Does $(1+\frac{x}{n})^n$ Uniformly Converge to $e^x$ on Different Intervals?

[evinda]

Hey again! (Blush)
I am looking at the following exercise:
We suppose $f_n(x)=(1+\frac{x}{n})^{n}, x \in \mathbb{R}$.
Check if $f_n$ converges uniformly at the intervals $(-\infty,a)$ and $(a,+\infty)$,where $a$ is a random real number.

$lim_{n \to +\infty} f_n(x)=e^{x}$,so $f(x)=e^{x}$
For the interval $(a,+\infty)$,we have:
$sup_{x>a}|f_n(x)-f(x)|=sup_{x>a}|(1+\frac{x}{n})^{n}-e^x|$
To find the supremum,we have to know if $ (1+\frac{x}{n})^{n}>e^x$ or $(1+\frac{x}{n})^{n}<e^x$ ,right? But how can I find which is greater?

I looked at the solution of the exercise and saw that at the interval $(a,+\infty)$,they take $e^x-(1+\frac{x}{n})^{n}>0$,and at $(-\infty,a)$ $(1+\frac{x}{n})^{n}-e^x>0$.. But why is it like that?How can I know it?? (Thinking)

 

[ThePerfectHacker]

Let us pick $a=0$ and examine the interval $(0,\infty)$, does $f_n$ converge to $\exp$ on this interval uniformly?

If it did it would mean that for $\varepsilon = 1$, there be an $n$ such that,
$$ |\exp(x) - f_n(x)| < 1 \text{ for all }x>0 $$

But given any $n$ the function $f_n$ is a polynomial function of degree $n$, while $\exp$ is the exponential function which goes to $\infty$ faster than any polynomial, no matter how big the degree. Thus, given any $n$ we can find $x_0$ large enough such that $\exp(x_0) > f_n(x_0) + 1$ which contradicts what we need to have uniform convergence.

 

[evinda]

Let us pick $a=0$ and examine the interval $(0,\infty)$, does $f_n$ converge to $\exp$ on this interval uniformly?

If it did it would mean that for $\varepsilon = 1$, there be an $n$ such that,
$$ |\exp(x) - f_n(x)| < 1 \text{ for all }x>0 $$

But given any $n$ the function $f_n$ is a polynomial function of degree $n$, while $\exp$ is the exponential function which goes to $\infty$ faster than any polynomial, no matter how big the degree. Thus, given any $n$ we can find $x_0$ large enough such that $\exp(x_0) > f_n(x_0) + 1$ which contradicts what we need to have uniform convergence.

I understand..And how can I check the uniform converge at the interval $(-\infty,a)$?

 

[ThePerfectHacker]

I understand..And how can I check the uniform converge at the interval $(-\infty,a)$?

Pick $a=0$ to get an idea of what is going on $(-\infty,0)$. We have a polynomial $p_n(x)$ of degree $n$ and exponential function $\exp(x)$. As $x\to -\infty$ the polynomial $p_n(x)$ will approach either $\pm \infty$ depending on the sign of its highest degree coefficient. While $\exp(x) \to 0$ as $x\to -\infty$. Therefore, $\exp(x)$ and $p_n(x)$ will not stay close to one another but will deviate as $x\to -\infty$.

 

[blue_raver22]

Hello! The supremum-inequality is a mathematical concept that states that the supremum of a set of numbers is always greater than or equal to any element in that set. In this case, we are looking at the supremum of the function $f_n(x)$ over the interval $(a,+\infty)$. To find the supremum, we can use the fact that $f_n(x)$ is a monotonically increasing function, meaning that as $n$ increases, so does the value of $f_n(x)$. Therefore, we can take the limit as $n \to +\infty$ to find the supremum of $f_n(x)$ over the given interval. In this case, we can see that the limit is equal to $e^x$, which is the supremum of $f_n(x)$ over the interval $(a,+\infty)$.

As for finding which is greater between $(1+\frac{x}{n})^{n}$ and $e^x$, we can use the fact that $e^x$ can be written as a power series, which converges for all real numbers. Therefore, we can approximate $e^x$ by taking the first few terms of the power series and compare it to $(1+\frac{x}{n})^{n}$. As $n$ increases, we can see that $(1+\frac{x}{n})^{n}$ gets closer and closer to $e^x$, meaning that eventually, $(1+\frac{x}{n})^{n}$ will be greater than $e^x$.

I hope this helps clarify things for you! Let me know if you have any other questions.