I'd say it's easier, depending on the contents of the class of course! If you understood probability you shouldn't have any issue with statistics since it's mainly applied probabilities and it's also very formulaic and methodic, so you can actually pass it without really understanding it that much haha.
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The Math Help Thread
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I'd say it's easier, depending on the contents of the class of course! If you understood probability you shouldn't have any issue with statistics since it's mainly applied probabilities and it's also very formulaic and methodic, so you can actually pass it without really understanding it that much haha.
So I'm having a lot of trouble with probability. How do I tell when I'm supposed to use which formula? Like, how do I know when I should use Addition, Multiplication, Complement, and the rest without it getting spelled out? I'm seriously lost on this.
Here's an example.
11. 90% of all people between the ages of 30 and 50 drive a car. If 4 people in that age group are chosen randomly what is the probability that:
a) Exactly 4 drive a car?
b) At least 1 drive a car?
c) At most, 1 drive a car?
I'm not sure which rule I'm supposed to use here.
Here's an example.
11. 90% of all people between the ages of 30 and 50 drive a car. If 4 people in that age group are chosen randomly what is the probability that:
a) Exactly 4 drive a car?
b) At least 1 drive a car?
c) At most, 1 drive a car?
I'm not sure which rule I'm supposed to use here.
I'll let somebody confirm this. I think it is
A- (9/10)^4
B- (9^4)/10
C- (1/10)^3
A- You multiply the odds of driving 4 times because you need 4 drivers. Whenever you are doing probability where it is done multiple times, the odds are multiplied.
B- Since you only need 1 driver, only the numerator is multiplied.
C- Since you need at most 1 driver, you multiply the odds of not being a driver 3 times. You only do it 3 times, because the fourth person doesn't matter if they are or are not a driver.
Stumpokapow
listen to the mad man
This is the probability of the following event:
The first person drives
The second person drives
The third person drives
The fourth person drives
The probability of each "draw" is independent (drawing one driver does not impact the probability of drawing another) and 0.9
0.9*0.9*0.9*0.9 = 0.6561
This is the probability of the following events:
- Four people drive a car = 0.9^4 = 0.6561
- Three people drive a car (either 123, 124, 134, or 234) = 0.9^3 * (1-0.9) * (4 choose 3) = 0.9^3 * 0.1 * 4 = 0.2916
- Two people drive a car (12, 13, 14, 23, 24, 34) = (0.9^2) * (0.1^2) * (4 choose 2) = 0.9^2 * 0.1^2 * 6 = 0.0486
- One person drives a car (1, 2, 3, 4) = (0.9) * (0.1^3) * (4 choose 1) = 0.9 * 0.1^3 * 4 = 0.0036
We add the probability of each event together:
0.6561 + 0.2916 + 0.0486 + 0.0036 = 0.9999
Alternatively, this can be defined with the complement operator:
1 - (Probability that no one drives a car)
1 - (0.1^4) = 1 - (0.0001) = 0.9999
- One person drives a car (1, 2, 3, 4) = 0.0036
- Zero people drive a car = 0.1^4 = 0.0001
0.0036 + 0.0001 = 0.0037
The general form of Binomial RV with n draws, i successes, and p probability of individual draw success:
Pr(x = i) = p^i * (1-p)^(n-i) * (n choose i)
(n choose i) is defined as n!/(k!)(n-k!) -- note that this is symmetric: 5 choose 3 = 5 choose 2. Why is this so? Look at the function, try to figure this out. Look at how it is applied above.
You need to practice trying to think, conceptually, about what's going on in the word problem.
You're looking at 4 people. Each of these people can be a driver (with p=0.9) or a non-driver (with p=0.1). I will write these as D for Driver, ND for Non-Driver.
What are the possible "draws" you can get from the bag? These
DDDD
DDDN
DDND
DNDD
NDDD
DDNN
DNDN
DNND
NDND
NDDN
NNDD
DNNN
NDNN
NNDN
NNND
NNNN
These are the permutations of your draws.
In this question, order doesn't matter. For example, if you get the draw NDDD, that's the same as getting the draw DDDN, right? Okay, so let's rewrite as combinations:
DDDD
DDDN (also DDND, DNDD, NDDD -- a total of four ways to make this combination)
DDNN (also DNDN, DNND, NDDN, NDND, NNDD -- a total of six ways to make this combination)
DNNN (also NDNN, NNDN, NNND -- a total of four ways to make this combination)
NNNN
Notice that the number of permutations of a given combination is the (n choose i) term above. So instead of having to add up all the probabilities for all the different permutations, you can just do all the combinations and multiply by the number of permutations of the given combination.
