Math thread! Come in and have some pi...

[ctownballer04]

You need to set bounds in place.

Assume |x-3| < 1/2. Then 5/2<x<7/2, so 1/2<x-2<3/2 implies 2/3 <1/|x-2| = 1/(x-2) < 2. Also, 5/2<x<7/2 implies -3/2 < x-4 < -1/2, and so |x-4| < 3/2.

Therefore |f(x) - 5| < (2)(3/2)|x-3| = 3|x-3|. Choose delta as the minimum of 1/2,epsilon/3.

You can verify that 3|x-3| - |f(x)-5| is nonnegative when 5/2<x<7/2 here: https://www.desmos.com/calculator/v90k95ad5j

I think I'm just going to wait for the lecture lol. I'm confused. Your prof's book is pretty nice though, I like it.

[CandyMane]

Patrickjmt has videos on statistics . Will look into tomorrow

[CandyMane]

Mathematician, are you a professor? If you mind me asking your line of work and degree

[ctownballer04]

Okay. Here's a bit more info though if it helps: For all factors present, in this case, |x-4| and 1/|x-2|, you need to replace them by a number to obtain a specific delta. You can only use the fact that |x-3| is controllable. By assuming that |x-3| is less than some fixed number, this allows you to get bounds of the other factors in most cases.

I needed to choose |x-3|<1/2 (or something smaller than 1) because if you assume |x-3|<1, you have 0<x-2<2. Since |x-2| is in the denominator, it doesnt help that your bound allows it to be close to 0 (so that 1/|x-2| can be as large as it wants). But by assuming |x-3|<0.1 for instance, this gives 9/10 < x-2 < 11/10, making sure that the denominator is away from 0, so that 1/|x-2| < 10/9

I get that stuff. I'm mostly just confused on our assumption. I don't really understand the less than 1/2, where that comes from and why we can assume that. Then I was also confused on choosing the minimum value between 1/2 and epsilon/3.

Don't worry about it though, it's late now and I'm not going to get anything really done tonight. I can probably just figure it out at lecture tomorrow, I don't want to waste your time if I'm not going to be sitting down trying to figure it out.

Thanks though

[DesEsseintes]

I couldn't find the original problem you have been quoting, but in general the idea is this:

Suppose we have a given sequence of real numbers (x(m)), and a real number b, such that for any c > 0 we can find a large enough N so that |x(n)-b| < c whenever n > N.

Now, we know that this is equivalent to the following: b - c < x(n) < b + c, for large enough n.

Furthermore, suppose b is non-zero. Then |b-0| > 0, so in particular set c to be |b-0|/2. The reasoning behind this is that if we take half the distance of b from zero then add or subtract this to b, the interval [b-c,b+c] will not contain the number 0. You can draw a diagram, in addition to reasoning loosely spelled out above, to make sure you understand the motivation.

However, this is good enough for our purposes: we now know that we can take the multiplicative inverse of x(n) where n > N, and have some bounds for it, by using the triangle inequality and our noted special choice of c.

This usually happens in analysis. The triangle inequality is best thought of as the ability to say the following: If two points, x and y, are each close to a third point, z, then x and y are both close to each other.

[Muckle_Ewe]

Ok brahs, riddle me this

Let's say I have a function f(x) such that 0 < f(0) < 1, f(T) = 0 and f'(x) = log(f)

Is it possible to prove that the integral of 1/f from 0 to T diverges? Seems like it's obvious but haven't done real math in ages so not sure how to rigorously prove it... Thought about constructing some other function g such that integral of 1/g with same limits diverges but with 1/g <= 1/f, however that proved to be more difficult than I though...

[Muckle_Ewe]

Hey there OP


u = f(x). du =log(f(x))dx

INT_0^T 1/f(x) dx = INT_[f(0)]^[f(T)] 1/(ulog(u)) du

since f(T) is 0, this integral is improper. So set up a limit:

lim_{t\to T} [log|log(f(t))| - log|log(f(0))|]

Oh hai there. Been away from this thread for ages, basically stopped enjoying maths a few years ago which is a shame since it used to fascinate me... All about dat dere software now.

