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The Math Help Thread
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It's probably too late to get an answer, but I might as well try.
So I know that the best way to do this is to use transformations. Set s = (x^2) + (y^2) and the bounds of s become 1 < s < 4. But what would the bounds of t be? I feel like I did it write on the exam, but studying over the question, I'm at a loss. It wouldn't make sense for t to be between 0 and 0, but I don't see (and can't remember) how else you can set bounds for t.
Okay, so the first bound can just be 0 to 1. The third remains 0 to that equation. What about the middle bounds? I think on the exam I just bullshit and put 0 < y < sqrt(1 - x^2).
I completely forgot Calc II and have no idea what's going on.
So I know that the best way to do this is to use transformations. Set s = (x^2) + (y^2) and the bounds of s become 1 < s < 4. But what would the bounds of t be? I feel like I did it write on the exam, but studying over the question, I'm at a loss. It wouldn't make sense for t to be between 0 and 0, but I don't see (and can't remember) how else you can set bounds for t.
Okay, so the first bound can just be 0 to 1. The third remains 0 to that equation. What about the middle bounds? I think on the exam I just bullshit and put 0 < y < sqrt(1 - x^2).
I completely forgot Calc II and have no idea what's going on.
FortuneFaded
Member
Sorry, I should've been clearer
Lets say I have an equation with sinx, am I allowed to multiply it by 2 to make it 2sinx? Or if its 2sinx am I allowed to divide the equation to make it sinx?
I'm studying for an exam.. hope I pass
Thanks for the help
Sorry, I should've been clearer
Lets say I have an equation with sinx, am I allowed to multiply it by 2 to make it 2sinx? Or if its 2sinx am I allowed to divide the equation to make it sinx?
I'm studying for an exam.. hope I pass
Thanks for the help
If the coefficient is out front and isn't part of the function you can do simple algebra on it.
(2 sin(x))/2 = sin(x) -- Valid algebra
(sin(2x))/2 ≠ sin(x)
boviscopophobic
Member
It's probably too late to get an answer, but I might as well try.
So I know that the best way to do this is to use transformations. Set s = (x^2) + (y^2) and the bounds of s become 1 < s < 4. But what would the bounds of t be? I feel like I did it write on the exam, but studying over the question, I'm at a loss. It wouldn't make sense for t to be between 0 and 0, but I don't see (and can't remember) how else you can set bounds for t.
If you set s=x^2 + y^2 and t = x^2/y, then the bounds run from s=1 to s=4 and from t=1 to t=1/2 (note the orientation is reversed), and the Jacobian determinant neatly cancels the integrand with a little bit left over. What you end up getting out is should be -3*ln(2).
Note that the question exhorts you to think first, which suggests that some human-comprehensible interpretation of the integral can be found. What familiar geometric figure, or tractable subset of a geometric figure, is this integral describing? Once you figure that out, the solution is easy.
Okay, so the first bound can just be 0 to 1. The third remains 0 to that equation. What about the middle bounds? I think on the exam I just bullshit and put 0 < y < sqrt(1 - x^2).
You can do an easy substitution to express this in terms of half of a Gaussian integral, int_0^(infinity) e^{-t^2} dt. The trick for computing the full Gaussian integral, int_(-infinity)^(infinity) e^{-t^2} dt, was probably taught when you learned about polar substitutions.
I completely forgot Calc II and have no idea what's going on.
Hypertrooper
Member
HeyI
I have to show that:
So I started with writing down the matrices A and B as a system of vectors:
Is it okay to show it that way?
I have to show that:
So I started with writing down the matrices A and B as a system of vectors:
Then I wrote down the product of both matrices:
After that, I transposed the matrix C:
The transposed matrix C is the product of the following systems:
Is it okay to show it that way?
To be honest I don't really follow that proof. I'd do it as follows:
Element i,j of A*B = row i of A * column j of B
So,
Element i,j of (A*B)t = row j of A * column i of B
On the other hand,
Element i,j of Bt*At = row i of Bt * columnj of At
Element i,j of Bt*At = column i of B * row j of A
And that's it!
spelltropy
Member
Wouldn't mind a hand with two questions (ε − δ continuity)
1. Got a function f
a,b)->R which is continuous for some c in (a,b) and f(c)=/=0. Trying to show there exists a delta such that |f(x)|>f(c)/2 - hint is triangle inequality, but not sure what I should be doing here, wouldn't mind a push in the right direction, I'm sure it must be very basic stuff in terms of Weierstrass as it's very early in the syllabus.
