User talk:Akshaysrinivasan

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Hello, Akshaysrinivasan, and welcome to Wikipedia! Thank you for your contributions. I hope you like the place and decide to stay. Here are a few good links for newcomers:

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I hope you enjoy editing here and being a Wikipedian! Please sign your name on talk pages using four tildes (~~~~); this will automatically produce your name and the date. If you need help, check out Wikipedia:Questions, ask me on my talk page, or place {{helpme}} on your talk page and someone will show up shortly to answer your questions. Again, welcome!  Oleg Alexandrov (talk) 16:31, 28 March 2006 (UTC)[reply]

I reverted your changes at derivative as they were incorrect. It is not true that

a function is differentiable at a point x if its derivative is continuous at x

Also, a function of many variables is not necessarily an implicit function.

One last thing: it is good if you take more care when you write things down, like a space after period, proper spelling, making variables italic: <math>x</math>, etc. Thanks. Oleg Alexandrov (talk) 16:31, 28 March 2006 (UTC)[reply]

I am sure to differentiate a functiona x its derivative must be continuous.that is what the formal definition says doesn't it?a function is differentiable if and only if lim h->0 (f(x+h)-f(x))/h exists.well if the limit exists the derivative has to be continuous. —This unsigned comment was added by Akshaysrinivasan (talk • contribs) .

That's not right. If the function is differentiable, it must be continuous, but not necessarily its derivative is continuous.

I would also suggest some typographical tips. I noticed that when you write in articles sometimes you confuse the comma (,) and the period (.) and don't leave a space after a period. I suggest you use correct typography, so that articles look nice. :) Oleg Alexandrov (talk) 16:00, 30 March 2006 (UTC)[reply]

I am not convinced what you've said is correct. If the function is differentiable the limit ${\displaystyle limh->0{\frac {f(x+h)-f(x)}{h}}}$0{\frac {f(x+h)-f(x)}{h}}}"/> has to exist right? The limit itself is the derivative of the function,so if it has to exist it must be continuous at x. Let us take an example,consider the function:-

                 f(x)=|x|

Now ${\displaystyle {\frac {d|x|}{dx}}=-1,x<0;+1,x>0}$0}"/> If you plot the derivative of the |x| you will certainly find a discontinuity at x=0 where the function is not differentiable. Unless you give me an example to the contrary i shall not be convinced to think i am wrong.

You are just saying that ${\displaystyle {\frac {f(x+h)-f(x)}{h}}}$ is continuous at ${\displaystyle h=0}$, but that is not the same as saying ${\displaystyle f'(x)}$ is continuous. Do you have any references? Oleg Alexandrov (talk) 16:53, 31 March 2006 (UTC)[reply]

But according to the definition of the derivative:- ${\displaystyle f'(x)=lim_{h\to 0}{\frac {f(x+h)-f(x)}{h}}}$ Well for f(x) to be differentiable, ${\displaystyle lim_{h\to -0}{\frac {f(x+h)-f(x)}{h}}=lim_{h\to +0}{\frac {f(x+h)-f(x)}{h}}}$ This is to say ${\displaystyle f'(x)}$ is continuous at x Obviously this is not applicable to the end points of the curve. This is mandatory for a curve to be differentiable at a point x,because if this condition is not satisfied the curve will have two values of slope at the same point and leads to the formation of a sharp point or a corner.you may refer thomas and finney although it hasn't been stated explicitly. Could you provide me examples of functions where these conditions are not satisfied?? --unsigned

You are right that the function

${\displaystyle {\frac {f(x+h)-f(x)}{h}}}$

is continuous at ${\displaystyle h=0}$, as a function of ${\displaystyle h}$ with ${\displaystyle x}$ fixed. As ${\displaystyle h\to 0}$, with ${\displaystyle x}$ fixed, this gives ${\displaystyle f'(x)}$. But as a function of ${\displaystyle x}$ this is not ${\displaystyle f'(x)}$ for all ${\displaystyle x}$ and with ${\displaystyle h}$ fixed!

