You are not logged in.
Something I just noticed about k+63. The same can be said for all odd numbers, no?
Is there any easy way to show the limit of a non-continuous function, that is, besides doing an epsilon-delta proof?
For example, how would you show:
The best I've seen (which isn't saying much) is either the ti-89 or ti-92 when it comes to simplifying. You can download vti and get an os from www.ti.org so that you can have it on your computer. If you wish to do this and need help, send me an email.
I'm sure many have heard this before:
Women take time and money: Women = time * money
However, time is money: time = money
So Women = money * money
But money is the root of all evil: money = √evil
So women = √evil * √evil
∴ women = evil. QED.
Ah, I see now. Guess no calculus is needed.
Ah, I forgot to do the second derivative test and found a maximum instead of a minimum. Then the only critical point you are left with is 0, but 0 isn't positive. So in this situation, you want to find a number closest to this minimum that is in your domain. In this case, it would be:
lim as x->0+ of x, which goes towards 0. Are you sure you didn't mean maximum? That's how this question is usually given.
John, I'm curious how you came up with that. Think you could show more work? I can't seem to find where that equation is coming from.
Yea, so expand on that fact:
a^2 + b^2 = 200
So a*b = √(200 - b^2) * b
So you want to find the minimum of that function. Take a derivative, find the critical points, and test each one.
Hint: You should come up with 2 critical points, only one will be within your domain (i.e. positive).
Start with the basics.
y = mx + b
We know this has to pass through the point (x, y) = (1, 2), so lets plug those in:
2 = m + b, or, m = 2 - b
That didn't seem to get us very far, did it? But at least we have a relationship for m and b, this might come in handy later.
So lets go back. What we want is a general equation for the area of the triangle. Well, we know of the formula 1/2 * base * height. So lets try to find those variables.
The height of the triangle is going to be the y-intercept, b. Easy enough.
The base of the triangle is going to be the x intercept. Since y = mx + b, and y must be 0 (definition of the x-intercept) 0 = mx + b. We want to solve this for x, so that would be x = -b / m.
So the area of the triangle is 1/2 * b * -b/m, or -b^2 / 2m. But wait, isn't this going to be negative? Negative area? Nope, remember the line that we are drawing has a negative slope, so m is negative, making -b^2 / m positive.
So we want to find the least area of the function -b^2 / 2m. Huh, two variables, that's going to be pretty tricky without multi variable calculus. But wait, doesn't m = 2 - b? Told you that would come in handy. So -b^2 / 2 * (2 - b) is the area, or -b^2 / (4 - 2b).
Try to find the minimum for that function. This will tell you what b is, then you can find m because m = 2 - b.
Edit:
And for extra credit, what kind of triangle does this make?
"also (i, -2i) and (-i, 2i), whatever these complex thingys mean, I don't remember."
Every nth degree equation has n solutions. Some solutions may be double roots (i.e. (x-1)(x-1) = 0). The graph may only pass through the x-axis (where y = 0, which would be a solutions) less than n times. When this occurs, you get an imaginary solution, which is what i is.
Ah, I see now. I mostly invented this method on the fly, and the part which I was talking about was the part which I had the most trouble understanding. So I guess I just thought that you would have the same.
Do you have a graphing calculator? If not, you can do all these by hand by just plugging in points.
1. Try graphing 2^x + 0, 2^x + 1, and 2^x - 1
2. 1 * (2^x), 2*(2^x), 4*(2^x)
3. 2^(1*x), 2^(2*x), 2^(4*x)
4. 2^x and 2^-x
Just try different ones, and you should very quickly begin to see a pattern emerge.
would not b / a = (4 / 3)^x be a suitable answer?
Yea, that is quite an equation, I don't believe I've ever had to solve one like it before.
My calculator comes up with 9^x = 16^(ln(3)x / 2*ln(2)) and 12^x = 16^(ln(12)x/4*ln(2))
This makes the equation:
16^x = 16^(ln(3)x / 2*ln(2)) + 16^(ln(12)x/4*ln(2))
That seems to be a step in the right direction, getting a common base. And can anyone derive 9^x = 16^... and 12^x = 16^....?
ryos, what's n?
And where can I look up commands for this code?
ryos, my c++ program agreed with my result to the 4th decimal place, and I'm pretty sure that's no cosmic coincidence. That being said:
"So, as we roll along, our chances of hitting a six on one of the dice doesn't decrease."
