Calculus & Analysis Marathon & Questions (2 Viewers)

seanieg89

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Re: First Year Uni Calculus Marathon

Since this is one of the easier questions ill solve this :





This is actually a bit circular. How are you defining e^x and log(x) if not through one of these limits? In which case you need to prove existence of the limits / differentiability etc before you can invoke something like L'Hopital.

To avoid such circularity in your argument it is best to avoid mentioning e^x or log(x) entirely.
 

Paradoxica

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Re: First Year Uni Calculus Marathon

By default, the expression is greater than zero for all real x, so a lower bound exists already.

For any real x, we can find a sufficiently large n such that x/n ≤ 2. Thus, the expression is bounded from above by 3^x.

For the sum, the ratio test is sufficient to establish convergence for all real x.

Therefore, both expressions converge for all real x. (not sure if legit)

Expanding the power using the binomial theorem and using the monotone convergence theorem should be sufficient to establish coincidence for all real x.

^is any of that legit lmao
 
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seanieg89

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Re: First Year Uni Calculus Marathon

By default, the expression is greater than zero for all real x, so a lower bound exists already.

For any real x, we can find a sufficiently large n such that x/n ≤ 2. Thus, the expression is bounded from above by 3^x.

For the sum, the ratio test is sufficient to establish convergence for all real x.

Therefore, both expressions converge for all real x. (not sure if legit)

Expanding the power using the binomial theorem and using the monotone convergence theorem should be sufficient to establish coincidence for all real x.

^is any of that legit lmao
Yep :), almost all of it is correct and exactly what I was looking for. Bounding that first expression above by 3^n does not achieve much since 3^n is unbounded, but the convergence of the first expression follows from the application of the monotone convergence theorem so you get this for free basically.
 

Paradoxica

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Re: First Year Uni Calculus Marathon

Yep :), almost all of it is correct and exactly what I was looking for. Bounding that first expression above by 3^n does not achieve much since 3^n is unbounded, but the convergence of the first expression follows from the application of the monotone convergence theorem so you get this for free basically.
cool, just that the way you worded it implied you wanted proof of convergence before proof of coincidence... :)
 

seanieg89

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Re: First Year Uni Calculus Marathon

Originally I did want that (and to achieve this you can prove that the first thing is monotonic in n and bounded, which establishes convergence), but there is nothing logically wrong with skipping this step though so I cannot fault a proof that does not mention this. Make sure you understand why it is okay to skip it though.
 

Paradoxica

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Re: First Year Uni Calculus Marathon

I didn't even get up to the last part of my quadrature question but I feel this last part is a good marathon question so I'll drop it here... I could not get it out though.

Too difficult to bound. Decided to go for the more direct route.

By continuity, it suffices to show that

Substitute k = 1/(3u) to transform the limit to the origin.



Split the logarithm apart and use the limiting secant definition of the first derivative on both terms.

This evaluates to 1, so the original limit goes to e, by continuity.
 
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Paradoxica

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Re: First Year Uni Calculus Marathon

Here's a fun exercise.



porcupinetree is not allowed to answer this
 

Paradoxica

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Re: First Year Uni Calculus Marathon



I actually need help with this one lol
 
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Paradoxica

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Re: First Year Uni Calculus Marathon

Evaluate:



You may need to familiarise yourself with trinomial coefficients.
 

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