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Calculus shortcut and Compression


ProgrammingGodJordan
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This is a clear explanation w.r.t. the "Trigonometric Rule Collapser Set", that may perhaps be helpful. (See source)

The above is not to be confused for u-substitution. (See why)

In the sequence: x = sin t, where dx = cost dt, and 1 − x2 = 1 − sin2 t = cos2 t ....(from problem: ∫ √1- x2)

..the novel formulation dx | dt · dx occurs such that the default way of working trigonometric equations is compressed, permitting the reduction of the cardinality of steps normally employable.

For example, in the video above, while evaluating ∫ √1- x2 dx, in a preliminary step, the instructor writes ∫ (√1-sin2θ/(√cos2θ) ) · cosθdθ. 

Using my trig collapser routine, this step (which may be a set of steps for other problems) is unnecessary, because applying my trig collapser set's novel form: dx | dθ · dx, we can just go right ahead and evaluate  ∫cosθ·cosθdθ.


The trigonometric rule collapser set may be a topic that may be an avenue which may spark further studies.

Edited by ProgrammingGodJordan
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9 hours ago, uncool said:

Walk me through your "collapser" for the integral of sin^3(x) dx?

The formulation works for many classes of integrals whose problems constitute some square rooted distribution.

Unfortunately, I don't know whether universal ways of collapsing are possible.

 

 

Edited by ProgrammingGodJordan
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On 9/7/2017 at 1:53 PM, uncool said:

Sorry; walk me through your "collapser" for the integral of x^2 sqrt(1 - x^2) dx, then?

Without TrigRuleCollapser:

Let x = sinϴ  dx = cosϴ

Preliminary Step (1): ∫ sin²ϴ · √1-sin²ϴ  · cosϴ

Preliminary Step (2): ∫ sin²ϴ · cosϴ  · cosϴ

Evaluation: ∫ sin²ϴ · cos²ϴ 

... ...

 

******************

 

With TrigRuleCollapser:

Let x = sinϴ  dx = cosϴ

Evaluation: ∫ sin²ϴ · cos²ϴ 

... ...

(No preliminary steps required, you Evaluate rule: x · dx/dϴ · dx)

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1 hour ago, uncool said:

So your collapser is just...doing the same multiplication, but skipping the explicit step of multiplication?

No.

Notice this preliminary Step (1): ∫ sin²ϴ · √1-sin²ϴ · cosϴ 

With my collapser, you can easily identify cosϴ from the initial substitution line; so instead of writing down the "√1-sin²ϴ" term, then finding cos²ϴ, then square rooting it, you go straight ahead and evaluate cosϴ in the integral.

As a result, you don't need to look for cosϴ from 1-sin²ϴ in the identity table, and you don't need to square root.

In the scenario above, three preliminary lines are replaced (excluding explicit multiplication) and in other problems, more preliminary lines may be replaced (also excluding explicit multiplication).

Either way, you avoid searching the identity table to begin evaluation, and you avoid square rooting.

 

 

 

 

 

Edited by ProgrammingGodJordan
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That makes it sound like all you're doing is consulting your memory instead of a lookup table. It's like saying you have a shortcut by replacing 8*7 with 56, instead of looking it up. 

 

I mean, you can make it look like it's a shortcut, but...

 

That's not even getting into pedagogical issues. 

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1 hour ago, uncool said:

That makes it sound like all you're doing is consulting your memory instead of a lookup table. It's like saying you have a shortcut by replacing 8*7 with 56, instead of looking it up. 

 

I mean, you can make it look like it's a shortcut, but...

 

That's not even getting into pedagogical issues. 

Yes, you can use memory to look on the merely three standard trig rule collapser forms. (Just like the memory you could use to memorize the many many more standard trig identities)

So, using my collapser is still cheaper than looking up the many more trig identities.

You gain shorter evaluations, and you also compute with far less lookup.

 

FOOTNOTE:

"Uncool", thanks for your questions. I have improved the "Clear explanation" section in the paper, and removed some distracting typos too. (In addition, in the original post, the term: "∫ (√1-sin2θ/(√cos2θ) ) · cosθ dθ". should have been: "∫ (√1-sin2θ) · cosθdθ" instead, based on the problem in the video)

Edited by ProgrammingGodJordan
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  • 1 month later...
On 9/9/2017 at 9:16 PM, ProgrammingGodJordan said:

No.

Notice this preliminary Step (1): ∫ sin²ϴ · √1-sin²ϴ · cosϴ 

With my collapser, you can easily identify cosϴ from the initial substitution line; so instead of writing down the "√1-sin²ϴ" term, then finding cos²ϴ, then square rooting it, you go straight ahead and evaluate cosϴ in the integral.

As a result, you don't need to look for cosϴ from 1-sin²ϴ in the identity table, and you don't need to square root.

In the scenario above, three preliminary lines are replaced (excluding explicit multiplication) and in other problems, more preliminary lines may be replaced (also excluding explicit multiplication).

Either way, you avoid searching the identity table to begin evaluation, and you avoid square rooting.

 

 

 

 

 

I couldn't find the paper in your OP, but I did find something on "research gate" at this location with the same title:
 
https://www.researchgate.net/publication/318722013_Trigonometric_Rule_Collapser_Set_qua_Calculus

It just looks like you're plugging in values from basic u-substitution!

 

Edit 1: I re-read it once more.

It's not u-substitution after all. It reminds me of it, but it's actually far from u-substitution. My bad.

Edit 2: I am not much of a mathematician, but I've passed Calculus I and II. From what I can see, I can't find it in any of the 3 2016 Calculus texts I own. 

I will look online and report if I find it.

Edited by thoughtfuhk
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