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[Rich Uncle Skeleton]

Can anybody help me with this problem?

Basically, I need to show that there exist two real-valued functions f1 and f2 with continuous partial derivatives in a neighborhood of (2,1,-1,-2) such that the above two equations are valid in a neighborhood of (2,1,-1,-2). Ideally I would come up with some concrete counterexample to refute the claim that two such functions don't exist, but I can't really think of a way to do that. =(

I think you can just plug the first of those two equations into the second and solve for f_1, check that this is C^1 on some neighborhood of (2,1,-1,-2), then go back to the first equation and define f_2 = root{29-f_1^2-x_4^2} and check that this too is C^1 near (2,1,-1,-2). (Of course, you shouldn't actually be taking all these partials, just argue that "This is a sum of powers of x_i so as long as the x_i are bounded away from zero...", "29-f_1^2-x_4^2 is positive and C^1 on some neighborhood, so f_2 is too", etc.)

 

[dinazimmerman]

I think you can just plug the first of those two equations into the second and solve for f_1, check that this is C^1 on some neighborhood of (2,1,-1,-2), then go back to the first equation and define f_2 = root{29-f_1^2-x_4^2} and check that this too is C^1 near (2,1,-1,-2). (Of course, you shouldn't actually be taking all these partials, just argue that "This is a sum of powers of x_i so as long as the x_i are bounded away from zero...", "29-f_1^2-x_4^2 is positive and C^1 on some neighborhood, so f_2 is too", etc.)

Great, this definitely seems like the right strategy! It seems so obvious now, and I'm an idiot for not just trying to solve for f1 from the start. I thought it was more complicated than that (I thought I had to do implicit differentiation), and wasted a lot of time looking for a complicated solution. T.T

 

[TimesEunuch]

Great, this definitely seems like the right strategy! It seems so obvious now, and I'm an idiot for not just trying to solve for f1 from the start. I thought it was more complicated than that (I thought I had to do implicit differentiation), and wasted a lot of time looking for a complicated solution. T.T

Have you covered the Implicit Function Theorem? If so, you can answer this question by just calculating the partial derivatives of the functions on the LHS wrt f1, f2, putting them in a matrix and showing the determinant isn't zero around the point (x1, x2, x3, x4, f1, f2) = (2, 1, -1, -2, 4, 3). That's usually how you approach such problems. This one is strange in that you can actually explicitly solve for f1, f2 in terms of (x1, x2, x3, x4), using simple algebra (and the choice of the correct square root). But I can't help thinking your lecturer wanted you to use the IFT here. Just a thought

 

[dinazimmerman]

Have you covered the Implicit Function Theorem? If so, you can answer this question by just calculating the partial derivatives of the functions on the LHS wrt f1, f2, putting them in a matrix and showing the determinant isn't zero around the point (x1, x2, x3, x4, f1, f2) = (2, 1, -1, -2, 4, 3). That's usually how you approach such problems. This one is strange in that you can actually explicitly solve for f1, f2 in terms of (x1, x2, x3, x4), using simple algebra (and the choice of the correct square root). But I can't help thinking your lecturer wanted you to use the IFT here. Just a thought

Yes, we've covered the Implict Function Theorem, and I thought about using it, but I didn't realize I could treat f1 and f2 as variables -- that's clever. So does this seem like the right approach:

Define a function g: R^6 -> R^2 with the coordinate functions:

• g_1(f1, f2, x1, x2, x3, x4) = f1^2 + f2^2 + x4^2 - 29
• g_2(f1, f2, x1, x2, x3, x4) = f1^2 / x1^2 + f1^2 / x2^2 + x4^2 / x3^2 - 17

g(4, 3, 2, 1, -1, -2)=0, and since the coordinate functions of g are polynomials, g is C^1. Then, as long as I can show that the derivative of g (a 2x6 matrix) has rank 2 locally, I can apply the Implicit Function Theorem and say that locally g=0 iff there exist two C^1 functions h_1 and h_2 from R^4 to R s.t. f1=h_1(x1, x2, x3, x4) and f2=h_2(x1, x2, x3, x4).

