Yes, you can, but keeping (6n+7) on the denominator (and the minus sign)
(so sum of -n(4n-7)/(6n+7) * x^(n-1) )
Or I don't understand the issue?
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The Math Help Thread
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(so sum of -n(4n-7)/(6n+7) * x^(n-1) )
Or I don't understand the issue?
When you differentiate you get
-n*(4n-7)*x^(n-1)/(6n+7)
Now, a power series has a general term with the form a_n*x^n, not a_n*x^(n-1), so you need to do k = n - 1 and you get:
-(k + 1)*(4k - 3)*x^k/(6k+13)
Link1110
Member
Thanks guys! I get it now!
Just one more week and a final for the sequences and series class, then I graduate from coursera and move onto the Calc 2 and 3 on Udemy! Bring it!
God I feel so dumb right now. I have following problem (awful translation, sorry)
and you can't use completeness theorem (don't think it is even helpful in this situation but whatever). I feel like this should be really obvious but I just can't do it. Tried to get inconsistency from either getting negation of B or just A but I just get to dead ends from those.
Anyone here have a clue?
and you can't use completeness theorem (don't think it is even helpful in this situation but whatever). I feel like this should be really obvious but I just can't do it. Tried to get inconsistency from either getting negation of B or just A but I just get to dead ends from those.
Anyone here have a clue?
Not Math but close enough!
I need help with this, I'm having so much trouble seeing the angels with these type of questions
I'm not sure, is 57 supposed to be between between the red F line and the X axis or the the entire angle from the 37 of the mg to the F red line?
The 57 degrees is the angle (cf. Hot Fuzz) between the vector F and the local x-axis.
On the left picture, we see that F acts at an angle of 20 degrees; that, plus the angle of 37 degrees from inclination, gave us 57 degrees.
Thanks for the quick answer!
If I understand correctly that means that now the 20 degrees are between the vector F and the Y axis in the second picture ( are those vertical and adjacent angle pairs or is that achieved just by virtue of moving vector F?), but wouldn't mean that for there to be 90 degrees in the fourth quadrant there would need to be 70 degrees between the F vector and the X axis and not 57? That what confuses the hell out of me, I'm looking at it incorrectly or at mixing up something
So I'm trying to prove this:
I figure I will have to prove by four cases, and for each prove the left hand side and then the right hand side. But I'm not even sure where to start to prove the left hand side. For case 1 the absolute value of x-y is the same as x-y, and since y is greater than or equal to zero, then absolute value of y is equal to y, but then how do we prove that as the difference between absolute valye of x-y and 3 times the absolute value of y?
I figure I will have to prove by four cases, and for each prove the left hand side and then the right hand side. But I'm not even sure where to start to prove the left hand side. For case 1 the absolute value of x-y is the same as x-y, and since y is greater than or equal to zero, then absolute value of y is equal to y, but then how do we prove that as the difference between absolute valye of x-y and 3 times the absolute value of y?
not psycho
Member
So I'm trying to prove this:
I figure I will have to prove by four cases, and for each prove the left hand side and then the right hand side.
If you have proved the triangle inequality already, it can be used to quickly do your current proof:
http://www.mathwords.com/t/triangle_inequality_with_absolute_value.htm
If you haven't, you may want to just prove that first so you can use it.
Thanks. So I've proven the triangle inequality and that made it easier to solve the problem without using proof by case.
But let's say I was asked to solve it by case on an exam. If I were to solve this problem by cases, could I first add +3|y| to both sides of the equation and still preserve the inequality to prove it or would this be considered a completely different proof?
I haven't finished all of the cases yet, but does this look valid?
There are some steps in your proofs that could use some explanation. In your triangle inequality proof, how do you conclude that |x|-2|y|<=|x+2y|? I think not psycho intended that you write |x-y|=|x+2y-3y| and use the triangle inequality on that.
Similarly, on your proofs by cases, you conclude that |x|+|y|<=|x+2y|+3|y|. Why? I think if you are considering cases, you would benefit from considering _all_ the cases, i.e. consider whether x+2y is positive or negative as well. (Sometimes there will be no choice, as in the first case you worked out.) Then you can make more obvious arguments to demonstrate the inequalities.
I've had two more tests since the last time I posted and have kept up the good grades! I got a 97 on the second one and 93 on the most recent! I thought I was going to bomb the most recent one but to my surprise, I didn't! We're now on Law of Sines and Cosines and started some calculator stuff today and this seems easier than the previous stuff. I should probably start looking back at some stuff to prepare for the cumulative final.
