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[Rich Uncle Skeleton]

This one feels like juvenile trickery showing that epsilon < 2epsilon/3 since we can pick epsilon to be whatever we want. But I suppose it works out, because it doesn't work when A > B, and it does when A = B and A < B.

The "for all epsilon" in the definition of the limit is not a juvenile trick. It's the whole point! Maybe my choice of notation is confusing. If you prefer, say A = B + d, and then apply the definition of the limit with, in the traditional notation, epsilon = d/3 > 0.

If the proof doesn't feel right to you, draw a number line and mark A and B with A to the right of B. Then certainly you can draw tiny, disjoint open intervals around those two points. For large enough n, all the A_n will be trapped in the interval around A and all the B_n will be trapped in the one around B (and thus, to the left of the A_n). Hence the contradiction.

Now go back and formalize this: Give a name to the distance between A and B. Decide how tiny those intervals should be. Refer to the rigorous definition of the limit and apply it to show that the terms must lie inside the intervals. If you go through it all carefully, you'll see that what I wrote before is simply a formal version of the picture on the number line.

 

[Government-man]

Can't seem to figure out this question which has to do with limits.

How do I show that

((sin(Nx))/sin(x))^2 --> N^2 when x --> k(pi), k being any integer.

 

[Sanskrit]

Damn, it's hard to understand the english equivalents of mathematical expressions even though I'm at differential equations :/

 

[dinazimmerman]

Can't seem to figure out this question which has to do with limits.

How do I show that

((sin(Nx))/sin(x))^2 --> N^2 when x --> k(pi), k being any integer.

Well, you have to show that the limit of ((sin(Nx))/sin(x))^2 as x approaches k(pi) is equal to N^2, so the problem is definitely related to limits, right?

There probably is a simpler way to do this, but this also works:

1) Use the multiple-angle formula on sin(Nx): http://upload.wikimedia.org/math/3/e/b/3ebc1b1c1345f431d718be7393ff8ac2.png. Let's use j instead of k here since we already are using k.
2) Use l'Hopital's rule on that expression.
3) Verify that the limits of the terms for which 0 =< j =< N-2 are equal to zero (because (sin(k(pi))^(N-j-1) =0 if (N-j-1)=/=0). Verify that the term for which j=N-1 goes to N (because (sin(x))^0 =1). At first, the limit of the term for which j=N is indeterminate (0/0), but use l'Hopital's rule to see that it is also equal to zero.
4) Since the limit of sums is equal to the sum of limits, (sin(Nx))/sin(x) --> (0 + 0 + ... + N +0) = N when x --> k(pi). Since the limit of a product is equal to the product of the limits, (sin(Nx))/sin(x))^2 --> N^2 when x --> k(pi)

 

[Althane]

So I'm in Computer Architecture class, and I need help learning probability and statistics.

Take, for example a problem from my last homework (which I did miserably on):

"We have a 1-bit predictor with 2 stages, T (the last branch was taken) and NT (The last branch was not taken). All branches have a uniform probability of P to be taken. Assuming our predictor starts in T, what is the probability that our predictor will be in T after a very long time running the program?"

I have no idea. I said something like "P, since each branch has an equal probability to be taken, it really only matters on what the last branch was".

Was this right? Was it even close?

So the next one, then, was "We have the same 1-bit predictor, except this time it has the probability Q of starting in T. What is the probability that our predictor will be in T after a very long time running the program?"

...

I had no idea where to even START. So, if someone wants to show me a resource where I can bone up on statistics rapidly, I'd be much obliged.

 

[poweld]

So I'm in Computer Architecture class, and I need help learning probability and statistics.

Take, for example a problem from my last homework (which I did miserably on):

"We have a 1-bit predictor with 2 stages, T (the last branch was taken) and NT (The last branch was not taken). All branches have a uniform probability of P to be taken. Assuming our predictor starts in T, what is the probability that our predictor will be in T after a very long time running the program?"

I have no idea. I said something like "P, since each branch has an equal probability to be taken, it really only matters on what the last branch was".

