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Equations for functions

Here are some equations I have come across over the years. Some are series equations for calculating trigonometric and numerous other types of functions. Hopefully your browser will not mess up these equations.
The term a for the following equations can be a real number or a complex number such as 2+3i. Have a look at my article on complex numbers too for alternate equations. They may also help you understand the results they give.
On Mathcad (and probably other programs like it), use (-1)r instead of -1r because -1r means raise 1 to the power of r and make it negative.
It is used to make alternate signs +2, -3, +4, -5 ... in series equations.

Trigonometry

sin(a) =  ∞
∑ 
r=0
 (-1)r a2r+1
(2r+1)!
  cos(a) =  ∞
∑ 
r=0
 (-1)r a2r
(2r)!
  tan(a) =  sin(a)
cos(a)


sin(a) =  eai - e-ai
2i
  cos(a) =  eai + e-ai
2
  tan(a) =  i(1-e2ai)
1 + e2ai  where i=√-1

Inverse Trigonometric functions

Note that sin-1(a) = arcsin(a) = asin(a) etc. Not to be confused with sin2(a) = sin(a) • sin (a)
arcsin(a) =  arctan (a )
√(1-a2)
arcos(a) = (π/2) - arctan(a)
√(1-a2)
arctan(a) =  ∞
∑ 
r=0
 (-1)r•a2r+1 where |a| ≤1
2r+1

arcsin(a) =  ∞
∑ 
r=0
 ( (2r)! ) a2r+1
22r(r!)2 2r+1  where |a| ≤ 1 which shows that it still can be a complex number
arccos(a) =  π  - arcsin(a)
2
arctan(a) =  ∞
∑ 
r=0
 (-1)ra2r+1  where |a| ≤ 1 a ≠ i, -i because the arctangent of i or -i does not have a solution
2r + 1

Leonhard Euler found a more efficient series for the arctangent, which is:
arctan(a) =  a ∞
∑ 
r=0
 r
∏
k=1		 2ka2
1+a2 (2k + 1)(1 + a2)
(Notice that the term in the sum for n= 0 is the empty product which is 1.)

Alternatively, this can be expressed:
arctan(a) =  ∞
∑ 
r=0
 22r(r!)2 a2r+1
(2r + 1)!(1 + a2)r+1
I tried it with some expressions like a = 0.3 - 0.5i on Mathcad and in 6 or so iterations, I got an accurate answer but there does appear to be some restriction in range when I used negative numbers and some larger numbers

arcsin(a) = -i•ln(ia + √(1-a2)
arccos(a) =  -i•ln(a + i√(1-a2) =  π - arcsin(a)
2
arctan(a) =  i (ln(1-ia) - ln(1+ia))
2

There exists infinite series equations for arcsecant etc. and logarithmic forms too which I have omitted.

Hyperbolia

sinh(a) =  ea - e-a
2
  cosh(a) =  ea + e-a
2
  tanh(a) =  e2a - 1
e2a + 1

arcsinh(a) = ln (a + √(a2+1))
arccosh(a) = ln (a + √(a2-1))

arctanh(a) =  ln(1+a) - ln(1-a)  =  ln((1+a)/(1-a))  a ≠ 1, -1 
2 2
 =  ∞
∑ 
r=0
 a2r+1  |a| < 1
2r + 1
The series operator has this restriction: |a| <1 but you can calculate a result when |a| >1 by using the equation with ln() in it but you will also need an equation from the complex number article I have written. There exists no solution when a = ± 1.

Secant, cosecant and cotangent

sec(a) =  1
cos(a)
  cosec or csc(a) =  1
sin(a)
  cot(a) =  1
tan(a)
These too have infinite series equations to define them

Logarithmic and exponential

en :=  ∞
∑ 
r=0
 nr
r!
  ln(n+1) :=  ∞
∑ 
r=0
 (-1)r•nr+1
r+1
 =  ∞
∑ 
r=1
 (-1)r-1 nr
r



For en, n≠0 but it can be a real number an imaginary number or even a complex number although for a complex number a + bi, you can use ea + bi = ea × ebi if you wish (although you are using more series equations). I have tested the series equation on Mathcad with complex numbers and it worked.
For ln(n+1),  -1 < n ≤ 1 so if you want ln(0.5) then set n = -0.5. This gives you a range for taking logarithms of a number greater than 0, up to <2 but for higher numbers, you can invert it and make the logarithm negative e.g. ln(10) = -ln(1/10).
If this is a little confusing, you can use a slightly modified form:
ln(n) :=  ∞
∑ 
r=1
 (-1)r-1•(n-1)r
r  0 < n ≤ 2
Although for small values of n, you need to calculate maybe a million iterations and for larger numbers, maybe just a thousand is enough.
You may think that taking a natural logarithm of a negative number is not possible because most calculators are not programmed to evaluate such things but in fact, you can and it gives a complex number as a result:
if n < 0 then ln(n) = ln|n| +  i•π and if you wish to use this, or other functions with complex numbers, I have written a page on this subject.
I have tried this also with complex numbers and it worked for a + bi = 1 + i but I have not worked out the range of numbers:
ln(a+bi) :=  ∞
∑ 
r=1
 (-1)r-1•(a-1+bi)r
r
but by using ln(n) = ln|n| +  i•π then you do not need to use complex numbers in the series equation. I just present it to you out of interest.

Combinations and permutations

nCr =  n!   nPr =  n!
r!(n-r)! (n-r)!

Miscellaneous equations that interest me

cosh(a) ± sinh(a) = e ±2a




π  =  ∞
Π
r=0		 4r2 + 8r + 4  =  ∞
Π
r=0		 (2r +2)2
2 4r2 + 8r +3 (2r + 1)(2r+3)

π  =  4•tan-1(1) = 4× ∞
∑ 
r=0 		(-1)r
2r +1

eiθ= cos θ +i•sin θ (Euler's formula)
eπi= -1 (Euler's identity)

n! ∼ √(2πn)• ( n ) n
e Stirling's formula
e  =  lim
n→∞ 		(1 + n-1)n
Modify an equation above (which converges quickly) gives e =  ∞
∑ 
r=0
 1
r!



sin-1(n)= π/2 + i•cosh-1(n) where I thought n>1 but it works on Mathcad for 0.1 etc. I'll investigate sometime.

To get any Fibonacci number, first calculate f given n where n≥0 (or more formally n ∈ ℕ0)
f=  1 (  1+√5 ) n
√5 2

Now take ⌊f⌉ = ||f|| = nint(f) = Round(f) = a rounded up integer
or you can also use int(f + 0.5) on a computer or some calculators which rounds up a positive number.
You can also use:
f=  1 ((  1+√5 ) n - (1-√5)n)
√5 2 2
(expect a few rounding errors on a computer)

These produce:
n ⌈ f ⌉
0 0
1 1
2 1
3 2
4 3
5 5
6 8
7 13
8 21
... ...





Have you found an error or do you want to add more information to these pages?
You can contact me at the bottom of the home page.

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