This maybe? Not really sure what difference there is between explaining and teaching.
Also what is the exact part you don't get? Maybe someone here could try to explain it.
Thanks for the link. I just want someone to explain it to me like a child. Sometimes instructors go through teaching the steps and assume you(the viewer)know a thing when you don't.
I dont understand any part of it. I would simply use the substitution or elimination methods everytime if I could.
so to find the number of hands in poker with 2 pairs, the equation is:
nCr(13,2) * nCr(4,2) * nCr(4,2) * nCr(44,1)
In the first term you choose 2 out of 13 different card denominations in the same step. What I don't understand is, why can't I choose one denomination for a pair, then choose another in 2 separate steps using this equation:
nCr(13,1) * nCr(4,2) * nCr(12,1) * nCr(4,2) * nCr(44,1)
The 2nd equation is twice that of the first, but I'm not seeing where the 2x comes from...?
nCr(13,2) * nCr(4,2) * nCr(4,2) * nCr(44,1)
In the first term you choose 2 out of 13 different card denominations in the same step. What I don't understand is, why can't I choose one denomination for a pair, then choose another in 2 separate steps using this equation:
nCr(13,1) * nCr(4,2) * nCr(12,1) * nCr(4,2) * nCr(44,1)
The 2nd equation is twice that of the first, but I'm not seeing where the 2x comes from...?
FortuneFaded
Member
The second equation is counting 2 of hearts, 2 of clubs, 8 of diamonds, 8 of hearts, 3 of spades as different from 8 of diamonds, 8 of hearts, 2 of hearts, 2 of clubs, 3 of spades. Hence the factor of 2.
The nCr(13,2) automatically removes order. However nCr(13,1) * nCr(12,1) = 2 * nCr(13,2) doesn't.
Need to double check on statistics problem. Its very simple but I feel like I missed an obvious thing.
If part 1 has an .85 reliability and part 2 .93, what is the probability the system will fail (both parts need to operate to be successful).
Is it as simple as multiplying them together and taking the inverse? Or is this a nonmutually exclusive event?
If part 1 has an .85 reliability and part 2 .93, what is the probability the system will fail (both parts need to operate to be successful).
Is it as simple as multiplying them together and taking the inverse? Or is this a nonmutually exclusive event?
Killer Queen
Banned
Fairly simple question but I'm having trouble with it as I haven't done math in a while.
((x^(4a))/(y^(2a)))^(3b^2)
((x^(4a))/(y^(2a)))^(3b^2)
Stumpokapow
listen to the mad man
Where's the question?
Stumpokapow
listen to the mad man
Assuming the failure or operation of both parts are independent, then the probability that neither fails is 0.85 * 0.93 = 0.7905, so the probability that the system will fail is 1 - 0.7905 = 0.2095. The events are clearly not mutually exclusive (if two events are mutually exclusive then their probability must sum to be less than or equal to 1 -- Kolmogorov axioms 2 and 3)
Killer Queen
Banned
Oh whoops, I just need to evaluate what I posted.
Edit: I'm having trouble figuring out the order in which I do things in addition to some of the rules.
Killer Queen
Banned
Whoops, it says: Expand and simplify the following powers. Give all of the answers in terms of positive exponents.
Sorry I kinda fucked it up. But thank you guys for the help.
Alright, so the question is about expanding (which is, to get rid of the parentheses). The goal is simply to get used to the rules relating to exponents in algebra.
The first rule you want to use is (a/b)^m = a^m/b^m. Applying the rule, you get this :
—(x^(4a))^(3b^2)
______________
(y^(2a))^(3b^2)
Next rule you'll want to use is (a^m)^n = a^(mn). Basically, multiply the exponents together.
After that, the problem is over because the bases (x and y) are different, so there's no simplication possible.
The first rule you want to use is (a/b)^m = a^m/b^m. Applying the rule, you get this :
—(x^(4a))^(3b^2)
______________
(y^(2a))^(3b^2)
Next rule you'll want to use is (a^m)^n = a^(mn). Basically, multiply the exponents together.
After that, the problem is over because the bases (x and y) are different, so there's no simplication possible.
Killer Queen
Banned
Ah how did I not see this. Thank you!
