I understand it in the sense that I know we're supposed to:
1)Prove our hypothesis is true for base case.
2)Assume it's true for some random case (k)
3)Prove it's true for (k+1) using (2)

But I really haven't been able to successfully show 3 that often. I think it's largely due to lack of algebraic intuition and creativity.

Do you understand why if the sum of the first k odd integers is k^2 for some k, then the sum of the first k+1 is k^2 + 2k + 1 = (k+1)^2, so that proves the sum of the first n is n^2 for all n >= k?

Not really.

I understand the bolded. What I don't understand is the unbolded.
It feels as though we are affirming the consequent/converse error.

I see how in other cases if we show its true for the base case, then we prove its true for (k+1), we get sort of a domino effect because if we've demonstrated it for one, and shown it holds for arbitrary (k+1), that means we get it for the 2nd case and on and on. I don't make that connection here though.

I suppose if it's phrased the other way get it.

If the sum of the (k+1st) integer is K+1=k^2+2k+1=(k+1)^2
then k^2 is the sum of the first k integers.

since we showed that if k^2 is sum of first n then K+1=k^2+2k+1=(k+1)^2 I suppose we've somehow managed to show them both.

It just feels really really circular to think in this way.

If the sum of the first k odd is k^2, and the next one is 2k+1, then when we add that one on too obviously we get k^2 + 2k + 1.

Example:

Sum first k = 3 odds: 1 + 3 + 5 = 9 = 3^2

The next odd is 2k+1 = 2*3+1 = 7. So the sum of the first k+1 = 4 odds has to be 3^2 + 7.

1 + 3 + 5 + 7 = 3^2 + 7 = 16 = 4^2 = (k+1)^2.

I follow this from the other night. What I don't understand is how this proves that the first n odd terms is k^2. To demonstrate my confusion:

(1)if it rains then it will be wet
(2)it is wet
therefore it rained.

Is this not what we are doing when we say

(1)if k^2 is the...then so and so is the case
(2)so and so is the case
therefore k^2 is the sum of first k odd integers.

Both of which are logical fallacies. I believe what you are saying is true, I just don't see how we've proved it.

Ya, I misread your question so I deleted

I see you solved it on your own. Another idea is motivated by sin x <= x

No. That would be logically invalid. We are doing this:

(1) 3^2 is the sum of the first 3 odd integers.

(2) If k^2 is the sum of the first k odd integers for some k, then (k+1)^2 is the sum of the first (k+1) odd integers for that k.

Therefore, n^2 is the sum of the first n odd integers for all n >= 3. That's because (1) and (2) implies that it must be true for n=4, and if it must be true for n=4, then it must be true for n=5, and if it must be true for n=5, then it must be true for n=6, and so on ad infinitum. That's a valid proof using the logical principle of induction, and all inductive proofs have this form.

Isn't this true for all n >=1 ? Is there a reason you made this distinction? Or are you just showing that since we used 3 for our base case then it follows for n >=3, but if we use 1 as base case it follows for all n >=1?

Yes, I used k=3 as the base case, so all you can logically conclude from that is that it's true for all n >= 3. I could have used k=2 as the base case and proved it was true for n >= 2. Even n >=1 if you consider 1 to be a sum of 1 number.

Do you see why it's logically valid now? Do you understand that's all we did? You need to understand this proof to be able to do any others. This is the simplest one, and the rest are all essentially the same just with more complicated algebra than noticing k^2 + 2k + 1 = (k+1)^2. In fact even if you didn't already know that, it's clear that is what we need to be true to establish (2), so then you'd set about showing that it is true, say by multiplying out (k+1)*(k+1). So you always know what you need to show to get the induction to work. Do you understand how we used that to establish that (2) is true?

True/false question: If the overall cascade of two LTI systems is causal, then each of the two systems must be individually casual.

It seems like it must be true. Like if you have a causal filter cascaded with an ideal low pass filter the system composed of the two cascaded filters seems like it would be non-causal. Right?

Suppose the non-causal system had an impulse response h1(t) = e^(t+4)*u(t+4). That's non-causal since it is non-zero for t < 0, so if you input an impulse at t=0, you get an exponential out that started 4 seconds before you put it in at t = -4. Clairvoyant. Now cascade that with an ideal delay h2(t) = delta(t-5) so it delays everything by 5 seconds. The overall impulse response will be the convolution of those impulse responses which will delay h1(t) by 5 seconds to produce h3(t) = e^(t-1)*u(t-1) which starts at t=1 and is causal.

Thank you.

