Undergrad Is There a Way to Override Reserved Symbols in Webpage Code?

Click For Summary
The discussion centers on the confusion surrounding the notation and calculation of derivatives, particularly the second derivative of functions like v(x) = a/x. Participants emphasize the importance of clear mathematical notation, criticizing the use of equal signs where implications should be used instead. There is a focus on the correct representation of derivatives, with some arguing that sloppy notation could lead to misunderstandings or deductions in academic settings. Additionally, a technical aspect regarding the overriding of reserved symbols in webpage code is mentioned, highlighting a method to prevent certain symbols from rendering incorrectly. Clear communication in mathematical expressions is deemed essential for effective understanding.
Anonymous_
Messages
10
Reaction score
1
TL;DR
I understand the formula f(x) =mx(dot) * x(double dot)
You basically just take the second derivative of the given function and multiply it by the original then multiple everything by m. I just don’t understand how the second derivative would be negative.
 

Attachments

  • FF14BD75-E926-41ED-B3D7-6D9ABCF2F3E0.jpeg
    FF14BD75-E926-41ED-B3D7-6D9ABCF2F3E0.jpeg
    18.5 KB · Views: 379
Mathematics news on Phys.org
Got it

(1/x)=(x^-1)=(-x^-2)=(-1/x^2)
 
Are you asking why the derivative of ##1/x## w.r.t. ##x## is ##-1/x^2##?
 
Anonymous_ said:
Summary:: I understand the formula f(x) =mx(dot) * x(double dot)
This is somewhat hard to read. This site supports LaTeX, which can be used to write equations formatted nicely.
##f(x) = m\dot x \ddot x##
The script I used for the above is ##f(x) = m\dot x \ddot x##
There's a link to our tutorial at the lower left of the text entry pane -- click the link titled LaTeX Guide.
Anonymous_ said:
You basically just take the second derivative of the given function and multiply it by the original then multiple everything by m. I just don’t understand how the second derivative would be negative.
You're given that v(x) = a/x, so v(x) is decreasing (assuming a > 0), which makes its derivative negative.
Anonymous_ said:
Got it
(1/x)=(x^-1)=(-x^-2)=(-1/x^2)
This makes no sense.
##\frac 1 x = x^{-1}##, but ##x^{-1} \ne -x^{-2}##
You apparently took the derivative to get from the 2nd expression to the 3rd, but you don't give any indication that that's what you did. If I were your instructor, I would mark your work down for this.
 
Mark44 said:
This is somewhat hard to read. This site supports LaTeX, which can be used to write equations formatted nicely.
##f(x) = m\dot x \ddot x##
The script I used for the above is ##f(x) = m\dot x \ddot x##
There's a link to our tutorial at the lower left of the text entry pane -- click the link titled LaTeX Guide.
You're given that v(x) = a/x, so v(x) is decreasing (assuming a > 0), which makes its derivative negative.
This makes no sense.
##\frac 1 x = x^{-1}##, but ##x^{-1} \ne -x^{-2}##
You apparently took the derivative to get from the 2nd expression to the 3rd, but you don't give any indication that that's what you did. If I were your instructor, I would mark your work down for this.

Not saying they are equal. The equal signs are steps => basically you have to multiple by negative one.
 
Anonymous_ said:
Not saying they are equal. The equal signs are steps => basically you have to multiple by negative one.
But you said they are equal:
Anonymous_ said:
(1/x)=(x^-1)=(-x^-2)=(-1/x^2)
I wouldn't have complained about this if you had showed what you are doing.
##\frac d {dx}(\frac 1 x) = \frac d {dx}(x^{-1}) = -x^{-2} = -\frac 1 {x^2}##
 
Mark44 said:
But you said they are equal:
I wouldn't have complained about this if you had written "implies" => instead of "equals" =.
Or better yet, showed what you are doing.
##\frac d {dx}(\frac 1 x) = \frac d {dx}(x^{-1}) = -x^{-2} = -\frac 1 {x^2}##
I would just assume you would know what I mean since this is a physics site...
 
Anonymous_ said:
I would just assume you would know what I mean since this is a physics site...
Because this is a physics site, you should write what you mean.

If I were to write something like this, would that be OK? ##2x + 1 = 5 = 2x = 4 = x = 2##
I hope not, since it is saying that 5, and 4, and 2 are all the same number, among other nonsensical statements.
 
Mark44 said:
Because this is a physics site, you should write what you mean.

If I were to write something like this, would that be OK? ##2x + 1 = 5 = 2x = 4 = x = 2##
I hope not, since it is saying that 5, and 4, and 2 are all the same number, among other nonsensical statements.
Yes, I can understand what’s going on. I hate when people overcomplicate things like this.
 
  • #10
Anonymous_ said:
Yes, I can understand what’s going on. I hate when people overcomplicate things like this.

Overcomplicating? As written, your post #2 is simply wrong and suggests you don't understand what an equal sign means.
 
  • #11

We try to help you optimally. I asked you a question in #3 which you ignored. Could you please answer so we can move on with the question?

We are just trying to tell you that using equal signs like that is wrong (although we get what you mean). Instead use ##\implies## arrows. Most people that mark exams would deduct points because this is very sloppy notation. By saying this to you, we hope that you can avoid this to happen.
 
  • #12
Mark44 said:
This is somewhat hard to read. This site supports LaTeX, which can be used to write equations formatted nicely.
##f(x) = m\dot x \ddot x##
The script I used for the above is ##f(x) = m\dot x \ddot x##
Where did you find the fake # sign, "#"? The result is pretty slick and I'd like to be able to borrow the trick.

Edit: found it here. Just apply a textcolor to the ##, making it black like so: ##\sqrt{x}##
 
Last edited:
  • #13
Now that you mention it, that means @Mark44 overrode the ## reserved symbol set somehow. Interesting...
 
  • #14
jbriggs444 said:
Edit: found it here. Just apply a textcolor to the ##, making it black like so: ##\sqrt{x}##

jim mcnamara said:
Now that you mention it, that means @Mark44 overrode the ## reserved symbol set somehow. Interesting...
Yes, @jbriggs444 hit it, but I did it to the first # symbol at each end. Doing so prevents the browser from rendering the expression between the # pairs.
 
Last edited:
  • Like
Likes jim mcnamara

Similar threads

  • · Replies 11 ·
Replies
11
Views
3K
  • · Replies 18 ·
Replies
18
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 10 ·
Replies
10
Views
1K
  • · Replies 10 ·
Replies
10
Views
1K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 5 ·
Replies
5
Views
4K
  • · Replies 2 ·
Replies
2
Views
2K