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[Feep]

So I've tried to get a hold of my math instructor a few times here in order to be prepped for my test Saturday. Basically I'm having issues knowing how to properly plug in numbers into a D=R*T formula.

Here's an example problem of which I'm having issues with:

Carol and Richard are travelling North in separate cars on the same highway. Carol is travelling at 65 miles per hour, Richard is travelling at 70 miles per hour. Carol passes Exit 102 at 1:30PM. Richard passes the same exit at 1:45PM. At what time will Richard catch up to Carol?

Another example would be:

In practicing maneuvers, two fighter jets fly toward each other. One flies east at 582 miles per hour and the other flies west at 625 miles per hour. If the two planes are 22 miles apart, how much time will it take for them to meet?


Once I can do these types of problems, then I'm prepared for my test. Thanks to anyone who can help out.

I'll repost my answers from your original thread:

In any case, in fifteen minutes, how much of a lead does Carol have? 65 miles per hour * .25 hours = 16.25 miles. The relative difference between the two speeds is 5 miles per hour, so this difference is made up in 16.25 / 5 = 3.25 hours.

Second part, relative rates, just combine the two. Their distance is closing at a combined rate of 582 + 625 = 1207 mph. 22 miles / 1207 mph = 0.018 hours, or 1.094 minutes.

 

[zlatko]

Thank you guys a bunch! Now to crack on some other problems myself and see if I'm doing it right on my own.

I don't know what I'd do without GAF some days.

 

[SniperHunter]

need some quick help, what's the base function of this? f(x) = -3sin2(x+pi/4)+1

I need to transform it

 

[Ya no]

Hey GAF, I need help with a calc problem that's driving me crazy...

What's the derivative of (sin(pi*x) + cos(pi*y))^9 = 13

I can get started by doing 9(sin(pi*x) + cos(pi*y))^8 and after that I know I have to take the derivative of (sin(pi*x) + cos(pi*y) [[I think?]] and then after tak the derivatives of pi*x and pi*y??

Anyways, after doing that first part, I'm getting stuck. Any help greatly appreciated!!

 

[Partial Gamification]

need some quick help, what's the base function of this? f(x) = -3sin2(x+pi/4)+1

I need to transform it

What sort of transform? I see the function sin(x) as the "building block" of that function, but I'm not sure what math this is needed for. The given function has the y-intercept at negative one, and the sine function is shifted left pi-over-four units, flipped, and "streched" by three. Apologies if this is way off from what you need.

Hey GAF, I need help with a calc problem that's driving me crazy...

What's the derivative of (sin(pi*x) + cos(pi*y))^9 = 13

I can get started by doing 9(sin(pi*x) + cos(pi*y))^8 and after that I know I have to take the derivative of (sin(pi*x) + cos(pi*y) [[I think?]] and then after tak the derivatives of pi*x and pi*y??

Anyways, after doing that first part, I'm getting stuck. Any help greatly appreciated!!

[edit:] Third equation should not be a partial derivative.

 

[JustProgress]

Hey guys, I have a question from my Discrete Math class:

"There are 999,999 natural numbers less than one million. For 1 <= K <= 6, determine how many of these have k distinct digits in their decimal representation. Leading zeroes count; treat 111 as 000111 and count it for k = 2."

The way I did it was when k=1 you have 9 choices for the first number, then 1 choice for all the others so its 9 choices. When k=2 you have 10 choices for the first number (can now use 0) and 9 choices for the second number, then 2 choices for the 4 remaning digits so it's 10*9*2^4. It goes in a pattern like this: k=3 10*9*8*2^3, etc.

Can anyone tell me if I did this right?

 

[Partial Gamification]

Hey guys, I have a question from my Discrete Math class:

"There are 999,999 natural numbers less than one million. For 1 <= K <= 6, determine how many of these have k distinct digits in their decimal representation. Leading zeroes count; treat 111 as 000111 and count it for k = 2."

The way I did it was when k=1 you have 9 choices for the first number, then 1 choice for all the others so its 9 choices. When k=2 you have 10 choices for the first number (can now use 0) and 9 choices for the second number, then 2 choices for the 4 remaning digits so it's 10*9*2^4. It goes in a pattern like this: k=3 10*9*8*2^3, etc.

