Calculate Upper and Lower Sums for Variable Intervals

[Uniquebum]

Let f:[0,1]->ℝ and f(x) = \frac{1}{n} when x=\frac{1}{n²}, n=1,2,... and 0 in other case.

Define such spacing/interval D that S_D-s_D < \frac{1}{100}. Here S_D refers to the upper sum and s_D the the lower sum.Now, I'm not sure how to approach this really since the sum is defined so that the values of x jump a different interval every time. Any help would be appreciated. Hopefully i managed to explain the problem clearly enough.

 

[SammyS]

Let f:[0,1]->ℝ and f(x) = \frac{1}{n} when x=\frac{1}{n²}, n=1,2,... and 0 in other case.

Define such spacing/interval D that S_D-s_D < \frac{1}{100}. Here S_D refers to the upper sum and s_D the the lower sum.

Now, I'm not sure how to approach this really since the sum is defined so that the values of x jump a different interval every time. Any help would be appreciated. Hopefully i managed to explain the problem clearly enough.

That looks like a typo.

Is \displaystyle f(x) = \frac{1}{x}, when \displaystyle x=\frac{1}{n^2}, n=1,2,... and 0 in otherwise.

... or perhaps ... \displaystyle f(x) = x, when \displaystyle x=\frac{1}{n^2}, n=1,2,... and 0 in otherwise.

Are you doing this with Riemann sums ? ... and ... Do the intervals need to be equal in size?

 

[Uniquebum]

It actually is f(x)=\frac{1}{n}. Yea, I'm trying to approach this with Riemann sums and the intervals don't need to be equal in size as far as i understand.

 

[Uniquebum]

Right, I gave it a wild shot and this is what i came up with.

The upper sum is defined as
\sum_{k=1}^{n}(x_{k+1}-x_k)*sup(f(x))

Now let's choose such width for the Riemann sum quadrilateral that
0.5*(\frac{1}{n} - \frac{1}{n+1})= \frac{1}{2n(n+1)} = r_n.

Lets place each quadrilateral so that the supremus (=the point 1/n) is in the middle. This way the infimum will always be 0 and the lower sum s_D=0.

Calculating the upper sum gives
\sum_{k=1}^{n}(x_{k+1}-x_k)*sup(f(x)) = \sum_{k=1}^{n}(2*r_n*\frac{k}{n})
because \frac{k}{n}, k=1,...,n goes through all the supremums.
So
S_D = \sum_{k=1}^{n}(2*r_n*\frac{k}{n}) = 2*r_n\sum_{k=1}^{n}k
which with a little work leads into
S_D = \frac{1}{2n} since \sum_{k=1}^{n}k=\frac{n(n+1)}{2}

Now S_D-s_d=\frac{1}{2n}-0=\frac{1}{2n}<\frac{1}{100} when n\geq 50.

Whether that's right or wrong is beyond me...

 

[Uniquebum]

That looks like a typo.

Is \displaystyle f(x) = \frac{1}{x}, when \displaystyle x=\frac{1}{n^2}, n=1,2,... and 0 in otherwise.

... or perhaps ... \displaystyle f(x) = x, when \displaystyle x=\frac{1}{n^2}, n=1,2,... and 0 in otherwise.

Are you doing this with Riemann sums ? ... and ... Do the intervals need to be equal in size?

I'm actually starting to wonder whether it might be a typo by my professor.

Since it seems the function would only get one value if defined the way i stated in the original post. \frac{1}{n} right?... or?