Hmmm, yep this definitely seems right to me. Thanks a lot! By the way, you're right; I think the professor did want us using the Implicit Function Thorem, but he would've given me full points anyway if I presented a concrete example instead. =P

 

[Ave22]

Hint for the bolded: You need to make ample use of |a| - |b| ≤ |a+b| ≤ |a|+|b|. The first part of the inequality allows you to bound |x| using your bound on |x-c|, and the second part will let you bound |x^(n-1) + cx^(n-2) +...| by breaking it apart.

I forgot to say thanks for this.

Now on to my issue this week: Determine all values of x for which f(x) = floor(x) + floor(-x) is continuous. Prove your answer.

I can see that it is continuous when x is a real number that is not an integer, but I'm not sure how to go about proving this. I know I can show a function is continuous by showing that the limit of the function exists. So, I'm trying to show |floor(x) + floor(-x) - (-1)| < epsilon for all x satisfying 0 < |x+1| < delta. Let delta < .5. Then -1.5 < x < -.5. Then either -1.5 < x < -1, x = -1, or -1 < x < -.5. If -1.5 < x < -1, then floor(x) + floor(-x) - (-1) = 0 < epsilon, so it's continuous on this interval. If -1 < x < -.5, then floor(x) + floor(-x) - (-1) = 0 < epsilon, so it's continuous on this interval. But if x = -1, then floor(x) + floor(-x) - (-1) = 1, which may be greater than epsilon.

Is this enough to show that the function will be continuous for all real numbers that are not integers, since I could have let delta be anything? This doesn't feel right to me.

Edit: Ah, just realized I messed up the |x - c| < delta part. Let me work on this some more.

Edit 2: Disregard this. I've got it.

 

[movie_club]

.

 

[innervision961]

I'm back! And this time, I'm much needier than before
I have my first Trig test coming up Tuesday and sadly I've had to miss a few days of class
due to problems at home. Can you guys help me get caught up so that I don't bomb this first test? If you can help with one or all of these problems, grateful I will be.

I'm not looking for answers, so much as directions please, thanks guys:

For the test I get to use my TI - 83 plus calculator, as well as 3 note cards, so if you have any calculator shortcuts or good notes, I'll take them!

1.) Convert -8+5i to polar form. & Convert (8,120) to rectangular form

(Yeah I really don't have a clue on this one, I semi understand what a polar coordinate but I must have missed this class.)

2.) Converting from pi to radians and radians to pi etc.

EX: 28.8° to pi form of a radian, or 128.7° to decimal form of a radian, or 8pi/5 to degrees

3.) 2^x+4 = 4^x-1

(I missed this day also, no clue where to begin on this one)

4.) Logarithms:

EX: log3 (x+4) - log3 (x) = 4
log8 of 30
write log2 (a^3/b^2c^4) with two or more logs

4.) e^x+2 = 18, solve for x

5.) pH= 13.8, find H3O, also: H3O+ = 2.74 * 10^-3 find pH


Man, if you guys could steer me in the right direction, or help me with any of this, I'd be grateful. I'm off to khanacademy and google, I'll keep checking back.

 

[close to the edge]

1) Complex numbers are either in Cartesian coordinates (a+b*i where a,b are rationals) or in Polar coordinates. Polar coordinates give you an angle and a length, which you measure from the point of origin to get to the number in the complex plane. To convert from Cartesian to Polar coordinates:
r = sqrt(a^2+b^2) (this is the length)

or

if x < 0. (this is the angle)

2.) You should be able to do this with your calculator by switching between the different modes for displaying angles (RAD, DEG are the relevant ones). I don't know the calculator that you're using but it should be on any scientific calculator.