2. Another function w:[0,infinity)->[0,infinity) which is continuous at x=0 with w(0)=0. Then theres another function f:R->R which satisfies
|f(h+c)-f(c)|<w(|h|) for all real h. Trying to show f is continuous at c - written down what it means for w(x) to be continuous at 0, not sure where to go from here?
1. Got a function f
a,b)->R which is continuous for some c in (a,b) and f(c)=/=0. Trying to show there exists a delta such that |f(x)|>f(c)/2 - hint is triangle inequality, but not sure what I should be doing here, wouldn't mind a push in the right direction, I'm sure it must be very basic stuff in terms of Weierstrass as it's very early in the syllabus.2. Another function w:[0,infinity)->[0,infinity) which is continuous at x=0 with w(0)=0. Then theres another function f:R->R which satisfies
|f(h+c)-f(c)|<w(|h|) for all real h. Trying to show f is continuous at c - written down what it means for w(x) to be continuous at 0, not sure where to go from here?
Wouldn't mind a hand with two questions (ε − δ continuity)
1. Got a function f
2. Another function w:[0,infinity)->[0,infinity) which is continuous at x=0 with w(0)=0. Then theres another function f:R->R which satisfies
|f(h+c)-f(c)|<w(|h|) for all real h. Trying to show f is continuous at c - written down what it means for w(x) to be continuous at 0, not sure where to go from here?
For the first one, this is obviously true when f(c)<0, so assume that f(x)>0. Now pick epsilon=f(c)/2 and see what happens.
For the second, it should just come down to that definition again. Let epsilon>0, so that there exists delta>0 such that |w(x)|<epsilon whenever 0<=x<delta. Or you could say that w(|h|)=|(w(|h|)|<epsilon for |h|<delta (justify this equality). Then apply the bound involving f(c) to finish.
You can consider that 1/x = x/x^2, as its clear that x != 0
1 = x/(x^2) + 1/(x^2)
1 = (x + 1)/(x^2)
x + 1 = x^2
x^2 - x - 1 = 0
spelltropy
Member
Forgot to say thank you for this! Cheers
Fishing for a hint with this one now:
Got a function f:R->R, x if x is rational, 0 if x irrational - trying to show it is only continuous at the origin. Anyone got a suggestion of where to start?
e: think I figured it out, wouldn't mind if someone checked: let delta=epsilon, if x is rational then |f(x)|<|x|<d=e, if x is irrational then |f(x)|=0<d=e - would this be enough? I've shown its continuous at the origin but what about other points?
Forgot to say thank you for this! Cheers
Fishing for a hint with this one now:
Got a function f:R->R, x if x is rational, 0 if x irrational - trying to show it is only continuous at the origin. Anyone got a suggestion of where to start?
e: think I figured it out, wouldn't mind if someone checked: let delta=epsilon, if x is rational then |f(x)|<|x|<d=e, if x is irrational then |f(x)|=0<d=e - would this be enough? I've shown its continuous at the origin but what about other points?
Never liked these problems, bleh. Your proof for continuity at the origin is almost correct: change the first strict inequality in the rational case to a "less than or equal to." You can combine both cases into one as follows:
Given a tolerance epsilon e > 0, the delta that we can set is d = e (or anything smaller than this). We see that, if |x - 0| < d (i.e. if x is within d-distance from the origin), then, by definition of f,
|f(x) - f(0)| = |f(x) - 0| <= |x| < d (= e).
Hence, f(x) is within e-distance from f(0).
For points other than origin, you'll likely have to invoke density of real numbers. Using the definition of continuity, think about what it means for a function to be not continuous at a point.
Forgot to say thank you for this! Cheers
Fishing for a hint with this one now:
Got a function f:R->R, x if x is rational, 0 if x irrational - trying to show it is only continuous at the origin. Anyone got a suggestion of where to start?
e: think I figured it out, wouldn't mind if someone checked: let delta=epsilon, if x is rational then |f(x)|<|x|<d=e, if x is irrational then |f(x)|=0<d=e - would this be enough? I've shown its continuous at the origin but what about other points?
I'm not sure if you're familiar with the concept of density, so I'll give a bit more of a hint.
