youngwerther
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nvm
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The Math Help Thread
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youngwerther
Banned
OK what am I doing wrong here
I get the function for linear approximation, and as far as I know I'm just supposed to plug in sqrt (36.1) into that and it's my answer. Apparently that isn't what I'm supposed to do. Anyone know what they are wanting?
I get the function for linear approximation, and as far as I know I'm just supposed to plug in sqrt (36.1) into that and it's my answer. Apparently that isn't what I'm supposed to do. Anyone know what they are wanting?
mugurumakensei
Member
OK what am I doing wrong here
I get the function for linear approximation, and as far as I know I'm just supposed to plug in sqrt (36.1) into that and it's my answer. Apparently that isn't what I'm supposed to do. Anyone know what they are wanting?
Your sqrt of 36.1 is wrong. The sqrt of 36 is 6. 36.1 > 36.1 so SQRT 36.1 > ( SQRT 36 = 6)
youngwerther
Banned
Dear God...
I feel like an idiot.
Thanks guys. Sure I'll be back in a few minutes.
I feel like an idiot.
Thanks guys. Sure I'll be back in a few minutes.
youngwerther
Banned
How do you find intervals over which a function is increasing or decreasing?
Take a function f(x). Find its derivative, f'(x). The intervals on which f'(x) is positive are the intervals on which f(x) is increasing. The intervals on which f'(x) is negative are the intervals on which f(x) is decreasing. Points at which f'(x) equals zero are called critical points.
thequickandthedead
Member
Haven't learned about complex roots of unity. What do you mean by "W_3"?
cpp_is_king
Member
W is the greek letter omega, and with 3 subscript it means "third root of unity". basically, the "nth rooths of unity" are a set of n distinct complex numbers which, when raised to the nth power, yield 1. W_2 = {1, -1} for example
Well if you haven't learned about complex roots of unity then you probably can't use this, but I'll try explaining it anyway if you're interested. (Note: e^i*theta = cos(theta) + i*sin(theta))
If n is a positive integer, the complex number W_n = e^i*2pi/n = cos(2pi/n) + i*sin(2pi/n) is called an nth root of unity since it is a root of the polynomial x^n - 1 (since (W_n)^n = e^i*2pi = cos(2pi) + i*sin(2pi) = 1 + i*0). (Also note that (W_n)^k = cos(2pi*k/n) + i*sin(2pi*k/n), where k is a positive integer, is also an nth root of unity. The polynomial x^n - 1 will always have n distinct complex roots, which are given by (W_n)^k for k = 1, ... , n). If you know that 'a' to the nth power equals b, then multiplying 'a' by an nth root of unity will give you something whose nth power still equals b. So in your example, you had that sigma is a real number satisfying (x-3)^3 - 2 = 0, so sigma - 3 = 2^1/3. But omega and gamma are complex numbers which also satisfies (x-3)^3 - 2 = 0, so omega - 3 has to equal 2^1/3 multiplied by the complex third root of W_3 (and gamma - 3 = 2^1/3 * (W_3)^2).
if you haven't learnt that, you should be able to use the fact that a polynomial with real coefficients only has real roots or pairs of complex conjugated roots. What this means is that since the polynomial only has real coefficients, and you know that β=P+iQ, you instantly gain the knowledge that γ=P-iQ, the complex conjugate of β.
thus, you know that your equation is of the form (x - α
(x -β(x - γ = 0.The complex conjugate roots can be multiplied together to a real form by (x -β
(x - γ = (x - P +iQ)(x - P -iQ) = (x-P)^2 + Q^2.This means that your equation is (x - α
((x-P)^2 + Q^2) = 0.From here, you can expand, multiply, and then just identify coefficients by equaling the symbolic and numeric coefficients for x^2, the ones for x, and the ones for the constant term, respectively. It should be easy to solve, since you get three equations and three unknowns.
youngwerther
Banned
I'm trying to find two critical numbers for a function.
I can't tell where I'm going wrong so could someone look over the algebra and see if they spot anything?
f(x) = 9x + 3x^-1
f'(x) = 9 - 3x^-2
0 = 9 - 3x^-2
-9 = -3x^-2
3 = x^-2
3 = 1/x^2
sqrt 3 = 1/x
1/sqrt3 = x
nvm that is right. Guess it took writing it out here to figure it out.