3.) Are you supposed to solve this? I'm lazy right now, so I'll let Wolfram Alpha help me
http://www.wolframalpha.com/input/?i=solve+2^x%2B4+%3D+4^x-1

4.) Take a look at the logarithmic identities (http://en.wikipedia.org/wiki/Logarithm#Logarithmic_identities), that should help.

5.) No clue about anything physics or chemistry related, get this sorcery away from me.

 

[poweld]

I'm back! And this time, I'm much needier than before
I have my first Trig test coming up Tuesday and sadly I've had to miss a few days of class
due to problems at home. Can you guys help me get caught up so that I don't bomb this first test? If you can help with one or all of these problems, grateful I will be.

I'm not looking for answers, so much as directions please, thanks guys:

For the test I get to use my TI - 83 plus calculator, as well as 3 note cards, so if you have any calculator shortcuts or good notes, I'll take them!

1.) Convert -8+5i to polar form. & Convert (8,120) to rectangular form

(Yeah I really don't have a clue on this one, I semi understand what a polar coordinate but I must have missed this class.)

2.) Converting from pi to radians and radians to pi etc.

EX: 28.8° to pi form of a radian, or 128.7° to decimal form of a radian, or 8pi/5 to degrees

3.) 2^x+4 = 4^x-1

(I missed this day also, no clue where to begin on this one)

4.) Logarithms:

EX: log3 (x+4) - log3 (x) = 4
log8 of 30
write log2 (a^3/b^2c^4) with two or more logs

4.) e^x+2 = 18, solve for x

5.) pH= 13.8, find H3O, also: H3O+ = 2.74 * 10^-3 find pH


Man, if you guys could steer me in the right direction, or help me with any of this, I'd be grateful. I'm off to khanacademy and google, I'll keep checking back.

For 2, in case you can't figure out how to make your calculator do it, or it's unable to, it's really very simple to do with basic math. Radians are measured in multiples (or fractions of) pi. Pi = 180 degrees. Therefore, 28.8 degrees in rads is (28.8/180)*pi. I'm guessing by "decimal form of radians" they just mean multiply out pi, so (128.7/180)*3.14159..., and (8pi/5) in degrees is [(8*180)/5]

Too lazy to get the calculator out, but you get the idea.

 

[innervision961]

Thanks guys! This is going into the note

 

[TimesEunuch]

Define a function g: R^6 -> R^2 with the coordinate functions:

• g_1(f1, f2, x1, x2, x3, x4) = f1^2 + f2^2 + x4^2 - 29
• g_2(f1, f2, x1, x2, x3, x4) = f1^2 / x1^2 + f1^2 / x2^2 + x4^2 / x3^2 - 17

g(4, 3, 2, 1, -1, -2)=0, and since the coordinate functions of g are polynomials, g is C^1. Then, as long as I can show that the derivative of g (a 2x6 matrix) has rank 2 locally, I can apply the Implicit Function Theorem and say that locally g=0 iff there exist two C^1 functions h_1 and h_2 from R^4 to R s.t. f1=h_1(x1, x2, x3, x4) and f2=h_2(x1, x2, x3, x4).

Know I'm pretty late with my response, but don't post on GAF much. Yes, that's the correct approach. Really what you're saying is that there exist functions h_1, h_2:R^4->R such that

g_1(h_1(x1 ,x2 ,x4 ,x4), h_2(x1, x2, x3, x4), x1, x2, x3, x4)=0
g_2(h_1(x1 ,x2 ,x4 ,x4), h_2(x1, x2, x3, x4), x1, x2, x3, x4)=0

for all (x1, x2, x3, x4) in the neighborhood of (2, 1, -1, -2). It's a little confusing in the question since f1, f2 serve the dual role there as variables and as functions, so you have the right idea in calling the variables f1, f2 and the functions h_1, h_2 to avoid ambiguity.

 

[Poyunch]

Okay I have this problem with quizzes and tests so I blanked out on how to do this problem even though I did a similar problem before.

