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Question 953598: Decreasing or increasing ? an=ne^-n
I need not only answer but explanation too.
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! an = ne^-n
I believe you might mean:
An = n * e^-n
An meaning the nth term in the progression.
The fact that the exponent is -n, I would assume that the sequence is decreasing.
You can take a few values of n and show what is happening.
since it doesn't tell you that n has to be > 0, you should assume that n can be negative or positive or 0.
so, let's take some values of n between -3 and 3 and see what happens.
the following table shows what happens:
n n * e^(-n)
-3 -3 * e^(-(-3)) = -3 * e^(3) = -60.25661...
-2 -2 * e^(-(-2)) = -2 * e^(2) = -14.77811...
-1 -1 * e^(-(-1)) = -1 * e^(1) = -2.71828...
0 0 * e^(0) = 0
1 1 * e^(-1) = .367879...
2 2 * e^(-2) = .270670...
3 3 * e^(-3) = .149361...
100 100 * e^(-100) = 3.720075... * 10^(-42)
it's clear that the function gets very small the further negative you go with the value of n and that the function stays positive and approaches 0 the further positive you get.
it does not appear that there is a vertical asymptote.
in order for there to be a vertical asymptote, the denominator of a rational equation needs to be equal to 0 at some point.
n * e^(-n) is the same as n / e^n
the denominator is e^n, but the denominator can never be equal to 0 because e^n can never be equal to 0.
when n = 0, e^n is equal to e^0 which is equal to 1.
so, there is no vertical asymptote, even though the function gets extremely negative as n gets more negative.
in fact, it gets so negative that you won't be able to see it on the graph after n becomes more negative than -10 or so unless you scale into the millions.
for example:
when n = -1000, n * e^(-n) becomes -1000 * e^1000 which becomes so large a negative value that the calculator can't handle it.
when n = -100, n * e^(-n) becomes -100 * e^100 which becomes -2.688117... * 10^45.
that's a very large negative number, but the function does not approach infinity as n approaches a certain number, so there is no vertical asymptote.
is there a horizontal asymptote?
that can be found by increasing n as large as it can get.
as n approaches infinity, the function approaches 0.
you get n * e^-n becomes n / e^n.
as n approached infinity, the function gets smaller and smaller.
you can see this clearer by dividing numerator and denominator by n to get:
(n/n) / ((e^n)/n) which results in 1 / ((e^n)/n)
(e^n)/n will get larger and larger as n gets larger.
the numerator stays the same at 1.
the function gets smaller and smaller and approaches zero as n grows larger and larger and approaches infinity.
for example, when n = 1000, n/e^n is equal to 1000 / e^1000 which is equal to 0.
it's not really equal to 0, but the calculator can't handle a number that small so the calculator shows 0.
when n = 100, n/e^n is equal to 100 / e^100 which is equal to 3.72007... * 10^(-42).
that's a very small number that's pretty close to 0.
as n gets larger, it gets close to 0 but never reaches 0.
so you have a function that approaches minus infinity as n gets more negative and you have a function that approaches 0 as n gets more positive.
a graph of the equation is shown below:
in the graph, n is replaced by x in order to satisfy the requirements of the graphing software.
otherwise the equation is the same and when i talk about n, i'm also talking about x because they represent the same thing.
there are 2 graphs to show you how the function behaves.
the first graph is a view from a distance to see what happens on the negative side of the graph as n gets more negative.
the second graph is a close up view to see what happens on the positive side of the graph as n gets more positive.
don't forget that x represent n in the graph.
so what is the answer to your question?
the question was:
Decreasing or increasing ? an=ne^-n
I need not only answer but explanation too.
the function increases to a maximum point and then decreases thereafter as it approaches the horizontal asymptote of y = 0.
how do you explain that?
good question.
i would say you can say that n * e^-n is the same as n / e^n and it is clear clear that e^n rises dramatically faster than n as the value of n gets larger.
as n approaches infinity, the function n / e^n therefore approaches 0.
the function is decreasing after it reaches it's maximum point at somewhere between n = 0 and n = 2.
in fact, it appears that the maximum value is as n = 1.
when n = 1, the function is equal .3678794412...
when n = .999, the function is equal to .3678792571
when n = 1.001, the function is equal to .3678792574...
I can't swear to it, but n = 1 does appear to be the maximum point of the function.
so the function reaches a maximum at n = 1 and then decreases thereafter and approaches 0 but never quite touches it.
the short answer.
the function increases for a while and then decreases.
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