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The Math Help Thread
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Need help with a simple calc problem.
For part A I would just make a piecewise function with each equation with their boundaries yeah?
I'm a little confused with the second part. I would use the mean value theorem for integrals to find the first average value but what would I do for the other part? Or do I have that backwards?
For part A I would just make a piecewise function with each equation with their boundaries yeah?
I'm a little confused with the second part. I would use the mean value theorem for integrals to find the first average value but what would I do for the other part? Or do I have that backwards?
nincompoop
Banned
Need help with a simple calc problem.
For part A I would just make a piecewise function with each equation with their boundaries yeah?
I'm a little confused with the second part. I would use the mean value theorem for integrals to find the first average value but what would I do for the other part? Or do I have that backwards?
Part a): f(x) = F'(x), so f is a piecewise constant function whose value on each interval is the slope of F on that interval
For part b) and c): Integrate f and F from 0 to 3 and divide by 3 (the length of the interval). That is your average value. You can integrate F by calculating the area underneath it geometrically.
Here's what I'm having trouble with.
I've done part a. Having trouble with part b. One of those stupid situations where I simply cannot think of an N to pick. I've been playing around with the formula from part a and trying to choose something in terms of epsilon to no avail.
I think I'm much closer with c. A couple nights ago I solved it (at least I remember thinking I solved it) but I'm having trouble reading my scratch. I think it's that I can solve the sequence down so that each element is equal to ((M^M)/(M!)) * ((a/M)^k), and since a is strictly smaller than M, a/M is less than one so it fits the geometric series.
I've done part a. Having trouble with part b. One of those stupid situations where I simply cannot think of an N to pick. I've been playing around with the formula from part a and trying to choose something in terms of epsilon to no avail.
I think I'm much closer with c. A couple nights ago I solved it (at least I remember thinking I solved it) but I'm having trouble reading my scratch. I think it's that I can solve the sequence down so that each element is equal to ((M^M)/(M!)) * ((a/M)^k), and since a is strictly smaller than M, a/M is less than one so it fits the geometric series.
nincompoop
Banned
No, you got the same answer because you integrated the same function twice, presumably because you wrote down a piecewise definition of the function F as your answer for part a. Your answer for part a needs to be the derivative of the function F.
CoffeeJanitor
Member
Okay can someone solve this problem for me and show that I'm not crazy? All I need are the eigenvectors for this matrix. I have the eigenvector for lambda = 0, but not for lambda = 3+- 4i (I only need the one for lambda = 3+4i, I think, because the other can be found from that answer).
Here is the matrix. I plugged it it online but my eigenvectors are coming back much, much messier than the ones the calculators/MATLAB are showing.
12, -7, 3
14, -4, 1
2, 3, -2
I seem to recall that complex eigenvectors can vary based on how you solve the problem, so maybe I'm right, I don't know
Here is the matrix. I plugged it it online but my eigenvectors are coming back much, much messier than the ones the calculators/MATLAB are showing.
12, -7, 3
14, -4, 1
2, 3, -2
I seem to recall that complex eigenvectors can vary based on how you solve the problem, so maybe I'm right, I don't know
Here's kind of a wonky question. The square root of variance is the standard deviation. With variance we take the average of the squared differences from the mean. What would it mean if we take the square root of the standard deviation (or raise it to the power of 2). I've looked through 2 statistics textbooks and it's never done or spoken about so I know that it probably does not lead to a number of any importance but I'm just trying to work something out.
close to the edge
Member
Well, stdev(X)^2 = Var(X) again. As for the square root of the standard deviation.. no idea. I guess the values of the standard deviation become more concentrated in one area and smaller, as the square root function tends to do. That's pretty much all I know. I only took one semester of stats, so I'm no expert.
im missing something in fundamental understanding of integrals..
so a problem like 15*((cosx)^1/4)*sinx*dx
i understand how it becomes 15 I (cos^1/4)*(-sinx)dx (I being integral sign)
and i understand how that become -12(cos^5/4)+C, mostly. what i dont get is why you carry over the negative sign from the derivative of cos, -sin.
are there any good explanations of the power rule? my book/teacher arent doing it for me right now
so a problem like 15*((cosx)^1/4)*sinx*dx
i understand how it becomes 15 I (cos^1/4)*(-sinx)dx (I being integral sign)
and i understand how that become -12(cos^5/4)+C, mostly. what i dont get is why you carry over the negative sign from the derivative of cos, -sin.
are there any good explanations of the power rule? my book/teacher arent doing it for me right now
The -1 there is to make up for the fact that you multiplied the original equation by -1 to get the -sinxdx in order to do the substitution for the integral (u = cosx and du = -sinxdx).