Okay, let's make this a little more simple for you. Let's say you just draw TWO people, so your possibilities are:
DD
DN
ND
NN
What's the probability for each? One way you can do this is to draw a tree. Start on the left and draw a branch going up for "First draw is a driver", and down for "First draw is not a driver". Now from each of the new nodes, draw a branch up for "Second draw is a driver" and down for "Second draw is not a driver". Look at the outcomes, and calculate their probabilities. What's the probability your first draw is a driver? 0.9, we know this right? What's the probability both draws are D? Here it's 0.9 * 0.9. What about DN? 0.9 * 0.1. What about ND? 0.1*0.9. What about NN? 0.1*0.1. To do the probability of a given outcome on your draws, you multiply. The total probability of all the possible outcomes has to add up 1, because these draws partition the probability space--they're everything that's possible.
When do you do addition? When you're adding different combinations together. Let's say I ask you in our simple example, what's the probability that you get at least one driver? Well, thats (0.9*0.9) + (0.9*0.1) + (0.1*0.9), right? Look at the outcomes above. Which outcomes give you at least one driver? What are their probabilities? Add them together.
When do you use complement? When it's easier to get the exact opposite of the answer you want than the answer you want. Imagine you draw 1000 people. What's the probability you draw fewer than 1000 drivers? Well you can add the probability of 999 drivers + 998 drivers + 997 drivers ... + 3 drivers + 2 drivers + 1 driver + 0 drivers.... or you can just take the total probability (1) and substract the complement of what you're looking for. The complement is the probability that you'll draw exactly 1000 drivers. (0.9^1000)
I mean, I would really recommend reviewing the basic concepts. List out the possible outcomes. Do a tree if it's possible.
Ok, so I figured out my last problem now I'm racking my brain too hard on this one. I'm not sure if I'm over thinking it. I have to find the area bounded by the graphs y=x^2, y=2-x, and y=0. I'm not sure how to proceed though. None of the examples we did had three graphs. 
Any help is appreciated. I'll keep seeing if I can figure it out in the mean time.
Any help is appreciated. I'll keep seeing if I can figure it out in the mean time.Shao Kahn Brewing a Stew
Banned
Ok, so I figured out my last problem now I'm racking my brain too hard on this one. I'm not sure if I'm over thinking it. I have to find the area bounded by the graphs y=x^2, y=2-x, and y=0. I'm not sure how to proceed though. None of the examples we did had three graphs.
Check this out: https://www.khanacademy.org/math/in.../area-between-curves-with-multiple-boundaries
This is your graph: https://www.wolframalpha.com/input/?i=y=0,+y=x^2,+y+=+2-x
This is the area you need to find i.e. highlighted in black:
You need to find out intersecting points, and set limits on it. Remember, y= 0 graph isn't doing shit so you don't have to worry about that.
Check this out: https://www.khanacademy.org/math/in.../area-between-curves-with-multiple-boundaries
This is your graph: https://www.wolframalpha.com/input/?i=y=0,+y=x^2,+y+=+2-x
This is the area you need to find i.e. highlighted in black:
Ok that's what I thought. I think I was letting the y=0 throw me off.
Edit : actually I think I got it Nvm
Shao Kahn Brewing a Stew
Banned
There you go. Good luck!
Edit: Alright, I'll provide steps lol. Give me a few minutes.
y = x^2
y = 2 - x
If Y is equal:
x^2 = 2 - x
x^2 + x = 2
x(x+1) = 2
http://www.wolframalpha.com/input/?i=x^2+++x+=+2
x = -2 and x = 1
This is your limit.
Shao Kahn Brewing a Stew
Banned
Hmm. I'm thinking and my brain is failing me at this hour. You MIGHT be absolutely right. You may have to add that up separately.
Or work it in to the solution. That little area looks like 2 integrals by itself, and you wind up with 3 integrals in total. BUT, you could take the original integral from -2 to 0, and then treat all of the remaining area as a triangle to make things ridiculously simple.