Thanks for the answer though, a lot easier than what I was trying to do...

[ctownballer04]

Okay. Here's a bit more info though if it helps: For all factors present, in this case, |x-4| and 1/|x-2|, you need to replace them by a number to obtain a specific delta. You can only use the fact that |x-3| is controllable. By assuming that |x-3| is less than some fixed number, this allows you to get bounds of the other factors in most cases.

I needed to choose |x-3|<1/2 (or something smaller than 1) because if you assume |x-3|<1, you have 0<x-2<2. Since |x-2| is in the denominator, it doesnt help that your bound allows it to be close to 0 (so that 1/|x-2| can be as large as it wants). But by assuming |x-3|<0.1 for instance, this gives 9/10 < x-2 < 11/10, making sure that the denominator is away from 0, so that 1/|x-2| < 10/9

I think I got the most basic ones figured out after lecture today.
Does this look alright (will reorganize, this is just scratch paper)?



Also, do you have any advice for higher powered problems? For example, lim x-->1 x^3-2x+3=2?

Approaching it similarly, you get |f(x)-L|=|x-1||x^2+x-1|

I'm not sure how to deal w/ that x^2+x-1. Bounding |x-1| doesn't seem to help as I'd have to scale the inequality through by x and that would leave me w/ an x-variable in the LB and UB.



EDIT: Wait, I think I was just going full retard earlier and overthinking this.
Let |x-1|<1
Then, -1<x-1<1
So, 0<x<2
and, 1<x+1<2

So, when |x-1|<1, we have:

|x-1||x^2+x-1|=|x+1||x*(x+1)-1|<|x+1||2*3-1=5|x+1|

Choose delta to equal the min(1,epsilon/5)
Then if 0<|x-3|<delta we have:
|f(x)-L|=|x-1||x*(x+1)-1|<5*|x+1|<5delta=5*(epsilon/5)=epsilon

[Slacker23]

(...) Also, do you have any advice for higher powered problems? For example, lim x-->1 x^3-2x+3=2?

Approaching it similarly, you get |f(x)-L|=|x-1||x^2+x-1|

I'm not sure how to deal w/ that x^2+x-1. Bounding |x-1| doesn't seem to help as I'd have to scale the inequality through by x and that would leave me w/ an x-variable in the LB and UB.

Your proof seems fine to me, at least at first sight, but double-check the numbers anyway just in case.

For the second question, why wouldn't bounding |x-1| work?

Suppose |x-1|<=1/4, i.e. $x\in[0.75,1.25]$. Show that x^2+x-1 is increasing if x>0 by showing (x^2+x-1)>(0^2+0-1) if x>0, which implies that (x^2+x-1)<=(1.25)^2+1.25-1, then use a regular limit argument.

As a general rule, if you have something of the form $(x-a)f(x)$ where $f(a)\neq0$ and $f$ is continuous, then for any $\epsilon>0$ you have that $f$ will have a minimum and maximum on $[a-\epsilon,a+\epsilon]$ and, by choosing $\epsilon$ small enough, you can have both the min and max to be positive (or both negative).

But I might be getting ahead of myself...

ETA : wrote before I saw the edit...

[DesEsseintes]

You've got the right idea good job brah, however here are some pointers in setting out a correct epsilon delta proof.

Step 0) Do some rough working to see how to proceed, and if your argument is long you may wish to identify some points to highlight before proceeding to your proof. In this case, you may want to write, before beginning your proof, that if it is the case |x-3| < 1/2, then 1/2 < |x-2| and |x-4| < 3/2.

Step A) State what you wish to demonstrate and introduce some notation, such as what is 'f' is and etc.

Step 1) In any analysis proof you should always begin with something like:

Let epsilon greater than 0 be arbitrary

Let e > 0

Suppose epsilon is a positive real number, etc.

Step 2) Write down a bound on the interval you wish to constrain x to in terms of epsilon: for instance, 0 < |x-3| < min(1/2,epsilon/3)

So far your proof is something like this:

Let e > 0 and suppose that x is in the domain of f such that 0 < |x-3| < min(1/2,e/3). (*)

Step 3) Then write down the chain of inequalities for |f(x)-L| to end up with something like this:

Let e > 0 and suppose that x is in the domain of f such that 0 < |x-3| < min(1/2,e/3). Then it follows that |f(x)-L| = ... series of inequalities with reference to any observations in Step 0 ... < e.

