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#26 Help Me ! » Estimating a sum of a series » 2013-03-17 10:55:35
- White_Owl
- Replies: 5
The problem: Calculate a partial sum for the first ten terms, estimate the error. Round answers to five decimal places.
Well, the partial sum is not a problem - a few minutes with a calculator and the answer is 1.04931. But the error estimation part gives me the trouble.
As far as I understand, I am supposed to solve:
And to solve this I need to find an integral of that scary function. But I have no idea how to approach this integral.
#27 Help Me ! » Summation of harmonic sequence » 2013-03-07 02:33:30
- White_Owl
- Replies: 1
Problem:
Here c is a positive constant, find is this sequence convergent or divergent and if it is convergent, what it converges to?
I see that the series is convegent.
And I also see that it converges to some value which is bigger than 1/c, but less than 2/c.
But what exactly is the value? How to reach it?
#28 Re: Help Me ! » Expected value of a function of a random variable » 2013-03-06 08:58:49
Ok, I tried it both ways and got:
if h(x)=1.5-x, then
if h(x)=x-1.5, then
Maybe I do not understand it at all, but: If pdf is defined for the [1,2] range, does that mean that we expect "the demand" to be at least 1, but no more than 2? but from this point of view the 1.5 at the beginning of the week will satisfy part of the demand and at the end of the week we would have a shortage. So negative amount should be the correct answer?
#29 Help Me ! » Expected value of a function of a random variable » 2013-03-06 08:44:55
- White_Owl
- Replies: 2
The problem states:
The weekly demand for propane gas (in 1000s of gallons) from a particular facility is an rv X with pdf
If 1.5 gallons are in stock at the beginning of the week and no new supply is due in during the week, how much of the 1.5 thousand gallons is expected to be left at the end of the week? [hint: let h(x)=amount left when demand =x.]
As far as I understand, to solve the problem I need to calculate:
And E(h(X)) would be the answer for the problem. But what is the h(x)??? Is it (1.5-x) or is it something else?
#30 Re: Help Me ! » Gabriel's Horn » 2013-03-03 17:11:13
Divergence should be proven or shown...
I think this solution can be used as a prove, but maybe there is an easier way?
And I repeat the question: Why Wikipedia uses incomplete formula?
#31 Re: Help Me ! » Gabriel's Horn » 2013-03-03 12:06:17
Another attempt:
Starting from here:
Let
Then:
We have a formula #24 in the table of integrals in the textbook:
So:
And here we have first limit is infinity divided by infinity, second limit is infinity and a constant.
Therefore we have an infinity in the final answer...
Did I make any mistakes?
#32 Re: Help Me ! » Gabriel's Horn » 2013-03-03 11:36:25
Now I wonder myself where did I got u^6
Thanks.Now I am not sure what to do next?
#33 Help Me ! » Gabriel's Horn » 2013-03-03 07:19:26
- White_Owl
- Replies: 9
According to the textbook, the surface area of the curve y=1/x for x>=1, rotated around x-axis is infinite.
According to my calculations it is finite. I suspect I have a mistake, but I cannot find it. Please help:
Surface area is:
Here we have a=1, b=\infty, and f(x)=1/x
Since one of the bounds is infinity, we have an improper integral and have to do it with a limit:
Looking at the description of Gabriel's Horn in Wikipedia, I see that they used for the surface a function:
Why is that? How did they manage to convert into 1?
#34 Re: Help Me ! » Chebyshev's Inequality » 2013-02-22 17:14:44
Oh! So inequality inside parenthesis in the probability arithmetic means sum of probabilities of all events which satisfy the inequality?
Well, it is kinda strange and unintuitive, but ok. I guess there is some logic in that and I would have to live with it...
But why not just write something like:
I think this would be more clear. Don't you think?
#35 Re: Help Me ! » Chebyshev's Inequality » 2013-02-21 08:09:58
Well? Does anyone have an answer?
#36 Re: Help Me ! » Chebyshev's Inequality » 2013-02-19 12:46:42
I am sorry, I do not understand your answer.
You said a=b=1? Where did 1 come from?
What exactly should I sum and why?
I am now even more confused...
#37 Re: Help Me ! » Chebyshev's Inequality » 2013-02-19 12:02:16
Ok, then lets take the k=2.
In that case we have two X which satisfy the inequality in parenthesis: 7 and 8.
