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May 7th, 2014, 06:55 AM   #1
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Question Differentiation 2

If y=(image)
then dy/dx=?
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May 7th, 2014, 08:14 AM   #2
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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,
$$\frac{dy}{dx} = \frac{1}{\arctan{\left(1+x+x^2\right)}}$$
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Last edited by v8archie; May 7th, 2014 at 08:22 AM.
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May 7th, 2014, 09:08 AM   #3
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Quote:
Originally Posted by v8archie View Post
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)]
also answer is 1/(1+(x+1)^2)

Last edited by Raptor; May 7th, 2014 at 09:12 AM. Reason: missed a point
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May 7th, 2014, 09:24 AM   #4
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Quote:
Originally Posted by Raptor View Post
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.
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May 7th, 2014, 09:28 AM   #5
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Quote:
Originally Posted by v8archie View Post
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 09:28 AM. Reason: grammer
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May 7th, 2014, 10:34 AM   #6
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I have solved this... if u don't know how to solve the equation, check the attached image...
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May 7th, 2014, 11:33 AM   #7
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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.
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