Question

Using LMVT (Lagrange's Mean Value Theorem) for solving algebra problems

Answer 1

@dan815

Answer 2

The square roots of two successive natural numbers greater than \(\large{N^2}\) differ by:

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Let n be a natural number greater than \(\large{N^2}\). Consider \(\large{f(x) = \sqrt{x}}\) where \(\large{x \in [n,n+1]}\).

Since, f(x) satisfies conditions for LMVT, thus:

\[\large{f'(c) = \cfrac{f(n+1)-f(n)}{n+1-n}}\] for some \(\large{c \in (n, n+1)}\tag{1}\)

and \(\large{f'(x) = \cfrac{1}{2\sqrt{x}}}\) \(\large{\implies f'(c) = \cfrac{1}{2\sqrt{c}}}\tag{2}\)

Thus, using (1) and (2):

\[\large{\cfrac{1}{2\sqrt{c}} = \sqrt{n+1}-\sqrt{n}}\tag{3}\]

Since, \(\large{n > N^2}\), thus:

\[\large{c > N^2 ~\implies~\cfrac{1}{c} < \cfrac{1}{N^2} ~\implies~\cfrac{1}{2\sqrt{c}} < \cfrac{1}{2N}}\]

Thus:

\[\large{\boxed{\sqrt{n+1}-\sqrt{n} < \cfrac{1}{2N}}}\tag{Ans}\]

Answer 3

@ganeshie8 @Kainui @Mimi_x3 @satellite73

Any more ideas ?

Answer 4

This was a question of sequences and series and I used LMVT :P ;)

Answer 5

Interesting, I'd like to see what kind of applications this could have. I've actually never heard of Lagrange's mean value theorem before but it looks familiar.

Answer 6

It is a generalization of `Rolle's theorem`

Answer 7

http://en.wikipedia.org/wiki/Mean_value_theorem

Answer 8

@ikram002p Any idea ???