How To Be More Impressive
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Suppose we want to publish something that is as simple as
1 + 1 = 2 (1) \tag{1} 1+1=2 1+1=2(1)
This is not very impressive. If we want our article to be accepted by IEEE reviewers, we have to more abstract. So, we could complicate the left hand side of the expression by using
l n ( e ) = 1 and s i n 2 x + c o s 2 x = 1 ln(e) = 1 \qquad\text{ and }\qquad sin^2x+cos^2x=1 ln(e)=1 and sin2x+cos2x=1
and the right hand side can be stated as
2 = ∑ n = 0 ∞ 1 2 n 2 = \sum^\infin_{n=0}\frac{1}{2^n} 2=n=0∑∞2n1
Therefore, Equation (1) can be expressed more scientifically as:
l n ( e ) + ( s i n 2 x + c o s 2 x ) = ∑ n = 0 ∞ 1 2 n (2) \tag{2} ln(e) + (sin^2x+cos^2x) = \sum^\infin_{n=0}\frac{1}{2^n} ln(e)+(sin2x+cos2x)=n=0∑∞2n1(2)
which is far more impressive. However, we should not stop here. The expression can be further complicated by using
e = lim z → ∞ ( 1 + 1 z ) z and 1 = c o s h ( y ) 1 − t a n h 2 y e=\lim_{z\to\infin}\Big(1+\frac{1}{z}\Big)^z \quad\text{ and }\quad 1=cosh(y)\sqrt{1-tanh^2y} e=z→∞lim(1+z1)z and 1=cosh(y)1−tanh2y
Equation (2) may therefore be written as
l n [ lim z → ∞ ( 1 + 1 z ) z ] + ( s i n 2 x + c o s 2 x ) = ∑ n = 0 ∞ c o s h ( y ) 1 − t a n h 2 y 2 n (3) \tag{3}ln\Big[\lim_{z\to\infin}\Big(1+\frac{1}{z}\Big)^z\Big] + (sin^2x+cos^2x) = \sum^\infin_{n=0}\frac{cosh(y)\sqrt{1-tanh^2y}}{2^n} ln[z→∞lim(1+z1)z]+(sin2x+cos2x)=n=0∑∞2ncosh(y)1−tanh2y(3)
Note: Other methods of a similar nature could also be used to enhance our prestige, once we grasp the underlying principles.