Derivative Process Using Fractal Indices K Equals One-Half, One-Third, and One-Fourth

Journal of Applied Mathematics and Physics, 2017, 5, 453-461 http://www.scirp.org/journal/jamp ISSN Online: 2327-4379 ISSN Print: 2327-4352

Derivative Process Using Fractal Indices k Equals One-Half, One-Third, and One-Fourth

Salvador A. Loria

College of Engineering & Graduate School, Nueva Ecija University of Science and Technology, Cabanatuan City, Philippines

How to cite this paper: Loria, S.A. (2017) Abstract Derivative Process Using Fractal Indices k Equals One-Half, One-Third, and One- In this paper, the researcher explored and analyzed the function between Fourth. Journal of Applied Mathematics integral indices of derivative. It is proved that getting half-derivative twice is and Physics, 5, 453-461. equivalent to first derivative. Also getting the triple of one-third derivative is https://doi.org/10.4236/jamp.2017.52040 equal to first derivative. Similarly, getting four times of one-fourth derivative Received: January 12, 2017 is equal to first derivative. Accepted: February 18, 2017 Published: February 21, 2017 Keywords

Copyright © 2017 by author and Fractal Index, Derivative, Gamma Function, Factorial Scientific Research Publishing Inc. This work is licensed under the Creative Commons Attribution International License (CC BY 4.0). 1. Introduction http://creativecommons.org/licenses/by/4.0/ Open Access Fractal is a general term used to express both the geometry and the procedures

which display self-similarity, scale invariance, and fractional dimension [1]. Geo- metry deals with shapes or objects described in integral dimension. A point has 0-dimension, a line having 1-dimension, a surface has 2-dimension and the solid has 3-dimensions [2]. However, there are phenomena that are suitably characte- rized their dimension between any two integral dimensions. A straight line has dimension of 1 and a zigzag has dimension n={ dstd. . ∈( 1 , 2)} . Here, the dimension is indicated as fractional dimension—a dimension whose value lies between integral values. Similarly, derivative process is an integral index in nature, such as first, second, third and up to nth derivative. The function obtained from derivative process is very useful in the field of physical science and technology. Thus, it is interesting to describe and analyze the function between integral indices of derivatives. Specifically, the study aimed to explore derivative process using fractal indices k that equals one-half, one-third, and one-fourth.

2. The Derivative Process

The derivative of y with respect to x is itself a function of x , and may in

DOI: 10.4236/jamp.2017.52040 February 21, 2017 S. A. Loria

turn be differentiated [3]. The derivative of the first derivative is called second d2 y derivative and is written or y′′ ; the derivative of the second called third dx2 d3 y derivative, or y′′′ , and so on… dx3 Let yx= n , then the derivatives are as follows:: dy = nxn−1 first derivative dx d2 y =nn( −1) xn−2 second derivative dx2 d3 y =−−nn( 12)( n) xn−3 third derivative dx3 d4 y =−−nn( 123)( n)( n −) xn−4 fourth derivative dx4 d5 y =−−nn( 1234)( n)( n −−)( n) xn−5 fifth derivative dx5 … … Repeating the process up to k times, we have dk y =nn( −123)( n −)( n −)( n −+ k 1) xnk− kth derivative (1) dxk

3. Derivatives Using Fractal Indices

Let us consider the function between 0 and first derivative or between first and second derivative. The index fraction indicates that the derivative process is called fractal. These can be denoted as follows: 12 d y ′ or y12 one-half derivative (between zero derivative and first deriva- dx12 tive) 32 d y ′ or y32 three-halves derivative (between first derivative and second dx32 derivative) and so on………. Let yx= n be the function, then 1 For kth derivative, where k is an element of sta. . ∈ N, using Equation (1). a d!k n ( xn) = nn( −12)( n −)( n −+ k 1) = xnk− (2) dxk (nk− )!