Can anyone tell me if I did this right?

I think that you have the right idea, but I'm not 100% on my approach either:
k=1: nine choices (can't have 000000)
k=2: ten choices, nine choices, two choices, two choices, two choices, two choices
k=3: ten choices, nine choices, eight choices, three choices, three choices, three choices
k=4: 10*9*8*7*4*4
k=5: 10*9*8*7*6*5 (last digit: five selected to choose from)
k=6: 10*9*8*7*6*5 (last digit: five distinct remaining to select from)
Add up the cases for final answer.
I hope this helps, or at least causes some fresh eyes to have a look.

 

[cpp_is_king]

Sum[k=1; k=6] 10! * k^(6 - k) / (10-k)!

this is basically just a generalization of the 6 explicit cases you've already worked out.

10! / (10 - k)! gives you 10*9*..., for a total of k elements in the product. So for k=1 it's 10, for k=2 it's 10*9, for k=3 it's 10*9*8, etc.

k^(6-k) gives you k*k*k*...*k, for a total of 6-k terms in the product. So when k=1, it gives you 1*1*1*1*1, when k=2 it gives you 2*2*2*2, when k=3 it gives you 3*3*3, etc.

 

[cpp_is_king]

Add up the cases for final answer.

Err, wait a second. Is the question asking for the sum of the cases? I thought so too at first, hence my answer involving a sum. After thinking about it some more, the sum is trivial, it's 999,999. I mean think about it. You've got 999,999 natural numbers less than 1,000,000. Obviously, for each of these numbers N, there is some number k such that N has k distinct digits. Furthermore, for any two numbers k and k', if N has both k distinct digits and k' distinct digits, then k = k'. So basically, every single number from 1 to 999,999 corresponds to exactly one number k. Also known as, you count every number exactly once.

So the sum of all cases k = 999,999. This leads to a rather interesting identity, namely that

x^p = x! * Sum[k=1; k=p] k^(p - k) / (x-k)!

For 0 <= p <= x. I wonder if this identity could be somehow useful in simplifying other complicated expressions, although it does seem rather specialized.

Here, p is the number of digits of the initial number (e.g. p = 6 for the case of 999,999), x is the numeric base (10 for this example).

But this means you could then apply the solution to any numeric base. Unfortunately I can't find an obvious way to generalize this to numbers with an arbitrary number of digits (in other words, removing the restriction that p <= x) without either writing it as a piecewise function or using the min() function, so a small amount of utility of this identity is lost. But if someone could find a way, that might be interesting.

 

[My Name Is Luke]

Yeah, I'm "learning" about logs and logarithmic functions in my College Algebra class but nothing about this is easy or sticking with me.

 

[Partial Gamification]

Err, wait a second. Is the question asking for the sum of the cases? I thought so too at first, hence my answer involving a sum. After thinking about it some more, the sum is trivial, it's 999,999. I mean think about it. You've got 999,999 natural numbers less than 1,000,000. Obviously, for each of these numbers N, there is some number k such that N has k distinct digits. Furthermore, for any two numbers k and k', if N has both k distinct digits and k' distinct digits, then k = k'. So basically, every single number from 1 to 999,999 corresponds to exactly one number k. Also known as, you count every number exactly once.

So the sum of all cases k = 999,999. This leads to a rather interesting identity, namely that

x^p = x! * Sum[k=1; k=p] k^(p - k) / (x-k)!

For 0 <= p <= x. I wonder if this identity could be somehow useful in simplifying other complicated expressions, although it does seem rather specialized.

Here, p is the number of digits of the initial number (e.g. p = 6 for the case of 999,999), x is the numeric base (10 for this example).

But this means you could then apply the solution to any numeric base. Unfortunately I can't find an obvious way to generalize this to numbers with an arbitrary number of digits (in other words, removing the restriction that p <= x) without either writing it as a piecewise function or using the min() function, so a small amount of utility of this identity is lost. But if someone could find a way, that might be interesting.

