Competition - post a photo you've taken of nature >>

Which convergence theorem?

I am trying to show that :

\int^1_0 (1-\frac{x}{n})^n dx

converges to 1-1/e as n tends to infinity.

How do I know whether to use the monotone or dominated convergence theorem?

Reply 1

ztibor

Original post by Big_Sam

I am trying to show that :

\int^1_0 (1-\frac{x}{n})^n dx

converges to 1-1/e as n tends to infinity.

How do I know whether to use the monotone or dominated convergence theorem?

Just integrate with substitution or using the following simple rule

\int \left (f(ax+b)\right )^n\ dx=\frac{\left (f(ax+b)\right )^{n+1}}{a(n+1)}

The linear ax+b is -1/n x+1 in this question
Calculate the limit of the terms you've got considering that
the integral itself is improper because subbing 0 you get

1^{\infty}

form, so here you should to use another limit as x->0
And it is good to know

\lim_{n \rightarrow \infty} \left (1+\frac{a}{n}\right )^n=e^{a}

for example a=-x and x->0

\lim_{x \rightarrow 0}\left (\lim_{n \rightarrow \infty}\left (1+\frac{-x}{n}\right )^n\right )=

=lim_{x \rightarrow 0}e^{-x}=\frac{1}{e^0}=1

(edited 13 years ago)

Reply 2

Big_Sam

Original post by ztibor

Just integrate with substitution or using the following simple rule

\int \left (f(ax+b)\right )^n\ dx=\frac{\left (f(ax+b)\right )^{n+1}}{a(n+1)}

The linear ax+b is -1/n x+1 in this question
Calculate the limit of the terms you've got considering that
the integral itself is improper because subbing 0 you get

1^{\infty}

form, so here you should to use another limit as x->0
And it is good to know

\lim_{n \rightarrow \infty} \left (1+\frac{a}{n}\right )^n=e^{a}

for example a=-x and x->0

\lim_{x \rightarrow 0}\left (\lim_{n \rightarrow \infty}\left (1+\frac{-x}{n}\right )^n\right )=

=lim_{x \rightarrow 0}e^{-x}=\frac{1}{e^0}=1

I was going to try something similar to that but I know that I need to use a convergence theorem rather than straight forward integration and comparison.

Reply 3

ztibor

Original post by Big_Sam

I was going to try something similar to that but I know that I need to use a convergence theorem rather than straight forward integration and comparison.

Using the monotone convergence theorem, the limit of the integral
will be the integral of the limit of f which limit is e^(-x)

Reply 4

Big_Sam

Original post by ztibor

Using the monotone convergence theorem, the limit of the integral
will be the integral of the limit of f which limit is e^(-x)

Do i need to show that

\int^1_0 (1-\frac{x}{n})^n dx \leq K

for all n?

Also if I changed it to

\int^{\frac{\pi}{2}}_0 sin(x)^n dx

tends to zero as n tends to infinity then would I need the dominated theorem rather than the monotone?

Reply 5

.matt

Original post by Big_Sam

I am trying to show that :

\int^1_0 (1-\frac{x}{n})^n dx

converges to 1-1/e as n tends to infinity.

How do I know whether to use the monotone or dominated convergence theorem?

It's fairly easy to find an integrable function that bounds the (absolute value of the) integrand, given that x is between 0 and 1, so dominated convergence should work.

Spoiler

Original post by ztibor

Just integrate with substitution or using the following simple rule

\int \left (f(ax+b)\right )^n\ dx=\frac{\left (f(ax+b)\right )^{n+1}}{a(n+1)}

This is false in pretty much all cases: taking the derivative of the right hand side gives f(ax+b)^n f'(ax+b). It's true in the case that f(x) = x, but if you take for example f(x) = sin(x), a = 1 and b = 0 it's clear that it doesn't always hold.

NB: I am quite drunk so there is a rather high probability that I'm talking nonsense

Reply 6

ztibor

[QUTE]

This is false in pretty much all cases: taking the derivative of the right hand side gives f(ax+b)^n f'(ax+b). It's true in the case that f(x) = x, but if you take for example f(x) = sin(x), a = 1 and b = 0 it's clear that it doesn't always hold.

NB: I am quite drunk so there is a rather high probability that I'm talking nonsense

THanks.
I miss the derivative of f(ax+b) as factor from the intergrandus, but without an 'a' factor.
So (sinx)'(sinx)^2=(sinx)^3/3

Reply 7

Big_Sam

Original post by .matt

It's fairly easy to find an integrable function that bounds the (absolute value of the) integrand, given that x is between 0 and 1, so dominated convergence should work.

Spoiler

So you mean for the first one I should use the dominated not the monotone? Or do you mean that for the sin question?

Reply 8

ztibor

Original post by Big_Sam

Do i need to show that

\int^1_0 (1-\frac{x}{n})^n dx \leq K

for all n?

You need to show

\displaystyle \lim_{n \rightarrow \infty} \int^1_0 \left (1-\frac{x}{n}\right )^n dx =K

and this K exists and finite.
This limit would be

\int^1_0 (\lim_{n \rightarrow \infty}(1-\frac{x}{n})^n) dx

and the limit function of f_n is e^(-x) so

\int^1_0 e^{-x} dx = [-e^{-x}]^1_0=-\frac{1}{e}+1

Also if I changed it to

\int^{\frac{\pi}{2}}_0 sin(x)^n dx

tends to zero as n tends to infinity then would I need the dominated theorem rather than the monotone?

I think so

(edited 13 years ago)

Reply 9

Big_Sam

I've plotted some of the fn and I can't seem to find a function that all of the fn are less too or that they converge to, so is it even possible to use one of the theorems?

(edited 13 years ago)

Reply 10

.matt

Original post by Big_Sam

I've plotted some of the fn and I can't seem to find a function that all of the fn are less too or that they converge to, so is it even possible to use one of the theorems?

Ah, I forgot about this topic, sorry :redface:

x is between 0 and 1, so x/n is going to be between 0 and 1 too. What does that say about 1-x/n? (1-x/n)^n? A constant is clearly integrable over a finite interval, hence if you can find a constant to bound (1-x/n)^n, you can use DCT to interchange the limit and integral.

Reply 11

ztibor

Original post by Big_Sam

I've plotted some of the fn and I can't seem to find a function that all of the fn are less too or that they converge to, so is it even possible to use one of the theorems?