Was this right? Was it even close?

So the next one, then, was "We have the same 1-bit predictor, except this time it has the probability Q of starting in T. What is the probability that our predictor will be in T after a very long time running the program?"

...

I had no idea where to even START. So, if someone wants to show me a resource where I can bone up on statistics rapidly, I'd be much obliged.

You are looking to solve Branch Prediction (http://en.wikipedia.org/wiki/Branch_predictor)

I've never done this particular work, but I have taken a class heavy in state machine logic. I believe since in the first problem the original state is T, then your state machine will begin in the Weakly Taken node. You need to calculate the probability that the state machine will not end up in the Weakly Taken or Strongly Taken nodes over a long period of time.

Given that the probability for both T and NT is P, with an odd number of steps there is a 0.5 chance that you end in T (Strongly Taken) and 0.5 chance you end in NT (Weakly Not Taken). If there is an even number of steps, there is a 0.5 chance of ending in T (Weakly Taken, by incrementing then decrementing, and vice-versa), a 0.25 chance of ending in T (Strongly Taken, double increment), and a 0.25 chance of ending in NT (Strongly Not Taken, double decrement). Combined, this gives you a probability of ending in T of 0.625.

Now, if Q is the probability of beginning in T then we also have to take into account the inverse probability, (1 - Q), of beginning in NT. Therefore, we will take our value for when we start in T and multiply it by the chance that we actually do start in T, and get 0.625Q.

We can take a shortcut for figuring out the odds of ending in T if we begin in NT, because it is the inverse of the odds of ending in T when beginning in T. Therefore, the odds of ending in T when beginning in NT are 0.375(1 - Q), or (0.375 - 0.375Q). The total odds of ending in T for this example is (0.375 + 0.25Q).

 

[zoku88]

^ You picture is actually a 2 bit predictor with 4 stages, instead of the problems 1 bit predictor. Weakly Not Taken and Weakly Taken don't exist.

 

[Althane]

^ You picture is actually a 2 bit predictor with 4 stages, instead of the problems 1 bit predictor. Weakly Not Taken and Weakly Taken don't exist.

Yeah. I have the math on how to do a 2-bit predictor in the book, but not a 1-bit predictor. Which was why I was confuzzled.

 

[poweld]

^ You picture is actually a 2 bit predictor with 4 stages, instead of the problems 1 bit predictor. Weakly Not Taken and Weakly Taken don't exist.

Yeah. I have the math on how to do a 2-bit predictor in the book, but not a 1-bit predictor. Which was why I was confuzzled.

Oh, huh. In that case, I'd have to agree with your original answer - it's always 0.5, but you already said that was an incorrect answer...

 

[Rich Uncle Skeleton]

Well, you have to show that the limit of ((sin(Nx))/sin(x))^2 as x approaches k(pi) is equal to N^2, so the problem is definitely related to limits, right?

There probably is a simpler way to do this, but this also works:

1) Use the multiple-angle formula on sin(Nx): http://upload.wikimedia.org/math/3/e/b/3ebc1b1c1345f431d718be7393ff8ac2.png. Let's use j instead of k here since we already are using k.
2) Use l'Hopital's rule on that expression.
3) Verify that the limits of the terms for which 0 =< j =< N-2 are equal to zero (because (sin(k(pi))^(N-j-1) =0 if (N-j-1)=/=0). Verify that the term for which j=N-1 goes to N (because (sin(x))^0 =1). At first, the limit of the term for which j=N is indeterminate (0/0), but use l'Hopital's rule to see that it is also equal to zero.
4) Since the limit of sums is equal to the sum of limits, (sin(Nx))/sin(x) --> (0 + 0 + ... + N +0) = N when x --> k(pi). Since the limit of a product is equal to the product of the limits, (sin(Nx))/sin(x))^2 --> N^2 when x --> k(pi)

Why not simply apply l'Hopital to sin(Nx)/sin(x)? You get Ncos(Nx)/cos(x), which tends to either N or -N depending on the parity of N and k. Either way the square of the ratio will tend to N^2.

