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# Tough problems on Calculus.

Here are some problems in calculus which i am not being able to solve:

$$1.$$ Evaluate: $\large{\lim\limits_{n \to \infty}\frac{1}{2n}log\binom{2n}{n}}$

$$2.$$ Consider the following sequence of positive integers regarding $${a_i}$$

$$a_1=1$$ and $$a_n=n(a_{n-1}+1)$$ for all $$n≥2$$.

Calculate: $\large{\lim\limits_{n \to \infty}(1+\frac{1}{a_1})(1+\frac{1}{a_2})\dots(1+\frac{1}{a_n})}$

$$3.$$Let $$f:\mathbb{R}\to \mathbb{R}$$ be function that is differentiable $$n+1$$ times for some positive integer $$n$$. The $$i^{th}$$ derivative is denoted by $$f^{i}$$. Suppose:

$f(1)=f(0)=f^{1}(0)=f^{2}(0)=\dots=f^{n}(0)=0$

Prove that: $$\large{f^{n+1}(x)=0}$$

$$4.$$Let $$f$$ be a real valued differentiable function on the real line $$\mathbb{R}$$ such that $$\large{\lim\limits_{x\to 0}\frac{f(x)}{x^{2}}}$$ exists, and is finite. prove that: $$\large{f'(0)=0}$$

Note by Aritra Jana
2 years, 9 months ago

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2) $$lim_{n \to \infty} ( \dfrac{a_{1} + 1}{a_{1}}\dfrac{a_{2} + 1}{a_{2}}...\dfrac{a_{n} + 1}{a_{n}}$$

$$a_{n-1} - 1 = \dfrac{a_{n}}{n}$$

Thus,

$$lim_{n \to \infty} (\dfrac{a_{2}}{2}\dfrac{a_{3}}{3}....\dfrac{a_{n+1}}{n+1}.\dfrac{1}{a_{1}a_{2}...a_{n}}$$

$$= lim_{n \to \infty} \dfrac{a_{n+1}}{(n+1)!} = lim_{n \to \infty}\dfrac{1+a_{n}}{n!}$$

$$= lim_{n \to \infty} \dfrac{1}{n!} + \dfrac{a_{n}}{n!} = lim_{n \to \infty} \dfrac{1}{n!} + \dfrac{1}{(n-1)!} \dfrac{a_{n -1}}{(n-1)!}$$

$$lim_{n \to \infty} \dfrac{1}{n!} + \dfrac{1}{(n-1)!} + \dfrac{1}{(n-2)!} + .... \dfrac{1}{(1)!} + \dfrac{a_{1}}{(1)!}$$

$$= e$$ as $$a_{1} = 1$$

- 2 years, 9 months ago

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Great :D

- 2 years, 9 months ago

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SUPERB! :D

- 2 years, 9 months ago

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In 1st use limit as a sum

We will first cancel one $$n!$$

Now it becomes

$$\displaystyle lim_{n \to \infty } \frac{1}{n} log\left( \dfrac{(n+1)(n+2)....(n+n)}{1.2.3...n} \right)$$

$$\displaystyle lim_{n \to \infty } \frac{1}{2n} \sum_{r = 1}^{n} log \left( \dfrac{n + r}{r} \right)$$

Changing summation into integration

$$\displaystyle \frac{1}{2} \int_0^{1} log\left( \dfrac{1+x}{x} \right) dx$$

Now this can be solved easily

4th is easy if limit exist then it is $$\dfrac{0}{0}$$ form Applying L-hospital $$\dfrac{f'(x)}{2x}$$ so this should also be $$\dfrac{0}{0}$$ because limit exist hence

$$f'(0) = 0$$

2nd looks interesting I'll try it

- 2 years, 9 months ago

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for the 4th one.. that is exactly what i thought.. but i was confused that it might be incorrect.. are you sure that method is right?

- 2 years, 9 months ago

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Yes :), what's the answer to 2nd?

- 2 years, 9 months ago

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the answer given is $$\LARGE{\boxed{e}}$$

- 2 years, 9 months ago

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1. Use the Stirling's approximation $n!\sim \sqrt{2\pi n}\bigg(\frac{n}{e}\bigg)^n$

- 2 years, 9 months ago

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3: I don't know if this is a valid proof:

$$f^{(n+1)}(x) = 0$$ iff $$f^{(n)}(x) = C_{n}$$ where $$C_{i}$$ are some arbitrary constants. However, $$f^{(n)}(0) = 0$$; because $$f^{(n)}(x) = f^{(n)}(0) = 0$$ for all $$x \in \mathbb{R}$$, therefore, $$f^{(n)}(x) = C_{n} = 0$$.

Repeating the same argument for $$n-1$$, $$n-2$$, and so on, we see that it must be true that $$f(x) = 0$$ and hence $$f^{(n+1)}(x) = 0$$.

Logically it seems sound:

$\forall n \in \mathbb{Z}, x \in \mathbb{R} \Bigg(\boxed{f^{(n+1)}(x) = 0 \wedge f^{(n)}(0) = 0} \longleftrightarrow \boxed{f^{(n)}(x) = C_{n} \wedge f^{(n)}(0) = 0} \longleftrightarrow \boxed{f^{(n)}(x) = 0} \Bigg)$

- 2 years, 9 months ago

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Using PMI

- 2 years, 9 months ago

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What's PMI?

- 2 years, 9 months ago

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Principle of mathematical induction

- 2 years, 9 months ago

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@U Z Yeah, it's pretty much induction but in reverse with base case $$n+1$$.

- 2 years, 9 months ago

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@megh choksi @Krishna Sharma @Sandeep Bhardwaj ..any idea as to how to approach the 3rd one?

- 2 years, 9 months ago

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I posted a solution just now, and I think it works. It seems too easy though, which is troubling :<

- 2 years, 9 months ago

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