Factorial is equivalent to gamma function [4] as nn!1= ⋅Γ( n) =Γ( n + ) (3)

Thus, in gamma function dk Г(n +1) ( xn) = xnnk− for > 0 (4) dxk Γ(nk −+1)

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For negative integer power, yx= −n , the derivative are as follows: Let yx= −n be the function, then dy =( −nx) −−n 1 first derivative dx d2 y =−−−nn( 1) x−−n 2 second derivative dx2 d3 y =−−−nn( 12)( −− n) x−−n 3 third derivative dx3 d4 y =−−−nn( 123)( −− n)( −− n) x−−n 4 fourth derivative dx4 d5 y =−−−nn( 1234)( −− n)( −− n)( −− n) x−−n 5 fifth derivative dx5 ...... Repeating the process to k times, we have dk y =−−−n( n123)( −− n)( −− n)( −− nk + 1) x−−nk kth derivative dxk k d y k −+(nk) or =−( 1) nn( + 123)( n +)( n +)( n +− k 1) x kth derivative dxk In factorial form: k d y k (nk+−1!) −+(nk) =( −1) x (5) dxk (n −1!) It is equivalent to gamma function as: k d y k Γ+(nk) −+(nk) =( −1) x (6) dxk Γ(n) Thus, k d − k Γ+(nk) −+(nk) xn =(−>1) xn for 0 (7) dxk Γ(n)

4. Gamma Function

∞ Definition 1. [4] [5] Gamma function is defined as Γ=(n) ∫ed−−xn xx1 , for 0 nZ∈ + , then, ∞ Γ+=(n1ed) ∫ −xn xx 0 ∞ ∞ Γ+=−(n1) xnx e− + nx n−−1 ed x x 0 ∫ 0 Γ(n +10) = +Γ nn( ) =Γ nn( ) (8) For Γ(12) ∞ Γ=(12) ∫ e−−x xx12 d 0 1 Let x= u2 , u = x2 , d x = 2 uu d ,

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ddxx = = = = −12 du 1 2dux d x 2u 2 2x ∞2 ∞∞ 22 Γ=(1 2) ∫ e−−uxx12 d = ∫∫ e− uu 2d u = 2 e− d u 0 00 we can express in terms of other variable. Thus,

∞ 2 Γ=(12) 2 e−v d.v ∫0

Getting the product of the two above equations, we have 2 ∞ ∞ ∞∞ 22 1 −−uv22 −+(uv) Γ= 2ed2edu v= 4 e dduv 2 ∫0 ∫0 ∫∫00  let ur= cosθ and vr= sinθ

2 22 r= uv + dduv= rr ddθ 2 π ∞ 1 −r2 Γ= 4∫∫2 err ddθ 2 00 1 π 2 ∞ 2 =4 ×− deθ −r − 2drr 2 ∫∫00 π =−−−2 001[ ] 2 = π Thus, Γ==(12) π 1.77245. The gamma function of fraction with multiple of one-half will be obtained as follows: Γ+=(n1!) n =Γ nn( )

3 1  1  11  Γ =!1 =Γ+=Γ    2 2  2  22  3 11 Γ=== ! π 0.88623 2 22 53 and Γ= (0.88623) = 1.32934. 22 Since the solution of gamma function using integral is complex, the Burn-

xx− 1 side’s approximate solution [6] [7], xx!= e2π x1+ , s. t. x∈ R﹨{ 0} 12x can be used. Table 1 shows the actual values of gamma function and the approximate Burnside’s solution. The absolute percentage of error was computed, which it is ranges between 0.06 to 1.5 percent, so Burnside’s equation found acceptable.  1 1  112 −  1 1 Using Burnside’s formula, != e22 π 1 +=0.88687, 22  21 12 2

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Table 1. Comparison of values of gamma function of R+.

Γ(k ) Actual Value Burnside’s Formula Absolute Percentage of Error

Γ(12) 1.77245 1.77374 0.07278

Γ(32) 0.88623 0.88557 0.07447

Γ(52) 1.32934 1.32884 0.03761

Γ(72) 3.32335 3.32263 0.02166

Γ(14) 3.62561 3.68104 1.52885

Γ(34) 1.22542 1.22431 0.09058

Γ(54) 0.90640 0.90559 0.08936

Γ(13 4) 2.54926 2.54863 0.02471

Γ(15 4) 4.42299 4.42214 0.01922

Γ(13) 2.67894 2.67897 0.00112

Γ(23) 1.35412 1.35321 0.06720

Γ(43) 0.89298 0.89222 0.08511

Γ(10 3) 2.77816 2.77751 0.02340

Γ(11 3) 4.01220 4.01140 0.01994

111 1 then, ! = Γ or we have Γ=1.77374 222 2  1 1  332 −  3 1 for != e22 π 1 +=1.32835 22  23 12 2 3 then, Γ= 0.88557. 2