 

[dinazimmerman]

Why not simply apply l'Hopital to sin(Nx)/sin(x)? You get Ncos(Nx)/cos(x), which tends to either N or -N depending on the parity of N and k. Either way the square of the ratio will tend to N^2.

Because I always like taking the complicated route! =(
Your method is infinitely simpler, I just didn't see it at the time.

 

[Althane]

Oh, huh. In that case, I'd have to agree with your original answer - it's always 0.5, but you already said that was an incorrect answer...

It'd be P, actually, whatever P is. I just wasn't sure that it'd be an incorrect answer.

And I had no idea how changing the starting point would change the effect of the long-running program.

So yeah. <<;;

 

[poweld]

It'd be P, actually, whatever P is. I just wasn't sure that it'd be an incorrect answer.

And I had no idea how changing the starting point would change the effect of the long-running program.

So yeah. <<;;

Right, I meant "always" within the scope of the question - from t=0 until t=infinity.

Maybe ask your prof or a TA.

 

[DEAD RABBIT]

I got a problem with something involving infinities.

Look at the following:

x = tan(a)/tan(b)

Now a & b are equal to pi/2. I thought, since the tangent functions grow to infinity at the same rate because a & b are just different variables limiting to the same value, x must equal 1. I had my chemistry professor say this:

tan(b) * x = tan(a) -> infinity * x = infinity -> x can be anything, because any number * infinity = infinity .......?

I protested, but couldn't find a convincing critique at the time. Now I am quite fucking pissed at this (it was homework). When dealing with infinities, there is no number that is equal to infinity. It's a continuous never ending operation of a function that has an unbounded value. I think I am right to say that x must 1. Is the chemistry professor's argument correct in the slightest?

 

[Cereal KiIIer]

I don't remember anything about trigonometry. COS SIN TAN are now stranger to me and I need to remember this ASAP for a job (That's what 5 years of humanities does to you).

Any good tutorial online?

 

[funk0ar]

I don't remember anything about trigonometry. COS SIN TAN are now stranger to me and I need to remember this ASAP for a job (That's what 5 years of humanities does to you).

Any good tutorial online?

http://www.khanacademy.org/


good for all levels of math

 

[Chris R]

Soh-Cah-Toa!

 

[Door2Dawn]

So I'm having trouble with a problem involving graphing an equation that has a absolute value in it.

For example: x = l y l - 2

I'm not really sure what I'm doing wrong. If you plug a number, say zero, into x, do you move y to the left to get y = -2?

 

[poweld]

So I'm having trouble with a problem involving graphing an equation that has a absolute value in it.

For example: x = l y l - 2

I'm not really sure what I'm doing wrong. If you plug a number, say zero, into x, do you move y to the left to get y = -2?

The graph will look like y = x + 2, but mirrored over the x axis at y = 0. So it will look like a < with the vertex landing on (-2, 0)

 

[Lonely1]

So I'm having trouble with a problem involving graphing an equation that has a absolute value in it.

For example: x = l y l - 2

I'm not really sure what I'm doing wrong. If you plug a number, say zero, into x, do you move y to the left to get y = -2?

First you graph when y>=0 and then when y<0.

 

[survivor]

Doing some linear algebra and I'm wondering if anyone can help me tell the difference between these 2 questions.

Say I have 3 vectors, v1, v2, and w and they are all in R3. And I'm asked these 2 questions

1) Does w belong to {v1, v2}
2) Does w belong to the subspace spanned by set {v1, v2}

Anyone mind telling me what's the difference between the 2 questions? I'm guessing for 2 I need to prove w is part of the span which I know how to do. But for 1 I got no idea what I'm asked to do.

 

[hemtae]

related rates problem, normally I'm pretty good with them but this has been giving me some trouble

A water trough is 10 m long and has a cross-section in the shape of an isosceles trapezoid that is 30 cm wide at the bottom, 90 cm wide at the top, and has height 60 cm. If the trough is being filled with water at the rate of 0.2 m^3/min how fast is the water level rising when the water is 30 cm deep?