The One Who Knocks
Member
Bit of a long shot, but has anyone here done any study on Bloch Groups and Dilogarithms? Would anybody have any recommendations for (essential) reading material on them?
Stumpokapow
listen to the mad man
Unless you are given BD = BX (where X is the point on the radius of the circle that intersects line BC), I do not believe you have any mathematical basis to assume that point B is the center of the segment you observe -- but by visual inspection it appears that it is, as you surmised.
That's the only information I was given and was asked to express AD using alpha and a, and find the ratio of AD/DB if alpha = 30
We're not supposed to determine data points such as BD=r by looking at the visuals, so maybe they just forget to add that data point?
Thanks for the answer!
BD=BX doesn't seem like it would imply that B is the center. If you were to form similar triangles by shifting B back and forth along the line bisecting angle B, you would still have BD=BX in any of the resulting diagrams.
I think the intention was to have it appear that the curve meets each line at a 90 degree angle, from which you could infer that B is the center.
Stumpokapow
listen to the mad man
Here is an ellipsoid [I guess more properly an ellipse, since it is in two dimensions] that has the property that the arc meets each line forming a 90 degree angle, but where BD and BX are not equal; by visual inspection, is the arc a quarter-circle, or a stretched ellipsoid?
(But it is equally trivial that my condition was insufficient; observe that the red and blue arcs clearly do not form a quarter-circle but BD = BX
Sure. I was assuming the curve was part of a circle as in the original post, and that the only question was where the center lies. If you assume nothing at all about the curve then the the problem is clearly nonsensical.Here is an ellipsoid [I guess more properly an ellipse, since it is in two dimensions] that has the property that the arc meets each line forming a 90 degree angle, but where BD and BX are not equal; by visual inspection, is the arc a quarter-circle, or a stretched ellipsoid?
(But it is equally trivial that my condition was insufficient; observe that the red and blue arcs clearly do not form a quarter-circle but BD = BX
When doing a partial differentiation with respect to x, just deal with y as if it was a constant.
For example,
d/dx( 3.x^2 + 4.y + x.y^2 + sin(x.y) )
is simply
6.x + 0 + y^2 + y.cos(x.y)
If you know how to solve du/dx + 3u = x.4^2 + 4, it's the same. Just treat y as if it was a constant.
That's the thing though, I don't. It's got du(x,y)/dx AND u(x,y) terms so I cant work out how to separate the variables; or if i want one of them substitutions with Greek letters or what.
so, as usual, you begin without the right part:
du/dx + 3u = 0
means u(x, y) = A
exp(-3x)
A can have a dependancy in y, since y is a constant when differenciating with respect to x (just substitute if you need to check)
Then, you look for a specific solution to du/dx + 3u = xy^2 + y
You look for an affine solution in x, so B x + C
Let's substitute, you get B + 3 B x + 3 C = xy^2 + y
By identification, 3 B = y^2, so B = (1/3).y^2
And B + 3 C = y, so C = (1/3).(y - (1/3).y^2)
This the solutions are f(x, y) = A exp(-3x) + (1/3).x.y^2 + (1/3).(y - (1/3).y^2)
(unless I've made a mistake, it's hard typing this kind of stuff)
A can be ANY function of y: e.g. 2, 14.y, sin or cos.exp(-y)
du/dx + 3u = 0
means u(x, y) = A
exp(-3x)A can have a dependancy in y, since y is a constant when differenciating with respect to x (just substitute if you need to check)
Then, you look for a specific solution to du/dx + 3u = xy^2 + y
You look for an affine solution in x, so B
x + CLet's substitute, you get B
+ 3 B x + 3 C = xy^2 + yBy identification, 3 B
= y^2, so B = (1/3).y^2And B
+ 3 C = y, so C = (1/3).(y - (1/3).y^2)This the solutions are f(x, y) = A
exp(-3x) + (1/3).x.y^2 + (1/3).(y - (1/3).y^2)(unless I've made a mistake, it's hard typing this kind of stuff)
A
can be ANY function of y: e.g. 2, 14.y, sin or cos.exp(-y)All Possible Ways
Member
He means that since your equation has no y derivatives in it, you can treat y as a constant. Just try a polynomial in x for your solution and match terms with equal powers of x on both sides of the equation, it will work out, you will see (you may also add an exponential in x for the complete solution).