5. Roots of Negative 1

The roots of −1 such as square roots, cube roots, fourth-roots, etc. can be obtained using the roots of complex numbers. θ Definition 2. [8] Let zC∈ , then z=+= x yi r(cosθθ + i sin) = r cis θ = r ei , ik(θ +2π ) sine θ is a multiple of 2π , then the general form is zr= e , where kZ∈ . ik(θ +2π ) Definition 3. [8] Let z0 be the nth root of complex number zr= e 1 n i(θ +2π kn) then zr0 = e for kn= 0,1,2,3 − 1 . The principal root is the root at k = 0, hence the principal nth-root is 1 ninθ i(π+2π kn) zr0 = e. Let z =−=1, 1e (π+2πk ) i ik(π 2+π ) Taking the square-root, we have 1e2 = 1e where k = 0 and 1. Thus, the roots are: i and −i and the principal root is i . (π+2πk ) i Taking the cube-roots, we have 1e 3 , where k = {0,1, 2} . Thus the roots are 12+ 3 2,ii −− 1,12 3 2 and the principal root is 12+ 3 2.i

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(π+2πk ) i Taking the fourth-roots, we have 1e4 , where k = {0,1,2,3} . Thus the roots are 22+ 22,i −+22 22, i −−22 22, ii 22 − 22, and the principal root is 22+ 22i .

6. Main Result

For k = 12, 11 let the function y=2 xn3 , = 3, suppose k = 12 then yy22′, ′′ are the first and second half-derivative. 15Γ+(n 14) Γ( ) 3! y22′ = 2 xnk−−= 22 xx3 12 = Γ(nk −+1) Γ( 4 − 12) 7 Γ 2 half-derivative 6 96 = 2 xx52 =52 = 3.61081 x52 15 π 15 π 8 Differentiate again the function, we have 1 51 96 Γ+(52 1) − 96 Γ(72) y2′′ = xx22= 2 15 π Γ(52 −+ 12 1) 15 π Γ(3) 15 π 96 = 8 xx22= 6 . 15 π 2 Thus, getting half-derivative twice is equivalent to first derivative

3 1 dd12 12yy d d2( x ) yx2′′ = = = = 6 2 ddxx12 1 2 ddxx

−3 12′ 12 ′′ ■ Let the function y=2 xn , = 3, suppose k = 1 2, then yy, are the first and second half-derivative. 1 k Γ+(nk) 1/2 Γ+(3 12) yx2′ =21(−=) −+(nk) 21(−) x−+(3 12) ΓΓ(n) (3)

17 − Γ(72) −(72) 3.32335 − y22′ = 2 i x= 2 i x 72 = 3.32335ix . Γ(32)

Differentiate again the function, we have

1 k Γ+(nk) yi2′′ =3.32335( − 1) x−+(nk) Γ(n)

12 Γ+(7 2 1/2) =3.32335ix( − 1) −+(72 12) Γ(72)

1 Γ(4) 3! y2′′ =3.32335 ii x−4 =−=3.32335( 1) x−−44−6 x ■ 7 3.32335 Γ 2

This completes the proof that twice of half derivative is equivalent to first derivative. For k = 1 3,

458 S. A. Loria

1 ′ ′′ ′′′ let y=2 xn3 , = 3, suppose k = , then yyy13 , 13, 13 , are the first, second, 3 and third 13 derivatives respectively. Γ+(31) Γ(4) y13′ = 2 x3− 13 = 2 xx83 = 2.99089 83 1 Γ(11 3) Γ31 −+ 3  8  11 Γ+1 81 Γ 7 7   −  13′′ 33   y= 2.99089 x33= 2.99089 xx3 = 4.31942 3 8 1  10 Γ −+1  Γ 33 3    10 Γ 71 − 13′′′ 3 2 y= 4.31942 xx33= 6 ■ Γ(3)

Thus, triple of one-third derivative is equal to first derivative.