Shao Kahn Brewing a Stew
Banned
Your graph makes it look more like 1/2 x 2 x 2.Essentially this:
Shao Kahn Brewing a Stew
Banned
Calculting a probability of an event is easy if you understand the logical flow, if you will, to reaching that event. Write down how the event can occur as a sentence and see if you encounter the words "and" (multiply), "or" (add), or "not" (complement, i.e. subtract from 1).
For example, the event that exactly 4 out of 4 people drive a car can be written as, "A drives a car and B drives a car and C drives a car and D drives a car." So the probability that this happens is (0.9) * (0.9) * (0.9) * (0.9).
The probability that at least 1 out of 4 people drives a car can be more easily found, as shown by Stumpokapow, by calculating the probability that no one drives a car, since (double negatives!)
at least 1 out of 4 people drives a car = not(not(at least 1 out of 4 ...)) = not(0 out of 4 ...).
These are correct, the combination operator ("choose") takes care for us the fact that ordering does not matter.
Therion is right. You should be calculating this region instead.
MoG EclipsE
Member
Hello, back with a stats question for you guys:
The strength of a certain type of rubber is tested by subjecting pieces of the rubber
to an abrasion test. For the rubber to be acceptable, the mean weight loss µ must
be less than 3.5 mg. Twenty five pieces of rubber that were cured in a certain way
were subject to the abrasion test. From these 25 observations, we computed a mean
3.27 mg and a standard deviation of 0.75 mg. Assume that weight loss is normally
distributed. Do we have significant evidence that this type of rubber is acceptable at
a level of significance of 5%?
(a) Use a 95% upper confidence bound for the mean weight loss to answer the question.
(b) Use a p-value to answer the question.
For (a) I got mew<=3.51675
(b) is where I am having trouble most. I believe I am supposed to compute Zo , but I am missing mew since I have only X bar for the formula of Zo. Once I compute Zo, the p-value is then P(Z<Zo)?? Also, how does the p-value prove that there is significant evidence that the type of rubber is acceptable at a level of significance of 5%?
Thank you in advance for the help!
The strength of a certain type of rubber is tested by subjecting pieces of the rubber
to an abrasion test. For the rubber to be acceptable, the mean weight loss µ must
be less than 3.5 mg. Twenty five pieces of rubber that were cured in a certain way
were subject to the abrasion test. From these 25 observations, we computed a mean
3.27 mg and a standard deviation of 0.75 mg. Assume that weight loss is normally
distributed. Do we have significant evidence that this type of rubber is acceptable at
a level of significance of 5%?
(a) Use a 95% upper confidence bound for the mean weight loss to answer the question.
(b) Use a p-value to answer the question.
For (a) I got mew<=3.51675
(b) is where I am having trouble most. I believe I am supposed to compute Zo , but I am missing mew since I have only X bar for the formula of Zo. Once I compute Zo, the p-value is then P(Z<Zo)?? Also, how does the p-value prove that there is significant evidence that the type of rubber is acceptable at a level of significance of 5%?
Thank you in advance for the help!
Shao Kahn Brewing a Stew
Banned
... I need to go back to high school.
So the limit is from 0 to 1 then, and you remove that from the triangle. Thanks for this!
Stumpokapow
listen to the mad man
First, although it's pronounced "mew", it's spelled "mu" when rendered in the English alphabet. Saving you embarrassment later in life
Before I post my answer for this problem, how did you get mu <=3.51675?
MoG EclipsE
Member
First, although it's pronounced "mew", it's spelled "mu" when rendered in the English alphabet. Saving you embarrassment later in life
Before I post my answer for this problem, how did you get mu <=3.51675?
I used alpha equal to 0.05 (for 95% confidence).
I used what I believe is the upper bound confidence interval formula which is:
mu<=Xbar + (sigma/sqrt
)*Zalphawhere Zalpha I found from the table to be 1.645 for alpha=0.05
so, mu<=3.27 + (0.75/sqrt(25))*1.645 = 3.51675
Thank you!
Would any of you happen to know a decent online tool as an aid to practice calculations for first year chemistry? I'm not looking for a calculator, but an online resource to practice calculations for unit analysis, stoichometry, molar mass, moles, and so on. I also don't mind paying for it.
spelltropy
Member
I got a simple SHM q, its been a while since i've done it. I dont know why, but Im getting half the answers they give at the back. We have a track where someones distance north of the centre is 60cos0.08t.