Step 4) Write something like 'as epsilon was arbitrary we have proven the theorem (or a variant thereof).

(*) N.B. you did not write this in your proof - there is a distinction between a limit point of a set and an element of a set. For instance, consider the function x -> 1/(x-1) on (0,1). Then 1 is not in the domain of this function, but 1 is a limit point of the domain (0,1), so it makes sense to talk about the limit of the function as x tends to 1. It may seem that the inclusion of 0 < |x-3| < stuff in the proof was unnecessary, especially when we specify x must be in the domain.

However, consider the following function:


f(x) = x for x=/=0, and f(0) = 1. Then using the idea of taking those values of x not equalling 0 when discussing the limit as x tends to zero, i.e. 0 < |x-0| < delta, gives the function tending to 0 as expected However, f(0) = 1 so we know it is not continuous. This is where we need to remember that we are dealing with limits (or limit points) which may or may not be in the domain, but we need to exclude them when talking about limits.

[iDontExistHere]

I'm in kind of a tight space with this summer class. In order for me to get a c in this class for it to transfer I need to get a 10 on my extra credit(1 point per question). If I posted one or two questions daily could you guys make sure it's 100% correct ? It's only 10 questions is the catch

[ctownballer04]

You've got the right idea good job brah, however here are some pointers in setting out a correct epsilon delta proof.

Step 0) Do some rough working to see how to proceed, and if your argument is long you may wish to identify some points to highlight before proceeding to your proof. In this case, you may want to write, before beginning your proof, that if it is the case |x-3| < 1/2, then 1/2 < |x-2| and |x-4| < 3/2.

Step A) State what you wish to demonstrate and introduce some notation, such as what is 'f' is and etc.

Step 1) In any analysis proof you should always begin with something like:

Let epsilon greater than 0 be arbitrary

Let e > 0

Suppose epsilon is a positive real number, etc.

Step 2) Write down a bound on the interval you wish to constrain x to in terms of epsilon: for instance, 0 < |x-3| < min(1/2,epsilon/3)

So far your proof is something like this:

Let e > 0 and suppose that x is in the domain of f such that 0 < |x-3| < min(1/2,e/3). (*)

Step 3) Then write down the chain of inequalities for |f(x)-L| to end up with something like this:

Let e > 0 and suppose that x is in the domain of f such that 0 < |x-3| < min(1/2,e/3). Then it follows that |f(x)-L| = ... series of inequalities with reference to any observations in Step 0 ... < e.

Step 4) Write something like 'as epsilon was arbitrary we have proven the theorem (or a variant thereof).

(*) N.B. you did not write this in your proof - there is a distinction between a limit point of a set and an element of a set. For instance, consider the function x -> 1/(x-1) on (0,1). Then 1 is not in the domain of this function, but 1 is a limit point of the domain (0,1), so it makes sense to talk about the limit of the function as x tends to 1. It may seem that the inclusion of 0 < |x-3| < stuff in the proof was unnecessary, especially when we specify x must be in the domain.

However, consider the following function:


f(x) = x for x=/=0, and f(0) = 1. Then using the idea of taking those values of x not equalling 0 when discussing the limit as x tends to zero, i.e. 0 < |x-0| < delta, gives the function tending to 0 as expected However, f(0) = 1 so we know it is not continuous. This is where we need to remember that we are dealing with limits (or limit points) which may or may not be in the domain, but we need to exclude them when talking about limits.

I'm confused about the bold. You said I didn't write this in my proof but what you're talking about appears to just be added insight. I am familiar w/ what you are saying because my professor made a point to explain that a limit point need not be in the functions domain, but I'm not sure where that applies to this proof.

Thanks a lot for the detailed assistance, it's very much appreciated.

[ctownballer04]

This is closer to what I'd write up for a HW assignment (format wise, I get sloppy and cut corners when working on scratch paper). Does this look better?

