Sevaji, please help

by Rishi Sun Apr 07, 2013 9:24 am

Can you please help me with the following problem? I am stuck.

Problem: Find the x coordinate of the point on the curve y=cos(x) that is close to (0,0)

My solution:

Minimize (x-0) **2 + (y-0)** 2

x **2 + (cos x)**2

Differentiating the above expression, we get

2x + (2cos(x)) * (-sin(x))

2x - 2sin(x) cos(x)

Find the critical point by setting 2x - 2 (sin(x)cos(x)) = 0

We get 2x = sin(2x)

x = sin(2(x)) /2

Then what?

First mid-term, I got 94. The second mid-term, 76

Not going well. Should work harder..

by MaxEntropy_Man Sun Apr 07, 2013 9:30 am

your solution so far is good. after this point you can use some iteration technique like picard's iteration or plot y=2x and y=sin(2x) and find where they intersect. the intersection point is the solution.

by MaxEntropy_Man Sun Apr 07, 2013 9:33 am

also by inspection the solution to your equation is x=0 and thus the answer (0,1). the picard's iteration technique is necessary for more difficult problems. not necessary on this one.

by Rishi Sun Apr 07, 2013 9:46 am

MaxEntropy_Man wrote:also by inspection the solution to your equation is x=0 and thus the answer (0,1). the picard's iteration technique is necessary for more difficult problems. not necessary on this one.

I think x=0 is correct because cos(0)=1

by Seva Lamberdar Sun Apr 07, 2013 9:51 am

Rishi wrote:Can you please help me with the following problem? I am stuck.

Problem: Find the x coordinate of the point on the curve y=cos(x) that is close to (0,0)

My solution:

Minimize (x-0) **2 + (y-0)** 2

x **2 + (cos x)**2

Differentiating the above expression, we get

2x + (2cos(x)) * (-sin(x))

2x - 2sin(x) cos(x)

Find the critical point by setting 2x - 2 (sin(x)cos(x)) = 0

We get 2x = sin(2x)

x = sin(2(x)) /2

Then what?

First mid-term, I got 94. The second mid-term, 76

Not going well. Should work harder..

Rishi, here is the method.

let r be the distance on any point on the curve (y = cos x) from origin (0,0).

Then, r = sqrt (x^2 + y^2),

or, r = sqrt {x^2 + (cos x)^2}

Then you can differentiate r with respect to x (to find x) such that r has the lowest or min value,

i.e. dr/dx = 0 and d/dx(dr/dx), or the second derivative, is positive.

by Rishi Sun Apr 07, 2013 9:51 am

Yes x=0 is correct.

by MaxEntropy_Man Sun Apr 07, 2013 9:51 am

i meant newton's method, not picard's iteration. i can never keep the names of these techniques straight. i know how they work, but never remember the names.

by MaxEntropy_Man Sun Apr 07, 2013 9:53 am

Seva Lamberdar wrote:
Rishi wrote:Can you please help me with the following problem? I am stuck.

Problem: Find the x coordinate of the point on the curve y=cos(x) that is close to (0,0)

My solution:

Minimize (x-0) **2 + (y-0)** 2

x **2 + (cos x)**2

Differentiating the above expression, we get

2x + (2cos(x)) * (-sin(x))

2x - 2sin(x) cos(x)

Find the critical point by setting 2x - 2 (sin(x)cos(x)) = 0

We get 2x = sin(2x)

x = sin(2(x)) /2

Then what?

First mid-term, I got 94. The second mid-term, 76

Not going well. Should work harder..

Rishi, here is the method.

let r be the distance on any point on the curve (y = cos x) from origin (0,0).

Then, r = sqrt (x^2 + y^2),

or, r = sqrt {x^2 + (cos x)^2}

Then you can differentiate r with respect to x (to find x) such that r has the lowest or min value,

i.e. dr/dx = 0 and d/dx(dr/dx), or the second derivative, is positive.

that is what rishi was also doing except he did not have the sqrt. it is unnecessary and adds needless complication to deal with the sqrt in this case, since minimizing the square of the distance is the same as minimizing the distance as rishi had correctly concluded.

by Seva Lamberdar Sun Apr 07, 2013 10:03 am

"that is what rishi was also doing except he did not have the sqrt. it is unnecessary and adds needless complication to deal with the sqrt in this case, since minimizing the square of the distance is the same as minimizing the distance as rishi had correctly concluded." Max

>>> I am also helping my wife doing her taxes, so handn't read Rishi's solution.

Anyway, he needs to calculate the second derivative of r (not r^2) to get the min. (corresponding to the second derivative of r being positive for particular x, at dr/dx =0).

by MaxEntropy_Man Sun Apr 07, 2013 10:39 am

i agree with you on the second derivative test. because the distance is not a monotonically varying function, one needs to be careful about correctly applying the second derivative test.

by Marathadi-Saamiyaar Sun Apr 07, 2013 2:05 pm

MaxEntropy_Man wrote:i meant newton's method, not picard's iteration. i can never keep the names of these techniques straight. i know how they work, but never remember the names.

That is what I thought but then may be I did not know any better.

by Marathadi-Saamiyaar Sun Apr 07, 2013 2:06 pm

Rishi wrote:Can you please help me with the following problem? I am stuck.

Problem: Find the x coordinate of the point on the curve y=cos(x) that is close to (0,0)

My solution:

Minimize (x-0) **2 + (y-0)** 2

x **2 + (cos x)**2

Differentiating the above expression, we get

2x + (2cos(x)) * (-sin(x))

2x - 2sin(x) cos(x)

Find the critical point by setting 2x - 2 (sin(x)cos(x)) = 0

We get 2x = sin(2x)

x = sin(2(x)) /2

Then what?

First mid-term, I got 94. The second mid-term, 76

Not going well. Should work harder..

Una Pergunta:

Are you also studying for your G E D ?

by Seva Lamberdar Sun Apr 07, 2013 5:28 pm

Seva Lamberdar wrote:

Rishi, here is the method.

let r be the distance on any point on the curve (y = cos x) from origin (0,0).

Then, r = sqrt (x^2 + y^2),

or, r = sqrt {x^2 + (cos x)^2}

Then you can differentiate r with respect to x (to find x) such that r has the lowest or min value,

i.e. dr/dx = 0 and d/dx(dr/dx), or the second derivative, is positive.

Anyway, I scribbled on a piece of paper and found the following,

the first derivative of r w.r.t x, dr/dx = (1/2)*(2x - sin 2x) / D^(1/2), ......... (1)

and second derivative, d/dx (dr/dx) = [(1/2)*{(cos x)^2 - x*x*cos 2x} + x*sin 2x] / D^(3/2) , ....... (2)

the denominator term D in Eqs. (1) and (2), D = {x*x + (cos x)^2},................. (3)

the optimum (max / min) condition in Eq (1) leads to, from numerator, 2x - sin 2x =0, which leads to the condition for optimum at lim x --> 0, such that sin 2x --> 2x, leading to 2x - sin 2x --> 0. In other words the optimum (max or min) is at / near x = 0.

If we substitute x = 0 in Eq. (2), we get the second derivative of r at x = 0 as positive (or greater than 0), indicating that optimum value at x = 0 corresponds to the minimum. Btw D is always positive from Eqn (3). Thus x = 0 is the required answer.

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