First Post!

[Dont Fear The Reefer]

Lol, I flunked calculus in college my freshman year. The final had 3 questions and the Professor wrote the book we used. I looked at it and walked out. I always threatened to take it again but never have. I'm too stupid to do it now!<G>

 

[ClamDigger]

im going to love this forum, give me a few days to get some threads together.

Topics (please add your suggestions)
Lab testing:
The Cannabinoids
The Terpenoids and Flavonoids
GCMS vs HPLC vs other

Full Spectrum Lab: why did they walk away from the industry?

Steep Hill Lab: the nations first testing facility.

Cannabinoid modulation:
Agonist vs Antagonist effects of terpenoids... why OG Kush tests low for THC but we all love it ... limonene acts as an agonist.
Other terpenoid modulation: myrcene

Neutralizing chlorine and chloramines with molasses by Tim Wilson

Calculating ppm from NPK and food labels.

YES!!
some intersting subjects.
count me in!

 

[cannabineer]

This is what I meant (sorry for forma/spelling etc. typing on phone):

Please omit prior comment about relative sizes I forgot it was with respect to r and the a and r in te denominator would be switched to get natural log anyway however:
shouldn't the definite integral in step 3 be :

r(arcSin r/a) for a>0, -a<0<r. ?

I'm pretty sure there's a screw up between lines 2 and 3...

This Q doesn't seem as hard as it initially looked I just hadnt seen these types in a while very interesting though I'll give it a go tomorrow anyone else with any input please post.

Cheers.

uuhhh... the answer is Three, right? cn

 

[potpimp]

Anybody for Chinese?

 

[cannabineer]

Anybody for Chinese?

Deal me in. I'll have Spring Rolls and General Tso's ...
cn

 

[chris75]

Ok yeh if someone could actually explain how to integrate with respect to phi (called it theta before my bad) when it isn't a variable in the function e.g. in last line that'd be appreciated.

 

[cannabineer]

Ok yeh if someone could actually explain how to integrate with respect to phi (called it theta before my bad) when it isn't a variable in the function e.g. in last line that'd be appreciated.

My memory of calculus is a bit fuzzy, but the indefinite integral of k d(phi) will be k(phi) + c. From the parameter's perspective, all the rest of the salad is treated as a constant, which i've denoted as k to avoid confusion with the constant c that needs to be assigned to any solved integral. cn

 

[chris75]

My memory of calculus is a bit fuzzy, but the indefinite integral of k d(phi) will be k(phi) + c. From the parameter's perspective, all the rest of the salad is treated as a constant, which i've denoted as k to avoid confusion with the constant c that needs to be assigned to any solved integral. cn

What I don't get though is in the last line there isn't even a constant in the integral. The only way how I can see that last operation would work if the integral was:

Integrate (1) between 2pi/5 with respect to phi.

In that case you could just say 1 = phi^0

Thus integrating you would just get phi^1 = phi.

Then you've got your definitie integral, you plug in the parameters, subtract, multiply the constant hanging out the front and then you get their stated final result of a^2(2pi/5).

But the integral isn't of 1. There isn't anything. Just the constant a^2 which doesn't have anything to do with the integral, and the integral which is just:

Integrate (nothing) between 2pi/5 and zero, with respect to phi. And I don't think integrating nothing can get you something. I'm pretty sure that'd just make the final result zero...

 

[Beansly]

My memory of calculus is a bit fuzzy, but the indefinite integral of k d(phi) will be k(phi) + c. From the parameter's perspective, all the rest of the salad is treated as a constant, which i've denoted as k to avoid confusion with the constant c that needs to be assigned to any solved integral. cn

What I don't get though is in the last line there isn't even a constant in the integral. The only way how I can see that last operation would work if the integral was:

Integrate (1) between 2pi/5 with respect to phi.

In that case you could just say 1 = phi^0

Thus integrating you would just get phi^1 = phi.

Then you've got your definitie integral, you plug in the parameters, subtract, multiply the constant hanging out the front and then you get their stated final result of a^2(2pi/5).

But the integral isn't of 1. There isn't anything. Just the constant a^2 which doesn't have anything to do with the integral, and the integral which is just:

Integrate (nothing) between 2pi/5 and zero, with respect to phi. And I don't think integrating nothing can get you something. I'm pretty sure that'd just make the final result zero...

People who actually knows calculus, nice.

 

[cannabineer]

What I don't get though is in the last line there isn't even a constant in the integral. The only way how I can see that last operation would work if the integral was:

Integrate (1) between 2pi/5 with respect to phi.

In that case you could just say 1 = phi^0

Thus integrating you would just get phi^1 = phi.

Then you've got your definitie integral, you plug in the parameters, subtract, multiply the constant hanging out the front and then you get their stated final result of a^2(2pi/5).