 

[zoku88]

So I'm having trouble with a problem involving graphing an equation that has a absolute value in it.

For example: x = l y l - 2

I'm not really sure what I'm doing wrong. If you plug a number, say zero, into x, do you move y to the left to get y = -2?

Graph, x=|y|and then shift the whole group left by two. The domain of the equation is from -2 to infinity.

It the same graph as x+2 = |y|, so if you plug 0 into x, you get 2 for |y|, and thus y=2 or -2. You'll basically get a sideways V when you're finished.

 

[zoku88]

Doing some linear algebra and I'm wondering if anyone can help me tell the difference between these 2 questions.

Say I have 3 vectors, v1, v2, and w and they are all in R3. And I'm asked these 2 questions

1) Does w belong to {v1, v2}
2) Does w belong to the subspace spanned by set {v1, v2}

Anyone mind telling me what's the difference between the 2 questions? I'm guessing for 2 I need to prove w is part of the span which I know how to do. But for 1 I got no idea what I'm asked to do.

For the second question, it would have to be, does w = a_1 * v_1 + b_2 * v_2.

For the first, I would guess, given the notation, it just means does w = v_2 or v_1... ? But I'm not sure...

 

[hitsugi]

New favorite thread on GAF. thanks in advance for all of the work that you guys have put into this! I recently discovered khan academy though my phone and have a tooon of catching up and relearning to do or my major (comp sci) so this thread is pretty much perfect

 

[survivor]

For the second question, it would have to be, does w = a_1 * v_1 + b_2 * v_2.

For the first, I would guess, given the notation, it just means does w = v_2 or v_1... ? But I'm not sure...

Ya for the first question I'm just gonna compare w with v1 and v2 since that's the only thing that seems to make sense. Thanks for the help.

 

[Therion]

x = tan(a)/tan(b)

The biggest problem I see is that tan isn't defined at pi/2. It's not infinity; it just doesn't exist.

If you're talking about limits as a,b->pi/2...well, those don't exist either. You need to use one-sided limits, which will approach positive or negative infinity, depending on the direction.

If you specifically have a=b->pi/2 from one side, then, yes, x->1.

 

[Rich Uncle Skeleton]

I got a problem with something involving infinities.

Look at the following:

x = tan(a)/tan(b)

Now a & b are equal to pi/2. I thought, since the tangent functions grow to infinity at the same rate because a & b are just different variables limiting to the same value, x must equal 1. I had my chemistry professor say this:

tan(b) * x = tan(a) -> infinity * x = infinity -> x can be anything, because any number * infinity = infinity .......?

I protested, but couldn't find a convincing critique at the time. Now I am quite fucking pissed at this (it was homework). When dealing with infinities, there is no number that is equal to infinity. It's a continuous never ending operation of a function that has an unbounded value. I think I am right to say that x must 1. Is the chemistry professor's argument correct in the slightest?

The problem is he's asking you about tan(pi/2)/tan(pi/2) and you're telling him about the LIMIT as a -> pi/2 of tan(a)/tan(a). If you drew a graph of tan(x)/tan(x), it'd be equal to one everywhere except where it's not defined. At those values of x, the limit will exist and be one, but the expression itself is undefined for the reasons your professor explained.

Ya for the first question I'm just gonna compare w with v1 and v2 since that's the only thing that seems to make sense. Thanks for the help.

Yes, this is what he's asking. {v_1,v_2} is just set notation; it means the set containing the elements v_1 and v_2. To ask whether w belongs to that set is simply to ask whether it's one of the two elements listed.

related rates problem, normally I'm pretty good with them but this has been giving me some trouble

A water trough is 10 m long and has a cross-section in the shape of an isosceles trapezoid that is 30 cm wide at the bottom, 90 cm wide at the top, and has height 60 cm. If the trough is being filled with water at the rate of 0.2 m^3/min how fast is the water level rising when the water is 30 cm deep?

