cdot dx$ So in general i can do whatever I want with dx's, df's and integrals in both sides of an equation like when I do with numbers or variables? Well the truth is that I'm using all these stuff but nobody told me that. By seeing how our professors are teaching different stuff in mathematics, I just noticed that these operations are actually matching the same techniques we use in algebra so I decided to treat them like that. I hope I'm doing things right. Thank you
 March 19th, 2019, 03:19 AM #2 Senior Member   Joined: Dec 2015 From: iPhone Posts: 486 Thanks: 75 $\displaystyle dx$ is the limit of delta $\displaystyle x$. For the integration part , you must use the word ”integration of equality” or ”add both sides the integral” and not multiply . Example : Since $\displaystyle e^x >x \; \Rightarrow \int_{D} e^x dx > \int_{D} xdx$. Let $\displaystyle D\in [0,1]$ , $\displaystyle e^x |_{0}^{1} > 1/2 x^2 |_{0}^{1}$. Which gives $\displaystyle e-1>1/2$ without calculating them . Last edited by idontknow; March 19th, 2019 at 03:22 AM.
 March 19th, 2019, 04:46 AM #3 Senior Member   Joined: Dec 2015 From: iPhone Posts: 486 Thanks: 75 About the differential equation : $\displaystyle y’=yx$, you lose some solutions if you divide by y. The general solution is found with the integrating factor which is $\displaystyle z(x)$ where $\displaystyle \frac{z’ }{z} = -x$ . When you substitute $\displaystyle -x$ to the equation the left side of equation is $\displaystyle zy’+yz’=(zy)’$ and in the right side of equation you have an easy integral .
 March 19th, 2019, 06:41 AM #4 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,640 Thanks: 2624 Math Focus: Mainly analysis and algebra It's complicated. Especially with regard to your third question. Differentials are things, but they aren't limits of $\Delta x$ because the limit value of $\Delta x$ is zero. Also, the symbol $\frac{\mathrm df}{\mathrm dx}$ is not a ratio of differentials, it is an operator $\frac{\mathrm d}{\mathrm dx}$ acting on the function $f$. I tend to think of differentials as infinitesimals after the manner in which Newton used $\epsilon$ and conversion between those and derivatives as just something that works. Wiki page on differentials Thanks from topsquark and babaliaris
March 19th, 2019, 06:53 AM   #5
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Quote:
 Originally Posted by v8archie It's complicated. Especially with regard to your third question. Differentials are things, but they aren't limits of $\Delta x$ because the limit value of $\Delta x$ is zero. Also, the symbol $\frac{\mathrm df}{\mathrm dx}$ is not a ratio of differentials, it is an operator $\frac{\mathrm d}{\mathrm dx}$ acting on the function $f$. I tend to think of differentials as infinitesimals after the manner in which Newton used $\epsilon$ and conversion between those and derivatives as just something that works. Wiki page on differentials
But if I treat $\displaystyle \frac{df}{dx}$ as a ration and do rearrangements or canceling will I have a problem? or it just works. Because in the class of differential equations we do that a lot.

 March 19th, 2019, 08:48 AM #6 Senior Member   Joined: Oct 2015 From: Greece Posts: 115 Thanks: 8 Also what If I do this: Let's say I have the function $\displaystyle f(x) = x$ then $\displaystyle \frac{df}{dx} = 1$ Is the below wrong? $\displaystyle df = dx \Leftrightarrow df \cdot dx = dx^2 \Leftrightarrow \int df \cdot dx = \int dx^2$
March 19th, 2019, 11:58 AM   #7
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Quote:
 Originally Posted by babaliaris But if I treat $\displaystyle \frac{df}{dx}$ as a ration and do rearrangements or canceling will I have a problem? or it just works. Because in the class of differential equations we do that a lot.
It works because $$\mathrm dy = \frac{ \mathrm dy }{ \mathrm dx } \mathrm dx$$

 March 20th, 2019, 08:35 AM #8 Newbie     Joined: Feb 2019 From: United Kingdom Posts: 24 Thanks: 3 The most embarrassing thing about this post is not on the OP but the glorified “Math Team” who don’t give a sh*t. Who can spot what’s wrong with the OP’s definition of derivative? How on Earth that slipped past by this so called “team” I don’t know. Either give up your gloves for the sake of your reputation or be active to correct people.
 March 20th, 2019, 08:58 AM #9 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,640 Thanks: 2624 Math Focus: Mainly analysis and algebra I'm comfortable with all the answers I gave. I never claimed to be correcting every error (that isn't my job) nor did I even claim to be answering every question. At least I was able, in the short time available to me, to contribute something positive and useful to the thread.
March 20th, 2019, 09:29 AM   #10
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Quote:
 Originally Posted by NineDivines The most embarrassing thing about this post is not on the OP but the glorified “Math Team” who don’t give a sh*t. Who can spot what’s wrong with the OP’s definition of derivative? How on Earth that slipped past by this so called “team” I don’t know. Either give up your gloves for the sake of your reputation or be active to correct people.
So let's see. You are complaining that we are overlooking a basic definition. So instead of correcting us for the benefit of the OP you are merely calling us out? Man up and tell the OP how to correct the mistake! That's why this site exists.

Personally I feel that it's been fixed "off to the side." As $\displaystyle \lim_{h \to 0} \Delta x = dx$ has been pointed out to be inaccurate I, personally, feel the problem has been resolved. But if you disagree then point it out.

-Dan

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