calculus

Huygens’ problem to Leibnitz: solution

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In the Feb 23 2018 blog problem, we posed the following question:

Sum the following infinite series:

1+\frac{1}{3} + \frac{1}{6} + \frac{1}{10} + \frac{1}{15}+ \ldots.

Solution:

The sum can be written as:

S=\sum_{n=1}^{\infty}P_{n}, where P_{n}=\frac{2}{n(n+1)}=2(\frac{1}{n}-\frac{1}{n+1}).

Thus, 2(1-\frac{1}{2} + \frac{1}{2} - \frac{1}{3} + \frac{1}{3} - \frac{1}{4} + \ldots)=2. This is the answer.

If you think deeper, this needs some discussion about rearrangements of infinite series also. For the time, we consider it outside our scope.

Cheers,

Nalin Pithwa.

RMO Training: taking help from Nordic mathematical contest: 1988

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Problem:

Let m_{n} be a smallest value of the function f_{n}(x)=\sum_{k=0}^{2n}x^{k}. Prove that m_{n} \rightarrow \frac{1}{2} when n \rightarrow \infty.

Proof:

For n>1,

f_{n}(x)=1+x+x^{2}+\ldots=1+x(1+x+x^{2}+x^{4}+\ldots)+x^{2}(1+x^{2}+x^{4}+\ldots)=1+x(1+x)\sum_{k=0}^{n-1}x^{2k}.

From this, we see that f_{n}(x)\geq 1 for x \leq -1 and x\geq 0. Consequently, f_{n} attains its maximum value in the interval (-1,0). On this interval

f_{n}(x)=\frac{1-x^{2n+1}}{1-x}>\frac{1}{1-x}>\frac{1}{2}

So, m_{n} \geq \frac{1}{2}. But,

m_{n} \leq f_{n}(-1+\frac{1}{\sqrt{n}})=\frac{1}{2-\frac{1}{\sqrt{n}}}+\frac{(1-\frac{1}{\sqrt{n}})^{2n+1}}{2-\frac{1}{\sqrt{n}}}

As n \rightarrow \infty, the first term on the right hand side tends to the limit \frac{1}{2}. In the second term, the factor

(1-\frac{1}{\sqrt{n}})^{2n}=((1-\frac{1}{\sqrt{n}})^{\sqrt{n}})^{2\sqrt{n}}

of the numerator tends to zero because

\lim_{k \rightarrow \infty}(1-\frac{1}{k})^{k}=e^{-1}<1.

So, \lim_{n \rightarrow \infty}m_{n}=\frac{1}{2}

auf wiedersehen,

Nalin Pithwa.

Reference: Nordic Mathematical Contest, 1987-2009.

 

Functions — “s’wat” Math is about !! :-)

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Reference: Nordic Mathematical Contest 1987, R. Todev:

Question:

Let f be a function, defined for natural numbers, that is strictly increasing, such that values of the function are also natural numbers and which satisfies the conditions f(2)=a>2 and f(mn)=f(m)f(n) for all natural numbers m and n. Define the smallest possible value of a.

Solution:

Since, f(n)=n^{2} is a function satisfying the conditions of the problem, the smallest possible a is at most 4. Assume that a=3. It is easy to prove by induction that f(n^{k})={f(n)}^{k} for all k \geq 1. So, taking into account that f is strictly increasing, we get

{f(3)}^{4}=f(3^{4})=f(81)>f(64)=f(2^{6})={f(2)}^{6}=3^{6}=27^{2}>25^{2}=5^{4}

as well as {f(3)}^{8}=f(3^{8})=f(6561)<f(8192)=f(2^{13})={f(2)}^{13}=3^{13}<6^{8}.

So, we arrive at 5<f(3)<6. But, this is not possible, since f(3) is an integer. So, a=4.

Cheers,

Nalin Pithwa.

Famous harmonic series questions

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Question 1:

The thirteenth century French polymath Nicolas Oresme proved that the harmonic series 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \ldots does not converge. Prove this result.

Question 2:

Prove that \frac{1}{2} + \frac{1}{4} + \frac{1}{6} + \ldots does not converge.

Question 3:

Prove that 1 + \frac{1}{3} + \frac{1}{5} + \frac{1}{7} + \ldots does not converge.

Even if you solve these all on your own, you won’t achieve the glory of that French polymath, but you will have “re-discovered” some “elements of truth of analysis” …You can give yourself a pat on the back!

Cheers,

Nalin Pithwa.