You need to relate volume, V, and water level, h. You know that the volume of water in the tank is just the length of the trough times the area of a cross section of the water. To find this draw a picture of the cross-section of the trough (trapezoid with height .9m), mark some water level, and shade in the area beneath that water level. That smaller trapezoid represents the cross-section of the water in the trough. You have to find the length of the top side of that trapezoid so you can find its area. Since the other sides of the trapezoid are lines the length of the top sides changes linearly, so since it's .3 at height 0 and .6 at height .9, it must be .3 + h/3. By the area formula for trapezoids, the cross-sections have area h(.3 + .3 + h/3)/2, so

V = 10h(.6+h/3)/2 = 3h + 5h^2/3.

By the chain rule, dV/dt = dV/dh dh/dt = (3 + 10h/3)dh/dt

Plugging in the values you have for dV/dt and h, you can solve for dh/dt.

 

[Sealda]

I am to decide the ellipse of this equation!

2x^2+(y-1)^2=1


My answer is

x^2+((Y-1)/Sqrt2)=1/2

which seems kinda off...


The ellipse is supposed to be centered around (0,1) and then have the widths 1/sqrt2 and 1. ...


Never mind i actually found someone else on teh internet asking about the exact same question. Apparently

2X^2 is written as (x/(1/sqrt2)^2

 

[Tornado Condor]

It's been a year and a half since i took calculus 2 so can someone help me integrate this?


Integral (x / ((L - d - x)^2)) dx. L and d are constants

Thanks!

edit: got the answer

 

[Minamu]

Would this plae be the best place to ask about polygon triangulation? I'm supposed to explain for an exam why it's not always obvious how to (effectively I assume) triangulate a polygonal shape. I can't seem to find any clear-cut answers on google regarding this :S Is it simply because there are usually so many vertices to start triangulating from, which make it hard to see how to draw the fewest triangles possible? Seems a bit of a pointless reason since it's not done manually by an artist in 3D graphics anyway

 

[harriet the spy]

Better Explained is a supurb site that tries to break out of the rote bullshit and actually explains what's going on.

Favorite one is the one about complex numbers, for example.

And I found this excellent article on that site.

http://betterexplained.com/articles/a-friendly-chat-about-whether-0-999-1/
this is pretty horrible, though.

 

[Father_Mike]

How do I do an inductive proof for this recurrence relation.

T = T(n/2) + n n > 1
= 3 elsewise

where n = 2^k


I think the answer is nlg2n + 3 but I can't figure out the proof and feel dumb.

 

[close to the edge]

How do I do an inductive proof for this recurrence relation.

T = T(n/2) + n n > 1
= 3 elsewise

where n = 2^k


I think the answer is nlg2n + 3 but I can't figure out the proof and feel dumb.

What do you mean with "where n = 2^k"? k doesn't show up anywhere else.

 

[Mudkips]

How do I do an inductive proof for this recurrence relation.

T = T(n/2) + n n > 1
= 3 elsewise

where n = 2^k


I think the answer is nlg2n + 3 but I can't figure out the proof and feel dumb.

What are you asking? What you have there is a function definition - nothing more.
T = something when n > 1, otherwise it's 3.
Also, n = 2^k, where k is some integer.


I think you're supposed to solve it so it isn't recursive, then prove it inductively forward. Assuming k is a non-negative integer, we can simply state that T = 2n + 1.

Why?

T(0) = Undefined (since k is a non-negative integer, n is 1, 2, 4, 8, 16, etc.)
T(1) = 3
T(2) = 5
T(4) = 9
T(8) = 17

If you expand T(8) out you get:
T(8)
= T(4) + 8
= T(2) + 4 + 8
= T(1) + 2 + 4 + 8
= 3 + 2 + 4 + 8

As you can see, the last expanded terms will always be +2 and +3.
You should recognize the is sequence as being similar to
1 + 2 + 4 + 8 + 16 + 32 ... + N = 2N - 1
So
3 + 2 + 4 + 8 + ... + N = 2N + 1.