This got me at first too. But then I thought about this example:
Keep flipping a coin till you get heads.
Well, the first time I filp it, I have a 1/2 chance of getting heads. The second time I flip it, I have a 1/2 chance of getting heads. So this would be 1/2 + 1/2, no? Does this mean that I have a 100% chance of getting heads if I flip a coin twice? Obviously, that's wrong. Heck, what happends if I flip it three times? I get a 150% chance of getting heads.
The thing is, you _only_ flip a second time when the first time was a bust. How often will that happen? 1/2 the time. So it's 1/2*1/2, or 1/4.
Think about it this way. You are asked to flip a coin two times. What are the possible outcomes?
Flip 1 Flip 2
h h
h t
t h
t t
And all of these are equally weighted. So what's the chance of flipping tails, then heads? 25%.
If you flip a coin till you get heads, what's the chance of flipping heads, and then heads? 0%
Since these are the only two possibilities of getting heads on the 2nd flip, its 25% + 0%, which is 25%.
2^(k+1) - 2^(k-1)
K = 1:
2^2 - 2^0 = 4 - 1 = 3
K = 2:
2^3 - 2^1 = 8 - 2 = 6
K = 3:
2^4 - 2^2 = 16 - 4 = 12
K = 4:
2^5 - 2^3 = 32 - 8 = 24
And so on.
Don't ask me how I got this, I couldn't tell you. Just a bunch of guessing and checking:
Summation of k = 1 to n of 2^(k+1) - 2^(k-1)
It's a standard trig identity yes.
Not a problem. Mostly all you have to do is work backwards:
cos²(5x) - cos²(x) = 1/2(2cos²(5x)-1+1 - (2cos²(x) - 1 + 1) )
= 1/2(cos²(10x) - cos²(2x) )
Since 10x = 6x - (-4x) and 2x = 6x + (-4x):
= sin(6x)sin(-4x)
= -sin(4x)sin(6x)
I normally do a lot more steps than needed, just so others can easily follow along.
http://www.pen.k12.va.us/Div/Winchester/jhhs/math/lessons/trig/ident2.html
sinAsinB = 1/2(cos(A-B) - cos(A+B))
-sinA = sin(-A)
cos(2A) = cos²A - sin²A
-sin(4x) = sin(-4x)
sin(-4x) * sin(6x) = 1/2(cos(-4x - 6x) - cos(-4x + 6x)) = 1/2(cos(-10x) - cos(2x))
1/2(cos(2 * -5x) - cos(2 * x)) = 1/2(cos²(-5x) - sin²(-5x) - ( cos²(x) - sin²(x) ) )
Since sin²(-5x) = 1 - cos²(-5x):
1/2(cos²(-5x) - (1 - cos²(-5x)) - (cos²(x) - (1 - cos²(x)) ) )
1/2( 2cos²(-5x) - 1 -2cos²(x) + 1) = 1/2(2cos²(-5x) - 2cos²(x)) = cos²(-5x) - cos²(x)
And finally, since cos(-A) = cos(A):
cos²(5x) - cos²(x)
Right again.
3/3, well done. They are all correct.
Took a few minutes and made a C++ program to run a little numerical analysis:
Runs: 1000
Hits: 64
Precent: 0.064
Runs: 10000
Hits: 785
Precent: 0.0785
Runs: 100000
Hits: 7566
Precent: 0.07566
Runs: 100000000
Hits: 7478819
Precent: 0.0747882
Source code:
#include <iostream>
#include <time.h>
#include <stdlib.h>
using namespace std;
int main()
{
srand(time(NULL));
const long NUM_RUNS = 100000;
long numHits = 0;
for (long x = 0; x < NUM_RUNS; x++)
{
int die1 = 0, die2 = 0, die3 = 0;
for (int y = 0; y < 3; y++)
{
if (die1 != 6) die1 = rand() % 6 + 1;
if (die2 != 6) die2 = rand() % 6 + 1;
if (die3 != 6) die3 = rand() % 6 + 1;
}
if (die1 == 6 && die2 == 6 && die3 == 6) numHits++;
}
cout << "Runs: " << NUM_RUNS << endl;
cout << "Hits: " << numHits << endl;
cout << "Precent: " << (double)numHits / NUM_RUNS << endl;
return 0;
}