1 , 2 2 1 , 3 3 1 , 4 4 1 , 5 5 1 , 6 6 1 , ⋯
Find the largest surd in the above series. Give your answer correct to three decimal places.
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I should've marked it as a calculus problem but in a bit of rush I accidentally chose Number Theory for this.
Anyways, a straight forward approach as always sir.
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It is okay to mark it as Number Theory. It is about numbers. It is just that the method to solve it is Calculus.
Tapas, don't use Huge. It makes the problem looks childish. Punctuation rules do not need a space before comma ",". There should be a space after comma and LaTex takes care of that. No need to assume that 1 is a surd because it does not affect the answer and it is not necessary that all the elements of the series must be a surd.
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Okay, I'll take care of this from now. Thank you sir. :)
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Let y = x x 1 . Then d x d y = x x 1 ( x 2 1 − ln x ) . And d x d y = 0 ⟹ ln x = 1 ⟹ x = e . We note that d x 2 d 2 y ∣ ∣ ∣ ∣ x = e < 0 , therefore, the maximum y , y m a x = e e 1 for all real x . The nearest term in the series to e e 1 is 3 3 1 ≈ 1 . 4 4 2 .