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**White_Owl**- Replies: 0

Is there a special name for a formula:

Or backward - differentiate the right side, the result should be equal to the expression under the integral.

Ok, I found another approach.

As we can see, the blue curve (3*cos(theta)) exists in the first and fourth quadrants. But the red curve (1+cos(theta)) exists in all four quadrants. So we can divide the region in question into four segments:

And mirror these two into Q3 and Q4.

If I calculate it like this, then

And the total area:

Well... It certainly looks plausible. But am I correct or not?

This is the image I am working with:

It is easy to see that they have three points of intersections: pi/3, pi, 2pi/3

(Note that the upper limits of integration are different.) Now calculate the half-area outside the first curve and inside the second curve:

Why??? And I do not understand where did pi/2 come from?

The half-area inside the first curve and outside the second is then given by

Sorry, but I do not think so....

But if we take that approach, then we can try:

So which answer is a correct one???

After sleeping on it, I realize that the area is bounded by

on the inside and on the outside.In other words:

while

Therefore:

Ok.... If the result is negative, I took the wrong order on of the limits... Not sure which one.

**White_Owl**- Replies: 6

Find the area of the region inside

and outsideI understand that this should be calculated by a double integral of

and . And I see that the area I am looking for can be calculated by halves (one half above x-axes and one below).So my current solution is:

But I do not understand what are the limits of ?

**White_Owl**- Replies: 1

We have a linear transformation from a matrix space, to a real space defined as a product of all elements in the matrix.

How can we define a **matrix** representing such transformation with respect to the standard basis?

The 'standard basis' here is a set of matrices:

But the result is not a matrix.

So I can write something like that:

But how to define a matrix representing such transformation?

**White_Owl**- Replies: 0

Question:

Here is what I understand:

1)

Since the matrix representing T is in "the most appropriate" basis and it is diagonal, it have to be

Correct?

2)

Correct?

So how do I find b2 an b3?

Vectors which undergoing the transformation.

hmm... Ok. It does look correct.

The next question is what if A is not invertible? Does it affect invertibility of T or not?

Oops. mistake.

So my nice set of equations is not correct

Homomorphism? I guess not. At least what I read in my lecture notes does not give me any clues on how to solve the problem.

Ok. Taking a step back. According to my notes:

- Let T:V->W be a linear transformation.

- If T is invertible, we call it an isomorphism.

So to prove that

is isomorphic, I need to show that it is invertible. Yes?And if T is invertible, that would mean that has a unique solution.

Well, maybe the trick is in the properties of invertible matrices?

It does look kinda nice...

Does that work as a proof, or did I prove something else?

**White_Owl**- Replies: 8

Problem:

Let

How to start the proof?

I am thinking about something like this: Let

, then by definition of matrix invertibility . So .But there is an issue of order, it does matter in the matrix product and I do violate it in this proof. So it must be wrong. But I do not see how to fix it.

you mean to draw a line through centers of the spheres, and look at this line as at a 'new' x-axis? Yes, this will work, thank you.

**White_Owl**- Replies: 3

We have two spheres with centers (x_1, y_1, z_1) and (x_2, y_2, z_2). And radii r_1 and r_2.

What is the volume of solid intersection of these two spheres?

I am not sure how to approach this question.

I know that a point will belong to the solid iff:

But how to find a volume limited by this surface?

It is not a curve, so I cannot use a straight 'solid of revolution' formulas, can I?

It should be something like a sum of all points which satisfy the inequality:

but how?

Is this correct or am I oversimplifying it?

**White_Owl**- Replies: 2

We have two three digit binary numbers. In one number all digits do invert every single step (does not matter what the initial value was, lets say we have a 000-111).

In another number, each digit inverts on random.

Each inversion requires some work to be done. So for the first number, in 10 steps we require 3*10=30 energy.

If none of the digits in the second number was inverted (this can happen), then in the same 10 steps we require 0*10=0 energy. If all of the digits in the second number was inverted (this also can happen), then we will require 3*10=30 energy.

But since the second number is under randomizer we will usually have less energy spent on it in comparison with the first number.

The question is: what is the average reduction of energy which would be required by the second number in comparison with the first one?

My current train of thought is:

There are four possible cases:

0 inverted - 2^3 combinations

1 inverted - (2^3)^3 combinations

2 inverted - (2^3)^3 combinations

3 inverted - 2^3 combinations

At each step one of those four cases can happen.

So the probability to spend get any single from combination A to combination B is

We have probability to spend 0 energy at one step

We have probability to spend 1 energy at one step

same for 2 and 3.

So on average we supposed to get

In other words, the second number will spend 1.5 energy to invert random number of bits and on average we will spend 1.5*10=15 energy on the second number.

Is this correct?

**White_Owl**- Replies: 1

How to find *n* in the equation:

Here cdf_binomial() is a distribution function of binomial random variable:

By trial and error I found that n=164. But is it possible to express *n* for a direct calculation? How did people solve such problems before the age of computers?

Yes, of course. Thank you.

anonimnystefy, thank you. I see the mistake now

So my new answer is:

Therefore, series converges for k>=2

I do not think there are mistakes:

Or are you talking about different equations?

**White_Owl**- Replies: 10

The problem is:

For which positive integers *k* is the following series convergent?

My Answer:

For series to be convergent the next inequality should be true (by the Ratio Test):

Since we know that both k and n are positive we can omit absolute bars.

And now I simplify:

But since *k* is a constant this limit will never be less than 1. Therefore the series divergent for all possible *k*.

Did I make a mistake somewhere? Textbook is looking for a convergent series...

Actually the approach you used is discussed in calculus - the Squeeze Theorem. I just did not think it can be used here as well.

Thank you.

I am reading Stewart's Calculus Early Transcendentals (7th edition) and there is only one method described (ch 11.3)