I Understanding "Terrible" Math Notation: A Calculus Guide

[BillhB]

Is there some standardized math text with "proper universal notation" I could read for calculus?

In one of my courses, $$\int\frac{dx}{x}$$ had a red mark through it, with a note that said "impossible" or something. I earned a zero on the question due to the above. In another instance $$\int(x^{-2}v)'dx$$ had a red-mark that said it was equal to zero and said terrible. In the same question $v'=yy'$ during a substitution had a mark that just said terrible. I got two points for the question, even though the answer matched others who had gotten full credit so I'm assuming I was just marked down for notation.

What's wrong with the above? Maybe this instructor just hates prime notation...

 

[jedishrfu]

I would ask your instructor about it. It may not be notation but perhaps Something more fundamental.

In the first, your teacher may not have liked you dividing the $dx$ by $x$ and would have preferred that you wrote $(1/x) dx$ instead. It doesn't make sense to divide $dx$as it's not a value but a kind of placeholder that indicates what variable to integrate over.

In the second, I'd say the prime In the integral means that $f(x)' dx$ integrates to $f(x)$. If that not what you meant then you can see your teachers concern.

 

[Mark44]

Is there some standardized math text with "proper universal notation" I could read for calculus?

In one of my courses, $$\int\frac{dx}{x}$$ had a red mark through it, with a note that said "impossible" or something.

What you show above is an indefinite integral. Was the integral that the instructor marked a definite integral?

I earned a zero on the question due to the above. In another instance $$\int(x^{-2}v)'dx$$ had a red-mark that said it was equal to zero and said terrible.

Again, was the actual integral a definite integral?

In the same question $v'=yy'$ during a substitution had a mark that just said terrible. I got two points for the question, even though the answer matched others who had gotten full credit so I'm assuming I was just marked down for notation.

What's wrong with the above? Maybe this instructor just hates prime notation...

What do v' and y' mean here? Prime notation doesn't show which variable the derivative is taken with respect to.

 

[Mark44]

In the first, your teacher may not have liked you dividing the dx by x and would have preferred that you wrote 1/x dx instead.

I don't think this is very likely. $\int \frac{dx}{x}$ is a wellknown integral that is often written this way.

 

[BillhB]

What you show above is an indefinite integral. Was the integral that the instructor marked a definite integral?
Again, was the actual integral a definite integral?

No. It was just a simple separable differential equation.. $$\frac{dy}{dx}=\frac{1}{x}$$ $$\int{dy}=\int \frac{dx}{x}$$ $$y=ln|x| + C$$

What do v' and y' mean here? Prime notation doesn't show which variable the derivative is taken with respect to.

It was just me writing out of a substitution for a Bernoulli Equation.. $$xy\frac{dy}{dx}+x^2-y^2=0$$ $$\frac{dy}{dx}-yx^{-1}=-xy^{-1}$$ so I had $v=y^2$, $v'=2yy'$ so I guess v' would be $\frac{dv}{dx}$ and y' is $\frac{dy}{dx}$ then I multiplied, found integrating factor, and subbed out to get to $$\int(vx^{-2})'dx=\int\frac{-2dx}{x}$$ $$vx^{-2}=-2ln|x|+C$$

I don't think this is very likely. $\int \frac{dx}{x}$ is a wellknown integral that is often written this way.

Yeah, we've wrote it that way in earlier courses, no one seemed to mind. Physics professor always puts the differential in the expression... but she's a physics teacher, and probably doesn't care as much if it is wrong. The instructor was trying to show how it was wrong in the very next class, but I didn't really follow the "proof" that well.

 

[Mark44]

No. It was just a simple separable differential equation.. $$\frac{dy}{dx}=\frac{1}{x}$$ $$\int{dy}=\int \frac{dx}{x}$$ $$y=ln|x| + C$$

I don't see anything wrong with this, speaking as a former college math teacher of 18 years. I would ask the instructor why it was marked off.

It was just me writing out of a substitution for a Bernoulli Equation.. $$xy\frac{dy}{dx}+x^2-y^2=0$$ $$\frac{dy}{dx}-yx^{-1}=xy^{-1}$$

You have a sign error in the 2nd equation.

so I had $v=y^2$, $v'=2yy'$ so I guess v' would be $\frac{dv}{dx}$ and y' is $\frac{dy}{dx}$ then I multiplied, found integrating factor, and subbed out to get to $$\int(vx^{-2})'dx=\int\frac{2dx}{x}$$ $$vx^{-2}=-2ln|x|+C$$
The instructor was trying to show how it was wrong in the very next class, but I didn't really follow the "proof" that well.

 

[BillhB]

You have a sign error in the 2nd equation.

Whoops, fixed it.

I don't see anything wrong with this, speaking as a former college math teacher of 18 years. I would ask the instructor why it was marked off.

Okay, but he seemed adamant that it was "wrong" notation. Terrible, terrible notation, garbage notation notes give that kind of vibe. Just a bit worried, have an exam in that class soon. Most of my quiz scores are abysmal due to notation I guess. To be fair, some it's certainty because I made errors like the above, missing a sign..etc. The problems are long, so I'm working on being more careful. Was kind of hoping there was some kind of one stop shop for "proper notation."

 

[Mark44]

What you have here -- $\int(vx^{-2})'dx=\int\frac{2dx}{x}$ isn't very good. An improvement would be $\int d(vx^{-2})=\int\frac{2dx}{x}$

If your instructor is dinging you for $\int \frac{dx} x$, s/he is being pedantic, IMO.

 

[BillhB]

What you have here -- $\int(vx^{-2})'dx=\int\frac{2dx}{x}$ isn't very good. An improvement would be $\int d(vx^{-2})=\int\frac{2dx}{x}$

Got it.

If your instructor is dinging you for $\int \frac{dx} x$, s/he is being pedantic, IMO.

Just worried about what other notation 'mistakes' I'm not aware of.

Thanks though, appreciate all the comments and time you've spent replying.

 

[mathman]

For your first question I believe he was looking for \int \frac{dy}{dx}dx=\int \frac{dx}{x}. Writing \int dy should not be used, until you are taking a differential equations course.