I don't know if I am reading this right, but for values of n = {111111,222222,...,666666} -> k=1 and therefore the sum will not be 999,999 for k={1,2,3,4,5,6}

Maybe I'm misreading what you have stated, but k=999,999 would be that many distinct digits, this would only occur in modulo greater than 999,999. Even so, is there an x! missing from the numerator in the given sum? I like your thinking on this and would like to better understand what you are getting at.

Yeah, I'm "learning" about logs and logarithmic functions in my College Algebra class but nothing about this is easy or sticking with me.

It is just practice and familiarity with the properties of this function. It helps me to remeber that it is the inverse of the exponential function but your probably at the point when they are just "torturing" you with exercises to drill the procedures into the fabric of your being. Maybe this will help, if they are giving you the slow reveal:

There is a really interesting history of these functions, youtube certainly has some good videos but I don't know of any off-hand.

 

[cpp_is_king]

I don't know if I am reading this right, but for values of n = {111111,222222,...,666666} -> k=1 and therefore the sum will not be 999,999 for k={1,2,3,4,5,6}

Maybe I'm misreading what you have stated, but k=999,999 would be that many distinct digits, this would only occur in modulo greater than 999,999. Even so, is there an x! missing from the numerator in the given sum? I like your thinking on this and would like to better understand what you are getting at.

Basically what I am saying is that, suppose you pick a number N in the interval [0, 10^6), and assume leading 0s are significant. Then N either has 1, 2, 3, 4, 5, or 6 distinct digits. So for each number N, there is exactly 1 k such that N has k distinct digits. Obviously such a number cannot have less than 1 distinct digit, nor can a number have more than 6 distinct digits. So we know that all numbers from 0 - 999,999 have an associated k.

On the other hand, for each number k from 1 to 6, you can find 1 or more numbers N such that N has k distinct digits. In other words, all values of k are covered by the numbers from 0 to 999,999. For this you just need to find one example for each k.

k=1 - 111111
k=2 - 111112
k=3 - 111123
k=4 - 111234
k=5 - 112345
k=6 - 123456


So given this information, assuming you come up with some formula f(k,p) such that f(k,p) accurately results in the number of numbers in the interval [0,10^p-1] with k distinct digits (with leading zeros significant), then it follows that

10^p = Sum[k=1;k=p] f(k,p)

The left side is tautologically equal to the total number of numbers in the interval [0,10^p-1]. And the right side is f(k,1) + f(k,2) + f(k,3) + ... + f(k,p). So as per the above discussion, they should be equivalent.

So let's use the case of p=6.

k=1: 10*1*1*1*1*1 = 10
k=2: 10*9*2*2*2*2 = 1440
k=3: 10*9*8*3*3*3 = 21870
k=4: 10*9*8*7*4*4 = 80640
k=5: 10*9*8*7*6*5 = 151200
k=6: 10*9*8*7*6*5 = 151200

So as you said, the sum is 403,930. However, I'm not sure where my logic fails.

The sum itself should be able to be expressed in simple English as "the number of numbers from 0 to 999,999 with any number of distinct digits". Well, that's kind of obviously equal to 1 million, right?

 

[Partial Gamification]

The sum itself should be able to be expressed in simple English as "the number of numbers from 0 to 999,999 with any number of distinct digits". Well, that's kind of obviously equal to 1 million, right?

There is a map that is onto but not one-to-one. I want to call it the set of Natural Numbers to the "tally" set. Because this is not bijective it is not invertible and the "tally" cannot map to the set of Natural Numbers. If you have a k=2 element with zeroes and ones, what six digit number is it?
I think that the problem asks for, and the sum can be considered: “the count of six digit numbers that have either; one, two, three, four, five, or six unique digits, where the order of the digits does not matter.” The combined cardinality of the sets k={1,2,3,4,5,6} does not equal the size of the set of Natural Numbers having a chosen size n. For instance, the element in k=2 : 000001 can also be represented: 001000 010000 … but only counts once in the “tally of Natural Numbers with two distinct digits.”

|N| = |k1|+|k2|+...+|k6|+|"repeat" elements|

 

[squidyj]

