For
$f_n(x) = \frac{nx + x^2}{n^2}$, this converges pointwise to the function $f(x) = 0$.
How would I prove this formally?
This is my attempt:
$$|f_n(x) - f(x)| = |\frac{nx + x^2}{n^2}| \leq |\frac{nx + nx^2}{n^2}| = |\frac{x + x^2}{n}|$$
So if I were to pick $N := \frac{|x + x^2|}{\epsilon}$ then by reversing the steps, we get the required inequality.
My question is: Is my choice of $N$ correct (is there allowance for the absolute values)? And if so, just to confirm for pointwise convergence, the allowance of having $x$ terms in the value for $N$ is O.K. right?
Asked
Active
Viewed 43 times
3
Natash1
- 1,379
-
"is there allowance for the absolute values": provided you find a suitable value of $N$ for any $\epsilon$, any expression is "allowed". – Jun 02 '17 at 09:54
-
Thanks, the main reason why I asked that is because in other epsilon delta proofs, I would always try to find a way to get rid of absolute values, but I didn't know why. – Natash1 Jun 02 '17 at 09:56
3 Answers
1
First, $N$ has to be an integer number.
Second, what you wrote $$|f_n(x) - f(x)| = |\frac{nx + x^2}{n^2}| \leq |\frac{nx + nx^2}{n^2}| = |\frac{x + x^2}{n}|$$
is incorrect the middle inequality does not hold necessarly ! For example: if $x=-1$ The RHS is zero, LHS is always positive for all $n \ge2.$
Red shoes
- 6,948
-
-
Regardingyour edit:
That's weird, because I thought $nx + nx^2 \geq nx + x^2$ for all $n \geq 1$, so I'm not sure why this doesn't cover that $x=1$ case. – Natash1 Jun 02 '17 at 09:55 -
-
Hmm, but if I have $n \geq 1 \iff n-1 \geq 0 \iff x^2(n-1) \geq 0 \iff nx^2 \geq x^2 \iff nx + nx^2 \geq nx +x^2$ – Natash1 Jun 02 '17 at 09:59
1
Why did you stop? :) From where you stopped, note that $N:= \lfloor \frac{|x+x^{2}|}{\varepsilon} \rfloor + 1 > \frac{|x+x^{2}|}{\varepsilon}$ and is an integer for sure.
Yes
- 20,719
-
Thanks! So absolute value signs do not matter right? Iirc, in other epsilon delta proofs like just the limit of a function to a value, I remember always trying to find ways to remove absolute value signs, but I wasn't always sure why – Natash1 Jun 02 '17 at 09:45
-
You only have to make sure $N$ is an integer; the absolute value thing does not matter here. – Yes Jun 02 '17 at 11:17
1
We may establish
$$\left|\frac{nx+x^2}{n^2}\right|<\left|\frac{x}{n}\right|+\left|\frac{x^2}{n^2}\right|<\epsilon.$$
An easy option is to make both terms smaller than $\epsilon/2$, which is achieved by
$$n>\left|\frac{2x}{\epsilon}\right|\text{ and }n>\sqrt{\left|\frac{2x^2}{\epsilon}\right|}$$ and we can take the maximum of the two.