Thus, it's looking pretty obvious that T = 2n + 1


Now we need to prove it.

T(n/2) + n = 2n + 1 (obeying the same restrictions on n and k).

Assume it is valid for some k (and we've shown that it is, for up to k = 3, yielding n = 8).
Now show it it is valid for k+1.

Substitute 2^(k+1) for n
T([2^(k+1)]/2) + 2^(k+1)
= T(2^k) + 2^(k+1)

From proving it valid for some k (we did up to 8), we can state that the T(2^k) term does in fact equal 2(2^k) + 1.

= 2(2^k) + 1 + 2^(k+1)

Since 2(2^k) = 2^(k+1), we have

= 2^(k+1) + 1 + 2^(k+1)

Combine those bad boys and you get

= 2 (2^(k+1)) + 1

Since by definition 2^(k+1) is our n in this inductive step...

T = 2n + 1 where n > 1, 3 otherwise, with n = 2^k+1, and k is some non-negative integer.

You've proven it for some k, and inductively for k+1. (For the lost and curious, inductive proofs mean you can apply the proof of a computed instance of k (I proved it for k=3) to prove it for k+1, and then apply that to prove for k+2, etc. etc., effectively proving it for all possible k.)

Thus indeedy:
T = 2n + 1 where n > 1, 3 otherwise, with n = 2^k, and k is some non-negative integer.

Edited in bolding and further explanation.

 

[Father_Mike]

What are you asking? What you have there is a function definition - nothing more.
T = something when n > 1, otherwise it's 3.
Also, n = 2^k, where k is some integer.


I think you're supposed to solve it so it isn't recursive, then prove it inductively forward. Assuming k is a non-negative integer, we can simply state that T = 2n + 1.

Why?

T(0) = Undefined (since k is a non-negative integer, n is 1, 2, 4, 8, 16, etc.)
T(1) = 3
T(2) = 5
T(4) = 9
T(8) = 17

If you expand T(8) out you get:
T(8)
= T(4) + 8
= T(2) + 4 + 8
= T(1) + 2 + 4 + 8
= 3 + 2 + 4 + 8

As you can see, the last expanded terms will always be +2 and +3.
You should recognize the is sequence as being similar to
1 + 2 + 4 + 8 + 16 + 32 ... + N = 2N - 1
So
3 + 2 + 4 + 8 + ... + N = 2N + 1.

Thus, it's looking pretty obvious that T = 2n + 1


Now we need to prove it.

T(n/2) + n = 2n + 1 (obeying the same restrictions on n and k).

Assume it is valid for some k (and we've shown that it is, for up to k = 3, yielding n = 8).
Now show it it is valid for k+1.

Substitute 2^(k+1) for n
T([2^(k+1)]/2) + 2^(k+1)
= T(2^k) + 2^(k+1)

From proving it valid for some k (we did up to 8), we can state that the T(2^k) term does in fact equal 2(2^k) + 1.

= 2(2^k) + 1 + 2^(k+1)

Since 2(2^k) = 2^(k+1), we have

= 2^(k+1) + 1 + 2^(k+1)

Combine those bad boys and you get

= 2 (2^(k+1)) + 1

Since by definition 2^(k+1) is our n in this inductive step...

T = 2n + 1 where n > 1, 3 otherwise, with n = 2^k+1, and k is some non-negative integer.

You've proven it for some k, and inductively for k+1. (For the lost and curious, inductive proofs mean you can apply the proof of a computed instance of k (I proved it for k=3) to prove it for k+1, and then apply that to prove for k+2, etc. etc., effectively proving it for all possible k.)

Thus indeedy:
T = 2n + 1 where n > 1, 3 otherwise, with n = 2^k, and k is some non-negative integer.

Edited in bolding and further explanation.

Yes this is right. I got it a few hours ago but thank you anyways.

 

[Ave22]

How do I prove the limit of a general function?

