How To Rewrite An Equality With An Equal Sign

Probability (statistics): Why is E [(E [X|Y]) ^2] equal to (E[X]) ^2?

It isn't always. Denote [math]Z[/math] as the random variable [math]E[X|Y][/math].Rewriting, we're asking if [math]E[Z^2] = E[X]^2[/math]. We can further rewrite [math]E[X][/math] as [math]E[E[X|Y]][/math] from the law of iterated expectations / tower property / whatever floats your boat.We get [math]E[X]^2 = E[E[X|Y]]^2 = E[Z]^2. [/math]The equality in question is whether [math]E[Z^2] = E[Z]^2. [/math]Equality only holds above when the Var(Z) is 0, else the equality you propose is not true.

Which number’s square root is equal to two times its cubic root?

To solve this, just set the square of the number equal to twice the cube of the number and solve.√ ̅x = 2(³√ ̅x)Take each side to the third power. We can do this, because each side is equal to each other, so it follows that their cubes are also equal to each other.(√ ̅x)³ = (2 * ³√ ̅x)³To cube the right side, we take the cube of 2 and also the cube of cubic root x and multiply them. For the left side, we can just move the exponent under the square root sign.√ ̅(x³) = 8xWhile the right side cancels out neatly, the left side doesn’t; however, it can still be simplified by taking the square root of x² * x.√ ̅(x²)(x) = 8xThen, we can rewrite the left side as two separate square roots.√ ̅(x²)√ ̅(x) = 8xAfter that, we can simplify √ ̅(x²) to x, neatly reducing the original square root function to x√ ̅x.x √ ̅x = 8xSquare each side to get rid of the square root. We don’t divide the x out just yet, because we want to keep 0 as a solution.x³ = 64x²Now, we can rewrite this equality.x²(x) = x²(64).As you can see, we have 0 as one of the solutions to the equation. Keep track of this as it is important and is one of the answers. However, for now, we can cancel out the x²’s.x = 64There is the other answer. Remember that 64 is just one of the answers, while 0 is another, so a more accurate solution set isx = {0, 64}By mentally checking, we know that the square root of 0 (0) is twice the cube root of 0 (0), as any number multiplied by 0 will remain 0.The square root of 64 is 8, which we know is twice its cubic root 4.I hope this answers your question.

If A + B = C, what does A equal?

The algebra required to solve this question is very simple.Remember that an equation consists of two expressions with an ‘'equal’ sign between them. This sign tells you that the expression on the left side of it has the same value as the one on its right.Also remember that an equation can only be manipulated by applying exactly the same mathematical process to the expressions on either side of the equal sign. If this is not done, you no longer have an equation.NOW TO APPLY THIS SIMPLE PROCESS TO THE GIVEN QUESTION.A + B = C, what does A equal?Subtract B from BOTH SIDES of the equation.The left hand side is A + B, subtracting B leaves the value: AThe right hand side is C, subtracting B results in the value: C - BAs the same operation has been done to both sides of the equation, it is still an equation, i.e., the new left hand side is equal to the new right hand side.So the equation is now: A = C - B , which is the answer to the question.

Why are two vectors, which have equal magnitude and direction but different starting points, equal?

1.28.2017 “Why are two vectors, which have equal magnitude and direction but different starting points, equal?”Magnitude and direction determine a vector. A vector does not have a starting point. In not having a starting point, a vector is like a number — numbers do not have starting points.In depicting a vector on an image, there is of course a point that corresponds to the “tail” and another point that corresponds to the “head”. But these are not part of the definition. However, since we use depictions, it gives us the impression that vectors have starting points.The definition is not arbitrary; the abstract definition in which a start point is not part of the concept of a vector is useful, especially in developing the notion of vector space in mathematics. The vector space is then not dependent on another underlying space and is therefore applicable more generally than if it were made to depend on an underlying space.However, that does not mean that the notion of point of application of a a physical quantity such as force is irrelevant. If there are two forces written as vector quantities F and -F acting at different points on a body, then the net force will be -F + F = 0. However there will be a couple acting on the body which is another vector, and is characterized by a net moment M, given by M = r x F where r is any vector joining the line of action of -F to the line of F. Even though the net force is zero and the center of mass will not accelerate, the moment results in angular acceleration about the center of mass. Thus in physical applications the point of application is important.

How do I find [math]x[/math]  satisfying the following equality:

Thanks  for  the A2A.  Based on the definition domains of [math]\log[/math]  we conclude [math]x \in \mathbb{R}_{\gt 0} \setminus \{ 1\}.[/math]So  [math]\log x^{k} =k \log x,\; k \in \mathbb{N}.[/math]Let   [math]q = \dfrac{1}{\log x}.[/math]   Then  we can rewrite the above equality as[math]2q +3q^2 +4q^3 + \ldots = 8.[/math]Or adding [math] 1[/math] to both sides as  [math]S(q)=1+2q +3q^2 +4q^3 +\ldots =9.[/math]So  you see that  we have a powers series in disguise and the problem reduces to finding [math]q[/math] for which it converges to [math]9.[/math]This  is not difficult in this case because  this powers series can be obtained by deriving the geometric series  [math] 1+q +q^2+ q^3 +\ldots = \dfrac{1}{1-q}.[/math]In particular, they both  have the same convergence radius 1  and [math]S(q)=\left(\dfrac{1}{1-q}\right)'= \dfrac{1}{(1-q)^2}.[/math]  Solving the  equation  [math]\dfrac{1}{(1-q)^2}=9[/math]  gives  [math]q=\dfrac{2}{3}[/math] or [math]q=[/math][math]\dfrac{4}{3}.[/math]  The latter one is not possible since this  [math]q[/math] lies outside the convergence interval of the series.So  [math]\dfrac{1}{q}=\log x= \dfrac{3}{2}.[/math]   The actual value of [math]x[/math] depends on the base of the logarithm which you're using.