Math Is Fun Forum

You’re on the right track, but sometimes you just need to be a little creative. By the triangle inequality

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Hence

so you can choose

such that

, i.e.

(note that

).

I didn’t realize this was supposed to be a computer question and solved it the hard way.

and

come from

; they can be proved by induction.

Note that there are

integers strictly between

and

, and also

integers strictly between

and

Hence, by strict-increasingness,

for

. Get it? Put another way:

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If there are m integers strictly between a and b and m integers strictly between f((a) and f(b) and f is strictly increasing, then f(a + r) = f(a) + r for 1 ≤ r ≤ m,[/*]
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Hence

.

Similarly you can show that if

then

– but you don’t need this bit to calculate

.

It did, thanks!

Yeah, I was just commenting on the statement – not referring to anyone in particular.

The formal definition of the rational numbers

is the field of fractions of the integral domain

of the integers.

The line

is perpendicular to

and so has gradient

; since it passes through the origin, its equation is

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The point

lies on it. Substitute in these co-ordinates and find the value of

.

Weight of m (taking

) is

.

So the frictional force required to keep m from slipping is

.

This can be supplied by a horizontal force of

.

Now the van is accelerating to the right, meaning m would be accelerating to the left relative to the van with a force of

.

Therefore the horizontal force required to pin m to M is

.

NB: The 150 N horizontal force consists of 25 N required to keep m stationary relative to the van and 125 N needed to generate the frictional force to balance its weight.

If you are given one particular solution to be

presumably the other linearly independent solution is of the form

. So

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Hence

.

No, we don’t. Such a mapping is not a linear transformation. Example: Let A be the matrix with all entries 1; then A+A is the matrix with all entries 2. However

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so the additivity property of a linear transformation is violated. Hence T is not a linear transformation.

From what I understand on Wikipedia, it’s the expectation of an expectation, the second expectation being treated like a random variable. For example, if X and Y are random variables and y is in the sample space of Y, we define the conditional expectation

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(where P(X = x | Y = y) denotes that probability that X = x given that Y = y). This is a function of y and so can be regarded as a random variable: Z = E(X|Y). The expectation of Z, E(Z), is then an iterated expectation:

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So an iterated expectation is just the expectation of an expectation. In fact, the theorem on Wikipedia states that E(Z) = E(X) for this particular iterated expectation.