Now what is the values of
Can you tell me the values for a and b?
What is the value of Pr(TRUE)???
#38 Re: Help Me ! » Chebyshev's Inequality » 2013-02-19 11:31:02
Yes, I read the definition, but it is not enough, I need a practical example.
Definition also says that it is true for all k>0.
So, if we have k=10:
Which X should I put here to see the truth of this inequality?
#39 Help Me ! » Chebyshev's Inequality » 2013-02-19 10:43:20
- White_Owl
- Replies: 9
I do not understand it.
Well, I kinda understand its meaning, but I am not sure how to read it or calculate it on a real numbers.
For example, if we have a pmf table:
Now, how to calculate it for lets say k=10? or for k=100?
What goes for X in the inequality and what is the result of Pr() function?
#40 Re: Help Me ! » Probabilities: pmf and cdf » 2013-02-19 02:46:16
oh.... yes.
Then I guess the problem is solved.
Thank you.
#41 Re: Help Me ! » Probabilities: pmf and cdf » 2013-02-18 15:22:00
Ok... Lets do it this way:
P(X)= Combination * (0.4^Total_Groups * 0.6^(5-Total_Groups))
And if I collapse rows with same number of people as a union of the two probabilities I receive:
Still no 1.0 in the total. What is wrong now?
#42 Re: Help Me ! » integrate (1+2e^x-e^(-x)^(-1) » 2013-02-18 11:22:44
mmmm.....
2u^2+u-1 does not have an (a*x+b)^2+c form or I cannot find it.
2u^2+u-1 = (2u-1)(u+1), and this I used in a "partial fractions" approach. But I cannot find any square form for this particular polynomial.
#43 Re: Help Me ! » Probabilities: pmf and cdf » 2013-02-18 11:15:50
I am not sure I understand. If we say that for the "exactly one group is late" we need to calculate: intersection of "one group is late" and "four groups are not late", and any of the five groups can be late. So the formula for the one group become:
But once I recalculate whole table like this - the sum become 1.98.
#44 Re: Help Me ! » integrate (1+2e^x-e^(-x)^(-1) » 2013-02-17 12:38:02
mmmm..... Still do not understand.
How do complete the square?
#45 Help Me ! » Probabilities: pmf and cdf » 2013-02-17 12:34:35
- White_Owl
- Replies: 7
I am reading and rereading the textbook and I still do not understand what exactly is the pmf and cdf?
Here is a problem from the last homework:
Suppose we have three couples and two individuals - five independent groups in total (eight people). Three couples are marked #1, #2, and #3, groups of one person are marked #4 and #5. Each of these five groups can be late for a meeting with a probability 40%. All groups are independent from each other.
Let X be a number of people who arrived late.
Determine pmf and cdf for X.
Well, first I tried to formalize X:
Since there are two possible combinations for 2, 4, and 6 people, we need to take a union of probabilities for these combinations and final P(X) become this:
And from here I need to find pmf
All examples in the textbook are just doing direct substitution p(x)=P(X=x) but if I summarize my P(X) I receive 1.4575616. And sum of p(x) for all x is supposed to be equal 1.0...
So, did I make a mistake somewhere or what am I missing?
#46 Re: Help Me ! » integrate (1+2e^x-e^(-x)^(-1) » 2013-02-17 12:12:16
In the denominator I have 2u^2+u-1. The trigonometric substitution requires to have just two elements in the polynom - a squared variable and a squared constant.
Here I have a third member - u, where do you propose it should go?
#47 Re: Help Me ! » integrate (1+2e^x-e^(-x)^(-1) » 2013-02-17 02:46:37
Your substitution will work just fine if you use the fact that the integral of 1/(a^2 + x^2) = (1/a)arctan(x/a).
Huh? How does that help here?
#48 Re: Help Me ! » integrate (1+2e^x-e^(-x)^(-1) » 2013-02-16 11:18:34
sorry, never mind, I got it.
"Partial fractions" approach is the key.
#49 Re: Help Me ! » integrate (1+2e^x-e^(-x)^(-1) » 2013-02-16 11:13:08
umm... by some mysterious reason, all plus signs disappeared from under the math tag.
#50 Help Me ! » integrate (1+2e^x-e^(-x)^(-1) » 2013-02-16 11:11:45
- White_Owl
- Replies: 11
I have:
using
And now I am stuck.
update: fixed plus signs in the equations