Your 2nd equation there should actually be this:
-15 I (cos^1/4)*(-sinx)dx
edit: or like spoo said above, you don't even need to multiply the equation by -1 as you can do:
15*((cosx)^1/4)*sinx*dx (let du = -sinxdx or -du = sinxdx)
so now you have 15*((u)^1/4)*-du or just -15*((u)^1/4)du
Vivalaraza
Member
Hi guys, this seems pretty simple but I can't figure out what I'm doing wrong.
I need to simply...
V(p-m) = m - p + theta (m - m) + delta
to
p = m + (1/1+V)delta
I've eliminated the 'theta(m-m)' bit which obviously is 0. Just can't figure out how I get the 'V(p-m)' bit over to the other side as '(1/1+V)'.
I need to simply...
V(p-m) = m - p + theta (m - m) + delta
to
p = m + (1/1+V)delta
I've eliminated the 'theta(m-m)' bit which obviously is 0. Just can't figure out how I get the 'V(p-m)' bit over to the other side as '(1/1+V)'.
hemtae
Member
Distribute the V so you get:
V*p-V*m = m-p+delta
Add V*m and p to each side then combine like terms so you get:
(V+1)p = (V+1)m+delta
divide each side by (V+1) and you get:
p = m + delta/(V+1)
-COOLIO-
The Everyman
^i have this problem where i know the deltas and k's but i need to find the o's in the right most matrix. im told i can do this by computing the pseudo inverse of a coefficient matrix. google isnt really helping me out here. what would i have to do? this isnt even for a math course, its a computer graphics course.
ThanksVision
Member
Very basic algebra problem here, but I just can't remember how to do it because it has been so long. We're integrating functions to find the area underneath the curve, and one of the functions was:
x = 2y-y^2
How do I solve for y in this? I barely remember how to factor. Isn't it completing the square or something?
x = 2y-y^2
How do I solve for y in this? I barely remember how to factor. Isn't it completing the square or something?
Multiply each side by -1
-x = (y^2) - 2y
add 1 to each side
1-x = (y^2) - 2y + 1
Right side can be factored.
1-x = (y - 1)^2
Square root of each side.
sqrt(1-x) = y - 1
Add one to each side.
1 + sqrt(1-x) = y
cpp_is_king
Member
This requires a little intuition. I would have jused used the quadratic equation
Earthstrike
Member
Hmm This is weird, because x = 2y - y^2 isn't a function. There exist values of x for which there is more than one value of y.
You can simplify the right side of the equation however. Just factor out the y to leave
x = y(2 - y) meaning x is 0 when y is 0 and when y is 2.
You can complete the squares or use the quadratic equation like the above 2 posters said.
If you really just want to integrate the curve, a more common tactic would be to graph the function for your limits of integration. Say you want to integrate from 0 to 1. The area under the curve to the x-axis is the same as the area of the square [0 1] x [0 1] minus the area under the curve to the y-axis.
So you can integrate the function with respect to y, so int(2y-y^2) from 0 to 1. This gives you the area under the curve to the y-axis.Then you take that area and subtract it from the square [0 1] x [0 1]. That would give you the area under the curve to the x-axis.
Does that make sense? It was kind of hard to explain without a picture.
ThanksVision
Member
Thank you!
I just needed it to be solved for Y so I could graph it in my calculator. Honestly, the calculus isn't bad, it's just stupid mistakes in the algebra.
I have a linear regression with 2 independent variables and n observations.
[Edit: Ill just add - Yi = b0 + b1Xi + b2Zi where i=1,2,...,n]
I have been given the OLS b_hat estimates for each of the independent variables plus the intercept in a 3x1 matrix, the variance covariance matrix as well as the standard error of the regression.
I need to work out the value of n, as well as RSquared.
Im guessing that either I need to work out SumSquaredResiduals, from which I am able to derive n, or I need to derive n from which I can work out the SSR.
Anyone able to point me in the right direction?
[Edit: Ill just add - Yi = b0 + b1Xi + b2Zi where i=1,2,...,n]
I have been given the OLS b_hat estimates for each of the independent variables plus the intercept in a 3x1 matrix, the variance covariance matrix as well as the standard error of the regression.
I need to work out the value of n, as well as RSquared.
