Start with the standard series for e^t:
e^t = Σ(n = 0 to ∞) t^n/n!.
Let t = -x^3/5:
e^(-x^3/5) = Σ(n = 0 to ∞) (-x^3/5)^n/n!
..............= Σ(n = 0 to ∞) (-1)^n * x^(3n)/(5^n * n!).
Integrate term by term:
∫ e^(-x^3/5) dx = [Σ(n = 0 to ∞) (-1)^n * x^(3n+1)/(5^n * (3n+1) * n!)] + C.
(Now, you can substitute bounds as needed.)
I hope this helps!
This was an answer to one of my posts. I don't know how to get the answer based on what the poster wrote. The e^(-x^3/5) is the equation for the survival function. I want the probability that someone lives past 5 years. I think I have to do an integral from 5 to infinity, but I don't know exactly how to using series. Do I have to plug in 5 for n & x and infinity for n & x, or do I plug in 1 - integral from 0 to 5? Please help.
e^t = Σ(n = 0 to ∞) t^n/n!.
Let t = -x^3/5:
e^(-x^3/5) = Σ(n = 0 to ∞) (-x^3/5)^n/n!
..............= Σ(n = 0 to ∞) (-1)^n * x^(3n)/(5^n * n!).
Integrate term by term:
∫ e^(-x^3/5) dx = [Σ(n = 0 to ∞) (-1)^n * x^(3n+1)/(5^n * (3n+1) * n!)] + C.
(Now, you can substitute bounds as needed.)
I hope this helps!
This was an answer to one of my posts. I don't know how to get the answer based on what the poster wrote. The e^(-x^3/5) is the equation for the survival function. I want the probability that someone lives past 5 years. I think I have to do an integral from 5 to infinity, but I don't know exactly how to using series. Do I have to plug in 5 for n & x and infinity for n & x, or do I plug in 1 - integral from 0 to 5? Please help.
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What you have written does not seem to be a probability density. The integral from 0 to infintiy is not 1. (Using Wolfram Alpha). Have you left out some factors involving t² perhaps?
Once you have an actual density f(x), just take
...... 5 ............ ∞
1 – ∫ f(x) dx = ∫ f(x) dx
..... 0 ............ 5
The function as you have gives rise to a Gamma function and, as I said, does not intgerate to 1 over the whole domain. Please check for typos (either your or in the book).
I hope these comments are helpful.
On the other hand if what you have given is not a den sity, b ut the Survvial function itself:
S(t) = P(T>t) where T is the random variable, then S(5) = P(T>5), so just compute f(5) = e^((-5³)/5)
In this case the deny is the negative of the derivative of what you gave and is s(x) = (3x²/5)e^(-x³/5)
-- I assume you left out parentheses and get simply S(5) = 1.38859 x 10^(–11)
Once you have an actual density f(x), just take
...... 5 ............ ∞
1 – ∫ f(x) dx = ∫ f(x) dx
..... 0 ............ 5
The function as you have gives rise to a Gamma function and, as I said, does not intgerate to 1 over the whole domain. Please check for typos (either your or in the book).
I hope these comments are helpful.
On the other hand if what you have given is not a den sity, b ut the Survvial function itself:
S(t) = P(T>t) where T is the random variable, then S(5) = P(T>5), so just compute f(5) = e^((-5³)/5)
In this case the deny is the negative of the derivative of what you gave and is s(x) = (3x²/5)e^(-x³/5)
-- I assume you left out parentheses and get simply S(5) = 1.38859 x 10^(–11)