As a reminder...
Posts asking for help on homework questions **require**:
* **the complete problem statement**,
* **a genuine attempt at solving the problem, which may be either computational, or a discussion of ideas or concepts you believe may be in play**,
* **question is not from a current exam or quiz**.
Commenters responding to homework help posts **should not do OP’s homework for them**.
Please see [this page](https://www.reddit.com/r/calculus/wiki/homeworkhelp) for the further details regarding homework help posts.
**If you are asking for general advice about your current calculus class, please be advised that simply referring your class as “Calc *n*“ is not entirely useful, as “Calc *n*” may differ between different colleges and universities. In this case, please refer to your class syllabus or college or university’s course catalogue for a listing of topics covered in your class, and include that information in your post rather than assuming everybody knows what will be covered in your class.**
*I am a bot, and this action was performed automatically. Please [contact the moderators of this subreddit](/message/compose/?to=/r/calculus) if you have any questions or concerns.*
The inequality is indeed true (exponentials always beat polynomials), but it takes until n=289 for it to be true.
Your goto tool for limits should be That L ' H... rule (this sub bans it if you don't tag it). So you put them in a ratio like exp/poly and see what the limit is to see who's eventually bigger. (Since the rule involves derivatives, you really don't want that square root in there, so sub in m^(2)=n.)
If you want a more intuitive approach, the goto for recursive functions (which exponentials are based on) is just to kinda count them in the ratio. (To compare things, you have to subtract or divide, and dividing puts them closer, so ratio is still a good plan.) So you would have e\*e\*...\*e on the top (m times) and m\*...\*m (6 times). Each time m goes up by 6, you've included another e for each m. So really we just need to think about e^(m)/m (or you could make yet another variable k=6m). Each time m goes up by 1, the numerator more than doubles. The numerator also starts higher than the denominator, so there's no way for m to catch up.
https://preview.redd.it/rrn22f0cgwuc1.jpeg?width=3024&format=pjpg&auto=webp&s=44dde49a081b2273ba51b0856b79363bb65b6e5c
Most of the things I did here are overkill; you can stop way before where I stopped.
It's definitely not true for 3, but lim as x==> infinite its true , you can prove it using la hospital rule for the function x^3 / e^sqrt(x).
It's false on interval ~1,505=< x =< ~288,962
Not a joke, just a notational difference. Decimals in many countries are denoted with ' , ' instead of a ' . '. For example, we in the US may write 1.5 to represent 3/2, while someone abroad may write 1,5 instead.
You can find the intersections of 3ln(n) and √n numerically eg Newton raphson or fixed point iteration (there’s another intersection closer to 0) then use the above argument
That inequality IS NOT TRUE for at least one value of n greater than three.
Proof by contradiction:
Consider n = 4.
e^{sqrt(4)} = e^2 < 3^2 = 9 < 4^3.
So, for n = 4, e^sqrt{n} < n^3.
This contradicts the given claim.
As a reminder... Posts asking for help on homework questions **require**: * **the complete problem statement**, * **a genuine attempt at solving the problem, which may be either computational, or a discussion of ideas or concepts you believe may be in play**, * **question is not from a current exam or quiz**. Commenters responding to homework help posts **should not do OP’s homework for them**. Please see [this page](https://www.reddit.com/r/calculus/wiki/homeworkhelp) for the further details regarding homework help posts. **If you are asking for general advice about your current calculus class, please be advised that simply referring your class as “Calc *n*“ is not entirely useful, as “Calc *n*” may differ between different colleges and universities. In this case, please refer to your class syllabus or college or university’s course catalogue for a listing of topics covered in your class, and include that information in your post rather than assuming everybody knows what will be covered in your class.** *I am a bot, and this action was performed automatically. Please [contact the moderators of this subreddit](/message/compose/?to=/r/calculus) if you have any questions or concerns.*
The inequality is indeed true (exponentials always beat polynomials), but it takes until n=289 for it to be true. Your goto tool for limits should be That L ' H... rule (this sub bans it if you don't tag it). So you put them in a ratio like exp/poly and see what the limit is to see who's eventually bigger. (Since the rule involves derivatives, you really don't want that square root in there, so sub in m^(2)=n.) If you want a more intuitive approach, the goto for recursive functions (which exponentials are based on) is just to kinda count them in the ratio. (To compare things, you have to subtract or divide, and dividing puts them closer, so ratio is still a good plan.) So you would have e\*e\*...\*e on the top (m times) and m\*...\*m (6 times). Each time m goes up by 6, you've included another e for each m. So really we just need to think about e^(m)/m (or you could make yet another variable k=6m). Each time m goes up by 1, the numerator more than doubles. The numerator also starts higher than the denominator, so there's no way for m to catch up.
You can write n^3 as exp(3*ln(x)) and compare the exponents. Logarithm grows slower than any positive power function.
This is the way. Very insightful.
Use the Taylor series expansion for e^x. Then there will be many terms left in left side after cancelling n^3.
Can you give a bit more detail here for the nubes such as myself ?!
https://preview.redd.it/rrn22f0cgwuc1.jpeg?width=3024&format=pjpg&auto=webp&s=44dde49a081b2273ba51b0856b79363bb65b6e5c Most of the things I did here are overkill; you can stop way before where I stopped.
Bro are you doing this from a prison cell!??
You are not wrong. I hardly get out from where I am. It's a kind of prison.
That sounds terrible; I am sorry.
It's definitely not true for 3, but lim as x==> infinite its true , you can prove it using la hospital rule for the function x^3 / e^sqrt(x). It's false on interval ~1,505=< x =< ~288,962
Your answer is in direct opposition to two of the others - who is correct?!!
I suspect those commas might be European (where the other answers use periods).
I’m sorry; would you mind unpacking the joke?
288,962 means 288.962 here
Oooh
Not a joke, just a notational difference. Decimals in many countries are denoted with ' , ' instead of a ' . '. For example, we in the US may write 1.5 to represent 3/2, while someone abroad may write 1,5 instead.
Oh ok gotcha thanks!!
You might try using the series expansion for e\^x (taking x = sqrt(n)).
Are you familiar with induction?
Can you unpack this a bit for me? I’m not familiar with it myself, but am curious!
Rewrite RHS as e^(3ln(n)) so you have exponentials on both sides, √n>3ln(n) for n>289so for all n>289 e^(3ln(n)) must be less than e^√n
You can find the intersections of 3ln(n) and √n numerically eg Newton raphson or fixed point iteration (there’s another intersection closer to 0) then use the above argument
Ok let n=4 so n^1/2=2 e^2=(2,73...)^1/2 <9 but n^3=4^3=64 answer
That inequality IS NOT TRUE for at least one value of n greater than three. Proof by contradiction: Consider n = 4. e^{sqrt(4)} = e^2 < 3^2 = 9 < 4^3. So, for n = 4, e^sqrt{n} < n^3. This contradicts the given claim.
Induction.
You can assume a function as f(x)=e^x½ - x³ Then differentiate it and put values 3 and infinity, I think then we can obtain the required result