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OCGuy

Lifer
Jul 12, 2000
27,224
37
91
Originally posted by: TruePaige
Originally posted by: eLiu
Originally posted by: OCguy
Its Friday evening...and you guys are discussing calculus. There are going to be some dry penises tonight.
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It's only 3pm on the west coast, and I'm bored at work. Dry my ass.
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Somebody pissed in OCguy's cheerios this morning, he's been shitting in every thread he can get his hands on.
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Maybe you cant read the tone of my posts....I guess I fail at being funny today. :(
 
sao123

sao123

Lifer
May 27, 2002
12,653
205
106
Originally posted by: DrPizza

(x²-2ax+a²)((x²-2bx+b²)=0

x^4 -2ax^3 -2bx^3 ...
You know that -2a -2b = 0, so a = -b (there's no x^3 term)
etc.

And, the last term, a²b² has to equal k

Should be trivial to figure out from here, so I'll leave that as an exercise for the reader. :p
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Originally posted by: eLiu
It's just a quadratic in x^2:
let y = x^2
then your fcn = y^2 - 4y + k

Now x = +/-sqrt(y), so y^2-4y+k needs to have one positive root of multiplicity 2. Complete the square for the obvious choice, k = 4, so you have (y-2)*(y-2) = y^2 - 4y + 4 and the roots are x={-sqrt(2),-sqrt(2),sqrt(2),sqrt(2)}

edit: DrPizza's way is a lot more general so you should understand what he did. I'm just relying on 2 obvious things: 1) the expression is quadratic, 2) if the quadratic doesn't have 1 root you'll get 4 different roots.

Edit2: OP, how would you do this WITH calculus?? I can't think of an efficient way to do it.
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both are great non-calculus ways to solve.
 
DrPizza

DrPizza

Administrator Elite Member Goat Whisperer
Mar 5, 2001
49,601
167
111
www.slatebrookfarm.com
Calculus method: since multiplicity is two, then the first derivative is equal to zero at each of the roots. (Each root is a relative min)
So
4x^3 - 8x = 0
4x ( x^2 - 2) = 0
x = 0 (relative max), and +/- sqrt(2)

Since each of those roots occurs twice,
(x - sqrt2)(x+sqrt2) (x-sqrt2)(x+sqrt2)
obviously, k = 4
 
E

eLiu

Diamond Member
Jun 4, 2001
6,407
1
0
Originally posted by: DrPizza
Calculus method: since multiplicity is two, then the first derivative is equal to zero at each of the roots. (Each root is a relative min)
So
4x^3 - 8x = 0
4x ( x^2 - 2) = 0
x = 0 (relative max), and +/- sqrt(2)

Since each of those roots occurs twice,
(x - sqrt2)(x+sqrt2) (x-sqrt2)(x+sqrt2)
obviously, k = 4
Click to expand...

NICE. That hadn't occured to me at all--that multiplicity 2 roots will be relative mins since (x-b)^2 will dominate as x->b. cool.
 
DrPizza

DrPizza

Administrator Elite Member Goat Whisperer
Mar 5, 2001
49,601
167
111
www.slatebrookfarm.com
Originally posted by: Regs
Originally posted by: JohnCU
this thread makes me horny:evil:
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I.


What's the square root of horny?
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Do you want me to calculate it using pre-calculus, or calculus? I was thinking that you could write a Taylor series to calculate it, but then again, that would be an infinite series & I know how sums of infinite series cause problems in OT.

sqrt(horny) = 1 + (horny -1)/2 - (1/8)(horny-1)² + (1/16)(horny-1)^3 - (5/128)(horny -1)^4 + (7/256)(horny-1)^5 - (21/1024)(horny-1)^6 + (33/2048)(horny-1)^7 - (429/32768)(horny-1)^8 + (715/65536)(horny-1)^9 - (2431/262144)(horny-1)^10 + . . .
 
Sea Moose

Sea Moose

Diamond Member
May 12, 2009
6,933
7
76
Originally posted by: schneiderguy
I'm taking a pre-calculus class in college (I took AP calc in high school but the college won't let me take the college calc class for some reason). We had a test today and the extra credit question was:

f(x) = x^4 - 4x^2 + k

For what value of k does the function have two real zeroes, each with a multiplicity of two?



So from there I figured out that since the multiplicity was 2, the function would just touch the x-axis at the zeroes, but not cross. So I took the derivative of the function and set y to 0 to find the extrema, which were +- sqrt(2). Then I plugged them back into the original equation, and set y to 0 to get k. I believe this is the correct solution, but how would you do this without derivatives? Basically, how do you get the extrema of an equation that has a degree greater than 2 without finding where the derivative of the function = 0?
Click to expand...

Step One: Quit math class.
Step two: Burn maths book
Step three: Buy beer
Step four: Buy Salted Nuts
Step Five: Nestle into couch and drink till you forget math
Step six: Go to step three and continue until effective









Profit
 
Rubycon

Rubycon

Madame President
Aug 10, 2005
17,768
485
126
Originally posted by: Sea Moose


Step One: Quit math class.
Step two: Burn maths book
Step three: Buy beer
Step four: Buy Salted Nuts
Step Five: Nestle into couch and drink till you forget math
Step six: Go to step three and continue until effective

Profit
Click to expand...

Abnormal program termination. Illegal function call in line 3.

Abort, Retry, or Epic Fail?

 
Sea Moose

Sea Moose

Diamond Member
May 12, 2009
6,933
7
76
Originally posted by: Rubycon
Originally posted by: Sea Moose


Step One: Quit math class.
Step two: Burn maths book
Step three: Buy beer
Step four: Buy Salted Nuts
Step Five: Nestle into couch and drink till you forget math
Step six: Go to step three and continue until effective

Profit
Click to expand...

