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Old 12-11-2012, 09:56 AM   #141
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I've seen about 4 methods taught on how to do partial fracs, 2 different ways in Calc2, 2 different ways in DiffEq and this last method I learned is uber easy.
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Old 12-11-2012, 10:40 AM   #142
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Partial Fracs with Laplace is easy - post problem and I'll respond. I have a way of doing Partial fractions that blows your hair back with ease.
No worries. I just got done with the final and had to decompose 2 problems but my teacher was nice and both were polynomials that were easily factored so it was straightforward.
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Old 12-11-2012, 11:02 AM   #143
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No worries. I just got done with the final and had to decompose 2 problems but my teacher was nice and both were polynomials that were easily factored so it was straightforward.
Must of been nice

I'm pretty sure I'll end up taking a victory lap in ODE
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Old 12-11-2012, 11:19 AM   #144
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I had a great teacher and the stuff actually makes sense to me so that helps. She normally teaches linear algebra and PDE too but not next semester
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Old 12-11-2012, 01:48 PM   #145
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I had a great teacher and the stuff actually makes sense to me so that helps. She normally teaches linear algebra and PDE too but not next semester
My ODE teacher only teaches Calc3 and lower, we only have 2 sections per semester of ODE.

PDE is typically a grad level course and not many people have to take it. Buddy of mine with a Ph.D in IE still has nightmares of PDE

Most of the class passed, I'm fairly sure I will end up with a D

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Old 12-15-2012, 10:38 AM   #146
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Old 12-15-2012, 11:12 AM   #147
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lol i hated this class. Its deff one of the toughest ive taken.
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Old 12-15-2012, 11:16 AM   #148
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Old 01-24-2013, 04:01 PM   #149
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Doing some review for PDE and I got thrown a curveball. I'm not even sure where to begin with this one. I've tried starting with an auxiliary equation but how the fvck do you factor a 4th degree polynomial?

y''''+4y'''+6y''+4y+4=0
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Old 03-03-2013, 11:06 PM   #150
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Hey,

Can any of you guys help me real quick with this little Calc quiz I need to finish. I got problems 1,4 and 5 done, those were simple. Problem 2 and 3 are throwing me off. I don't really remember the theory rule for the high number derivatives like number 2. I'm sure this is really easy for some of you DE guys. lol

Thanks

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Old 03-04-2013, 12:48 AM   #151
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f(x)=x^(-1/2)
f'(x)=(-1/2)x^(-3/2)
f''(x)=(-1/4)x^(-5/2)

continue that. I got F^(50)(x)=(-1/1125899906842624)x^(-101/2)

#3 would suck to type
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Old 03-04-2013, 02:48 AM   #152
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^You did #2 wrong, the coefficient in front isn't right. For #2, let's look at the first 4 derivatives:
f(x)=x-1/2
f'(x)=(-1/2)x-3/2
f''(x)=(3/4)x-5/2
f'''(x)=(-15/8)x-7/2
f''''(x)=(105/16)x-9/2
etc.

So we have to come up with a way to figure out what f(a)(x) is.

First let's do the easy part, the negative/positive sign in front. It just changes every time, so we need a (-1)a in front
f(a)(x)=(-1)a...

Then, we'll notice that in the denominator we have 2, 4, 8, 16, 32... so we need to put a (1/2)a also
f(a)(x)=(-1)a(1/2)a.....
Simplifying:
f(a)(x)=(-1/2)a.....

The numerator of the coefficient is a bit tricky, it's 1, 3, 15, 105.... but we know that this is just 1, 1*3, 1*3*5, 1*3*5*7.... so we have a hint that a factorial is needed. The second derivative has 1*3, the third has 1*3*5, the fourth has 1*3*5*7 etc so let's put (2a-1)! in there. That will give us the right numbers, but we need to get rid of the evens, e.g. for the fourth derivative, (2a-1)! will give you 1*2*3*4*5*6*7 but we need to get rid of the 2*4*6. You'll notice that 2*4*6 is just (1*2*3)*2*2*2, so if we take (2a-1)! and divide it by 2(a-1)(a-1)!, we should have the right coefficient. Let's add this whole part:

f(a)(x)=(-1/2)a(2a-1)!/(2(a-1)(a-1)!).....

Finally we need to get the power of x correct, we see that it's -3/2, -5/2, -7/2... so that's just -(2a+1)/2. Let's add that in

f(a)(x)=(-1/2)a(2a-1)!/(2(a-1)(a-1)!)x-(2a+1)/2

And that's the general formula for the a-th derivative. Then just put 50 in place of the a's and you're done (I think the actual numbers are way too big to calculate with a calculator).
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Old 03-04-2013, 03:07 AM   #153
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your post was last edited before your post? impressive.

