Warframe And Taylor's Expansion (With Valkyr's Behind) (Part One)

[Renegade343]

So, a few days ago, I made this thread, saying that I would be doing something with Warframe and Talyor's expansion. I am here to say that I am posting the results now. The content is below (please note the contents may offend some members, so the contents will have spoilers hiding it in case they do not want to read further): 

 

As most Warframe gamers may know, most players would perceive Valkyr's behind as, for lack of a better word, attractive. Then, I was bored on a Thursday morning, and decided to dust off one of the mathematics books I have, then flipped to negative binomials, Manclaurin series and Taylor series, and started re-learning how to do it again (since I have done it before, but left it to collect cobwebs in my mind). While I was doing the practice questions, a thought hit me: 

 

Is there a way to approximate a function for the curve of Valkyr's behind?

 

So, being the person who would go and search for answers, I did a few more questions, then started planning in order to answer that thought. And so, I first decided to get a Valkyr Warframe, go to the Arsenal, then turned her approximately π/4 radians to her clockwise direction, accounting for her positioning relative to the camera (image 1): 

 

 

Thus, from that image above, we will have two (and maybe a third) restrictions: 

 

- The approximate function obtained would only be for viewing Valkyr at approximately π/4 radians to her clockwise direction, in the Arsenal screen. 

- Valkyr must have no animation sets. 

- (Not yet tested with bows) Valkyr must be holding any Primary weapon that is not a bow. 

 

Also, from that image above, I have highlighted what this thread would be analysing. I cropped it out, then turned it π/4 radians anti-clockwise in order to have a function, since with the original image, I would be getting a one-to-many equation, which does not fit under the definition of a function. By rotating it in that direction, I will now be getting a many-to-one equation, which does fit under the definition of a function. Afterwards, I pasted the cropped image into GeoGebra (with the bottom left corner of the image being the origin), then scaled both axis so that it would accurately represent the size of the image in meters (the image size is 0.021652m x 0.028102m [with a laptop that has PPI of 110.27]). And thus, the image (image 2) of that result is below (ignore the point E for that image, as I was just testing with trying to manipulate the line): 

 

 

Once I made the image scale whenever I zoomed both axis in or out (so that I would not have headaches and frustration with it), I then started to use trial and error to find the tangent points of both ends of the limits, then used trial and error to find out the point where the gradient of the tangent = 0. The image (image 3) of that result is below: 

 

 

Once I was satisfied with the gradients of the tangents needed, I plotted a dy/dx vs. x graph to find the derivative of the curve (i.e.: f'(x)) (image 4): 

 

 

Then, I differentiated that line again to obtain f''(x) = -53.55.

 

Once I had all the data I needed, I used the Taylor series formula in order to get an approximation of the curve. For those who do not know what it is, the Taylor series represents a function with a sum of infinite terms for a certain point. The general formula for this is: 

 

f(x) = f(a) + (f'(a)/1!)*(x-a) + (f''(a)/1!)*((x-a)^2) + ... 

 

Using this formula, I first set a = 0 (meaning at point D in image 3), since that would be less tedious to calculate. And since differentiating f''(x) will result in 0, I omitted all other powers higher than x^2. Thus, the values for f(0), f'(0) and f''(0) are below: 

 

f(0) = 0.01985

f'(0) = 0.5077

f''(0) = -53.55

 

And so, the approximate function using all the data I have collected and manipulated is: 

 

f(x) ≅ (-26.775x^2) + 0.5077x + 0.01985

 

I then plotted this approximate function to see if it works, and the result is in the image: 

 

 

While it does fit with the curve for some values after x = 0, it deviates from it starting from x = 0.005. What I think may be the case is that I have set the limits too narrow, and thus did not have considered any possible points of inflexion (which would then indicate that the approximation can be written to higher powers of x). Next time, when I have the time to re-do this again, I will test to see if the curve can be approximated to a cubic function.

 

And so, to conclude, this is my first and failed attempt to combine Warframe and Taylor's expansion. But at least it was rather fun to do so.

Edited September 5, 2014 by Renegade343

[Gammaion]

I... I don't understand. Something about Valkyr's booty?

[ElHefe]

Brilliant application of this technique for "curve fitting" ... I'ill give it a 1 + 1 + 1 + ...

Although I have to stick to my point that there was too much "read my mind" contained in the question

 

Estimates of remainders could work for any sort of problem as would be finding a tangent or the value of a function at a point

[(PSN)JoeyTwoShoes]

You certainly know how to make high-level mathematics interesting.

[Renegade343]

Estimates of remainders could work for any sort of problem as would be finding a tangent or the value of a function at a point

Still practicing that part. 

[Renegade343]

Announcing that part two of this series will be posted over the weekend. Stay tuned!

[unknow99]

Oh, the maths guy strikes ag....Butt that's interesting! :3

[Utarious]

And I'm just sitting here.

 

No idea what is going on.

[Snydrex]

You.

 

Are.

 

Brilliant.

Edited September 6, 2014 by Snydrex

[RejectionOfFate]

And I'm just sitting here.

 

No idea what is going on.

It is the usage of math to figure out just how shapely Valkyr's rear is.

[Mesyra]

I am confuse.