Let n be a natural number, and let c >0. Show lim as x -> c of x^n = c ^n. I'm only allowed to use the definition of a limit of a function, and in the previous exercise I proved that (x^n - c^n)/x-c = x^(n-1) + cx^(n-2) + c^2 x^(n-3) + ... + x^2 c^(n-3) + x c^(n-2) + c^(n-1). So I work through it and end up with |x-c| < epsilon / [x^(n-1) + cx^(n-2) +...]. I let delta = epsilon / [x^(n-1) + cx^(n-2) +...], but then I don't know what to do from here, because delta isn't a constant.

 

[Rich Uncle Skeleton]

How do I prove the limit of a general function?

Let n be a natural number, and let c >0. Show lim as x -> c of x^n = c ^n. I'm only allowed to use the definition of a limit of a function, and in the previous exercise I proved that (x^n - c^n)/x-c = x^(n-1) + cx^(n-2) + c^2 x^(n-3) + ... + x^2 c^(n-3) + x c^(n-2) + c^(n-1). So I work through it and end up with |x-c| < epsilon / [x^(n-1) + cx^(n-2) +...]. I let delta = epsilon / [x^(n-1) + cx^(n-2) +...], but then I don't know what to do from here, because delta isn't a constant.

As long as delta < c, |x-c| < delta implies |x^(n-1) + cx^(n-2) +...| < |2c|^{n-1}n. Since n and c are fixed, we can just take delta = epsilon/|2c|^{n-1}n (actually, you want delta to be the minimum of this with c, so that what I said above will be true). Then |x-c| < delta implies |x^n-c^n| = |x^(n-1) + cx^(n-2) +...||x-c| < |2c|^{n-1}n delta = epsilon.

In words, the quantity |x^{n-1}+cx^{n-2}+...| is bounded above for x close to c, so even though the delta you defined isn't a constant, you can replace the non-constant part (the denominator) with this bound and then you're ok.

 

[Ave22]

As long as delta < c, |x-c| < delta implies |x^(n-1) + cx^(n-2) +...| < |2c|^{n-1}n. Since n and c are fixed, we can just take delta = epsilon/|2c|^{n-1}n (actually, you want delta to be the minimum of this with c, so that what I said above will be true). Then |x-c| < delta implies |x^n-c^n| = |x^(n-1) + cx^(n-2) +...||x-c| < |2c|^{n-1}n delta = epsilon.

In words, the quantity |x^{n-1}+cx^{n-2}+...| is bounded above for x close to c, so even though the delta you defined isn't a constant, you can replace the non-constant part (the denominator) with this bound and then you're ok.

Ah, I understand the idea now. I've been working on this for a couple hours now, but I'm struggling to verify that |x^(n-1) + cx^(n-2) +...| < |2c|^{n-1}n using the fact that |x-c| < c.

I appreciate the help.

 

[Rich Uncle Skeleton]

Ah, I understand the idea now. I've been working on this for a couple hours now, but I'm struggling to verify that |x^(n-1) + cx^(n-2) +...| < |2c|^{n-1}n using the fact that |x-c| < c.

I appreciate the help.

Hint for the bolded: You need to make ample use of |a| - |b| &#8804; |a+b| &#8804; |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.

 

[Lagspike_exe]

I need help with 2 problems. One is easier and one is a bit tougher.

1) Basis of a prism is a rhombus with d1 = 18, d2 = 24, while the diagonal of the side is 39 (I don't know the exact word for this one. Imagine a line connecting A1 to B2 in a prism).
Calculate P (P = 2B + M in prism).

2) A 4 side pyramid with a basic edge of 5&#8730;2 cm and s = 13 (s is the line that connect A, B, C or D with the peak of the pyramid). Calculate the edge of the cube that is written inside that pyramid, so it's 4 upper vertexes are located at the side edges of the pyramid.

 

[Feep]

I need help with 2 problems. One is easier and one is a bit tougher.