Time for one lap:
I did 60cos0.08t = 0, got the two values (300pi/16, 100pi/16), and took the smaller one away and got 200pi/16... answer at the back is 400pi/16.
Length of track:
2r=60, 2pir = 188.4..., answer at the back is twice this again, (120pi)
Any ideas what I'm doing wrong? I feel like I've missed a beat here... In my defense, I've not done SHM for 3 years now, I'm just helping someone out and need help myself
Time for one lap:
I did 60cos0.08t = 0, got the two values (300pi/16, 100pi/16), and took the smaller one away and got 200pi/16... answer at the back is 400pi/16.
Length of track:
2r=60, 2pir = 188.4..., answer at the back is twice this again, (120pi)
Any ideas what I'm doing wrong? I feel like I've missed a beat here... In my defense, I've not done SHM for 3 years now, I'm just helping someone out and need help myself
I believe you have to use t-test since you don't know the population standard deviation but do know the sample sd instead.
So for (a), to get a 95%-upper confidence bound for the population mean, you would calculate,
sample mean + t_critical * sample sd / sqrt(sample size)
= 3.27 + 1.711 * 0.75 / sqrt(25)
= 3.527
The critical t value I have was obtained from a table for df = 24 and alpha = 0.05 (one-tailed).
----
For (b), you would set up the null hypothesis that mu < 3.5 and the alternative hypothesis that mu >= 3.5, calculate your t value (call it t*), and find the probability P(t >= t*) that mu >= 3.5. If P(t >= t*) < .05 (the level of significance), then you would reject the null hypothesis; if not, accept.
Now,
t* = (3.27 - 3.5) / (0.75 / sqrt(25)) = -1.533.
My table doesn't consider negative t values, so I instead consider P(t > 1.533) and will subtract that from 1. In other words, due to symmetry of the t distribution curve,
P(t >= -1.533) = 1 - P(t < -1.533) = 1 - P(t > 1.533),
and the table shows that, for df = 24,
P(t > 1.318) = .10
P(t > 1.711) = .05.
Since 1.533 lies in-between, I know that .05 < P(t > 1.533) < .10. Hence,
.9 < P(t >= -1.533) < 0.95.
Since the probability is at least .9 (way higher than .05), we can safely accept the null hypothesis that mu < 3.5.
MoG EclipsE
Member
Thank you very much! For part a though, would we say there is not enough evidence since mu<=3.527 and we need it to be strictly less than 3.5?
boviscopophobic
Member
Imagine a point going one lap around a circular track and look at its distance north of the center. Does it reach 0 more than once?I got a simple SHM q, its been a while since i've done it. I dont know why, but Im getting half the answers they give at the back. We have a track where someones distance north of the centre is 60cos0.08t.
Time for one lap:
I did 60cos0.08t = 0, got the two values (300pi/16, 100pi/16), and took the smaller one away and got 200pi/16... answer at the back is 400pi/16.
Length of track:
2r=60, 2pir = 188.4..., answer at the back is twice this again, (120pi)
Any ideas what I'm doing wrong? I feel like I've missed a beat here... In my defense, I've not done SHM for 3 years now, I'm just helping someone out and need help myself
MoG EclipsE
Member
Trying to understand which way to go about this one, I am having a lot of trouble withthis chapter:
Specifications call for the wall thickness of two-liter polycarbonate bottles to be 4 mils
on average. (A mil is one thousandth of an inch.) A quality control engineer samples
7 two-liter polycarbonate bottles and measures the wall thickness in each bottle. The
results are:
3.990 4.037 4.125 4.101 4.062 3.969 3.955.
Assume that the wall thickness is normally distributed.
a) Is there significant evidence at a level of significance of 5% that the wall thickness
does not meet the specifications?
I know I calculate the sample mean and S.D first to calculate t*...
Now my question is: Would I find the P-value using t* and see if its higher than 0.05 or would I find the t value for 0.025 on the table and compare it to what I calculated for t*?My problem with these questions is how to go about accepting or rejected a hypothesis. I know there is more than 1 way, but my teachers notes are very unclear for this chapter.
Thank you for any clarification!
Specifications call for the wall thickness of two-liter polycarbonate bottles to be 4 mils
on average. (A mil is one thousandth of an inch.) A quality control engineer samples
7 two-liter polycarbonate bottles and measures the wall thickness in each bottle. The
results are:
3.990 4.037 4.125 4.101 4.062 3.969 3.955.