If x>0 is rational, to prove that the function is not continuous you need to prove that, given e > 0, there exists at least one irrational number y such that |x-y|<e. (Why? I guess I'll leave that to you)
As a hint for that, x - sqrt(2) is irrational.
For x irrational, you need to prove the opposite, that is, that there's a rational y at distance e or less. As a hint for that, every irrational is a number with infinite decimals, so you can "approach" it with numbers that have finite decimals (for example, 3.14 is similar to Pi).
spelltropy
Member
Cheers for the help, managed to do it in the end w/ density of rationals!
In linear algebra, if I have three vectors in R^5, and am asked to find a basis containing these vectors, I should be looking for 2 more? Because the standard basis in R^5 has 5 and basis theorem? I have a feeling I'm completely off the mark here..)
Also, given two vectors in R^2 which are scalar multiples of each other, is it sufficient to give a counter example to show they don't span R^2?
In linear algebra, if I have three vectors in R^5, and am asked to find a basis containing these vectors, I should be looking for 2 more? Because the standard basis in R^5 has 5 and basis theorem? I have a feeling I'm completely off the mark here..)
Also, given two vectors in R^2 which are scalar multiples of each other, is it sufficient to give a counter example to show they don't span R^2?
Awesome, linear algebra is my favorite.
If the given three vectors are linearly independent, then yes, you just need to find two more and check that the five vectors are linearly independent. It is enough to check for linear independence to conclude that the five vectors form a basis for R^5. This is because the space of R^5 has a dimension of five and we have five vectors, i.e. it is possible to span/cover R^5. With any fewer vectors, spanning R^5 is not possible; with any more vectors, linear independence is not possible.
I'm assuming the question asks you to show that the two vectors do not form a basis for R^2. Yes, you can come up with a vector that cannot be written as a linear combination of the two vectors. From the definition of linear independence, we can also show that, if one vector is a multiple of the other (in R^2), then the two vectors are linearly dependent (the converse is also true).
I'm doing a "Math Foundations" class that's basically a bunch of proof-writing. I know what direct/contrapositive/contradiction etc is, but I'm having a really hard time even getting started. I figure if I can get the first step, the rest should be easy, but I'm having a really hard time figuring out that first step for my asignment questions.
Any tips on this?
Any tips on this?
Maybe post the assignment question.
Well, it's not really so much this one specific question, but all of them. I'm having trouble doing starting proofs in general.
But here's the first question, supposed to be done by direct proof.
For all x in the real numbers, if 0 < x - 2, then 3 < 2x.
So I started with the "Let x be in the real numbers" and "Assume 0 < x-2" and then I just get stuck messing around with "0 < x-2" and "0 < 1" trying to find some way to use transitivity to get 3 < 2x.
boviscopophobic
Member
I'd look at what you start with and what you want to prove. You want to prove 3 < 2x, which is a statement about 2x. What you have is 0 < x - 2. How can you convert that into a fact about 2x? Well, first you could get x by itself:
2 < x (by adding to both sides)
Then you could multiply both sides by 2, which preserves the inequality:
4 < 2x
Now see if you've gotten where you need to be. Well, we wanted 2x > 3, but we have 2x > 4, which is even better! Since 4 > 3, it follows by the transitive property that 2x > 3.
Learning partial derivatives.
Can anyone help me with this one?
f(x,y) = sin(xcosy)
I know the answer but I don't know how they got it.
For fx, to me it looked like a chain rule problem with a product rule on the inside.
fx = (cosy - xsiny)cos(xcosy).
Apparently I'm way off. Any help?
Can anyone help me with this one?
f(x,y) = sin(xcosy)
I know the answer but I don't know how they got it.
For fx, to me it looked like a chain rule problem with a product rule on the inside.
fx = (cosy - xsiny)cos(xcosy).
Apparently I'm way off. Any help?
Like Bovi pointed out, in general when doing proofs, know where you want to go. It's simple, but immensely helpful.
Question about Laplace transforms.
I need to take the transform of this business
f(t) = 2u(t) - 5e^(-2t)u(t) - 3cos(2t)u(t) + 2sin(2t)u(t)
Totally lost.
I'm assuming u(t) is the step function, the L.T. of which is 1/s.
If that's so, then for the first term I know the coefficient just carries through the transform.
For the other 3 terms I have no idea.