Im guessing that either I need to work out SumSquaredResiduals, from which I am able to derive n, or I need to derive n from which I can work out the SSR.
Anyone able to point me in the right direction?
Long shot but here I go..
I have to show
δ(a/b-c/d) = bdδ(ad-bc)
where δ is the dirac delta function.
I know
∫f(x)δ(x-a)dx = f(a)
So thus I start with
∫δ(a/b-c/d)f(x)dx
But I'm not sure how to start...my equation doesn't even have an x in it. Can't figure out the correct substitution to start with
I have to show
δ(a/b-c/d) = bdδ(ad-bc)
where δ is the dirac delta function.
I know
∫f(x)δ(x-a)dx = f(a)
So thus I start with
∫δ(a/b-c/d)f(x)dx
But I'm not sure how to start...my equation doesn't even have an x in it. Can't figure out the correct substitution to start with
Been a long time but do you have the equation of the line? Could you find a vector that the dot product of it and your known point equal zero?
http://www.math.washington.edu/~king/coursedir/m445w04/notes/vector/normals-planes.html
http://www.math.washington.edu/~king/coursedir/m445w04/notes/vector/normals-planes.html
-COOLIO-
The Everyman
i just realized that i have to approximate the curve first. this is going to get ugly :s
cpp_is_king
Member
Do you know in advance that it's a polynomial? if so then it becomes quite easy.
That's what I got too, and then I tried to simplify it to (r-z)(r+13)(z-13). The computer gave me (r+z)(13+r-z), and I'm not sure how it got that.
cpp_is_king
Member
a^2-b^2 = (a+b)(a-b)
Always remember, and any time you see something with two different variables each squared in the equation, pull this trick out of your bag and see if it makes things simpler.
r^2 - z^2 + 13r + 13z
= (r+z)(r-z) + 13(r+z)
= (r+z)(r-z+13)
You can tell your original solution is wrong just by inspection, because if you multiply out the first item in each parens, you get 13r^2 which is not in your original equation
Hey I know this falls into physics but I'm not sure of my answers here. It involves simple harmonic motion and collisions.
A 4.00kg block is suspended from a spring with k = 500N/m. A 50.0g bullet
is fired into the block directly from below with a speed of 125m/s and becomes
embedded in the block.
(a) Find the amplitude and period of the resulting simple harmonic motion.
For the period I found omega (angular frequency) by omega = root k/m. m being the sum of the masses of the bullet and the block.
I did not know how to get A the amplitude. I used conservation of momentum to find the velocity of the block after the bullet has been embedded because I thought that might be needed. I think I also need to new equilibrium position so I so I set -kx = -mg. Not sure if that is how you find it though. Then I took my velocity and used it in the formula v = omega root A^2 - x^2. x being the new equlibrium position. I'm not sure if that was the way to go about it. I ended up with A = 0.1599 metres. Does anyone know if I'm on the right track here?
A 4.00kg block is suspended from a spring with k = 500N/m. A 50.0g bullet
is fired into the block directly from below with a speed of 125m/s and becomes
embedded in the block.
(a) Find the amplitude and period of the resulting simple harmonic motion.
For the period I found omega (angular frequency) by omega = root k/m. m being the sum of the masses of the bullet and the block.
I did not know how to get A the amplitude. I used conservation of momentum to find the velocity of the block after the bullet has been embedded because I thought that might be needed. I think I also need to new equilibrium position so I so I set -kx = -mg. Not sure if that is how you find it though. Then I took my velocity and used it in the formula v = omega root A^2 - x^2. x being the new equlibrium position. I'm not sure if that was the way to go about it. I ended up with A = 0.1599 metres. Does anyone know if I'm on the right track here?
Pleaaase help me math geniuses! This is a finite math (business I think) problem...
If $500 is deposited each quarter into an account paying a quaterly interest rate of 3%:
A. How long (in years) will it take to earn $15000?
I figured this one out using the future value formula already, but I may as well show my work.
I put the future value program into my calculator so these are the variables I used.
F = $15000
S = $500
I = .03
N = 21.7/4 which rounds to about 5.5 years (This is the answer of part A)
B. How much interest was earned during the first year? This is the part where I'm really stuck.
Any help would be just awesome, thanks so much!
If $500 is deposited each quarter into an account paying a quaterly interest rate of 3%:
A. How long (in years) will it take to earn $15000?
I figured this one out using the future value formula already, but I may as well show my work.
I put the future value program into my calculator so these are the variables I used.
