The Importance of Limits in the Development of Mathematics

[titaniumx3]

For nonegative integers n ,consider the limit:

(a) L=\sqrt{n+\sqrt{n+\sqrt{n+...}}}

(If a_{1}=\sqrt{n},a_{k+1}=\sqrt{n+a_{k}} that's
another way of notation we are interested in
L=\lim_{k\to\infty}a_{k})

Looking at the above, given the definition of a_{1} and a_{k+1}, we are told,

\lim_{k\to\infty}a_{k} = \sqrt{n+\sqrt{n+\sqrt{n+...}}}

But, this to me seems to be either false or slightly misleading. Shouldn't it actually be,

\lim_{k\to\infty}a_{k} = ...\sqrt{n+\sqrt{n+\sqrt{n}}}

Are those two limits equivalent or different?

 

[sadhu]

to me it seems to be same expression written in two forms,
there is no reason for their to be different , is there??

 

[titaniumx3]

to me it seems to be same expression written in two forms,
there is no reason for their to be different , is there??

Well, in the first (original) limit, for n=0 say, it's not clear what value is contained in the first square root, where as in the second limit I posted, it is clear that the value contained in the first square root is 0; hence the limit would definitely be 0.

 

[sadhu]

but if you see you are trying to find the infinite term of a series
a_{k}=\sqrt{n+a_{k-1}}

but also you defined in your post that
a_{1}=\sqrt{n}

thus it is clear that what is there in first term \sqrt{n},n=0,=0
hence limit=0

what is so unclear about it

sorry for the trouble with latex format

 

[titaniumx3]

Hmmm, I think I'm probably getting myself confused lol.

Say the two equations I posted have limits as k tends to infinity, L_{1} and L_{2} respectively and we consider the case where n = 0.

Then,

L_{1} = \sqrt{0+\sqrt{0+\sqrt{0+...}}} = \sqrt{\sqrt{\sqrt{...}}}

and,

L_{2} = ...\sqrt{0+\sqrt{0+\sqrt{0}}} = ...\sqrt{\sqrt{\sqrt{0}}}

Here, it is clear that L_{2} = 0 but L_{1} = 0 or L_{1} = 1.

Hmmm, this is really confusing, lol. To me L_{2} is a more accurate representation of the limit of the sequence a_{k} as k tends towards infinity.

 

[sadhu]

how can you say that L^{1}=1

unless you show it maths wise

 

[titaniumx3]

how can you say that L^{1}=1

unless you show it maths wise

I'm not entirely sure how you'd show it rigourously and in all honesty I'm not sure if it makes any sense at all with regards to L_1 but,

1 = sqrt(1) = sqrt(sqrt(1)) = sqrt(sqrt(sqrt(1))) = ...

That said, the above is using only the representation of L_1 in my previous post, not in the a_1 = sqrt(n), a_(k+1) = sqrt(n+a_K) representation (which I claim equals L_2, not L_1).

Maybe it is just poorly written notation, but the way L_1 is definined in my last post doesn't make any sense at all because:

L_{1} = \sqrt{0+\sqrt{0+\sqrt{0+...}}} = \sqrt{\sqrt{\sqrt{...}}}

implies there is no end term, so how on Earth do you know it is equal to 0? Personally I think it is best to define it simply as a_{1}=\sqrt{n},a_{k+1}=\sqrt{n+a_{k}} and leave it there.

 

[titaniumx3]

May I ask, when someone discusses "limiting processes", what exactly does that mean? Does it simply refer to things like the sum of an infinite series, calculation of differentials/integrals? What about taking the limit of a function or sequence, is that also a limiting process?

BTW, I'd like to thank whoever recommended the book, The Origins Of Cauchy's Rigorous Calculus; it's an interesting read, easy to comprehend and invaluable to the topic at hand.