I see that the L.T. of 1 is also 1/s. Does that mean I can treat u(t) as a 1 and just multiply it with the other stuff in each term?
I need to take the transform of this business
f(t) = 2u(t) - 5e^(-2t)u(t) - 3cos(2t)u(t) + 2sin(2t)u(t)
Totally lost.
I'm assuming u(t) is the step function, the L.T. of which is 1/s.
If that's so, then for the first term I know the coefficient just carries through the transform.
For the other 3 terms I have no idea.
I see that the L.T. of 1 is also 1/s. Does that mean I can treat u(t) as a 1 and just multiply it with the other stuff in each term?
Also, while I'm hear, another partial differentiation question
I need to find all the second derivatives for this business
v = e^(xe^y)
To get the first derivatives, I just took the natural log of both sides, followed through with implicit differentiation, and subbed back in v.
So vx = e^(xe^y)e^y
and
vy = e^(xe^y)(xe^y)
Now I need vxx, vxy, vyx, and vyy
I can get vxx easily enough, but the other ones I'm unsure of. Should I start by taking the ln again and using the product properties of logs to get a sum?
I need to find all the second derivatives for this business
v = e^(xe^y)
To get the first derivatives, I just took the natural log of both sides, followed through with implicit differentiation, and subbed back in v.
So vx = e^(xe^y)e^y
and
vy = e^(xe^y)(xe^y)
Now I need vxx, vxy, vyx, and vyy
I can get vxx easily enough, but the other ones I'm unsure of. Should I start by taking the ln again and using the product properties of logs to get a sum?
Not really a help question, but just kind of a planning one.
Should I take Linear Algebra right after I take Calc I?
I just don't want to end up taking Linear Algebra 2 years from now and forget everything from Calc and/or Discrete.
I'm a CS major, all I have to take is Calc I, Discrete Math, Linear Algebra, and Statistics I.
EDIT: I am in Calc I and Discrete Math at the moment.
Should I take Linear Algebra right after I take Calc I?
I just don't want to end up taking Linear Algebra 2 years from now and forget everything from Calc and/or Discrete.
I'm a CS major, all I have to take is Calc I, Discrete Math, Linear Algebra, and Statistics I.
EDIT: I am in Calc I and Discrete Math at the moment.
Not a fully educated opinion here, but I'm in Linear Algebra right now and thus far we haven't used anything from calculus at all.
I've heard from my professors that linear algebra is mostly quite separate from calculus.
That said, I don't know how it is at your school, but at my school it's still an upper-division math class and is kind of proof based, so whenever you feel like dealing with that.
Alright cool thanks, suppose I'll just wait and see who is teaching it and decide from there.
You won't need calculus for linear algebra. Linear algebra and calculus being important to each other happens in higher math level. That said, I would just take linear algebra and get it out of the way.
Well, since L.T. goes from 0 to infinity, doesn't that make u(t)=1? The second term seems to be a simple integral of 5*e^(-(2+s)*t). As for the third and fourth terms, why not turn the sin and cos in terms of e^ix?
My thoughts while reading that.
yea
uh...sure...
uhhhhhh.................(......)
Goodnight!
I apparently have no idea what I'm doing. I'm just not seeing the logic of your statements. It's midnight, so I guess I'll just turn it in like this. It's amazing to me that 4 (four!) of my classes right now are focusing on Laplace transforms and I still have no idea what I'm doing. I must be really stupid. Thanks for trying to help though.
yea
ibyea said:
uh...sure...
ibyea said:
uhhhhhh.................(......)
ibyea said:
Goodnight!
I apparently have no idea what I'm doing. I'm just not seeing the logic of your statements. It's midnight, so I guess I'll just turn it in like this. It's amazing to me that 4 (four!) of my classes right now are focusing on Laplace transforms and I still have no idea what I'm doing. I must be really stupid. Thanks for trying to help though.
I want to explain that part further.
e^(ix) = cos(x) + i*sin(x)
e^(-ix) = cos(x) - i*sin(x)
Therefore:
cos(x) = [e^(ix) + e^(-ix)]/2
sin(x) = [e^(ix) + e^(-ix)]/i2
The third term then turns into the integral of from 0 to infinity:
3*[e^((i-s)x) + e^(-(i+s)x)]/2
Although you could always try integration by parts.
