My Math Forum Differentiation 2

 Calculus Calculus Math Forum

May 7th, 2014, 07:55 AM   #1
Newbie

Joined: May 2014
From: India

Posts: 12
Thanks: 0

Differentiation 2

If y=(image)
then dy/dx=?
Attached Images
 Capture.JPG (9.4 KB, 16 views)

 May 7th, 2014, 09:14 AM #2 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,598 Thanks: 2583 Math Focus: Mainly analysis and algebra Surely $y(x)$ is a step function, so therefore has $$\frac{dy}{dx}=\begin{cases}0 & x \notin \mathbb{Z} \\ \text{undefined} & x \in \mathbb{Z} \\ \end{cases}$$ If it's not a step function, then it would seem more sensible to write it as $$y(x) = \int_1^x \frac{1}{\arctan{\left(1+r+r^2\right)}}dr$$ In which case, by the second fundamental theorem of calculus, $$\frac{dy}{dx} = \frac{1}{\arctan{\left(1+x+x^2\right)}}$$ Thanks from Raptor Last edited by v8archie; May 7th, 2014 at 09:22 AM.
May 7th, 2014, 10:08 AM   #3
Newbie

Joined: May 2014
From: India

Posts: 12
Thanks: 0

Quote:
 Originally Posted by v8archie Surely $y(x)$ is a step function, so therefore has $$\frac{dy}{dx}=\begin{cases}0 & x \notin \mathbb{Z} \\ \text{undefined} & x \in \mathbb{Z} \\ \end{cases}$$ If it's not a step function, then it would seem more sensible to write it as $$y(x) = \int_1^x \frac{1}{\arctan{\left(1+r+r^2\right)}}dr$$ In which case, by the second fundamental theorem of calculus, $$\frac{dy}{dx} = \frac{1}{\arctan{\left(1+x+x^2\right)}}$$
no no tan inverse is not in denominator,
its arctan[1/(1+r+r^2)]

Last edited by Raptor; May 7th, 2014 at 10:12 AM. Reason: missed a point

May 7th, 2014, 10:24 AM   #4
Math Team

Joined: Dec 2013
From: Colombia

Posts: 7,598
Thanks: 2583

Math Focus: Mainly analysis and algebra
Quote:
 Originally Posted by Raptor no no tan inverse is not in denominator, its arctan[1/(1+r+r^2)] also answer is 1/(1+(x+1)^2)
Oops. I don't really see how you get that answer though. The question seems poorly defined.

May 7th, 2014, 10:28 AM   #5
Newbie

Joined: May 2014
From: India

Posts: 12
Thanks: 0

Quote:
 Originally Posted by v8archie Oops. I don't really see how you get that answer though. The question seems poorly defined.
the equation is a telescoping series (hint given by my teacher)
in our country being a good proble-solver is better than being practical
if u ever find out how to solve it pls let me know

Last edited by Raptor; May 7th, 2014 at 10:28 AM. Reason: grammer

May 7th, 2014, 11:34 AM   #6
Newbie

Joined: May 2014
From: India

Posts: 12
Thanks: 0

I have solved this... if u don't know how to solve the equation, check the attached image...
Attached Images
 Capture.JPG (34.0 KB, 4 views)

 May 7th, 2014, 12:33 PM #7 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,598 Thanks: 2583 Math Focus: Mainly analysis and algebra The problem is that, for the sum to make sense, x must be restricted to integer values. But inn that case, the concept of a limit breaks down and there is no derivative. If x is not an restricted to integers, you get either a step function (as I said before) or the telescoping series fails to telescope (over the reals) and the sum is in fact an integral. So you result for the series is right (and very good), but the differential is wrrong and/or meaningless. Subscribing meaning to meaningless things is mysticism, not maths. Thanks from Raptor

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post helloprajna Economics 0 December 8th, 2012 07:55 AM helloprajna Calculus 3 December 8th, 2012 07:30 AM helloprajna Calculus 3 November 15th, 2012 03:25 AM Pumpkin99 Calculus 8 February 24th, 2012 01:54 AM watuwanz Calculus 1 March 22nd, 2009 11:21 AM

 Contact - Home - Forums - Cryptocurrency Forum - Top