[JimFromRaleigh1]

Doing optimization word problems, the problem reads "an open box is made from a tin sheet that is 8 inches^2...if a square is cut from each corner..."

When I googled the problem to figure out how to set it up, everybody has it as the length and width as 8-2x. My question is why use the 8? How do we know the length is 8 inches? Doesn't 8 inches squared of the tin sheet mean the length could be 4-2x and 2-2x? Hope I was clear.

[lambo4bkfast]

I'm having a lot of trouble integrating this integrand (e^x)^2. I tried substitution and ended up with this as the antiderivative (1/e^x)[((e^x)^3)/3]. I don't know if I am having a complete brain fart, but I stared at this problem for approx 15min and couldn't come up with anything. I derived the integral by evaluating a solid of revolution, but I am certain that part is correct. please respond

[Kobe81BakeShow]

Disregard this being a new account pls. My regular account is temporarily banned

I'm writing proofs of Laplace transformations from a table. The last proof is of a Bessel function (Idek what it is tbh, this is pretty much a bonus question for our assignment). Here is what I need to prove: i.imgur.com/TE3lfZt.jpg

I tried following a video of proving J0(t) but the teacher was Indian and I got too confused.

[fsujoseph]

Im fukken crying right now brahs. I got an A on all 4 tests on Calc 1 so I didn't study as hard on the final and now I think I fukked it all up. Needed a 75 on the final to get an A, don't know if im gonna make it

Maybe he'll curve the final a few points, it was extremely hard

[charity4thepoor]

Im fukken crying right now brahs. I got an A on all 4 tests on Calc 1 so I didn't study as hard on the final and now I think I fukked it all up. Needed a 75 on the final to get an A, don't know if im gonna make it

Maybe he'll curve the final a few points, it was extremely hard

Brah, always assume the next test will be harder than the previous in a math class. Especially if you do really well.

[JimFromRaleigh1]

Can you restate the whole problem word for word? Small things can make a difference.

If an open box is made from a tin sheet 8in. square by cutting out identical squares from each corner and bending up the resulting flaps, determine the dimensions of the largest box that can be made.

Every answer I see online is that the length/width are 8-2x. However if a tin sheet is 8 inches squared, isn't that the area? If so, shouldn't the length/w be 2 radical 2 on each side? Not sure why we use 8 just need some light.

[Slacker23]

If an open box is made from a tin sheet 8in. square by cutting out identical squares from each corner and bending up the resulting flaps, determine the dimensions of the largest box that can be made.

Every answer I see online is that the length/width are 8-2x. However if a tin sheet is 8 inches squared, isn't that the area? If so, shouldn't the length/w be 2 radical 2 on each side? Not sure why we use 8 just need some light.

You start with a square sheet of tin of sides 8 and cut out equal squares from each corner, giving you something that looks like this and is folded along the dotted lines:

   __
 _|..|_
| :  : |
|_:..:_|
  |__|

No comments about how potato my drawing is.

If the squares had side x, then the flaps have width x and you know your cross has width 8 at its widest point. The length of the flaps is the largest width of the cross without the two sides of squares that were cut off, giving you 8-2x.

It follows that, once you fold the dotted lines, you get a square base of side 8-2x and a height of x. I'll leave it to you to find the volume and maximize it.

[JimFromRaleigh1]

Exponential Equations

(3^2x)-12*3(^4)+27=0

Halp? I know 27=3^3

[juiceboxx03]

Im fukken crying right now brahs. I got an A on all 4 tests on Calc 1 so I didn't study as hard on the final and now I think I fukked it all up. Needed a 75 on the final to get an A, don't know if im gonna make it

Maybe he'll curve the final a few points, it was extremely hard

Would have had an A in cal 2 if not for this very mistake

[DesEsseintes]

Let y = 3^x: can you write down an equation in terms of y and then use logarithms to solve?

[ctownballer04]

I'm trying to prove a problem very similar to one I posted a week or two ago. It's basically just a proof I did for sequences changed to more general functions.

