Approximation of a function with this condition

Question

I'm wondering if I have the expression

$$f = y(ln(x) +1)$$

and I'm given $x \gg 1$. Is it valid to approximate

$$f \approx yln(x)$$

Answer 1

The symbol used is $\sim$ not $\approx$ when taking about asymptotic behaviour of functions.

Assuming $g(x)\neq 0$ then $f(x)\sim g(x)$ is a notation for $\lim\limits_{x\to x_0}\dfrac{f(x)}{g(x)}=1$

Most of the time the context (i.e. $x_0$ which is often $0$ or $\infty$) is implicit and is omitted, though you can use $f(x)\sim_{x_0} g(x)$ to specify it explicitly.

This notation goes along with Landau $o(\ )$ notation to denote something negligible.

Where we use $x\ll 1$ for numbers, or in the context of $x\to 0$, we can use $f(x)=o(g(x))$ to signify $f(x)\ll g(x)$ with definition $\lim\limits_{x\to x_0} \dfrac{f(x)}{g(x)}=0$

It is immediate that $$f(x)=g(x)+o(g(x))\iff f(x)\sim g(x)$$

Rem: you noticed we ask for $g(x)\neq 0$, but we can also rewrite this in epsilon-delta as $|f(x)|\le \epsilon|g(x)|$ equivalently to get rid of the division, and keep the same meaning.

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In the present case $f(x)=y\ln(x)+y$

Since $y$ is constant $\frac y{\ln(x)}\to 0$ at infinity since $\ln(x)\to+\infty$

It means $y=o(\ln(x))$ ($y$ is small/negligible compared to $\ln(x)$) and we have $f(x)\sim y\ln(x)$.

Answer 2

There is no strict rule for "approximating". So your question should perhaps be something like "How good or bad is this approximation?" I also cannot answer this question with certainty, since you did not provide any further context. It really depends on what your definition of "good" is and what you want to do with the approximation.

Generally speaking, this is a good approximation (again - it depends on what your understanding of "valid" and "good" is):

Define $f(x)=y(\log(x) +1)$ and $g(x)=y\log(x)$, where $y$ is a real constant. The functions $f$ and $g$ have the same time complexity, so $f=O(g)$ and $g=O(f)$. If you're unfamiliar with the big $O$ notation, here are two equivalent definitions of it for $x\to\infty$:

Def 1. $$\exists\ C > 0\ \exists\ x_0 > 0\ \forall\ x > x_0: |f(x)| \le C\cdot|g(x)|$$

Def 2. $$\limsup_{x \to \infty} \left|\frac{f(x)}{g(x)}\right| < \infty$$

I will leave the proof for $f=O(g)$ and $g=O(f)$ to you. In this case I would use the first definition but try to prove it with both. If you still have question ask in the comments.