This might not be precisely the right thread but I figure there are going to be a lot of people in here who will know what to look for

		\begin{align*}
			T(n) &= T(n - 1) + 1 \\
			&= T\left(\frac{n}{2} - \frac{1}{2}\right) + 1 + 1 \\
			&= T\left(\frac{n}{2} - \frac{1}{2} - 1\right) + 1 + 1 + 1 \\
			&= T\left(\frac{n}{4} - \frac{1}{4} - \frac{1}{2}\right) + 1 + 1 + 1 + 1 \\
			&= T\left(\frac{n}{2^2} - \frac{1}{2^2} - \frac{1}{2^1} - \frac{1}{2^0}\right) + 5 
		\end{align*}
		From this we should be able to guess the general form for the $k$th recursion.
		\begin{align*}
			T(\left\frac{n}{2^i} - \sum_{j=0}^{m} \frac{1}{2^p}\right) + k \\
			\text{Where}\\
			i =& \lfloor\frac{k}{2}\rfloor \\
			m =& \lceil \frac{k}{2}\rceil - 1 \\
			[INDENT][/INDENT]p =& \lfloor\frac{k}{2}\rfloor - \lceil \frac{k}{2}\rceil + j + 1 
----------->\end{align*}

The arrow obviously isn't part of the LaTeX but it's giving me a missing delimiter error on that line, and I'm just not seeing it anywhere in the above section, not bloody anywhere.

 

[harriet the spy]

First line of your second align, i think you want \left( instead of (\left

 

[TimesEunuch]

The arrow obviously isn't part of the LaTeX but it's giving me a missing delimiter error on that line, and I'm just not seeing it anywhere in the above section, not bloody anywhere.

You're missing a bracket after the first \left in the second align environment.

 

[squidyj]

Well... still doesn't work but I fixed that one error.

Ah, got it, thank you

 

[barnone]

Can anyone show me how the following conversion for logarithms takes place?

a^(log_b) = n^(log_b(a))

What's the intuition behind this? I am in college and logarithms have often felt non-intuitive.

 

[Fou-Lu]

Any ideas how to deal with a calc teacher that takes off ridiculous amounts of marks for everything you do that is not how he would do it? Even if it leads to a correct answer.

It's frustrating, I was good with calc until this class, and now I'm getting 61% on a midterm because I lose five marks for factoring out a negative one when finding a limit.

 

[Troll]

2. Give a closed plane curve C with K(kappa, or curvature) > 0 that is not convex.


3. Draw closed plane curves with rotation indices 0, 2, -2, and 3, respectively.

 

[Partial Gamification]

2. Give a closed plane curve C with K(kappa, or curvature) > 0 that is not convex.


3. Draw closed plane curves with rotation indices 0, 2, -2, and 3, respectively.

For the second one, consider how the curvature is equal to the magnitude of the cross product, usually written: |v x a| / |v|^3
where if the degree of the curve is one, linear, then the "velocity" is constant and the "acceleration" is zero. This doesn't seem to qualify as a "curve" though. Alternatively, if v and a are parallel, then their cross-product is also zero. The second derivative of the plane curve will determine curvature.


For the third, the rotation index measures how many times a curve turns with respect to a certain direction, so be sure to indicate the specific direction on the drawing. Examine the following Pascal Snail (or limaçon), note that depending on the direction its rotation index is 2 or -2.

...and for just one more bit on logarithms.
consider: y=a^x has the inverse: y= log_a(x)
examine: 100= 10^2 has the inverse 2 = log(100)
Given log of base "a" of "x," what "a to the y" equals "x?" This just seems confusing.

 

[TimesEunuch]

2. Give a closed plane curve C with K(kappa, or curvature) > 0 that is not convex.

There's a theorem that a simple (i.e. non-self-intersecting) closed plane curve with positive curvature everywhere has to be convex. So the curve you're looking for must self-intersect. The Pascal Snail from Partial Gamification's post above appears to work as an example.

 

[Minamu]

Probability!

A skewed coin has a head probability of 0,9. Another coin is normal with 0,5. By randomly choosing which coin to use, I want the probability to be 0,6 to get heads. So what should the probability be to get the skewed coin?

 

[rjc571]

Probability!