Assume that the wall thickness is normally distributed.
a) Is there significant evidence at a level of significance of 5% that the wall thickness
does not meet the specifications?
I know I calculate the sample mean and S.D first to calculate t*...
Now my question is: Would I find the P-value using t* and see if its higher than 0.05 or would I find the t value for 0.025 on the table and compare it to what I calculated for t*?My problem with these questions is how to go about accepting or rejected a hypothesis. I know there is more than 1 way, but my teachers notes are very unclear for this chapter.
Thank you for any clarification!
dr3upmushroom
Banned
Okay, so just started getting into Trig, and we've learned how to interpret graphs of cos and sin, but this hw problem has me confused.
Okay, so just started getting into Trig, and we've learned how to interpret graphs of cos and sin, but this hw problem has me confused.
You have to match how many As is a cycle, how many Bs the center is above from 0, and how much As the phase shifted. I'll do the first one, for cos:
y=B/2 + (B/2)*cos((3x/4A) - (2A/3))
dr3upmushroom
Banned
Sorry man, I plugged this in and I got he answer wrong, and I couldn't replicate the graph on desmos.
What I can gather is that:
Amp=7/2
Avg=7/2
Phase Shift=-(7pi/3)
Period=45/pi ?
Sorry, I got the phase shift wrong. Phase shift is pi, since that is a 180 degrees shift.
The idea is the same as before. Here, the null hypothesis is that mu (the wall thickness) = 4, and the alternative hypothesis is that mu does not equal to 4.
Since our alternative hypothesis involves an inequality, rather than a one-sided inequality > like in your previous problem, we look at when the area under the curve from t = t_critical to t = infinity is equal to 0.05/2 = 0.025 (assuming your table only considers area on the right end).
Note that, by symmetry, we are at the same time considering the area from t = -infinity to t = -t_critical and requiring this to be 0.025, so that the total area is 0.05.
So compute the t value (call it t). If |t| > t_critical, then we can reject the null hypothesis. Otherwise, we would accept it.
Sorry I don't explain the theory behind and made this sound like a mechanical process; it has been many years since I looked at any of this. I had studied stats using this book in high school, and had thought it was pretty good. You might find it useful as an extra resource.
MoG EclipsE
Member
edit: Thank you! So when do we use p-value as a decision factor rather than this approach?
MoG EclipsE
Member
Sorry for all the questions lately, but I have another problem with understanding the null hypothesis testing.
When formulating a null hypothesis, my understanding is that you cannot have an inequality in the null hypothesis. In other words, the null hypothesis consists always of an equality and the alternate hypothesis is of an inequality. Now, I was taught that the null hypothesis is also always the original claim.
Now, if I have an original claim which consists of an inequality, what do I do?
Example: A particular type of gasoline is supposed to have a mean octane rating greater than 90%. Five measurements (in %) are taken of the octane rating:
90.1 88.8 89.5 91.0 92.1.
We want to verify that the mean octane rating (in %) is greater than 90.
Here I would say that the null hypothesis is the original claim (mu>90) and the alternate is (mu<=90). But this contradicts the rules that a null hypothesis always has an equality. So what would I do in cases such as this?
Thank you!
When formulating a null hypothesis, my understanding is that you cannot have an inequality in the null hypothesis. In other words, the null hypothesis consists always of an equality and the alternate hypothesis is of an inequality. Now, I was taught that the null hypothesis is also always the original claim.
Now, if I have an original claim which consists of an inequality, what do I do?
Example: A particular type of gasoline is supposed to have a mean octane rating greater than 90%. Five measurements (in %) are taken of the octane rating:
90.1 88.8 89.5 91.0 92.1.
We want to verify that the mean octane rating (in %) is greater than 90.
Here I would say that the null hypothesis is the original claim (mu>90) and the alternate is (mu<=90). But this contradicts the rules that a null hypothesis always has an equality. So what would I do in cases such as this?
Thank you!
Wow. I didn't realize that we can't have an equality in the alternative hypothesis until I saw your question. (Please ignore the edits I've made, as they were not correct. I'm starting to know that I will need to review inferential stats again some time.)
Edit: Read this here, starting p.187 (p. 175 on print).
MoG EclipsE
Member
hmm, so wouldn't your edit from earlier where 0.9<p-value<0.95 just be the opposite, 0.5<p-value<0.1 and we accept the null hypothesis (where the null is now what H1 was)?
