No, add 5 not 1 to get 35. That's because the sum of 5 ones is 5.

sum(2i+1) = 2*sum(i) + sum(1) for i = 1 to 5

= 2*5*6/2 + 5

= 35.


You should be able to check for yourself that this is

3 + 5 + 7 + 9 + 11 = 35.


Also, the sum of the first n odd integers is n^2, but we're starting from 3 instead of 1, so this is the sum of the first 6 odd integers minus 1 or 6^2 - 1 = 35.


You can derive the general formula for an arithmetic series:

a + (a + k) + (a + 2k) + (a + 3k) + ... + [a + (n-2)k] + [a + (n-1)k]

where in your case

a = 3, the starting value
k = 2, what you have to add to each value to get the next value
n = 5, the number of items to sum.

Call the sum S, and write it forwards and backwards.

S = a + (a + k) + (a + 2k) + (a + 3k) + ... + [a + (n-2)k] + [a + (n-1)k]

S = [a + (n-1)k] + (a + (n-2)k] + ... + (a + k) + a

adding these term by term gives n terms that are all the same value 2a + (n-1)k so

2S = [2a + (n-1)k] * n

S = a*n + n(n-1)k/2

S = 3*5 + 5*4*2/2

S = 35.

Note that when a = 1 and k = 1, this simplifies to the n(n+1)/2 that you know for 1+2+3+...+n.

It's possible to simplify this by writing it in terms of the first term a and the last term, call it b. Since b = a + (n-1)k, we have

S = a*n + (b - a)n/2

S = n(a+b)/2.

So for any arithmetic series, no matter where it starts or what you have to add to get each term, the sum is the number of terms times the average of the first and last term. In our case

5*(3+11)/2 = 35.

My instinct is that there's a more fundamental misunderstanding going on -- like what the left hand side means. And what the statement Sum_{i=1}^n i = n(n+1)/2 means.

But I'm totally prepared to be wrong.

You're sort of right. I know what the left side means, and know what Sum_{i=1}^n i = n(n+1)/2 means. But because we have only been given Sum_{i=1}^n i = n(n+1)/2 (as well as i^2, i^3) as tools to help us evaluate limits, but haven't spent any time proving them, and because this last few weeks is the first time I've really worked with sums I didn't know that simply subbing the "identity" wouldn't work. Had I not known of Sum_{i=1}^n i = n(n+1)/2 , my instinct would have been to do it in the way that Bruce and you demonstrated (just sub 1-5 for I and sum them).

He said he stuck the identity n(n+1)/2 in for i and replaced n with 5. That takes care of the 2i, but you can't just add the + 1 to that. We have

sum(i) = 1 + 2 + 3 + ... + n = n(n+1)/2 = 5*(5+1)/2

sum(2i) = 2*sum(i) = 2n(n+1)/2 = 2*5*(5+1)/2

But

sum(2i+1) IS NOT sum(2i) + 1 = 2*sum(i) + 1 = 2*5*(5+1)/2 + 1

instead it is

sum(2i+1) = sum(2i) + sum(1) = 2*sum(i) + 1 + 1 + 1 + 1 + 1

= 2*5*(5+1)/2 + 5

= 35.

The key here is that the summation is a linear operation. That means 2 things. It means for any constant c we have

sum(c*i) = c*sum(i)

and for constant d we have

sum(i + d) = sum(i) + sum(d).

So putting them together we have

sum(c*i + d) = c*sum(i) + sum(d)

and sum(d) is just n*d so

sum(c*i + d) = c*sum(i) + n*d.


Linear operations will be fundamental to your future studies of electrical engineering.

Yeah, it makes sense; subbing in for i and calling it good neglects the fact that I needed to add 1 to each 1-5.

Essentially I forgot that the step in purple was necessary.

Right. Put the parentheses around the 2i+1 in the first line. If you leave them off, it's a different problem which has as its answer what you gave before.

Ha, yes. I'm on a roll

If asked to express the area in region c as a positive amount, would the three integers I've written satisfy this?

All are good but they aren't necessarily integers, are they?

I think he mean integrals

Ah, autocorrect prob

Right, integrals. Thanks for the confirmation.

Anyone here familiar with control systems engineering? Having difficulties grasping what exactly I am doing when determining the frequency response of a system. These Bode plots just makes no sense to me.

We're looking at the steady state response of a transfer function when the input is a sinusoid, right? But say the system is something like e^(-x)... Wouldn't oscillations from such a system die down in the steady state, ALWAYS? I don't see how the frequency of the input sin wave would make a difference...

Brainfart! Think I got it now.

No, if the impulse response is e^-t * u(t), then the response to a sine wave will be the convolution of the impulse response with a sine wave which will be a periodic sine wave in the steady state, but it's amplitude and phase will be different from the input sine wave. The Laplace transform of the impulse response is

H(s) = 1/(1 + s) for Re(s) > -1.

The frequency response is the Fourier transform

H(jω) = 1/(1 + jω).

The magnitude of the frequency response is

|H(jω)| = |(1 - jω)/[(1 + jω)(1 - jω)]| = |1 - jω|/(1 + ω^2)

= sqrt(1 + ω^2)/(1 + ω^2)

= (1 + ω^2)^(-1/2).

So

20 log(|H(jω)| = -10*log(1 + ω^2).

This is approximately 0 dB until we get close to the corner frequency of ω = 1. At that frequency the response is attenuated by 3 dB, and it continues to decrease by 10 dB for every decade of frequency, or 6 dB for every doubling of frequency (octave). So this is a first order lowpass filter. The angle of H(jω) or the phase response will be about 0 until we are about 0.1*ω, at which point it starts to roll of, becoming -pi/4 at ω=1, and about -pi/2 for frequencies about a decade above that and higher. For exponential impulse responses that decay more rapidly, the corner frequency will be higher, and for impulses that decay less rapidly, it will be lower.

Find a matrix A s.t. A^2 =/= 0 but A^3 = 0.

What have you tried?

I've mostly tried playing around on my calculator-- fwiw part a of the question was to find A (not zero) s.t. A^2 = 0 and I came up with
[-1 1 0
-1 1 0
0 0 0]

What do you know? Eg. eigenvalues, minimal polynomial, etc.

Here is my thought process:

The matrix can't be 2x2. (Why?)
The eigenvalues are 0. (Why?)
If the matrix is 3x3, it suffices to consider upper triangular ones. (Why?)
The main diagonal must be all zeroes. (Why?)

Then, consider matrices of the form:

You can get rid of the 0's and do 2x2 for that part.

Experimental mathematics solution in R for 2nd part:

A^2 is not 0 for the matrix you gave. Thanks for the thought process though, it was still helpful.

Thanks Bruce!

No, but A^3 is...which is what you were looking for.

Are you smarter than a 9th Grader?
This came from my son's math competition.
I spent an hour working on it and could not solve it!
I thought I was smart. Oh how the mighty have fallen.