A skewed coin has a head probability of 0,9. Another coin is normal with 0,5. By randomly choosing which coin to use, I want the probability to be 0,6 to get heads. So what should the probability be to get the skewed coin?

Assuming I understand your question correctly...

x(0.9) + (1 - x)(0.5) = 0.6
x(0.4) + 0.5 = 0.6
x(0.4) = 0.1
x = 0.25

 

[Minamu]

Assuming I understand your question correctly...

x(0.9) + (1 - x)(0.5) = 0.6
x(0.4) + 0.5 = 0.6
x(0.4) = 0.1
x = 0.25

Sorry, it's a literal translation of the practice exam. That seems to work! Thanks. There's four of us here, we've been at it for three hours :lol We forgot that you can do the 1-x part...

 

[awm8604]

Probability!

A skewed coin has a head probability of 0,9. Another coin is normal with 0,5. By randomly choosing which coin to use, I want the probability to be 0,6 to get heads. So what should the probability be to get the skewed coin?

C1 - Coin 1
C2 - Coin 2
H - Heads
T - Tails

P(H|C1) - 0.9
P(H|C2) - 0.5

P(H) - 0.6

P(C2|H) - P(HnC2)/P(H) = P(H|C2)P(C2)/P(H) = (0.5)(0.5)/(0.6) = 0.25/0.6 = 0.4167

Edit: Disregard, misunderstood the problem.

 

[bjb]

Calc related question if anyone is willing to help:

http://i.imgur.com/HRW0b.png

 

[Leezard]

Calc related question if anyone is willing to help:

http://i.imgur.com/HRW0b.png

try substitution, s = sqrt(u) might do it.

 

[Partial Gamification]

Calc related question if anyone is willing to help:

http://i.imgur.com/HRW0b.png

I tried to find a recursive definition of F(x) but I don't think I found the right substitution to use, with integration by parts. There is surely a nicer way to get a more explicit solution, but the following is what I came up with, deciding to go with a trignometric substitution after my algebraic gymnatics failed (errors possible):

Might be able to find generating functions, it would be easier to have a frame of reference from what topics you are covering. I had to look up the integral of tan^n(alpha), maybe someone else can see if and where I went wrong, or provide a different approach. Good luck on the test.

edit: I swapped the limits of integration. fixed.
[edit2:] The exponent in the series ought to be -(2n+1)/2

 

[cpp_is_king]

I had this crazy idea where I set s^2 = i * sin(u), where i = Sqrt[-1]. It worked out extremely cleanly, because the clever use of i made the denominator of the fraction cancel out after the substitution. (Later I did another substitution with v = sin(u)).

But for some reason, it leads to the wrong answer. Normally the use of complex numbers as an algebraic aid to simplify expressions is sound and works fine, so I wonder why it doesn't work in this case.

That being said, I punched this into Wolfram Alpha, and based on the answer I strongly suspect that the problem was copied down incorrectly.

 

[cpp_is_king]

Ok, I think I've finally got it. Jesus what a hard integral.

The basic idea is to split the integral into two separate integrals, each of which is easier to integrate than the original. I realize it's highly not obvious how I ended up at this separation, but it really just took a shitload of experimentation. Whenever you've got some power of 2 plus or minus 1, you should start thinking about how to factor the item, or write it as a difference of 2 squares. For example, x^4 + 1 = (x^2+1)^2 - (Sqrt[2] x)^2

That isn't what was needed here, but I think that was the idea that first got me down this path. I also needed a way to reduce the power of 4 to a power of 2, and the power of 2 to a power of 1. One way to do this is to make a substitution, but that doesn't lead anywhere useful here, because if you say u = s^2, then du = 2s ds = 2 sqrt ds. So you introduce a sqrt power, which is very difficult to work with.

Another clever trick, however, is to divide both the numerator and denominator by a power of 2. What this does is give you both a positive and negative power (either a +1/-1 in the case of the numerator, or a +2/-2 in the case of the denominator). This is often easier to factor, and the introduction of negative powers can have surprising results when it comes to choosing a u-substitution.