How does ones solve 5^x = 4^(x-1)

I got to here xlog5 = (x-1)log4

but I don't know how to isolate x and I know it's extremely simple but I'm seriously blanking and I'm pissed off because I needed an A on this quiz but I probably got a C because of this.

 

[jepense]

Okay I have this problem with quizzes and tests so I blanked out on how to do this problem even though I did a similar problem before.

How does ones solve 5^x = 4^(x-1)

I got to here xlog5 = (x-1)log4

but I don't know how to isolate x and I know it's extremely simple but I'm seriously blanking and I'm pissed off because I needed an A on this quiz but I probably got a C because of this.

How do you solve
a x = b x + c ?

(Answer: x = c / (a-b))

 

[Poyunch]

Okay so ax = (log5)x but what is bx + c ?

 

[Fjolle]

Okay I have this problem with quizzes and tests so I blanked out on how to do this problem even though I did a similar problem before.

How does ones solve 5^x = 4^(x-1)

I got to here xlog5 = (x-1)log4

but I don't know how to isolate x and I know it's extremely simple but I'm seriously blanking and I'm pissed off because I needed an A on this quiz but I probably got a C because of this.

xlog5 = xlog4 - log4

log4 = xlog4 - xlog5

log4 = x(log4-log5)

x = log4/(log4-log5)

 

[Kenka]

I've just seen something while browsing the web :

When you get a second degree equation like x^2+6x+25 = 0, it's often awkward to find the roots without a calculator, especially if they are complex.

I've been given the trick to cut back the constant to a certain value that can allow me to write the equation in a simple form. Like in this case : (x+3)^2 + 16 = 0

And all of a sudden, everything you have to do is to to take the root of the stand-alone constant and make it the imaginary part of the solution :

x = -3-4*i
x = -3+4*i

Now the form written earlier is often used to describ the circle in planes and it looks similar, barring the signs and a second square component :

(x &#8722; a)^2 + (y &#8722; b)^2 = r2 or (x &#8722; a)^2 + (y &#8722; b)^2 - r^2 = 0

I find it funny, is there any relation linking the two fields ?

 

[jepense]

I find it funny, is there any relation linking the two fields ?

Links between complex numbers and analytic geometry? Lots. Complex numbers are very often represented as a 2D plane where the axes are the real and imaginary parts. It's often also convenient to give complex numbers in polar representation (x+iy -> r=sqrt(x^2+y^2), tan theta = y/x). In the complex plane, the exponential function with an imaginary argument draws a circle (x+iy = r*e^(i theta)), etc. See wikipedia.

 

[Ydahs]

I think that part should be dy/dx = (y^2-9)/3y unless you wrote the problem wrong

as for the last part when you have log|y| = x + c,
you have to raise e to the (x+c) not just x using the laws of exponents you have (e^x)*(e^c). e^c is a constant that can be replaced with A

A few weeks late, but thanks for this explanation. Was doing revision for my exam in two days (PANIC) and I remembered I posted this here. Now it all makes sense.

 

[Domino Theory]

Could anyone help me out with this bonus problem?

My professor gave our group a proposition: if our group can demonstrate to the class how to do a typical linear programming problem WITHOUT going through the process of finding the 5 corners on a graph and plotting the points into the problem then we get 20 points extra credit.

 

[dr3upmushroom]

I'll be eternally grateful to anyone who can clear this up for me: I'm taking an economics class and the teacher expects that we know how to do calculus. I don't. We don't need to do anything too crazy though, mostly just derive equations.

So, I'm pretty sure I'm figured this out, but I don't think I'll be able to see a tutor to confirm before our next test. If you have an equation like 50 - p squared (don't know how to actually do the squared sign), if you want to use calculus to derive a linear equation from that, you just make the exponent of p a coefficient instead, so it become 50 - 2p?

And then if p already had a coefficient, like 90 - 1.5p squared, you first decided the equation by the coefficient of p squared, so 60 - p squared/ 1.5, then make the coefficient of p its exponent and divide by 1.5?