You're right, we should look at Ho: mu >= 3.5 and Ha: mu < 3.5, and calculate the probability P(t < -1.533) (= P(t >1.533)), and this is greater than 0.05. Sorry about that.
MoG EclipsE
Member
No it's ok, I didn't even know how to go about the problem in the first place, so you still helped me a lot!
Is it normal to not remember a large portion of what you learn (although reviewing or "relearning" being very easy)? I keep imagining math masters in here being able to answer any question despite not having had a math class in many years.
I totally spaced how to do integration by parts when I was in my physics class today and I'm guessing I've done at least one integration by parts problem in the last few months. I probably had to review it the last time I ran into it months ago too.
I totally spaced how to do integration by parts when I was in my physics class today and I'm guessing I've done at least one integration by parts problem in the last few months. I probably had to review it the last time I ran into it months ago too.
Typically you'll only remember things if you review it some time. Each time you review it you can wait a longer time until you review it next time, generally.
Helping people in this thread is one way to review things.
In my Linear Algebra book here's a question I do not understand...
Q: The row reduction algorithm applies only to augmented matrices of a linear system.
A: False. The algorithm applies to any matrix, whether or not the matrix is viewed as an augmented matrix for a linear system.
Q: The row reduction algorithm applies only to augmented matrices of a linear system.
A: False. The algorithm applies to any matrix, whether or not the matrix is viewed as an augmented matrix for a linear system.
The answer is somewhat right, and I don't think you were asked a good question in the first place.
We have to realize that a matrix A is a linear map between two vector spaces, i.e. the matrix represents in numbers a linear relationship between two quantities that we view as vectors. Hence, inherently, there is always an equation Ax = b that the matrix allows (a matrix just doesn't exist by itself).
So the question is right in the sense that row reduction can always be applied to any matrix A since there is an underlying equation for A, but in my view, it's also wrong since that equation can be viewed as an augmented matrix (really, a double array of numbers).
I'm not sure I understand what you're getting at here. If A is a matrix that satisfies Ax=b, then you can look at an augmented matrix [A | b], but A itself is still not an augmented matrix and I don't see how considering [A | b] addresses the question.
The row reduction algorithm can be applied to any matrix because it is just a series of steps performed on the entries. The algorithm itself says nothing about what the matrix represents; that only comes up when interpreting the algorithm's results. If it is being applied to an augmented matrix of a linear system, then you can use the result to find that system's solution. If it is applied to some arbitrary matrix, it could be used to find other properties, like the rank for example. In particular, it can be applied to a column vector, which clearly is not the augmented matrix of any linear system.
Sorry, I guess my explanation wasn't clear, and I am probably overthinking here.
I have a problem with the question and the answer that Timedog posted, because (1) a matrix is always associated with an equation (because it represents a linear relation between two vector spaces), and (2) whenever the row reduction is applied, there is an inherent assumption that it's to reduce a matrix equation. For example, to find a solution to Ax = b given a vector b, or to find the rank of the matrix, which is also finding a solution, to Ax = 0. We wouldn't apply row reduction to a column vector, unless it was really a matrix with one column.
So even though we can apply the reduction to only the matrix A for a homework problem, behind the scenes, we are in actuality doing the reduction for the augmented matrix [A | b] for some b that may or may not have been specified in the problem. In that sense, I think the statement in the question should be a true statement.
Particulam
Neo Member
Hello Gaf!
I'm currently very annoyed as I can't seem to solve a problem which I believe is very simple. I would be forever grateful if you helped me out!
I am applying overlap to a signal (input) and I am trying to calculate the length of the resulting signal (output). I will explain the problem with two examples.
The variables to consider is the input signal length, a package length, as well as the number of elements overlapping.
Example 1:
Length of package = 8
Length of the signal is 19.
We are overlapping with 4 elements.
Signal (input) =
[1 2 3 4 5 6 7 8;
9 10 11 12 13 14 15 16;
17 18 19 0 0 0 0 0]
Signal (output) =
[1 2 3 4 5 6 7 8;
5 6 7 8 9 10 11 12;
9 10 11 12 13 14 15 16;
13 14 15 16 17 18 19 0 ]
Note that each row represents one package (which is fixed to 8 elements).