You want a function which is differentiable but whose derivative is not continuous. Here you are:

${\displaystyle f(x)=x^{2}\sin {\frac {1}{x}}}$ for ${\displaystyle x\neq 0}$ and ${\displaystyle f(0)=0.}$

Its derivative is

${\displaystyle 2x\sin {\frac {1}{x}}-\cos {\frac {1}{x}}}$

which is not continous at 0. Oleg Alexandrov (talk) 03:05, 1 April 2006 (UTC)[reply]

Well i think we have a misunderstanding here.You see i said a function is differentiable at x=a if its derivative is continuous at x=a.And your function is no doubt differentiable at all points except at x=0 where f'(x)is discontinuous and varies randomly.I think i forgot to mention that.I never meant the variable to be h.If i did i would never have mentioned f(x)=|x| which itself is differentiable at all points except x=0 i.e its dervative is discontinuous at x=0.Akshay Srinivasan22:51,April 1 2006 IST

OK then, so you say that if the derivative is continuous at x=a, then it is differentiable at x=a. This statement says nothing, as if you assume that the derivative is continuous at x=a, that obviously implies that the derivative exists (if the derivative exists and is continuous, then it obviously exists).

What you should have said that if the derivative f'(x) exists in a neighbourhood of x=a (not including a), and the limit as x->a of f'(x) exists, then it follows that f'(a) also exists. That would have been accurate. :) Oleg Alexandrov (talk) 17:48, 1 April 2006 (UTC)[reply]

OK now can i put in the article derivative?Akshay Srinivasan23:23,April 1 2006

Now we agree on the correct statement. But frankly, I don't think it belongs in the derivative article. It is an interesting observation, but not as crucial. And that article is already big enough. Oleg Alexandrov (talk) 20:50, 1 April 2006 (UTC)[reply]

I reverted your edits to that article. What you wrote is not at all a generalization, because it is not even a series, it has a finite number of terms. Besides, that observation is described right below, in the "pitfalls" sections.

And some style tips. You should write "generalization" and "series", rather than "Generalization" and "Series", that is, small letters. Also, you should use a space afer comma (,) and after period (.) the way I do in this paragraph. Thanks. Oleg Alexandrov (talk) 08:51, 2 April 2006 (UTC)[reply]

Alright I ahall do that from next time on.Akshay 11:46, 2 April 2006 (UTC)[reply]

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Hi. I see you have created a nice page for the Hoysala temple there. I shall soon be doing some work on this page (I am medieval architecture buff) and adding more pictures. I shall be sourcing useful info about the temple structure and design from a book I have Complete guide to Hoysala temples by Gerard Foekemo. All this over the next month or so.EnjoyDineshkannambadi 18:37, 1 November 2006 (UTC)[reply]

Hi Akshay,

Namaskaara! :) I see that you had written message to Dinesh Kannambadi in his user page, User:Dineshkannambadi. In general it is not recommended to edit other user's page or putting a message in other user's page. We should be using talk page for those purposes. I have just moved that message from user page to Dinesh's talk page, here: User talk:Dineshkannambadi. Thank you and happy editing. - KNM ^Talk 16:28, 6 November 2006 (UTC)[reply]

Thanks for your reply. I have a few good pictures too which I planned to upload today showing doorway decor, minor shrine at coutryard corners etc. I shall however retain your existing one. Feel free to just upload your other file(s) into wiki and give me its .jpg file name so I can access it. Upload as a GDFL self image. While I have visited over 15 Hoysala monuments and have images for all of them, If you have some for other temples that I dont, then we can better use your resources. Do you have any for Belavadi ( I have uploaded mine in the page for this town but at that time I was not using my digital camera so I did not get the whole trikuta temple into one image). Do you have pictures from other Hoysala temples or any other architectural idioms that I could use.?Dineshkannambadi 16:58, 6 November 2006 (UTC)[reply]

Hi,

Regarding your correction of the hindi version of 'Technical' from 'प्रौद्योगिकी' to 'पौधोगिकी' in the BIT Mesra article 'http://en.wikipedia.org/wiki/Birla_Institute_of_Technology'. Please note the earlier version reflects the correct hindi translation. You may refer the link http://www.iitkgp.ac.in/ or http://www.iitd.ernet.in/ for details.

Thanks Pray4sid 11:24, 8 June 2007 (UTC)[reply]

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