F = $15000
S = $500
I = .03
N = 21.7/4 which rounds to about 5.5 years (This is the answer of part A)
B. How much interest was earned during the first year? This is the part where I'm really stuck.
Any help would be just awesome, thanks so much!
mysticwhip
Banned
this is prolly really easy but I suck.
O,A,B,C and D are such that B is the midpoint of OD and O is the origin. The vectors a & b are two position vector relative to O such that OA=a,OB=b and OC=3b-a.
1)Express AC and CD in terms of a & b.
AC would be 3b right? and CD would be a?
O,A,B,C and D are such that B is the midpoint of OD and O is the origin. The vectors a & b are two position vector relative to O such that OA=a,OB=b and OC=3b-a.
1)Express AC and CD in terms of a & b.
AC would be 3b right? and CD would be a?
AC = OC − OA = (3b−a) − a = 3b − 2a
CD = OD − OC = 2b − (3b−a) = -b + a
You can create a simple example and check that it's true.
mysticwhip
Banned
Im not sure I quite understand due to the dots not being connected but what program is that?AC = OC − OA = (3b−a) − a = 3b − 2a
CD = OD − OC = 2b − (3b−a) = -b + a
You can create a simple example and check that it's true.
Need advice. I'm pretty solid with all of Calc so far. Only places I really fuck up on is knowing previous match knowledge, especially in geometric properties and other such things.
This is why Optimization is the hardest thing for me in Calc I right now. It's essentially using previous knowledge in order to set up a problem. Any tips on optimization other than practice?
This is why Optimization is the hardest thing for me in Calc I right now. It's essentially using previous knowledge in order to set up a problem. Any tips on optimization other than practice?
Visualize the vectors in your head. AC is a line from point A to point C.
I used Paint.NET to make the image btw.
Alright I'm confused on how to represent a function as its Taylor series.
The problem I'm doing is x^(1/2) with the center at 4. I took the first four derivatives of it and plugged them into the theorem but I'm not seeing any pattern. The first problem I did was pretty obvious but I'm not sure if there's a trick/step I need to go through to make these easier.
The problem I'm doing is x^(1/2) with the center at 4. I took the first four derivatives of it and plugged them into the theorem but I'm not seeing any pattern. The first problem I did was pretty obvious but I'm not sure if there's a trick/step I need to go through to make these easier.
nincompoop
Banned
When you take the nth derivative of f(x) = x^1/2 you get x^((1 - 2n)/2) times some coefficient which is obtained by multiplying by the exponents of f(x), f'(x), f''(x), and so on, up to the exponent of the (n-1)st derivative. So the coefficient is 1/2 * -1/2 * -3/2 * ... * (1 - 2(n-1))/2. So when you multiply all those numbers together, you get ((-1)^n-1) times (1/2^n) times (the product of the first n-1 odd integers). The product of the first n-1 odd integers can be written using this formula (just replace n with n-1). The rest you can figure out.
OK, so it's been a few years since I've done any sort of maths and I'm going back over my old work, brushing up and relearning stuff. I've come to one part which I'm completely failing to understand. Quadratic inequalities.
I get most of it, but I don't know how to determine whether the range of variables is an 'and' or an 'or' (so, whether the answer is '#<x<#' or 'x<# or x>#')
http://www.purplemath.com/modules/ineqquad.htm
"My knowledge of graphing, together with the zeroes I found above, tells me that that I want the intervals on either end, rather than the interval in the middle:"
I have no idea how the author came to this conclusion. I'm sure it's something really simple and I'm just overcomplicating it in my head but I'm stumped.
I get most of it, but I don't know how to determine whether the range of variables is an 'and' or an 'or' (so, whether the answer is '#<x<#' or 'x<# or x>#')
http://www.purplemath.com/modules/ineqquad.htm
"My knowledge of graphing, together with the zeroes I found above, tells me that that I want the intervals on either end, rather than the interval in the middle:"
I have no idea how the author came to this conclusion. I'm sure it's something really simple and I'm just overcomplicating it in my head but I'm stumped.
It seems really obvious thinking about it a bit more. Thanks.
nincompoop
Banned
The first urn has 2 red balls and 3 white balls, if you pick a red ball from the first urn and put it in the second urn then the second urn has 5 red balls and 2 white balls, and if you pick a white ball from the first urn and put it in the second urn then the second urn has 4 red balls and 3 white balls. So before you put anything in the second urn it has 4 red balls and 2 white balls.
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