I'm pretty sure this is how to do this, but I don't understand the math behind it, so I'm not feeling great about it. Note that I don't need to know the concept behind why I'm doing this, I just need to know if I'm doing it right. Huge thanks to anyone who replies.

 

[dinazimmerman]

The derivative of 50-P^2 is -2P. Why?

The derivative of bP^n (where b & n are any real numbers) is bnP^(n-1).

The derivative of a standalone real number - in other words, a constant - is just 0.

The derivative of a finite sum of (differentiable) functions is equal to the finite sum of the derivatives of each function.

So the derivative of a + bP^n (where a is any real number) is just bnP^(n-1).

EDIT: For better understanding, check out these notes on derivatives: http://tutorial.math.lamar.edu/Classes/CalcI/DerivativeIntro.aspx

 

[dr3upmushroom]

The derivative of 50-P^2 is -2P. Why?

The derivative of bP^n (where b & n are any real numbers) is bnP^(n-1).

The derivative of a standalone real number - in other words, a constant - is just 0.

The derivative of a finite sum of (differentiable) functions is equal to the finite sum of the derivatives of each function.

So the derivative of a + bP^n (where a is any real number) is just bnP^(n-1).

EDIT: For better understanding, check out these notes on derivatives: http://tutorial.math.lamar.edu/Classes/CalcI/DerivativeIntro.aspx

:lol You went a little more in-depth than I need, but thanks.

So I was right, right? We're never going to deal with any exponent larger than squared, so I'll always just multiply the exponent and coefficient of a variable to derive its new coefficient?

 

[poweld]

:lol You went a little more in-depth than I need, but thanks.

So I was right, right? We're never going to deal with any exponent larger than squared, so I'll always just multiply the exponent and coefficient of a variable to derive its new coefficient?

Don't forget to reduce the exponent by 1 as well. My calc teacher called it "Hit and Dip", the "Hit" is multiplying the coefficient by the exponent, and the "Dip" is reducing the exponent by 1.

Also, remember that constant values become 0, and a variable with a power of 1 (e.g. x) becomes 1.

Some examples to reinforce:
d/dx 2x = 2
d/dx x^2 = 2x
d/dx (x^2 + 300) = 2x
d/dx 5x^5 = 25x^4
d/dx (x^2 - 5x^5) = (2x - 25x^4)

 

[Door2Dawn]

So I'm having a really hard time trying to combine this into a single logarithm

ln5+3lnx-lny

I keep thinking about how

 

[dinazimmerman]

So I'm having a really hard time trying to combine this into a single logarithm

ln5+3lnx-lny

I keep thinking about how

I'm pretty sure

ln5+ 3lnx - lny = ln ( 5x^3 / y).

That's because:
ln(x) + ln = ln(xy)
ln(x) - ln = ln(x/y)
and aln(x)=ln(x^a)

 

[Sealda]

Hello

I want to do multiple linear regression, but one of the requirements is the residuals to be normally distributed, and I can check that with QQplots but then the QQ plot shows it is about 95% of data fit into the normal line, but 5% is way off!

can I still proceed ?*or do I have to find a way to transform the data ?*

 

[Feep]

I'm normally pretty good at probability, but I'm not sure how to figure this one out.

Let's say you repeatedly press a button, and each press results in either a "pass" or a "fail". You keep hitting the button until you fail.

If the probability of a "pass" is p%, how can one find the average number of button presses before a fail?

 

[Therion]

I'm normally pretty good at probability, but I'm not sure how to figure this one out.

Let's say you repeatedly press a button, and each press results in either a "pass" or a "fail". You keep hitting the button until you fail.

If the probability of a "pass" is p%, how can one find the average number of button presses before a fail?

Sounds like you want the expected value of a geometric distribution.

 

[Mudkips]

I'm normally pretty good at probability, but I'm not sure how to figure this one out.

Let's say you repeatedly press a button, and each press results in either a "pass" or a "fail". You keep hitting the button until you fail.

If the probability of a "pass" is p%, how can one find the average number of button presses before a fail?