I totally overlooked that the coeficient in front was multipied by the exponent, which was changing for each derivitive. I just assumed it was (1/2) ^50 which made thing much more simple haha.

I completely do not belong in this thread. ill just go ahead and show myself the way out now.
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Old 03-04-2013, 03:09 AM   #154
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Old 03-04-2013, 04:43 AM   #155
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^You did #2 wrong, the coefficient in front isn't right. For #2, let's look at the first 4 derivatives:
f(x)=x-1/2
f'(x)=(-1/2)x-3/2
f''(x)=(3/4)x-5/2
f'''(x)=(-15/8)x-7/2
f''''(x)=(105/16)x-9/2
etc.

So we have to come up with a way to figure out what f(a)(x) is.

First let's do the easy part, the negative/positive sign in front. It just changes every time, so we need a (-1)a in front
f(a)(x)=(-1)a...

Then, we'll notice that in the denominator we have 2, 4, 8, 16, 32... so we need to put a (1/2)a also
f(a)(x)=(-1)a(1/2)a.....
Simplifying:
f(a)(x)=(-1/2)a.....

The numerator of the coefficient is a bit tricky, it's 1, 3, 15, 105.... but we know that this is just 1, 1*3, 1*3*5, 1*3*5*7.... so we have a hint that a factorial is needed. The second derivative has 1*3, the third has 1*3*5, the fourth has 1*3*5*7 etc so let's put (2a-1)! in there. That will give us the right numbers, but we need to get rid of the evens, e.g. for the fourth derivative, (2a-1)! will give you 1*2*3*4*5*6*7 but we need to get rid of the 2*4*6. You'll notice that 2*4*6 is just (1*2*3)*2*2*2, so if we take (2a-1)! and divide it by 2(a-1)(a-1)!, we should have the right coefficient. Let's add this whole part:

f(a)(x)=(-1/2)a(2a-1)!/(2(a-1)(a-1)!).....

Finally we need to get the power of x correct, we see that it's -3/2, -5/2, -7/2... so that's just -(2a+1)/2. Let's add that in

f(a)(x)=(-1/2)a(2a-1)!/(2(a-1)(a-1)!)x-(2a+1)/2

And that's the general formula for the a-th derivative. Then just put 50 in place of the a's and you're done (I think the actual numbers are way too big to calculate with a calculator).
Excellent! Thank you!

I hate it when my professor throws us these curve ball questions that we hardly cover in class. I understand finding 2nd, 3rd and 4th derivatives, however anything past that begins to recur and basically you need to determine what derivative f(a) lands on.

Any idea whats going on with the theory question #3?
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Old 03-04-2013, 08:11 AM   #156
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I was able to do #3, but it would be really hard to write it out here in a way that's understandable. Maybe I can write it out and take a picture...
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Old 03-04-2013, 08:25 AM   #157
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Ok, here it is. Let me know if you have any questions:


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Old 03-04-2013, 09:13 AM   #158
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I just wanted to say good luck to anyone taking PDEs. Never have I been so interested in a class and had such little understanding of what was going on. It is by far the most interesting class I've taken, qualitatively, and the most difficult quantitatively.
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Old 03-04-2013, 09:49 AM   #159
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I just wanted to say good luck to anyone taking PDEs. Never have I been so interested in a class and had such little understanding of what was going on. It is by far the most interesting class I've taken, qualitatively, and the most difficult quantitatively.
I never actually took a PDE class, although many aspects of physics uses them. My favorite class by far was Complex Analysis. It was a little weird at times, but not too difficult, and it opens up a whole new realm when it comes to mathematics. Really interesting stuff, I highly recommend it!
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Old 03-04-2013, 09:58 AM   #160
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But isn't QM mostly PDEs? There's an intro to it at the end of my PDE book. How did you manage to get through it?

I know there's a Real Analysis class offered and I've heard all the math majors bitching about that one; no one seems to enjoy it. I don't know if there is a complex analysis class offered or not. Topology is offered and I've heard good things about it, but I'm pretty sure I would have to go through Real Analysis in order to take it and I'm not sure I'm willing to risk my GPA to take a class I don't need. Plus, if I can't see how it applies to the real world then there's a better chance I'll get lost. That's pretty much how linear algebra is going, but luckily it's easy enough that I've given up trying to find a physical implication for all the theorems and just follow the rules to get the right answers.
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