The padded numbers has now increased the signal length from 19
to 19 + 4*3 = 31
Example 2:
Length of package = 8
Length of the signal is 21.
We are overlapping with 3 elements.
Signal (input) =
[1 2 3 4 5 6 7 8;
9 10 11 12 13 14 15 16;
17 18 19 20 21 0 0 0]
Signal (output) =
[ 1 2 3 4 5 6 7 8;
6 7 8 9 10 11 12 13;
11 12 13 14 15 16 17 18;
16 17 18 19 20 21 0 0]
The padded numbers has now increased the signal length from 21
to 21 + 3*3 = 30
My question:
How can I calculate the signal output length without drawing it. I'm looking for a relation between the variables stated above.
Hope someone can help me! Ask if something is unclear. Thanks!=)
I'm currently very annoyed as I can't seem to solve a problem which I believe is very simple. I would be forever grateful if you helped me out!
I am applying overlap to a signal (input) and I am trying to calculate the length of the resulting signal (output). I will explain the problem with two examples.
The variables to consider is the input signal length, a package length, as well as the number of elements overlapping.
Example 1:
Length of package = 8
Length of the signal is 19.
We are overlapping with 4 elements.
Signal (input) =
[1 2 3 4 5 6 7 8;
9 10 11 12 13 14 15 16;
17 18 19 0 0 0 0 0]
Signal (output) =
[1 2 3 4 5 6 7 8;
5 6 7 8 9 10 11 12;
9 10 11 12 13 14 15 16;
13 14 15 16 17 18 19 0 ]
Note that each row represents one package (which is fixed to 8 elements).
The padded numbers has now increased the signal length from 19
to 19 + 4*3 = 31
Example 2:
Length of package = 8
Length of the signal is 21.
We are overlapping with 3 elements.
Signal (input) =
[1 2 3 4 5 6 7 8;
9 10 11 12 13 14 15 16;
17 18 19 20 21 0 0 0]
Signal (output) =
[ 1 2 3 4 5 6 7 8;
6 7 8 9 10 11 12 13;
11 12 13 14 15 16 17 18;
16 17 18 19 20 21 0 0]
The padded numbers has now increased the signal length from 21
to 21 + 3*3 = 30
My question:
How can I calculate the signal output length without drawing it. I'm looking for a relation between the variables stated above.
Hope someone can help me! Ask if something is unclear. Thanks!=)
For example one, 3*(8-4) + 7 = 19, so 3*8 + 7 = 31.
For example two, 3*(8-3) + 6 = 21, so 3*8 + 6 = 30.
So, I have a basic algorithm. Step one is to subtract overlap element from package length. Step two is to keep adding until the difference between signal length and the addition is less than package length, and then find out what that difference is. Step three is to multiply the number of addition you had to apply by package length and then add in the difference calculated from step two.
Particulam
Neo Member
spelltropy
Member
Got two simple qs here, thinking about method more than anything. Finding the limit of the following two sequences.
For the first, I just divided the top and bottom by the first term of the denominator, and got (1/2n^2)/(1+2^n/8n) + 2. Is it sufficient to then say 1/(2n^2) -> 0 by the product rule, and so the first term here tends to 0(by product rule once again), and so the entire seq tends to 2?
For the second, I let the sequence (an, say) equal e^log(5n^3+6n^4)^(1/n), simplified it down, said lim an = e^lim(log 6n^4+5n^3)/n, then by L'Hopital's, this tends to 0, so an -> e^0 = 1. Is this right/is there a simpler way of looking at it? Have been away from Analysis I for a while, and turns out it makes up most of the first q for the Analysis II paper in the summer, so need to catch up on it.
For the first, I just divided the top and bottom by the first term of the denominator, and got (1/2n^2)/(1+2^n/8n) + 2. Is it sufficient to then say 1/(2n^2) -> 0 by the product rule, and so the first term here tends to 0(by product rule once again), and so the entire seq tends to 2?
For the second, I let the sequence (an, say) equal e^log(5n^3+6n^4)^(1/n), simplified it down, said lim an = e^lim(log 6n^4+5n^3)/n, then by L'Hopital's, this tends to 0, so an -> e^0 = 1. Is this right/is there a simpler way of looking at it? Have been away from Analysis I for a while, and turns out it makes up most of the first q for the Analysis II paper in the summer, so need to catch up on it.
dr3upmushroom
Banned
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