1/(1-p) | 0 <= p < 1
you'll never get a fail | p = 1

If the first press is a fail, then it took 1.
Otherwise, it's 1 wasted press, plus the average number of times to get a fail (since all presses are discrete and separate).

Average = odds of fail, + odds of pass * (Average + 1 wasted press)
Average = (1-p) + (p)(1 + Average)
Average = 1 - p + p + pAverage
Average = 1 + pAverage
1 = 1/Average + p
1 - p = 1/Average
1 / (1 - p) = Average

Try it out with p = 0, 1/5, and 1.
0 = always fail = 1/1 = 1 press.
1/2 = coin toss = obviously 2 = 1/(1/2) = 2 presses.
1 = never fail = 1/0 = we better write that condition separately.
.99999999999999 = almost never fail = 1/.0000000000001 = 100000000000000 presses.

 

[Manos: The Hans of Fate]

It's awesome that everyone is helping people out in this thread, but man did scrolling though this page remind me of how math scared the crap out of me in school.

 

[poweld]

I'm normally pretty good at probability, but I'm not sure how to figure this one out.

Let's say you repeatedly press a button, and each press results in either a "pass" or a "fail". You keep hitting the button until you fail.

If the probability of a "pass" is p%, how can one find the average number of button presses before a fail?

My guess would be 1/(1-p) excluding when p = 0 or 1, cause then the odds are 1 and 0, respectively.

edit: beaten badly! had typed out my response a while ago but forgot to press post :lol

 

[Feep]

Thanks, guys.

 

[Feep]

Alright, since you guys helped me out with the expected value of a geometric distribution, I was able to run a fairly complex analysis on the exact values one should use if that want to maximize their utility in my upcoming game. (You pay more gold for an increased chance of success.)

After a lot of work, the optimal value to use for p is, depending on the difficulty of the recipe, d:

d^2 [ pi*p/2*sec^2(pi*p/2) - tan(pi*p/2) ] - 8d = 0

Unfortunately, this is not solved for p. Can I get a confirmation that there is no real way to isolate p? It doesn't REALLY matter, my calculator can solve it, but...you know...it's not as pretty. = (

 

[Ave22]

I need to prove that any unbounded positive function on a closed interval is not Riemann integrable with the norm of a partition, ||P|| = (b-a)/n, but I have no idea how to begin.

Negating the definition of Riemann integrable function tells me that I need to show there exists epsilon > 0, such that for all delta, |Rsum - Rintegral| > epsilon. But what does this mean? I really hate how this class is taught (or isn't taught); we are given problems to work on with no knowledge how how these things work.

 

[Therion]

I need to prove that any unbounded positive function on a closed interval is not Riemann integrable with the norm of a partition, ||P|| = (b-a)/n, but I have no idea how to begin.

Negating the definition of Riemann integrable function tells me that I need to show there exists epsilon > 0, such that for all delta, |Rsum - Rintegral| > epsilon. But what does this mean? I really hate how this class is taught (or isn't taught); we are given problems to work on with no knowledge how how these things work.

I'm a little rusty, but this seems like the right way to go...
Choose your n so that ||P|| is less than delta. Then say that since the function is unbounded, there is some c in the interval such that f(c)>(epsilon + L)/||P||, so Rsum>f(c)||P||>epsilon+L. Therefore |Rsum-L|>epsilon-L+L>epsilon.

I think that'll do trick. You should probably actually multiply f(c) by the size of the particular interval within the partition and then say that's greater than ||P||, just to be clear. Also note where necessary that having a positive function makes the inequalities work nicely.

 

[Lonely1]

Aren't all functions defined in (topologically) closed intervals bounded? Also, Who are a and b? if a and b are the minimum and maximum of your interval in R then {(b-a)/n}_n is not a partition. Sorry, but I believe you have your problem wrong.