I can't tell where I'm going wrong so could someone look over the algebra and see if they spot anything?
f(x) = 9x + 3x^-1
f'(x) = 9 - 3x^-2
0 = 9 - 3x^-2
-9 = -3x^-2
3 = x^-2
3 = 1/x^2
sqrt 3 = 1/x
1/sqrt3 = x
nvm that is right. Guess it took writing it out here to figure it out.
opticalmace
Member
Almost, don't forget it could be +- when you take a square root.
So x = +- 1/sqrt(3).
ThanksVision
Member
doing practice probs for a test and this one is stumping me
so I know how to convert it to trig where
s = sin
ds = cos
the denominator will convert to cos and then be cancelled when multiplied by ds. So I get the integral of
(1/6)sin^3 + 3sin
and I can find the integral of those pieces, but I can't figure out how to find the polynomial Q from that. Any ideas on what I'm missing?
so I know how to convert it to trig where
s = sin
ds = cos
the denominator will convert to cos and then be cancelled when multiplied by ds. So I get the integral of
(1/6)sin^3 + 3sin
and I can find the integral of those pieces, but I can't figure out how to find the polynomial Q from that. Any ideas on what I'm missing?
The antiderivative is a polynomial in cos(theta), but theta = sin^-1(s), and cos(sin^-1(s)) = sqrt(1-s^2)doing practice probs for a test and this one is stumping me
so I know how to convert it to trig where
s = sin
ds = cos
the denominator will convert to cos and then be cancelled when multiplied by ds. So I get the integral of
(1/6)sin^3 + 3sin
and I can find the integral of those pieces, but I can't figure out how to find the polynomial Q from that. Any ideas on what I'm missing?
ThanksVision
Member
got my tablet so I could show you my work:
I see that cos(theta) works out to equal sqrt(1-s^2). So I can add the cos(theta) together to get (19/6)cos(theta) - (1/3)cos^3(theta)
then I convert both of them to sqrt like you said:
(19/6)sqrt(1-s^2) - (1/3)sin^2(theta)sqrt(1-s^2)
factor out the sqrt to make it look like the problem statement
sqrt(1-s^2)[(19/6)-(1/3)sin^2(theta)]
that makes the constant 19/6 it seems? But the answer is 28/9. Did I make a dumb mistake somewhere?
got my tablet so I could show you my work:
I see that cos(theta) works out to equal sqrt(1-s^2). So I can add the cos(theta) together to get (19/6)cos(theta) - (1/3)cos^3(theta)
then I convert both of them to sqrt like you said:
(19/6)sqrt(1-s^2) - (1/3)sin^2(theta)sqrt(1-s^2)
factor out the sqrt to make it look like the problem statement
sqrt(1-s^2)[(19/6)-(1/3)sin^2(theta)]
that makes the constant 19/6 it seems? But the answer is 28/9. Did I make a dumb mistake somewhere?
The constant term is 3 + 1/6 -1/18 = 28/9. Since -1/18*cos^3(theta) = -1/18*(1-s^2)*sqrt(1-s^2), so you get an extra constant term of -1/18 when you factor out sqrt(1-s^2).
ThanksVision
Member
oh, duhhh. Thanks for the help!
doing practice probs for a test and this one is stumping me
s
I think I am just dumb here, but how does the integration by parts work as explained in the question's hint, specifically, how does d(1/sin(s)) = (1-s^2)^(-1/2)ds?
sin^-1(s) = arcsin(s), not 1/sin(s)
So 1/sin(s) = sin(s)^(-1). Thanks.
I'm trying to get the determinant of this matrix using Cramer's rule.
2| -2 1 -3 |
| -5 2 3 |
| 3 -4 -4 |
I get:
2(-2| 2 3| - | -5 3 | -3 |-5 2 |
| -4 -4| | 3 -4 | | 3 -4 |
= 2 (-2(-20) - 29 - 3(26))
= 2(40-29-78)
= -134
The answer should be -122.
Can someone see where I'm going wrong? Sorry it looks so clumsy.
2| -2 1 -3 |
| -5 2 3 |
| 3 -4 -4 |
I get:
2(-2| 2 3| - | -5 3 | -3 |-5 2 |
| -4 -4| | 3 -4 | | 3 -4 |
= 2 (-2(-20) - 29 - 3(26))
= 2(40-29-78)
= -134
The answer should be -122.
Can someone see where I'm going wrong? Sorry it looks so clumsy.
gracias.
SailorCosmos
Member
I am studying boolean algebra and am having trouble figuring this question out.
Show algebraically that x'yz + xyz' + x'yz' = x'y + yz'
Could anyone help me?