Stumpokapow
listen to the mad man
In general, the derivative of an exponential function:
d/dx e^u = e^u du/dx
So you repeat the function and then multiply it by the derivative of the function in the exponent.
First partial with respect to x
dv/dx = e^(xe^y) e^y = e^(xe^
+yWhat's going on here? Well, first you know d/dx e^(xe^y) = e^(xe^y) * d/dx xe^y. e^y is a constant, x is a variable. Take the constant out, differentiate x to 1, you get e^(xe^y) * (1) e^y = e^(xe^y)e^y. Sum the exponents because you have like bases.
Second partial XX:
dv^2/dxdx = e^(xe^
+y) e^y = e^(xe^+2y)Second partial cross:
Here we switch and treat y as the variable and x as the constant.
dv^2/dxdy = e^(xe^
+y) (xe^ + 1)(Remember that cross partials are the same* regardless if you do dxdy or dydx * in most functions you're dealing with--all continuous functions IIRC)
First partial with respect to y
dv/dy = e^(xe^
) (xe^) = e^(xe^ + y) (x)Second partial YY:
dv^2/dydy = e^(xe^
+y) (xe^y + 1) (x)Why does (x) stick around? It's a constant here, so we can take it out before differentiating.
Because this is a mess, here's everything cleaned up:
You do not need any kind of natural logging. The derivative of e^x is e^x dx. So the derivative of, say, e^(4x) = e^(4x) d/dx (4x) = 4e^(4x). With partials, you treat the variable you are not differentiating with respect to as a constant and the one you are as a variable. So d/dx e^(xy) = e^(xy) d/dx(xy) = e^(xy) y. d/dy e^(xy) = e^(xy) d/dy(xy) = e^(xy) x. In general with e^ functions, continued differentiation just gets a longer and longer equation, the original equation will still be there.
du/dx sin(u) = cos(u) du/dx
u = xcosy
du/dx sin(u) = cos(xcosy) cos
Why?
d/dx xcos
= cos d/dx x <-- cos is a constant, take it out before you differentiate.To your question, it doesn't matter. Linear Algebra I is very separate from Calc I-III. There are no real topics common to both. I also didn't end up using Linear Algebra in my CS degree at all until I got to a few of the upper division specialization courses You should definitely consider doing more math than is required for your program, though. Calc II and Calc III especially. When you get to upper division CS courses, particularly if you do image processing or any kind of low-level processor stuff that requires signals analysis, you'll be very happy if you go to Calc III and beyond.
I don't know what you're permitted to use in proofs in terms of properties of elementary operators or numbers.
Assume 0 < x - 2 -- Given in problem
Therefor 2 < x-2+2 (Add two to both sizes, 2 < x, property of addition)
Therefor 4 < 2x (Multiple both sides by two, 2(2) < 2(x), property of positive multiplication)
Therefor 3 < 2x-1 (Subtract one from both sides, property of subtraction)
2x-1 < 2x for all x (Property of subtraction)
3 < 2x -1 < 2x so by transitivity 3<2x
Do take the linear algebra--it's an awesome topic. However, to really understand it, you will want to be comfortable with imagining abstract concepts like vector spaces and with proving simple statements. If you haven't written proofs before, you may want to wait till after calculus 2/3; by then you will have had more exposure to proofs. Since you are a CS major, I'd look into a numerical analysis or a numerical linear algebra course as well.
Linear algebra and calculus are related to an extent, but you may not see it in your study. There used to be a subject called vector calculus; while you can still find books with this title, I don't think calculus is taught in terms of vectors anymore. We briefly see the divergence, gradient, and curl operators when we consider multivariate functions in calculus 3; on the other hand, continuity and differentiability still play an important role when we work with matrices. In the field that I work in, understanding both subjects is crucial.
Second partial cross:
Here we switch and treat y as the variable and x as the constant.
dv^2/dxdy = e^(xe^
(Remember that cross partials are the same* regardless if you do dxdy or dydx * in most functions you're dealing with--all continuous functions IIRC)
IIRC, cross partials are the same only if all second partial derivatives are continuous
dr3upmushroom
Banned
An open rectangular box is to be made from a piece of cardboard 26 inches wide and 26 inches long by cutting a square from each corner and bending up the sides.
(a) Express the volume V of the box as a function of the size x of the cutout.
(b) Approximate the dimensions of the box with the largest volume. (Round your answers to one decimal place.)