"Let D be a subset of R, f: D-->R, and xbar be a limit point of D. Suppose lim x-->xbar f(x)=L where L>0. Prove that there is an open interval (a,b) containing xbar such that f(x)>0 for x in {(a,b) intersect D}, x=/=xbar"

I'm having a hard time wording this since my understanding of these open balls is iffy, but I think I mostly have it.


Edited Attempt:
Suppose lim x->xbar f(x)=L>0
By the definition of limit, for any epsilon>0, there exists a delta>0 such that |f(x)-L|<epsilon for all x in D for which 0<|x-xbar|<delta. The statement all x in D for which 0<|x-xbar|<delta is equivalent to all x in {(xbar-delta,xbar+delta)intersect D\xbar}
Since delta>0, we certainly have an xbar in the open interval of (xbar-delta,xbar+delta)
Let epsilon=L/2
Choose a corresponding delta such that |f(x)-L|<epsilon for all x in {(xbar-delta,xbar+delta)intersect D} where x does not equal xbar
So, for all x not equal to xbar in the open set {(xbar-delta,xbar+delta) intersect D}, we have:
|f(x)-L|<epsilon=L/2 <=> -L/2<f(x)-L<L/2 <=> L/2<f(x)<3L/2
Since L>0, 0<L/2<f(x)

Therefore, we have proven there exists an open interval containing xbar, namely {(xbar-delta,xbar+delta), such that f(x) is greater than 0 for all x not equal to xbar in the open set {(xbar-delta,xbar+delta) intersect D}



I think this is where I'm done lol? What say you misc?

[Slacker23]

I'm trying to prove a problem very similar to one I posted a week or two ago. It's basically just a proof I did for sequences changed to more general functions.

"Let D be a subset of R, f: D-->R, and xbar be a limit point of D. Suppose lim x-->xbar f(x)=L where L>0. Prove that there is an open interval (a,b) containing xbar such that f(x)>0 for x in {(a,b) intersect D}, x=/=xbar"

I'm having a hard time wording this since my understanding of these open balls is iffy, but I think I mostly have it.


My attempt:
Given e=L/2 where L>0
By the definition of limit, there exists a delta>0 such that for all x in D such that 0<|x-xbar|<delta which is equivalent to all x in {(xbar-delta,xbar+delta)intersect D\xbar}
implies |f(x)-L|<epsilon=L/2 <=> -L/2<f(x)-L<L/2 <=> L/2<f(x)<3L/2
Since L>0, 0<L/2<f(x)

I think this is where I'm done lol? What say you misc?

I don't think this works.

The epsilon-delta definition is only for sequences, i.e. for any {xn} such that xn->X, you have that for every epsilon there exists delta_epsilon, bla bla bla...

However, while this is true for any sequence {xn} in D converging to X, there could be an infinity of sequences in D converging to X and the delta_epsilon could, a priori, be different for each and not necessarily boundable.

I would use a proof by contradiction.
Suppose that there does not exist any d such that 0<|x-X|<d (x in D) => f(x)>0.

This means that, for any d>0 there exists some x_d in D such that 0<|x_d-X|<d and f(x_d)<=0.

Define the sequence {xn} by xn=x_(1/n) and derive a contradiction.

Edited to emphasize where the weak spot is, IMO.

[ctownballer04]

I don't think this works.

The epsilon-delta definition is only for sequences, i.e. for any {xn} such that xn->X, you have that for every epsilon there exists delta_epsilon, bla bla bla...

However, while this is true for any sequence {xn} in D converging to X, there could be an infinity of sequences in D converging to X and the delta_epsilon could, a priori, be different for each and not necessarily boundable.

I would use a proof by contradiction.
Suppose that there does not exist any d such that 0<|x-X|<d (x in D) => f(x)>0.

This means that, for any d>0 there exists some x_d in D such that 0<|x_d-X|<d and f(x_d)<=0.

Define the sequence {xn} by xn=x_(1/n) and derive a contradiction.

Edited to emphasize where the weak spot is, IMO.

I'm really confused by this. You obviously know way more about this subject than I do but let me post my thoughts so maybe you or someone else can detect the disconnect between us.

I went to my professors office today where he gave some hints into this problem. He pulled out our homework problem from last chapter (see quotes below) and said this weeks hw problem is done the same way by picking an appropriate epsilon.