A partition would be {a+i*(b-a)/n}_i with 0<=i<=n. And the solution would be, like stated above, the must be an i such that, for a c in [p_i,p_{i+1}], f(c)>(epsilon + L)/||P||, so Rsum>f(c)||P||>epsilon+L. Therefore |Rsum-L|>epsilon-L+L>epsilon.

Note: [p_i,p_{i+1}] must contain an element where your function f is undefined. Hence the domain of the function can't be a closed set.

 

[Ave22]

Hah, I go through the same confusion every time this professor gives us a problem to work on. The only part of the problem I left out is that the function is defined as f:[a,b] -> R. Also, a function that has an asymptote in a closed interval is unbounded.

Thanks for the attempts at helping. I'll try working through it now to see if I understand what you're saying.

 

[Lonely1]

Hah, I go through the same confusion every time this professor gives us a problem to work on. The only part of the problem I left out is that the function is defined as f:[a,b] -> R. Also, a function that has an asymptote in a closed interval is unbounded.

Thanks for the attempts at helping. I'll try working through it now to see if I understand what you're saying.

Well, if your function has an asymptote, then domain must be in the form of:

f:[a,d)(d,b]-> R or something like it. which is quite (topologically) different than f:[a,b]-> R. If its indeed like it, the c you are looking for is c=d+(a-b)/n or c=d-(a-b)/n. As n becomes smaller, f(c) approachs to infinite, making Rsum - Rintegral bigger than epsilon, since Rsum contains f(c) as a factor.

 

[Lonely1]

Edit:Sorry, I made a mistake. A function defined in a closed can be unbounded. It just have to be of the form:

F:[a,d)[d,b]->R but the idea of the proof is the same. Sorry for the confusion. I was thinking on differentiable functions.

 

[close to the edge]

Okay, need some help, we're doing ordinary differential equations right now and I'm slightly confused with this particular one:

[;x*y' + y = 5x^2 + 6x + 2;]

Okay, so I divide by x and rewrite y' as dy/dx. I can now separately integrate both sides, which looks about like this:

[;\int{\frac{1}{y} dy} = \int{-\frac{1}{x} dx};]
[;ln y = -ln x + C;]

Good, now is the part where I get confused, the solution looks like this:

I don't get the 6th step. Why is suddenly e^c in the numerator? Shouldn't it look like this:
e^(ln y) = e^-(ln x) + e^c

It seems like such a stupid question as well..


edit: Okay, I got it. It's really simple, like I thought. e^(x+y) = e^x * e^y. Oh man..

 

[ant_]

You have to raise the whole expression to the exponent, not each term individually.

So you would get e^(-lnx+c) which is the same as (e^c)(e^(-lnx)), and i think you can get it from there

 

[close to the edge]

You have to raise the whole expression to the exponent, not each term individually.

So you would get e^(-lnx+c) which is the same as (e^c)(e^(-lnx)), and i think you can get it from there

Yes, I found the solution already. Thanks anyways.

 

[Haly]

Basic math fail. I'm not sure what I'm doing wrong but I can't seem to get the right answer.

l1 is the line that passes through (4, 5) and (1,1)
l2 is the line that passes through (3, -3) and (0, 4)
a is the angle from l1 to l2
Find a

Relevant equation:
tan(a) = (m2 - m1)/(1 + m1m2)

m1 and m2 are the slopes of l1 and l2, respectively.

Book says answer is 60 degrees. I keep getting something like 23 degrees. This is not the kind of equation you can mess up after 4 or 5 tries unless I made a fundamental error or the book is wrong. Someone please do this for me and show me my mistake.

(For the record when I plot out the lines the angle does appear to be 60 degrees)

 

[poweld]

Basic math fail. I'm not sure what I'm doing wrong but I can't seem to get the right answer.

l1 is the line that passes through (4, 5) and (1,1)
l2 is the line that passes through (3, -3) and (0, 4)
a is the angle from l1 to l2
Find a

Relevant equation:
tan(a) = (m2 - m1)/(1 + m1m2)

m1 and m2 are the slopes of l1 and l2, respectively.