But the integral isn't of 1. There isn't anything. Just the constant a^2 which doesn't have anything to do with the integral, and the integral which is just:

Integrate (nothing) between 2pi/5 and zero, with respect to phi. And I don't think integrating nothing can get you something. I'm pretty sure that'd just make the final result zero...

Think of the (nothing) as a 1. Then you have the indefinite integral of d(phi) which is phi + c, and you can then solve the definite formulation. Work for you? cn

<edit> To clarify, the indefinite integral of 1 dx is the same as the integral of dx. It's notation ... the 1 isn't stated and only looks like (nothing).

 

[chris75]

Think of the (nothing) as a 1. Then you have the indefinite integral of d(phi) which is phi + c, and you can then solve the definite formulation. Work for you? cn

<edit> To clarify, the indefinite integral of 1 dx is the same as the integral of dx. It's notation ... the 1 isn't stated and only looks like (nothing).

Ahh... Yes now I understand, it was that last sentence I was unaware of.

You're sure of that? I.e.

The integral of (nothing stated) with respect to x = integral of 1 with respect to x

I couldn't recall that, though as throughout maths quite often a 1 isn't specified and you can't integrate zero it makes sense in terms of notation etc.

Thanks for that.

 

[cannabineer]

I found this ... cn

<edit> You can integrate zero but it doesn't really get you anywhere ... (int) zero dx = c

 

[chris75]

Any idea what was done between step 2 and 3? The integration performed. I.e:

Line 2, the fraction, parameter between a and 0, integrated with respect to r.

You can see line 3 their definite integral... However I've got no idea how the hell they got there.

See:
http://www.sngweb.com/images/formula3.jpg

The third last integral, that certainly would assist in the procedure, though the only issue is the numerator is r which is also what we're integrating with respect to, so it'd obviously require some product or quotient rule...

Any idea?

 

[cannabineer]

Any idea what was done between step 2 and 3? The integration performed. I.e:

Line 2, the fraction, parameter between a and 0, integrated with respect to r.

You can see line 3 their definite integral... However I've got no idea how the hell they got there.

See:
http://www.sngweb.com/images/formula3.jpg

The third last integral, that certainly would assist in the procedure, though the only issue is the numerator is r which is also what we're integrating with respect to, so it'd obviously require some product or quotient rule...

Any idea?

I do not recall how to actually solve that integral analytically. However if you go to this table, specifically entry 34, you'll find a result consistent with the integral in question. cn

<edit> Found a table that I can post in parts.

contrast with:

 

[chris75]

I found this ... cn

<edit> You can integrate zero but it doesn't really get you anywhere ... (int) zero dx = c

Ha ha,thanks: Yes the importance of that god damn constant is what I'd keep forgetting.
Makes much more sense.

I remember even back in high school doing pages and pages of indefinite integrals, everything was fine when it was marked except after every function the teacher had to write " + C " as I'd do the integration and forget the constant...

Cheers.

 

[cannabineer]

Ha ha,thanks: Yes the importance of that god damn constant is what I'd keep forgetting.
Makes much more sense.

I remember even back in high school doing pages and pages of indefinite integrals, everything was fine when it was marked except after every function the teacher had to write " + C " as I'd do the integration and forget the constant...

Cheers.

What finally cured me of the same problem was the old math joke:

What is the indefinite integral of d(cabin) / (cabin)?

Answer: A houseboat! (log (cabin) plus "Sea")
cn

 

[chris75]

I do not recall how to actually solve that integral analytically. However if you go to this table, specifically entry 34, you'll find a result consistent with the integral in question. cn

<edit> Found a table that I can post in parts.

contrast with:

Nice! Great find/post. None of the tables I found were that detailed. Do you have the direct link to that one?

Several +reps and a promotion I believe are in order.

 

[cannabineer]

Nice! Great find/post. None of the tables I found were that detailed. Do you have the direct link to that one?

Several +reps and a promotion I believe are in order.

Good thing polar bears can't blush ... ~grin~
Link to the page in question: http://www.sosmath.com/tables/integral/integ13/integ13.html
cn

 

[tyler.durden]

What finally cured me of the same problem was the old math joke:

What is the indefinite integral of d(cabin) / (cabin)?


Answer: A houseboat! (log (cabin) plus "Sea")
cn

Nice! Great find/post. None of the tables I found were that detailed. Do you have the direct link to that one?

Several +reps and a promotion I believe are in order.

Oh, yeah. I'm going to enjoy this forum

 

[chris75]

Ha ha ha that was a good one. Hadn't heard that, probably would've helped remembering to add the sea...

Here's one:

Why do mathematicians always confuse Halloween and Christmas?

Because 31Oct = 25Dec.

(different numerical bases).




I've got to pull out the old math book go over some of this stuff again, brilliant area of study.


Anyway don't drink and derive,

Take it easy.