So that being said, the answer is:

F(x) = Integral[Sqrt[x], 1] [ -atanh(k + 1/k) + atan(k - 1/k) ] / (4 Sqrt[3])

where k = (4/3)^(1/4) s

Note that the first term is the inverse hyperbolic tangent, and the second term is the inverse normal tangent.

Going to show my work later, after I have a chance to write it up in a presentable format.

 

[Partial Gamification]

Ok, I think I've finally got it.
Going to show my work later, after I have a chance to write it up in a presentable format.

Nice, looking forward to viewing it; you are putting my mind to ease. I was chasing a better series, than above, and its closed form expression that I believe was there due to the limits of integration. With repsect to the complex substitution, was the lack of a homomorphism the issue? I'm thinking that the codomain was not equal the domain.

Again, great work!

 

[cpp_is_king]

Alright, here we go. Hand written, sorry about that but it's been a long time since I did any serious math, and it would probably take me a while to remember how to format this in Latex.

 

[cpp_is_king]

I just noticed a slight mistake in the definitions of G(k) and H(k).

I wrote:

G(k) = (k + 1/k)^2 - 1
H(k) = (k - 1/k)^2 - 1

It should be:

G(k) = (k + 1/k)^2 - 2
H(k) = (k - 1/k)^2 - 2

Luckily, this doesn't change much. You still use the same substitution, but you end up with:

G(k): Integral[1 / (u^2 - 2)] du
H(k): Integral[1 / (v^2 + 2)] dv

G(k): - atanh(u / sqrt(2)) / sqrt(2)
H(k): atan(u / sqrt(2)) / sqrt(2)

So this will change some of the computation that follows, but everything else is the same.

It seems I also left out a factor of s in the inside of the atan and atanh. The numerator should have had an s^2 and the denominator should have had an s. After the corrections, I get


atan[ (2s^2 - Sqrt(3)) / (12^(1/4) * Sqrt(2) * s) ]

and the argument to atanh is the same but with a + in the numerator. I may have messed up the details the same, but assuming you do the substitutions correctly you should end up with a correct answer.

 

[Partial Gamification]

^Brilliant insight. Consider me in your fanclub, whether you like it or not.

 

[cpp_is_king]

^Brilliant insight. Consider me in your fanclub, whether you like it or not.

Heh, thanks. Stuff like this makes me miss math. Although I'm not sure whether I'm proud of myself or not, lol. It reminds me of this time in college where I was taking a course in point set topology. It was me (senior undergrad), and then a bunch of PhD students.

The teacher put this problem on the board and said it was homework, I think it was to prove the Cauchy-Schwarz inequality for generalized R^n

I worked on it for probably 20 hours straight, and it just felt impossible. I had finally gotten it down to this really small inequality with just 3 or 4 variables, which I even wrote a program to verify was correct for a few million numbers, but it wouldn't budge. I spent a few hours trying to learn about various theorems related to inequalities which might be useful, and I stumbled across Holder's Inequality.

If 1/p + 1/q = 1, then

And that was exactly what I was looking at. So I practically shat myself, wrote it up, and presented it in class. Apparently he gives that problem every year as a troll, and everyone knows when you get the Cauchy Schwarz inequality problem, not to even bother with it because he's just trying to troll you.

Felt pretty weird after that, lol.

Anyway, long story short, I spent WAY more time than I should have on this, and now I feel like taking a nap

 

[TyphoonStruggle]

My first time posting in the math thread... and I have a few numerical analysis questions:

1. I can't figure out how my instructor solved for roundoff error. He got "~ O(eps_machine) / h^2"

2. I need to compute the relative perturbations in y for x=1+eps(machine) for:
a) relative condition number = 1 / ln(x)
b) absolute condition number = 1 / x

------------------------------------------------------------------------------------------------------

Comments:
The book I have is really bad, and I'm having trouble finding examples for these cases.
I'm assuming that [ relative condition number * eps(machine) ] = the answer for (2a) but I'm not sure if absolute condition number is the same way.