1 ′ ′′ ′′′ Let y=2 xn−3 , = 3, suppose k = , then yyy13 , 13, 13 are the first, second, 3 and third ⅓ derivative respectively. 1 Γ+3 10 Γ+nk  − 13′ 13 ( ) −+(nk) 13 3 3 y=21(−=) x2+ ix ΓΓ(n) 22( 3) 10 − 10 =(1.38908 + 2.40596ix) 3 ; Γ= 2.77816 3 11 Γ(11 3) − y1 3′′ =(1.38908 + 2.40596ix) 3 Γ(10 3) 11 − =−+( 2.00609 3.47466ix) 3 11 1 Γ+ 12 − 1 3′′′ 13 33 3 y=−+( 2.00609 3.47466ii) + x ■ 22 11 Γ 3 =1.49544( − 4.01219 +=− 0ix) −42 6 x

This completes the proof that the triple of one-third derivative is equal to first derivative For k = 14 1 ′ ′′ ′′′ ′′′′ Let y=2 xn3 , = 3, suppose k = , then yyy14 , 14, 14 , y14 are the first, 4 second, third and fourth 14 derivatives respectively. Γ+(31) Γ(4) y14′ = 2 x3− 14 = 2 xx114 = 2.71310 114 1 Γ(15 4) Γ31 −+ 4  Γ(15 4) y1 4′′ = 2.71310 xx5 25= 3.61082 2 Γ(72)

Γ(72) y14′′′ = 3.61082 xx94 = 4.70726 94 Γ(13 4)

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′′′′ Γ(13 4) y14 = 4.70726 xx22= 6 ■ Γ(3)

This completes the proof that four times of one-fourth derivative is equivalent to first derivative. ′ ′′ ′′′ ′′′′ Let y=2 xn−3 , = 3, suppose k = 1 4, then yyy14 , 14, 14 , y14 are the first, second, third and fourth 14 derivatives respectively. 1 Γ+3 1 Γ+  −+3 14′ k (nk) −+(nk) 22 4 4 y=21(−=+) x2 ix ΓΓ(n) 22( 3) =(1.8026 + 1.8026ix) −13 4

7 Γ 15 − 1 4′′ 22 2 y=++(1.8026 1.8026i) ix4 22 13 Γ 4 = (2.54926i)( 1.30365) x−14 4 = 3.23334ix−144

15 Γ 15 − 14′′′ 22 4 y= 3.23334 i+ ix4 22 7 Γ 2 =−+( 3.12752 3.12752ix) −15 4

14′′′′ 22Γ(4) −4 y=−+( 3.12752 3.12752i) + ix 2 2Γ( 15 4) ■ =−=4.42298( 1.35655) xx−42− 6

This completes the proof that getting the one-fourth derivatives four times is equivalent to one whole or first derivative.

7. Conclusion

The study explored and analyzed the function between integral indices of derivative based on the theoretical deduction of the gamma function. The above solu- tions and proofs confirmed that derivatives using fractal indices exist every- where. Derivatives contributed significantly to the field of physical science. It is very interesting to describe and analyze the behavior of functions obtained through derivative process using fractal indices. Likewise, the process being used in this paper can be extended to analyze derivatives of different transcendental functions.

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Ontario. [4] Dulay, V.N. (1996) Reviewer/Text in Advanced Engineering Mathematics. Rex Book Store, Manila, Philippines. [5] Chen, B. (2012) Some Properties on the Function Involving Gamma Function. Ap- plied Mathematics, 3, 587-589. https://doi.org/10.4236/am.2012.36090 http://www.scirp.org/journal/PaperInformation.aspx?PaperID=20283 [6] Lu, D. (2014) A Generated Approximation Related to Burnside’s Formula. Journal of Number Theory, 136, 414-422. https://doi.org/10.1016/j.jnt.2013.10.016 http://www.sciencedirect.com/science/article/pii/S0022314X13002850 [7] Shi, X., Liu, F., and Qu, H. (2010) The Burnside Approximation of Gamma Functio. Analysis and Application, 8, 315-322. https://doi.org/10.1142/S0219530510001643 [8] Brown, J.W., Churchill, R.V. (1996) Complex Variables and Applications. 6th Edi- tion, McGraw-Hill, Inc., New York.

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