Calculus & Analysis Marathon & Questions (1 Viewer)

Drsoccerball · Oct 9, 2016

Re: First Year Uni Calculus Marathon

seanieg89 said:
Related:

$Prove that $\lim_{n\rightarrow \infty }\left(1+\frac{x}{n}\right)^n$ and $\lim_{n\rightarrow \infty }\left(\sum_{k=0}^n \frac{x^k}{k!}\right)$ both exist and coincide for all real numbers $x$.$

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

$\lim_{n\rightarrow \infty }\left(1+\frac{x}{n}\right)^n \ <=> \lim_{n\rightarrow \infty } e^{n\ln{(1+\frac{x}{n})}}$

$$ Applying L'hopitals rule we get $ e^x = \lim_{n\rightarrow \infty }\left(1+\frac{x}{n}\right)^n$

$e^x $ has a taylor series $ \lim_{n\rightarrow \infty }\left(\sum_{k=0}^n \frac{x^k}{k!}\right) $ which can be deduced through his formula. $$

seanieg89 · Oct 9, 2016

Re: First Year Uni Calculus Marathon

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

$\lim_{n\rightarrow \infty }\left(1+\frac{x}{n}\right)^n \ <=> \lim_{n\rightarrow \infty } e^{n\ln{(1+\frac{x}{n})}}$

$$ Applying L'hopitals rule we get $ e^x = \lim_{n\rightarrow \infty }\left(1+\frac{x}{n}\right)^n$

$e^x $ has a taylor series $ \lim_{n\rightarrow \infty }\left(\sum_{k=0}^n \frac{x^k}{k!}\right) $ which can be deduced through his formula. $$

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 · Oct 9, 2016

Re: First Year Uni Calculus Marathon

seanieg89 said:
Related:

$Prove that $\lim_{n\rightarrow \infty }\left(1+\frac{x}{n}\right)^n$ and $\lim_{n\rightarrow \infty }\left(\sum_{k=0}^n \frac{x^k}{k!}\right)$ both exist and coincide for all real numbers $x$.$

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

seanieg89 · Oct 9, 2016

Re: First Year Uni Calculus Marathon

Paradoxica said:
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 · Oct 9, 2016

Re: First Year Uni Calculus Marathon

seanieg89 said:
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 · Oct 9, 2016

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 · Oct 9, 2016

Re: First Year Uni Calculus Marathon

leehuan said:
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.

$$e) Check that $e=\lim_{k\rightarrow \infty}{{\left(\frac{2+\sqrt{3+9k^2}}{3k-1}\right)}^k}$

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

By continuity, it suffices to show that $\lim_{k \to \infty} k \log{\left(\frac{2+\sqrt{3+9k^2}}{3k-1}\right)} = 1$

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

$\lim_{u \to 0} \frac{1}{3u} \log{\left( \frac{2u + \sqrt{3u^2+1}}{1-u} \right)}$

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.

Paradoxica · Oct 10, 2016

Re: First Year Uni Calculus Marathon

Here's a fun exercise.

$\lim_{n \to \infty} \int_0^1 \sqrt[n]{x^n+(1-x)^n} \text{d}x$

porcupinetree is not allowed to answer this

Paradoxica · Oct 15, 2016

Re: First Year Uni Calculus Marathon

$\noindent Let $H_{x}$ be the analytical continuation of the Harmonic Number to all real numbers excluding the non-positive integers, defined as:$ \int_0^1 \frac{1-t^x}{1-t} \, \text{d}t \\ $By using the fact that $\lim_{x \to \infty} \left(H_{x+z} - H_{x}\right) = 0$ for any real $z$, or otherwise, prove that: $\int_0^1 H_{x} \, \text{d}x = \gamma \\ $You may assume that $H_0 =0$ without proof.$

I actually need help with this one lol

seanieg89 · Oct 15, 2016

Re: First Year Uni Calculus Marathon

Paradoxica said:
$\noindent Let $H_{x}$ be the analytical continuation of the Harmonic Number to all real numbers excluding the non-positive integers, defined as:$ \int_0^1 \frac{1-t^x}{1-t} \, \text{d}t \\ $By using the fact that $\lim_{x \to \infty} \left(H_{x+z} - H_{x}\right) = 0$ for any real $z$, or otherwise, prove that: $\int_0^1 H_{x} \, \text{d}x = \gamma \\ $You may assume that $H_0 =0$ without proof.$

I actually need help with this one lol

$\int_0^1 H(x)\, dx\\ \\ =\int_0^1\left(\int_0^1 (1-t^x)\sum_{k\geq 0}t^k\, dt \right)\, dx\\ \\ =\lim_{n\rightarrow\infty}\int_0^1\left( \int_0^1 (1-t^x)\sum_{k= 0}^n t^k \, dt\right) \, dx\quad (\star)\\ \\ = \lim_{n\rightarrow\infty}\left(h_{n+1}-\int_0^1 \sum_{k=0}^n\frac{1}{x+k+1}\,dx\right)\\ \\ =h_{n+1}-\log(n+1)\rightarrow \gamma.\\ \\ \\ $where $(\star)$ follows from the monotone convergence theorem. $$