"Prove that if the lim n--> a_n=L>0, then there exists n_0 in N such that a_n>0 for all n greater than or equal to n_0"

So he sort of pushed us down this route that I have taken. Not that it wouldn't work, but I'm certain he isn't expecting a proof by contradiction based on his office hour.

In regards to the bold in your post, I just don't follow this part at all. W/ sequences, we had the format of "for any epsilon>0, there exists an N in the naturals such that for all n greater than or equal to N, |a_n-a|<epsilon". The N was sort of our delta for sequences.

Now with more generalized functions w/ domains not restricted to the naturals, we have the limit definition w/ epsilons and deltas. More specifically, the text my professor wrote defines limits of functions as: "We say that f has a limit at xbar if there exists a real number L such that for every epsilon>0, there exists a delta>0 with |f(x)-L|<epsilon for all x in D for which 0<|x-xbar|<delta. In this case we write lim x-->xbar f(x)=L"

Please clarify if I misunderstood you, I'm just really trying to get this figured out because I thought I had until Thursday but apparently my HW is due tomorrow this week. Also, please see my edited post above. I re-read it and needed to reword some parts of it. It still could surely be incorrect but that is what I got based on what I'm reading in the text and the direction the professor push me.

[DesEsseintes]

I don't think this works.

The epsilon-delta definition is only for sequences, i.e. for any {xn} such that xn->X, you have that for every epsilon there exists delta_epsilon, bla bla bla...

However, while this is true for any sequence {xn} in D converging to X, there could be an infinity of sequences in D converging to X and the delta_epsilon could, a priori, be different for each and not necessarily boundable.

I would use a proof by contradiction.
Suppose that there does not exist any d such that 0<|x-X|<d (x in D) => f(x)>0.

This means that, for any d>0 there exists some x_d in D such that 0<|x_d-X|<d and f(x_d)<=0.

Define the sequence {xn} by xn=x_(1/n) and derive a contradiction.

Edited to emphasize where the weak spot is, IMO.

Hhhhmmmm. I don't mean to sound like I am attacking you here, but you're completely wrong/confused about analysis in general and need to find a proper text to study before giving out more advice.

ctownballer: A limit point - call it x - of some set D is characterised as follows:

For any e > 0, the 'open-ball' (I prefer the term epsilon neighbourhood) B(x;e) intersects A at points other than x (or, more succinctly, B(x;e) intersect A\{x} is not the empty set for any epsilon greater than zero). Here B(x;e) = {y in R: |x-y|<e}.

Now, in the real numbers, it so happens that an 'open-ball' is just an interval, as you rightly characterise. Your proof should be as follows:

As L > 0, set epsilon (=e) to be L/2 > 0. Then you know there exists a delta (=d) greater than zero such that for all y in D with 0 < |y-x| < d, we have |f(y)-L|<e.
Then note as x is a limit point for D, B(x;d) contains points of D other than x (in fact, it contains infinitely many points of D - a good exercise is to prove this). Furthermore, the points of D it does contain, let us take an arbitrary point y, are such that |f(y)-L|<e, so necessarily f(y) > 0 [because of our choice of e].

Now, open balls in R are actually intervals, so it follows (x-d,x+d) is the interval you require. Can you re-organise this guide into a tightly formatted style?

[JimFromRaleigh1]

Exponential Equations

(3^2x)-12*3(^4)+27=0

Halp? I know 27=3^3

Let y = 3^x: can you write down an equation in terms of y and then use logarithms to solve?

Not sure what you mean.

[DesEsseintes]

Not sure what you mean.

You should get something like:

y^2 = positive stuff, if you remember that (3^(2x)) - if this is what you intended to write - is the same as (3^x)^2. Then take square roots and then logarithms.

[JimFromRaleigh1]

You should get something like:

y^2 = positive stuff, if you remember that (3^(2x)) - if this is what you intended to write - is the same as (3^x)^2. Then take square roots and then logarithms.

Ok, so what am I trying to do? I do not think we are supposed to use Logs yet because this is a 1 section before it is introduced. I want to get the same bases, correct?