Having trouble understanding this problem. Halp!
(a) Express the volume V of the box as a function of the size x of the cutout.
(b) Approximate the dimensions of the box with the largest volume. (Round your answers to one decimal place.)
Having trouble understanding this problem. Halp!
Regret Upon Approval
Member
Draw out a rectangle.
Then draw out a square in each corner.
Then imagine folding it up to form a rectangle with the top missing.
This should help you visualize it, a key first step.
Do you understand that so far?
Stumpokapow
listen to the mad man
Draw the flat sheet as a rectangle. Draw the squares in the corners with size x. Draw lines connecting the inside edges of the squares to show what you are going to fold up to make the box. Measure the dimensions.
So we see from this picture that the bottom of the box is
26-2x wide by 26-2x long
x high
Volume of a rectangular prism is length * width * height.
The formula for the volume is
(26-2x)(26-2x)(x) = 4x^3 - 104x^2 + 676x
How do you optimize (find the highest value of a function)? To optimize, you
set the derivative of the formula to 0 (note: this might get you a minimum or a maximum, do a second derivative check to confirm but in this case it's a maximum). Why is this the case? The derivative of the Volume function with respect to X tells you how much V, volume, changes per unit change of X. We look for where it's 0 because that's the point where positive changes in X no longer make a difference in Volume, and subsequently lower total volume.
The derivative is
12x^2 -208x + 676.
4(3x^2-52x+169) = 4(3x-13)(x-13) So, then, x=13/3 or x=13. One of these is a minima, one of these is a maxima. You can do a second derivative test to figure out which is which or just plug both values in and see which is the maximizing value. f''(x) = 24x-208, plug in x=13 and get 104, which means x=13 is a minima; plug in x=13/3 and get -104 which means x=13/3 is a maxima; alternatively, the volume at x=13 is 0 which should trivially tell you it's a minima, since you'd literally be cutting out the entire sheet of cardboard.
Solution:
For a sanity check, check values of X just below and above the value you found to ensure that the volume is lower.
Khan Academy: Optimizing Box Volume
https://www.khanacademy.org/math/di...imization/v/optimizing-box-volume-graphically
https://www.khanacademy.org/math/di...mization/v/optimizing-box-volume-analytically
Solution:
X=13/3 is x=4.333, so you can round to 4.3
X~=4.3
Dimensions ~= 17.4x17.4x4.3
Volume ~= 1301.868
X~=4.3
Dimensions ~= 17.4x17.4x4.3
Volume ~= 1301.868
For a sanity check, check values of X just below and above the value you found to ensure that the volume is lower.
Khan Academy: Optimizing Box Volume
https://www.khanacademy.org/math/di...imization/v/optimizing-box-volume-graphically
https://www.khanacademy.org/math/di...mization/v/optimizing-box-volume-analytically
spelltropy
Member
Forgot to say thanks for the help with the linear algebra last week kgtrep, really appreciate it
I'm kind of struggling with Analysis right now, and wouldn't mind some help with these two questions: the first one seems obvious, but not sure how to go about it, whereas I have no idea what I'm doing for the second...
I'm kind of struggling with Analysis right now, and wouldn't mind some help with these two questions: the first one seems obvious, but not sure how to go about it, whereas I have no idea what I'm doing for the second...
boviscopophobic
Member
For part i) of the first, raise both terms (x^p)^(1/q) and (x^(1/q))^p to the q'th power. To simplify, you can apply any laws of exponents that you know for integer exponents (it would be begging the question to apply them to rational exponents), and cancellation of functions with their inverses. You should end up getting the same answer for both terms, which shows that the original terms were equal due to injectivity of the x |-> x^q function. A similar idea applies for part ii).Forgot to say thanks for the help with the linear algebra last week kgtrep, really appreciate it
I'm kind of struggling with Analysis right now, and wouldn't mind some help with these two questions: the first one seems obvious, but not sure how to go about it, whereas I have no idea what I'm doing for the second...
For the second exercise, what do you know about the completeness of the real numbers?
triplestation
Member
um does anyone here have essential calculus second edition by stewart?
is it much different than the first?
itd be great if someone could post a photo of 3 or 4 pages of the problem sets so i can compare it with the first edition
is it much different than the first?
itd be great if someone could post a photo of 3 or 4 pages of the problem sets so i can compare it with the first edition
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