EDIT: Ah nevermind, it's all worked out above
Allonym
There should be more tampons in gaming
Can anyone help me solve this problem? I'm doing an algebra udemy course and the instructor doesn't specifically tell you how to solve the subsequent problems and the way I go about doing them, doesn't give me the correct answer. These are links to the three problems
http://imgur.com/iN9fR3v
http://imgur.com/i0dSRiN
http://imgur.com/HSv5Hf4
Any and all help is appreciated.
http://imgur.com/iN9fR3v
http://imgur.com/i0dSRiN
http://imgur.com/HSv5Hf4
Any and all help is appreciated.
Allonym
There should be more tampons in gaming
Hmmm, sorry I can enlarge them. I'll give my solutions in a few minutes.
http://imgur.com/iN9fR3v
http://imgur.com/i0dSRiN
http://imgur.com/HSv5Hf4
The just provided the links to the images, it makes it easier and you can maximize them there
Now those are easier to read. That second one is a pain to count in the head lol. I try to write detailed answers.
Done. Will see if I can get a good image of it. As for advices before giving you straight answers, change the roots into exponents for easier calculation. Just need to remember the rules on how to calculate them. This will work well with the second and third problem. First problem I kinda want to see what you have done since you have probably done it right and the problem is just with presentation.
Done. Will see if I can get a good image of it. As for advices before giving you straight answers, change the roots into exponents for easier calculation. Just need to remember the rules on how to calculate them. This will work well with the second and third problem. First problem I kinda want to see what you have done since you have probably done it right and the problem is just with presentation.
so, as usual, you begin without the right part:
du/dx + 3u = 0
means u(x, y) = A
A can have a dependancy in y, since y is a constant when differenciating with respect to x (just substitute if you need to check)
Then, you look for a specific solution to du/dx + 3u = xy^2 + y
You look for an affine solution in x, so B
Let's substitute, you get B
By identification, 3 B
And B
This the solutions are f(x, y) = A
(unless I've made a mistake, it's hard typing this kind of stuff)
A
Dude, thank you! This totally makes sense now! Solve the homogeneous part first, duh!
Thanks so much!
Allonym
There should be more tampons in gaming
For the first problem I got √(1/3) with both the numerator and denominator being raised to the 3rd power and subsequently getting √1/27. Who knows if that's right though
FortuneFaded
Member
For the first one, it is teaching you to perform the power inside the bracket to both the numerator and denominator. So you get
1^(3/2) / 3^(3/2) = 1 / 3√3 and times both sides by √3 to get √3 / 9
The second teaches you to have the same number being raised so it is easier to convert everything to fractional powers and then add them
So it is converted to 2^(1) x 2^(4/3) x 2^(1/15) = 2^(36/15) = 2^(12/5)
The third is the same as the second just using 6 instead of 2.
Himynameischris
Member
Didn't know we had a math thread! I need serious help with this,
How does e^x=e^-x e^2x? (I know x=x)
The rule they use is highlighted in the picture but it doesn't make sense to apply it here. I guess my question is how did they know to apply that rule here
How does e^x=e^-x e^2x? (I know x=x)
The rule they use is highlighted in the picture but it doesn't make sense to apply it here. I guess my question is how did they know to apply that rule here
Stumpokapow
listen to the mad man
"How does e^x = e^2x e^-x"
a^b a^c = a^(b+c)
2x - x = x
"How did they know to apply it"
Because it allow them to factor out a term that could resolve the denominator of the fraction.
a^b a^c = a^(b+c)
2x - x = x
"How did they know to apply it"
Because it allow them to factor out a term that could resolve the denominator of the fraction.
I have a question pertaining to math courses in general as a computer science major. My school requires a weirdly low amount of math courses to graduate in computer science. All that is required is elementary statistics, calc 1 and discrete mathematics. Would it be beneficial for me to take higher level calc classes, perhaps even proofs or linear algebra?
I spoke to my advisor and he just gave me some wishy washy "well it's not required but if you'd like to that's fine" type answer. I also really enjoy math and have loved my calc class and discrete math class and did really well in them so it's not something I dread doing or anything like that.
Any insight would be greatly appreciated!
I spoke to my advisor and he just gave me some wishy washy "well it's not required but if you'd like to that's fine" type answer. I also really enjoy math and have loved my calc class and discrete math class and did really well in them so it's not something I dread doing or anything like that.
Any insight would be greatly appreciated!
Himynameischris
Member
Hmm ok I think I got it, thanks!
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