Translate sum and long variables to Julia (VS Code)

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1

Hi, I want to write code to transform the mathematical expression of:
y=sum from k=0 to k=infinite of x^k/k!
where x and y are long real var.
I try to write the code such this:

function taylor(k)
k = (0:Inf)
x = long(x)
y = long(y)
if n == 0
end
if n>0
for i in 1:n
f = k*i
y= x^k/f
end
@show x
@show y
end
end
(k)=taylor(3)

or:

function taylor(x)
u=0
v=1
k=0
a=1
while ((v-u)/u)>=(1/10^10)
u=u+((x^k)/factorial(big(k)))
k=k+1
v=v+((x^a)/factorial(big(a)))
a=a+1
end
return u
end
(u,v)=taylor(6)
@show u,v

both didn’t work. I am new to Julia and have no experience in coding before. How to represent series of numbers from 1 to infinite to the x function, let say x= int64. And also how to write proper function e.g function taylor(k). What variable should I put inside taylor( ).

Thank you!!!

2

Calculating the factorial and the power for each iteration is going to be very slow. BigFloats are also slow, but you don’t need them.

For each iteration you add a new term to your sum. Notice that

x_k = x^k/k! \hspace{1cm} x_{k+1} = x^{k+1}/(k+1)!

So x_{k+1} = x_k\,\cdot\, x/(k+1)

Use this in your algorithm:

function mysum(x, K)
    xₖ = one(x) / 1  # zeroth term x^0/0! equals one
    s = xₖ  # initialize sum
    for k in 1:K
        xₖ *= x / k  # update term as shown above
        s += xₖ  # update sum
    end
    return s
end

You should probably check for convergence, that is, occasionally check whether s + xₖ == s, at that point you can stop summing. I’ll leave adding that as an exercise for the reader :grin:

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