Show algebraically that x'yz + xyz' + x'yz' = x'y + yz'
Could anyone help me?
I haven't touched boolean algebra in years but I'll give it a try. It may be wrong so check it over!
We have
x'yz + xyz' x'yz'
Factor out the y (since boolean follows distributive properties)
y(x'z + z'x + x'z')
Factor the x' from above (the first and third terms in the brackets)
y( x'(z + z') + z'x)
In boolean a + a' = 1, so our z + z' = 1. So we get
y(x' + z'x)
Now follow the rule that: a + bc = (a+b)(a+c) for our part in the bracket.
y(x'+z')(x' + x)) Once again something equals 1.
y(x' + z') Distribute
yx' + yz' Can rearrange multiplication in boolean
x'y + yz'
And this is what we were looking for.
SailorCosmos
Member
Thank you. I had gotten up to y(x' + z'x) but didn't know what to do afterwards. Thank you again.
Yeah, it uses a tricky rule unique to boolean algebra. Although it is just standard distribution and then simplfied using special boolean rules.
(a + b)(a + c) = a^2 + ac + ab + bc
a^2 = a and ac + ab = a(c + b) so we get
a + ac + ab +bc
a(1+c+b) +bc and in boolean 1+ x = x, so for our 1 + c + b we get 1.
So we get: a +bc
-Pyromaniac-
Member
Holy shit I feel like a huge idiot. I'm brushing up on sequences and whatnot because I'm about to do an online test of sorts for an internship I applied to.
Anyway I found some questions online to practice finding patterns and for the life of me I cannot figure out this one: 4, 3, 5, 9, 12, 17, ___
What's the last number? The answer is apparently
but what's the pattern? I can't seem to get this one while the others were no problem.
Anyway I found some questions online to practice finding patterns and for the life of me I cannot figure out this one: 4, 3, 5, 9, 12, 17, ___
What's the last number? The answer is apparently
26
4,3,5 are the initial numbers. After that, you get 9 = (5 + 4), 12 = (9 + 3), 17 = (12 + 5), 26 = (15 + 9), 38 = (26 + 12), etc.
A recursive algorithm would be y(n+1) = y
+ y(n-2), for all n > 2. So y(1) = 4, y(2) = 3, y(3) = 5.Looks to me kinda like a_n=a_(n-1) + a_(n-3)
the first three terms (4, 3, 5) would have to be given so I'm not sure it's the best example but other than that it works, I believe. EDIT: beat
-Pyromaniac-
Member
4,3,5 are the initial numbers. After that, you get 9 = (5 + 4), 12 = (9 + 3), 17 = (12 + 5), 26 = (15 + 9), 38 = (26 + 12), etc.
A recursive algorithm would be y(n+1) = y
I honestly wouldn't have thought to look for such a thing. Is there a way you knew that it was this type of pattern? TBH I'm rusty as hell on this stuff so I checked for a normal pattern, then a threaded pattern (simple one as in 5-4, 9-3, etc..), but after that I didn't know what to check for. Is this type of pattern you guys pointed out the natural progression after that? Or is there something else that tells you...
Maybe you just know from experience of doing these that this type of thing is a possibility, I haven't done one like it.
Sometimes you just have to examine whether sums of earlier terms can give you a new term. I don't really have a rule when I figure out series, it's just looking for patterns for me.
Experience is always a good thing though. You can always find out a new pattern.
Like this one:
1,11,21,1211,1231
Experience is always a good thing though. You can always find out a new pattern.
Like this one:
1,11,21,1211,1231
youngwerther
Banned
If I get a C in Calculus 1 should I go ahead and change majors to like, art history or something?
-Pyromaniac-
Member
lol these are killing me. I don't even know that one, I have an idea but I can't bring it together. Same with these: 121, 232, 353, 474, 5115, ______
I notice something going on there but I can't figure out the last number. I notice the first 2 give the third number. But the 4th number is made up from the first and third. And then it all falls apart. I'm honestly screwed for this lol. Do you know that one? And is the next number in yours 1251?
The next one in the one I posted is 131221. Try saying the digits out loud, one after one.
In the one you posted, notice that the outer digits seem to be increasing separately from the inner digits. I would guess that the next one would be 6156.
-Pyromaniac-
Member
answer is 6136 but you were close. Christ I didn't even think to notice the actual digits in each number. I swear every time I see one there is a new possibility I discover. Yours is just too much for my brain atm.