Book says answer is 60 degrees. I keep getting something like 23 degrees. This is not the kind of equation you can mess up after 4 or 5 tries unless I made a fundamental error or the book is wrong. Someone please do this for me and show me my mistake.

(For the record when I plot out the lines the angle does appear to be 60 degrees)

Sounds like you are graphing it wrong, and getting the slopes incorrect. An eyeball estimate confirms 60 degrees. m1 = 4x/3, m2 = -7x/3

edit: Ugh, graph I posted failed. Still, the slopes are correct.

 

[malsumis]

Book says answer is 60 degrees. I keep getting something like 23 degrees. This is not the kind of equation you can mess up after 4 or 5 tries unless I made a fundamental error or the book is wrong. Someone please do this for me and show me my mistake.

(For the record when I plot out the lines the angle does appear to be 60 degrees)

well, I got this.

l1: y=4/3x-1/3, l2: y=-7/3+4

so 180deg - arctan(4/3) - arctan(7/3) = 60deg. I took arctan(-7/3) with a positive because the angle resulting from it is negative (what I mean here, is that I simply took the second vertical angle). Just remember that you can't simply subtract the angles to get the answer.

 

[Zeppu]

Here is how I did it

m1 = Δx1/Δy1 = (4 - 1)/(5 - 1) = 4/3
m2 = Δx2/Δy2 = (3 - 0)/(-3 - 4) = -7/3

if tan(a) = (m2 - m1)/(1 + m1m2)
 tan(a) = [ (m2 - m1)/(1 + m1m2) ]
        = [ (-7/3 - 4/3)/(1+[4/3*-7/3]) ]
        = [ -3.66 / -2.11 ]
      a = tan⁻¹( 1.735 )  
      a = 60.03

 

[f0rk]

Anyone have any experience with Maple?
I have to make a procedure to find the number of real roots of a polynomial using Sturm's Chain. I can produce the chains of positive/negative numbers, but I'm having a couple of problems.
How do I count the number of sign changes? I can obviously do it by hand but that's not the point!
And does anyone have any tips on how to take into account common factors? The question says to use quo and rem, but I'm not sure how to integrate it into the procedure.

 

[itxaka]

Just curious because I'm pretty bad at math.

Will it be possible to mathematically calculate where a ball was gonna fall in a roulette knowing the force of the ball and roulette?

Or is the possibility of the ball hitting a piece of the roulette and changing it's course radically a big problem?

 

[poweld]

Just curious because I'm pretty bad at math.

Will it be possible to mathematically calculate where a ball was gonna fall in a roulette knowing the force of the ball and roulette?

Or is the possibility of the ball hitting a piece of the roulette and changing it's course radically a big problem?

Because the function is highly sensitive to its initial conditions, we can't reliably determine where the ball will fall. For example, the minute scoring on the ball may affect the way it rolls slightly, which may vastly change the outcome.

 

[KajunW]

Hello

I want to do multiple linear regression, but one of the requirements is the residuals to be normally distributed, and I can check that with QQplots but then the QQ plot shows it is about 95% of data fit into the normal line, but 5% is way off!

can I still proceed ?*or do I have to find a way to transform the data ?*

I'm probably way too late here, but...

(1) Your residuals & QQ plots suggest you have highly kurtotic residuals. One way to deal with this is to use a transformation like y_new:=arctan((y-mean)/stdev), but this will sort of ruin your model coefficient interpretation. Non-normality, however, is never that big of a problem in practice, when compared to...

(2) heteroskedasticity. Your residuals vs fitted plot is alarmingly bad. Seems like the higher the fitted value, the more you overestimate with your model (i.e. residuals are more negative). Maybe this part will be solved by using above mentioned transformation, but I have a feeling it won't be enough. If that is the case, then considering correcting for heteroskedasticity by using (a) Weighted Least Squares, or (b) Standard OLS coefficient estimates but with standard errors calculated by using White's correction.