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OK I've done this and I get the inequality I need. But can I just check, I don't understand how I have used the equation I need to in the OP? For the second part if I know that -x<= sinx <= x then -1<=nx(sin(1/nx)) <= 1 but then I'm a bit stuck

Homework Statement I am attempting to show that -x \leq sin(x) \leq x for x>0 and thus \int^1_0 nxsin(\frac{1}{nx})dx converges to 1. Homework Equations I know that I need to use the fact that I have shown that the limit as T tends to infinity of \int^T_1 \frac{cos(x)}{\sqrt{x}}dx...

Do I need to use a summation somewhere?

Homework Statement Show from definition that if f is measurable on [a,b], with m<=f(x)<=M for all x then its lebesgue integral, I, satisfies m(b-a)<=I<=M(b-a) Homework Equations The Attempt at a Solution I know that the definition is that f:[a,b]->R is measurable if for each t...

Thank for all your help, it's very much appreciated!

I can't believe it! So can I just check I have understood it. So to find the integral I need to say how the limits change depending on whether x>0 or x<0 and then calculate the integral for each case?

1. so the integral would be between -1 and x+1 of 1dy which would be (x+1)-(-1)=x+2? 2.If x>0 then the integral is between 1 and x-1 of 1dy which would be (x-1)-(1)=x-2

1. This is the bit I'm struggling with i think? 2. I think so, is it because when x < 0 the overlap is given by 2+x and when x<0 the overlap is 2-x? 3. Because then x-y is not in (-1,1) and so the indicator function is zero and so the integral is zero

Erm to be honest I just used the previous question :uhh:, I thought that was all I needed to do? Do you mean that is it because the overlap is like a triangle function?

I'm not sure but is the measure of the overlap I worked out not the answer to the integral? So the integral will be equal to 2-|x| i.e 2+x when x is negative and 2-x when x is positive?

If x were to move to the right would it then be between x-1 and 1 which would then give 2-x?

Yeah it's being very slow with me too. Erm would ?=x+1 so the integral is equal to (x+2)?

I think I understand what the overlap is. Is the overlap not the values the make the integral non-zero. So to do the integral instead of the limits being 1 and -1 they will become -1 and ?

I think I have been a bit stupid here, there is a question before this that I think answers what you are asking. I had to measure the overlap of the intervals (-1,1) and (-1+x,1+x) for various values of x. I found that for values of x not in (-2,2) the overlap is empty and equals zero. For...

Sorry I think I am getting confused. I have done the graph like you said, so y must be between -1 and 1. So if x=-1.5 then I(x-y) = 1 for y between -0.5 and -1?

I understand what you mean, so I have y in (-1,1) and x in (2,-2) but I don't see a formula between the two that will make sure x-y is always in (-1,1)?

will the integral not just be 2 because the indicator function if triggered so it's the integral of 1 which is x between 1 and -1 which is 2?

So does x have to be between (2,-2) is that what you mean?

Homework Statement Calculate f*f where f is the indicator function (-1,1) Homework Equations The convolution f*g of functions f and g is defined by: f*g(x)=\int^{\infty}_{-\infty} f(x-y)g(y)\ dy The Attempt at a Solution I haven't really done convolution before as I am teaching myself, so...

Homework Statement calculate the deviance for the poisson model and show that the i'th component of the poisson deviance is non-negative Homework Equations Dev = 2[L(y)-L(m)] where L is the log-likelihood and m is the mle The Attempt at a Solution I have calculated the log...

Homework Statement Define a sequence (fn) from n=1 to infinity of functions on [0,1] by fn(t)=t^n does the sequence converge in (CL^2[0,1],||.||2) Homework Equations The Attempt at a Solution I am struggling on where to start. I am fairly new to the L2 space and so would just...

Is it that easy?! I was expecting more work than that to be honest ha. thanks for the help guys

sorry I forgot we had centred the ball at a, I was thinking a was any point. So now we can say that dist(a,B)>=r as well

b does not intersect the ball and so the distance between a and b will be >= r-b ?

Such that (x in X : ||x-a|| < r)?

Sorry yes the inf part should be (y in B) not y in a. So if B is closed the complement must be open. So the point a is in an open set?

Homework Statement Prove that a point,a, not belonging to the closed set B has a non-zero distance from B. I.e that dist(a,B)=inf(y in a) ||a-y||>0 Homework Equations I have no idea how to start this. It is only worth a few marks and I have been told it is fairly easy but I have always...

Hey guys, I have a function y=(x+2)/(x+1) and I have performed iterations to show that for any initial value other than -sqrt{2} the sequence converges to sqrt{2}. So I have found that sqrt{2} is a stable fixed point and -sqrt{2} is an unstable fixed point. Now I have to prove my...