 

[cpp_is_king]

My first time posting in the math thread... and I have a few numerical analysis questions:

1. I can't figure out how my instructor solved for roundoff error. He got "~ O(eps_machine) / h^2"

2. I need to compute the relative perturbations in y for x=1+eps(machine) for:
a) relative condition number = 1 / ln(x)
b) absolute condition number = 1 / x

------------------------------------------------------------------------------------------------------

Comments:
The book I have is really bad, and I'm having trouble finding examples for these cases.
I'm assuming that [ relative condition number * eps(machine) ] = the answer for (2a) but I'm not sure if absolute condition number is the same way.

I've never taken a numerical analysis course and so I'm just taking a wild guess here, but does it help that a common rounding method is to add 0.5 and truncate?

For example, round(3.3) = trunc(3.8) = 3
round(3.6) = trunc(4.1) = 4

Since you apparently already know how to get the truncation error, the rounding error of f(x) should be the truncation error of f(x+0.5)

 

[TyphoonStruggle]

I'm not sure it's possible to relate truncation and roundoff error.

The truncation error is derived from cutting off the Taylor Series of a function.
see http://mathworld.wolfram.com/TaylorSeries.html

The Taylor Series runs infinitely. Then wherever we choose to stop, the tail end that's lost is considered the truncation error.

 

[Partial Gamification]

You might already be aware but that is (basically) the Second Derivative Midipoint formula.

Exact = Approx. + Error

perturbation, in mathematics, method for solving a problem by comparing it with a similar one for which the solution is known. Usually the solution found in this way is only approximate.

 

[Troll]

A. If a(s) is an oval, prove that T''(second derivative of the tangent vector) is parallel to T(the tangent vector) at at least 4 points.

-So far I have that an oval has four vertices and that each vertex represents a critical point of K(curvature). And that the critical points represent the max, min and two inflection points...

B. Find the vertices of r(t)=1-2cos(t).

Hint: Need Four Vertex Theorem for both of them.

 

[big ander]

Hope this is clear despite me typing it in rather than using proper symbols, I'm on a train.

5) Prove the following are true for sequences in the complex numbers [I will denote a sequence by "zn"]:

a) For the sequence {zn}=(z^n), (z^n)–>0 if abs(z)<1, while (z^n) diverges if abs(z)>1.
b) If a sequence zn converges to zero and cn is any sequence in the set of complex numbers, then the sequence zn*cn converges to zero.​

Think I'm just too in my own head once again. For part a I've easily shown using the polar form that (z^n)=(abs(z)^n)(e^(in(theta))). It's obvious to me that abs(z)^n approaches zero since abs(z)<1 but it seems like, to show the final result, I'll actually need part b first. And I'm not sure how to do that.

 

[Partial Gamification]

A. If a(s) is an oval, prove that T''(second derivative of the tangent vector) is parallel to T(the tangent vector) at at least 4 points.

-So far I have that an oval has four vertices and that each vertex represents a critical point of K(curvature). And that the critical points represent the max, min and two inflection points...

B. Find the vertices of r(t)=1-2cos(t).

Hint: Need Four Vertex Theorem for both of them.

I don't know if this is the rigth approach but this is what I see.
Think about the standard equation for an ellipse, (x^2)/a + (y^2)/b = 1
Solving for y: sqrt( 1 - (x^2)/a ) or (-1)sqrt( 1 - (x^2)/a )
The first derivative of either of these functions will give you the tangent vector. Due to the symmetry, showing at least two points where the "velocity is parallel the to the jerk" in one of either halves will satisfy part A. So, what you want to evaluate is the acceleration and the snap at: x = -a, a, 0. If you are dealing with vectors, you would be looking for a zero cross-product between the pairs of derivatives of the position function vector.

position = int(Tangent) = int(velocity) = int(int(accel)) = int(int(int(jerk))) = int(int(int(int(snap))))

The same applies for part B, which is a Pascal Snail, BTW. The vertices are where the graph crosses the major and minor axes (of the "ellipse"), so you just need to consider the shape of the graph.

Think I'm just too in my own head once again. For part a I've easily shown using the polar form that (z^n)=(abs(z)^n)(e^(in(theta))). It's obvious to me that abs(z)^n approaches zero since abs(z)<1 but it seems like, to show the final result, I'll actually need part b first. And I'm not sure how to do that.