Note the reason you can't do that with an ideal low pass filter is that the impulse response is a sinc function that is non-zero out to minus infinity, so you can't delay it enough to make it non-causal. We have to truncate the impulse response at some point so it's finite (hence the term finite impulse response or FIR filter) and then we can produce a causal filter. It won't be ideal, but it will be as close as we want if we make it long enough so we only exclude points where the sinc function is small. This is called windowing, and windows that taper off the impulse response gradually will generally have more desirable frequency responses than ones that simply chop it off abruptly.

In general, if a non-causal system has h1(t) = 0 for t < -T, we can produce a causal system if we cascade it with a causal system that has an impulse response h2(t) which is 0 for t < T. You can draw the picture to see why that is by convolution.

Yeah, when I stopped thinking about a system that had values for all positive time and started thinking about finite non-causal systems it became easy to see why the statement is false.

Yes. I see it.

Here's an example that shouldn't be hard, but that I don't get anywhere with:

given a_1=1 and a_(n+1)=3-(1/a_ n)
I have an intuition that a_n < 3 for all n <=1
We know it's true for n=1
We assume it is true that a_k < 3
and need to show that the following is true
a_(k+1) < 3

My initial strategy is to manipulate the statement from the inductive hypothesis, while maintaining the equality, in order to derive the statement we want to prove is true.

a_k < 3
1/a_k > 1/3
3 - 1/a_k > 3 - 1/3

So at this point I've derived the statement I wanted to, but in doing so I wasn't able to prove what I thought was true.

It's 3 - 1/a_k < 3 - 1/3. The > changes to < when you change 1/a_k to -1/a_k. Or just think about it, if you're subtracting something bigger than 1/3 from 3, then it must become less than 3 - 1/3. You're done. 3 - 1/a_k is a_(k+1), and you showed that is less than 3.

Wow. Well that's exciting .

I notice that when I assume that a_n < 8/3 for all n, I end up showing that, but also showing that a_(n+1) < 21/8. That would lead me to believe that induction isn't very good at doing something like telling me what the sequence converges to. Am I correct there?

Induction is a technique for proving things for all n. It won't help you find the limit, but you know that the limit is [3+sqrt(5)]/2. That's because it's the solution to

x = 3 - 1/x.

Multiply through by x and solve the quadratic for the positive root since a_n is increasing. This works because you can replace the x on the left with the limit of a_(n+1), and the x on the right with the limit of a_n. But these limits are equal, so you can solve for the limit.

I'm having trouble seeing this trig relationship. I'm assuming it is because I really only know right angle trig and, unless I'm not seeing this correctly, they aren't using right angle trig.



The equation implies that tan=|bc|/2d_2 (d_2 is distance from water surface to mirrored bottom) but in terms of soh-cah-toa, I don't see which side they are referring to as the adjacent side that equals 2d_2.

Draw a vertical line through point D to the water surface, and you get two right triangles with legs d_2 and BC/2

See it. Thanks.

For the alternating series test my book tells me the following for the alternating series where the function is b_n

if

1) b_(n+1) <= b_n for all n
2) lim n->infinity of b_n = o

then the series converges. Fine.

Then I'm given the following problem

Σ (-1)^(n+1) * n/(n^2+9) (n=1 to infinity)

so the first thing I do is check condition 1.
Unless I am doing something stupid it is not the case that b_(n+1) <= b_n for all n because when n=1 b_n+1 > b_n. So I move on try other tests and they all are inconclusive.

I go to the solution manual and it says that if f(x)=n/(n^2+9) then f'(x) is decreasing for x greater than 3 which means that it is eventually decreasing and since the limit is 0 as n -> then, by the alternating series test, the series converges.

So am I missing something? I get why the function being eventually decreasing is important, but this isn't what the definition says, is it?

For the convergence of a sequence (or series) it doesn't matter what happens in the beginning; you can change the first elements without changing the convergence. So all convergence criteria should include the words "eventually" or "for n sufficiently large".

You need to stop blindly following tests and think about what is going on. b_1 - b_2 + b_3 is some finite constant C, and b_n for n > 3 converges, so the sum for n > 3 converges to some number S, so the whole series converges to C + S.

I don't think you fully appreciate the inductive proof for the sum of all odds. It should be a major event in your life when you realize you can add the next term and get (k+1)^2. Like getting your first car, or your first kiss. It also involves adding something to a series. Let's organize it this way:

We know the sum of the first k odds for some k is

1 + 3 + 5 + ... + (2k-1) = k^2

For the sum of the first k+1 odds, add 1 more on (not 1 moron):

1 + 3 + 5 + ... + (2k-1) + (2k+1) = k^2 + (2k+1)

But we recognize k^2 + (2k+1) is (k+1)^2, and that completes the proof by induction.

Since you didn't have this thrill yet for that series, try proving by induction that the sum of the first n positive integers is n(n+1)/2.