Paradoxica · Oct 15, 2016

Re: First Year Uni Calculus Marathon

seanieg89 said:
$\int_0^1 H(x)\, dx\\ \\ =\int_0^1\left(\int_0^1 (1-t^x)\sum_{k\geq 0}t^k\, dt \right)\, dx\\ \\ =\lim_{n\rightarrow\infty}\int_0^1\left( \int_0^1 (1-t^x)\sum_{k= 0}^n t^k \, dt\right) \, dx\quad (\star)\\ \\ = \lim_{n\rightarrow\infty}\left(h_{n+1}-\int_0^1 \sum_{k=0}^n\frac{1}{x+k+1}\,dx\right)\\ \\ =h_{n+1}-\log(n+1)\rightarrow \gamma.\\ \\ \\ $where $(\star)$ follows from the monotone convergence theorem. $$

I'd like the way that uses the hint...

seanieg89 · Oct 16, 2016

Re: First Year Uni Calculus Marathon

Paradoxica said:
I'd like the way that uses the hint...

Sure.

$$For $x\in [0,1]$ $\\ \\H(x+n)-H(n) \leq H(n+1)-H(n)\\ \\ \Rightarrow \lim_{n\rightarrow\infty}\int_0^1 H(x+n)-H(n)\, dx = 0\\ \\ \Rightarrow \lim_{n\rightarrow\infty}\left(\int_0^1 H(x)+\sum_{k=1}^n \frac{1}{x+k}\, dx - H(n)\right)=0\\ \\ \Rightarrow \int_0^1 H(x)\, dx =\lim_{n\rightarrow\infty} \left(H(n)-\log(n+1)\right)=\gamma.$

Paradoxica · Oct 22, 2016

Re: First Year Uni Calculus Marathon

Evaluate:

$\lim_{n \to \infty} \prod_{k=1}^n \left(1+\frac{k}{n^2}\right)$

Hint:

$1-x< \frac{1}{1+x} < 1 \quad \forall x \in \mathbb{R}^+$

Paradoxica · Oct 22, 2016

Re: First Year Uni Calculus Marathon

Evaluate:

$\lim_{n \to \infty} \sqrt[n]{\binom{3n}{n,n,n}}$

You may need to familiarise yourself with trinomial coefficients.

Paradoxica · Oct 24, 2016

Re: First Year Uni Calculus Marathon

$$\noindent Let $(a_n)_{n>0}$ be a convergent sequence. Discuss the convergence of: $\\\\ $a) $ \lim_{n \to \infty} \prod_{k=1}^n (a_{n+1} - a_{k}) \\\\ $b) $ \lim_{n \to \infty} \prod_{k=1}^n (1 + a_n - a_{k})$

InteGrand · Oct 25, 2016

Re: First Year Uni Calculus Marathon

Paradoxica said:
$$\noindent Let $(a_n)_{n>0}$ be a convergent sequence. Discuss the convergence of: $\\\\ $a) $ \lim_{n \to \infty} \prod_{k=1}^n (a_n - a_{k}) \\\\ $b) $ \lim_{n \to \infty} \prod_{k=1}^n (1 + a_n - a_{k})$

Is there a typo in the first one? The last term in the product is 0, so that product is just 0 for all n.

Paradoxica · Oct 25, 2016

Re: First Year Uni Calculus Marathon

InteGrand said:
Is there a typo in the first one? The last term in the product is 0, so that product is just 0 for all n.

Ah yes... mismatched indices, sorry.

InteGrand · Nov 3, 2016

Re: First Year Uni Calculus Marathon

$\noindent Let $\mathbf{F}\left(x,y\right)$ be a continuous vector field defined on the plane with $\left\|\mathbf{F}\left(x,y\right)\right\| = 1$ at all points $(x,y)$. Let $S$ be the unit square (vertices $(0,0), (0,1),(1,0),(1,1)$) and suppose that $\mathbf{\overline{F}}:= \iint_{S}\mathbf{F}\left(x,y\right) \, \mathrm{d}x\,\mathrm{d}y$ has unit length. Must $\mathbf{F}(x,y)$ be a constant vector field? Justify your answer.$

$\noindent (The vector obtained by integrating a vector field is that where each component is obtained by integrating the vector field's corresponding components.)$

Paradoxica · Nov 13, 2016

Re: First Year Uni Calculus Marathon

$$\noindent Let $(a_{n})_{n>0}$ be defined by the following: $ \\\\ a_1 = 1, a_{n+1} = \frac{n}{a_n} \\\\ $Evaluate: $ \lim_{n \to \infty} \frac{1}{\sqrt{n}} \left( \sum_{k=1}^n \frac{1}{a_k} \right)$

Paradoxica · Nov 13, 2016

Re: First Year Uni Calculus Marathon

$$\noindent Evaluate: $ \lim_{n \to \infty} \frac{e^{\mathcal{H}_n}}{\sum_{k=1}^n \sqrt[k]{k}} $ where $\mathcal{H}_n$ is the $n^{\text{th}}$ Harmonic Number.$

You are given the limit exists, so do not prove it.

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