1) Basis of a prism is a rhombus with d1 = 18, d2 = 24, while the diagonal of the side is 39 (I don't know the exact word for this one. Imagine a line connecting A1 to B2 in a prism).
Calculate P (P = 2B + M in prism).

2) A 4 side pyramid with a basic edge of 5&#8730;2 cm and s = 13 (s is the line that connect A, B, C or D with the peak of the pyramid). Calculate the edge of the cube that is written inside that pyramid, so it's 4 upper vertexes are located at the side edges of the pyramid.

I can't help you with 1), because I have no idea what you're talking about.

2) It helps to visualize this as a two-dimensional problem. Take a isosceles triangle with base 5&#8730;2 and the others sides 13. We need to draw a square in that triangle. So, we need to figure out when the horizontal distance between the two sides is equal to the height.

Looking at only the left-half of the triangle, we can find the height at a given point from the horizontal position along the base (x) with y = (13/(2.5&#8730;2))x. We're looking for the point where y = 2 (2.5&#8730;2 - x), as we want the height to be twice the width (since we cut the triangle in half). Solving this pair of equations gives us x = 1.2455 and y = 4.58. So, the length of a side of this square is 4.58.

This is also the length of the side of the cube, in three dimensions. The volume of this cube is 4.58^3 = 96.066.

 

[paparazzo]

Stupid question: how do I input a negative on my TI-84 calculator? I know how to get a negative exponent but I can't for the life of me figure out how to do a negative and simply using minus doesn't work and yields an error.

 

[dabig2]

Stupid question: how do I input a negative on my TI-84 calculator? I know how to get a negative exponent but I can't for the life of me figure out how to do a negative and simply using minus doesn't work and yields an error.

On the bottom left corner there should be a (-) that's on the same button as Ans(when using the 2nd fuction). That's the negative button.

 

[paparazzo]

Wow, i feel dumb, should've looked more closely. Thanks.

 

[Lagspike_exe]

I can't help you with 1), because I have no idea what you're talking about.

2) It helps to visualize this as a two-dimensional problem. Take a isosceles triangle with base 5&#8730;2 and the others sides 13. We need to draw a square in that triangle. So, we need to figure out when the horizontal distance between the two sides is equal to the height.

Looking at only the left-half of the triangle, we can find the height at a given point from the horizontal position along the base (x) with y = (13/(2.5&#8730;2))x. We're looking for the point where y = 2 (2.5&#8730;2 - x), as we want the height to be twice the width (since we cut the triangle in half). Solving this pair of equations gives us x = 1.2455 and y = 4.58. So, the length of a side of this square is 4.58.

This is also the length of the side of the cube, in three dimensions. The volume of this cube is 4.58^3 = 96.066.

Thank you.

 

[Ydahs]

Just want to check if my logic is right:

find y(x): dy/dx=(y/3)-(3/y)

(expansion) dy/dx = (y^2 -4)/2y

(transpose to integrate with respect to y and x) integral( 2y/(y^2 -4) ) dy = integral(1) dx

let u = y^2, therefore du/dy = 2y

integral(1/(u-4)) du = x + C

log|y^2 -4| = x

(rearrange) y = sqrt( e^x + 4)

Now with other examples similar to this, the Constant from the integral seems to disappear and be replaced by an A before the e^x. Does that apply in this case and if so, why?

 

[hemtae]

Just want to check if my logic is right:

find y(x): dy/dx=(y/3)-(3/y)

(expansion) dy/dx = (y^2 -4)/2y

(transpose to integrate with respect to y and x) integral( 2y/(y^2 -4) ) dy = integral(1) dx

let u = y^2, therefore du/dy = 2y

integral(1/(u-4)) du = x + C

log|y^2 -4| = x

(rearrange) y = sqrt( e^x + 4)

Now with other examples similar to this, the Constant from the integral seems to disappear and be replaced by an A before the e^x. Does that apply in this case and if so, why?

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

 

[dinazimmerman]

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. =(

 

[dinazimmerman]

Nobody has a suggestion at least?