Abnormal program termination. Illegal function call in line 3.

Abort, Retry, or Epic Fail?
Click to expand...

I suck at math, i am jealous of those that are good at it. I bought a nintendo DS so i can do brain training. A little japanese character tells me I suck at math :(
 
H

hypn0tik

Diamond Member
Jul 5, 2005
5,866
2
0
Originally posted by: schneiderguy
I'm taking a pre-calculus class in college (I took AP calc in high school but the college won't let me take the college calc class for some reason). We had a test today and the extra credit question was:

f(x) = x^4 - 4x^2 + k

For what value of k does the function have two real zeroes, each with a multiplicity of two?



So from there I figured out that since the multiplicity was 2, the function would just touch the x-axis at the zeroes, but not cross. So I took the derivative of the function and set y to 0 to find the extrema, which were +- sqrt(2). Then I plugged them back into the original equation, and set y to 0 to get k. I believe this is the correct solution, but how would you do this without derivatives? Basically, how do you get the extrema of an equation that has a degree greater than 2 without finding where the derivative of the function = 0?
Click to expand...

It's a bit messy but it can be done. Since you have 2 real zeroes, each with multiplicity of two, you have the form:

f(x) = (x-a)^2 (x-b)^2
So that when you set f(x) = 0, x = a and x = b are your only roots.

Now expand:

(x^2 - 2ax + a^2) (x^2 - 2bx + b^2) = [x^4 - 4x^2 + k]

(The quantity in the [] = RHS)

Expand further:

x^4 + x^3 (-2a -2b) + x^2 (a^2 + b^2 + 4ab) + x (-2ab^2 - 2a^2b) + a^2*b^2 = RHS

Now, you have 2 equations and 2 unknowns to solve:
The coefficient of x^3 = 0 and the coefficient of x^2 = -4

-2a - 2b = 0 ---> a = -b ----------- eqn (1)

a^2 + b^2 + 4ab = -4
Sub in (1)
(-b)^2 + b^2 + 4(-b)(b) = -4

b^2 + b^2 -4b^2 = -4
-2b^2 = -4
b^2 = 2

b = sqrt(2)
a = -sqrt(2)

You can pick b = -sqrt(2) to be your solution, in which case a = +sqrt(2). The symmetry is preserved since f(x) = (x-a)^2 (x-b)^2

So, comparing your LHS to your RHS, your constant k = a^2*b^2 = sqrt(2)^2 (-sqrt(2)^2)
so k = 2*2 = 4.
 
Sea Moose

Sea Moose

Diamond Member
May 12, 2009
6,933
7
76
Originally posted by: hypn0tik
Originally posted by: schneiderguy
I'm taking a pre-calculus class in college (I took AP calc in high school but the college won't let me take the college calc class for some reason). We had a test today and the extra credit question was:

f(x) = x^4 - 4x^2 + k

For what value of k does the function have two real zeroes, each with a multiplicity of two?



So from there I figured out that since the multiplicity was 2, the function would just touch the x-axis at the zeroes, but not cross. So I took the derivative of the function and set y to 0 to find the extrema, which were +- sqrt(2). Then I plugged them back into the original equation, and set y to 0 to get k. I believe this is the correct solution, but how would you do this without derivatives? Basically, how do you get the extrema of an equation that has a degree greater than 2 without finding where the derivative of the function = 0?
Click to expand...

It's a bit messy but it can be done. Since you have 2 real zeroes, each with multiplicity of two, you have the form:

f(x) = (x-a)^2 (x-b)^2
So that when you set f(x) = 0, x = a and x = b are your only roots.

Now expand:

(x^2 - 2ax + a^2) (x^2 - 2bx + b^2) = [x^4 - 4x^2 + k]

(The quantity in the [] = RHS)

Expand further:

x^4 + x^3 (-2a -2b) + x^2 (a^2 + b^2 + 4ab) + x (-2ab^2 - 2a^2b) + a^2*b^2 = RHS

Now, you have 2 equations and 2 unknowns to solve:
The coefficient of x^3 = 0 and the coefficient of x^2 = -4

-2a - 2b = 0 ---> a = -b ----------- eqn (1)

a^2 + b^2 + 4ab = -4
Sub in (1)
(-b)^2 + b^2 + 4(-b)(b) = -4

b^2 + b^2 -4b^2 = -4
-2b^2 = -4
b^2 = 2

b = sqrt(2)
a = -sqrt(2)

You can pick b = -sqrt(2) to be your solution, in which case a = +sqrt(2). The symmetry is preserved since f(x) = (x-a)^2 (x-b)^2

So, comparing your LHS to your RHS, your constant k = a^2*b^2 = sqrt(2)^2 (-sqrt(2)^2)
so k = 2*2 = 4.
Click to expand...

wall of numbers ... worse than wall of text. I can at least understand text if i try to read. You fucking math people make that shit up as you go along i swear to the mods!
 
G

gdextreme

Member
Nov 8, 2008
127
0
0
I don't know what you mean by multiplicity of 2, but if you want to solve it, you substitute 'x^2' as 'z'. Then you have a quadratic in the variable 'z'. You can use the quadratic formula to solve it.
 
M

marketsons1985

Platinum Member
Apr 15, 2000
2,090
0
76
Originally posted by: gdextreme
I don't know what you mean by multiplicity of 2, but if you want to solve it, you substitute 'x^2' as 'z'. Then you have a quadratic in the variable 'z'. You can use the quadratic formula to solve it.
Click to expand...

This. Super easy FTW. Then use the discriminant or b^2 - 4ac.
 
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