I'm just briefly stabbing at this; but for the first part, has it something to do with a notion of the inverse of the sequence and its radius of convergence?

The complex sequence, {C}, has elements a+bi and {C} will converge if and only if the sequences {a} and {b} both converge. I am thinking that you need to show bounded and decreasing but do not have the answer.

 

[Troll]

I don't know if this is the rigth approach but this is what I see.
Think about the standard equation for an ellipse, (x^2)/a + (y^2)/b = 1
Solving for y: sqrt( 1 - (x^2)/a ) or (-1)sqrt( 1 - (x^2)/a )
The first derivative of either of these functions will give you the tangent vector. Due to the symmetry, showing at least two points where the "velocity is parallel the to the jerk" in one of either halves will satisfy part A. So, what you want to evaluate is the acceleration and the snap at: x = -a, a, 0. If you are dealing with vectors, you would be looking for a zero cross-product between the pairs of derivatives of the position function vector.

position = int(Tangent) = int(velocity) = int(int(accel)) = int(int(int(jerk))) = int(int(int(int(snap))))

The same applies for part B, which is a Pascal Snail, BTW. The vertices are where the graph crosses the major and minor axes (of the "ellipse"), so you just need to consider the shape of the graph.

Okay so I took the tangent and second derivative of tangent cross product in the hopes of getting zero and it is not working out. I don't think I quite understand...could you elaborate a bit? Btw I'm talking about part A still.

 

[Partial Gamification]

Okay so I took the tangent and second derivative of tangent cross product in the hopes of getting zero and it is not working out. I don't think I quite understand...could you elaborate a bit? Btw I'm talking about part A still.

Note in the original reply, I forgot to distribute the b and just dropped it (error).
I need to checkout the theorem you listed to see what's the question asking for. The cross-product is implying they are always parallel... I'll try to flush-out what is being asked to identify from the Theorem kater today. Maybe someone else can chime in (with a little complex analysis too).

[Edit:] The image was updated, corrected.

 

[Partial Gamification]

Blaaargh! r(t) out to be < t, sqrt(b-(b*x^2)/a), 0 >
and then r'(t) = <1, ...>

I still need to revisit this later today.

 

[cpp_is_king]

Blaaargh! r(t) out to be < t, sqrt(b-(b*x^2)/a), 0 >
and then r'(t) = <1, ...>

I still need to revisit this later today.

I'm not sure why it says "at at least 4 points". I wrote the equation of an ellipse in parametric form:

F(t) = (X(t), Y(t))

where

X(t) = a cos t
Y(t) = b sin t

Then I computed the first, second and third derivatives of each one:

You can do this yourself, but it's easy to verify by inspection that

X'''(t) = -X'(t)
Y'''(t) = -Y'(t)

Thus,

F'''(t) = -F'(t)

Hence the second derivative of the tangent vector should be parallel to the tangent vector at all points.

It's early in the morning though, and I can't guarantee that I didn't do something retarded.


Edit: Apparently ellipse != oval. And looking into a little more, there are many different kinds of ovals and there is not a clear mathematical definition of the broad term "oval".

I guess I would ask the OP what definition of "oval" he was given.

 

[Partial Gamification]

The image above was updated corrected but I am not sure how to approach this. I have been assuming that the instructions went with oval to mean ellipse. It might be more to due with a circumscribed circle around the oval and relating Theorem 3 [pages 4-5], but I fail to divine how to associate all this information such that we can show "the oval" has anything but parallel T and T''.

Takes me back to the first time I encountered disc versus circle...

 

[Tha Robbertster]

Can someone give me a long (and maybe an easy to follow ) explanation on how I can prove a function has an inverse function?

I know you should prove that the function is injective and surjective but when I look at previous exams, they always take the derivative of the function. The thing I don't understand is what they do after that.

 

[Leezard]

Can someone give me a long (and maybe an easy to follow ) explanation on how I can prove a function has an inverse function?

I know you should prove that the function is injective and surjective but when I look at previous exams, they always take the derivative of the function. The thing I don't understand is what they do after that.

They probably examine whether the function is strictly increasing or strictly decreasing. If the derivative has a zero, there might be trouble finding an inverse.