Tough problems on Calculus.

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

1.1. Evaluate: limn12nlog(2nn)\large{\lim\limits_{n \to \infty}\frac{1}{2n}log\binom{2n}{n}}

2.2. Consider the following sequence of positive integers regarding ai{a_i}

a1=1a_1=1 and an=n(an1+1)a_n=n(a_{n-1}+1) for all n2n≥2.

Calculate: limn(1+1a1)(1+1a2)(1+1an)\large{\lim\limits_{n \to \infty}(1+\frac{1}{a_1})(1+\frac{1}{a_2})\dots(1+\frac{1}{a_n})}

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

f(1)=f(0)=f1(0)=f2(0)==fn(0)=0f(1)=f(0)=f^{1}(0)=f^{2}(0)=\dots=f^{n}(0)=0

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

4.4.Let ff be a real valued differentiable function on the real line R\mathbb{R} such that limx0f(x)x2\large{\lim\limits_{x\to 0}\frac{f(x)}{x^{2}}} exists, and is finite. prove that: f(0)=0\large{f'(0)=0}

Note by Aritra Jana
4 years, 10 months ago

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

Abhishek Sinha - 4 years, 10 months ago

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

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

Repeating the same argument for n1n-1, n2n-2, and so on, we see that it must be true that f(x)=0f(x) = 0 and hence f(n+1)(x)=0f^{(n+1)}(x) = 0.

Logically it seems sound:

nZ,xR(f(n+1)(x)=0f(n)(0)=0f(n)(x)=Cnf(n)(0)=0f(n)(x)=0)\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)

Jake Lai - 4 years, 10 months ago

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Using PMI

U Z - 4 years, 10 months ago

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What's PMI?

Jake Lai - 4 years, 10 months ago

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@Jake Lai Principle of mathematical induction

U Z - 4 years, 10 months ago

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

Jake Lai - 4 years, 10 months ago

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

Aritra Jana - 4 years, 10 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 :<

Jake Lai - 4 years, 10 months ago

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2) limn(a1+1a1a2+1a2...an+1anlim_{n \to \infty} ( \dfrac{a_{1} + 1}{a_{1}}\dfrac{a_{2} + 1}{a_{2}}...\dfrac{a_{n} + 1}{a_{n}}

an11=anna_{n-1} - 1 = \dfrac{a_{n}}{n}

Thus,

limn(a22a33....an+1n+1.1a1a2...an 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}}

=limnan+1(n+1)!=limn1+ann! = lim_{n \to \infty} \dfrac{a_{n+1}}{(n+1)!} = lim_{n \to \infty}\dfrac{1+a_{n}}{n!}

=limn1n!+ann!=limn1n!+1(n1)!an1(n1)! = 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)!}

limn1n!+1(n1)!+1(n2)!+....1(1)!+a1(1)!lim_{n \to \infty} \dfrac{1}{n!} + \dfrac{1}{(n-1)!} + \dfrac{1}{(n-2)!} + .... \dfrac{1}{(1)!} + \dfrac{a_{1}}{(1)!}

=e = e as a1=1a_{1} = 1

U Z - 4 years, 10 months ago

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Great :D

Krishna Sharma - 4 years, 10 months ago

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SUPERB! :D

Aritra Jana - 4 years, 10 months ago

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In 1st use limit as a sum

We will first cancel one n!n!

Now it becomes

limn1nlog((n+1)(n+2)....(n+n)1.2.3...n)\displaystyle lim_{n \to \infty } \frac{1}{n} log\left( \dfrac{(n+1)(n+2)....(n+n)}{1.2.3...n} \right)

limn12nr=1nlog(n+rr)\displaystyle lim_{n \to \infty } \frac{1}{2n} \sum_{r = 1}^{n} log \left( \dfrac{n + r}{r} \right)

Changing summation into integration

1201log(1+xx)dx\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 00\dfrac{0}{0} form Applying L-hospital f(x)2x\dfrac{f'(x)}{2x} so this should also be 00\dfrac{0}{0} because limit exist hence

f(0)=0f'(0) = 0


2nd looks interesting I'll try it

Krishna Sharma - 4 years, 10 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?

Aritra Jana - 4 years, 10 months ago

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

Krishna Sharma - 4 years, 10 months ago

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@Krishna Sharma the answer given is e\LARGE{\boxed{e}}

Aritra Jana - 4 years, 10 months ago

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