It's prime numbers, enclosed by their rank in ascending order.
121, 232, 353, 474, 5115, 6136, 7177, 8198, etc...
I see, the increase being 4 in the inner digits in the last number led me to think that the next inner digits would also increase by 4. How peculiar that it would increase by 1,2,2,4,2
edit: Ah, prime numbers. I never made that connection.
-Pyromaniac-
Member
FUBAR McDangles
Member
Hey guys, I had a test today and the final question kind of stumped me.
It was: Find a polynomial that has the roots 2,3,2i, -2i so that f(4)=80.
So after a bit of thinking, I figured that you'd just go (x-2)(x-3), FOIL that to get X^2-5x+6, then do (x-2i)(x+2i) to get x^2 -4(i)^2, or X^2 + 4 (-i^2 is equal to -1, no?)
Then do X^2-5x+6 * X^2 +4 and get:
X^4 - 5(x)^3 + 10(x)^2 - 20(x) +24
That being the polynomial. I figured I was good to go, so to final check I'd look to make sure that putting 4 in for X would result in the final answer being 80.
But I kept getting 40.
(4)^4 - 5(4)^3 + 10(4)^2 - 20(4) +24
256 - 320 + 160 - 80 + 24
440-400 = 40
Where did I go wrong? Anybody care to help me out?
Also, I was kind of stumped on another question and just had to guess. Not even going to bother with my guess. Solve for X: 7^X = 2^(2X+2), how do you do that?
It was: Find a polynomial that has the roots 2,3,2i, -2i so that f(4)=80.
So after a bit of thinking, I figured that you'd just go (x-2)(x-3), FOIL that to get X^2-5x+6, then do (x-2i)(x+2i) to get x^2 -4(i)^2, or X^2 + 4 (-i^2 is equal to -1, no?)
Then do X^2-5x+6 * X^2 +4 and get:
X^4 - 5(x)^3 + 10(x)^2 - 20(x) +24
That being the polynomial. I figured I was good to go, so to final check I'd look to make sure that putting 4 in for X would result in the final answer being 80.
But I kept getting 40.
(4)^4 - 5(4)^3 + 10(4)^2 - 20(4) +24
256 - 320 + 160 - 80 + 24
440-400 = 40
Where did I go wrong? Anybody care to help me out?
Also, I was kind of stumped on another question and just had to guess. Not even going to bother with my guess. Solve for X: 7^X = 2^(2X+2), how do you do that?
You had the right approach, but missed a constant.
All polynomials with the given roots can be denoted:
k(x-2)(x-3)(x-2i)(x+2i), where k is an arbitrary constant (though k is not equal to 0).
For sanity, multiply the complex roots so we have:
k(x-2)(x-3)(x^2 + 4).
When you insert x = 4 you get k*2*1*20 = 40*k.
As such, choose k to be 2, which will make the expression have the value 80.
You only missed that you could multiply the whole polynomial with an arbitrary constant and still have the same roots. As such, you would only have to choose the constant such that you would get the answer you wanted.
FUBAR McDangles
Member
Darn. I guess I didn't realize that. Well thanks guys. Hopefully I get at least partial credit on the question.
And can you guys help me on my edit? Or here it is again:
Also, I was kind of stumped on another question and just had to guess. Not even going to bother with my guess. Solve for X: 7^X = 2^(2X+2), how do you do that?
-Pyromaniac-
Member
edit: nvm figured it out, getting the hang of things.
I'm taking a math placement test next week, and I'm doing fine on the algebra stuff, but this one question from the basic math part(of the practice quiz) is giving me a nightmarish time.
Three of four numbers have a sum of 22. If the average of
the four numbers is 8, what is the fourth number?
A. 4
B. 6
C. 8
D. 10
I don't even know where to start. Save me, GAF. Oh and I have the answer, but I don't know how to work it out properly.
Three of four numbers have a sum of 22. If the average of
the four numbers is 8, what is the fourth number?
A. 4
B. 6
C. 8
D. 10
I don't even know where to start. Save me, GAF. Oh and I have the answer, but I don't know how to work it out properly.
Here's a hint:
You're given
w+x+y = 22
(w+x+y+z)/4 = 8
You need to solve for z. Is there any way you could combine or substitute the above equations to isolate z?
Oh snap, now I see it.
Set up a proportion where 8*4= w+x+y+z
Which becomes
32= w+x+y+z and we know we can sub in 22 for w+x+y so we get
32=22+z
and the rest is cake.
Thanks!
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