*mutter* Stupid calculus... has to be dependant on crap I never bothered to learn in high school....
How would you evaluate this:
(4/5)(1/(k^5/4) if k = infinity?
Step by step instructions would be nice... so I don't have to come and ask you guys every time I see one of these... heh.
Sarvis
Math question...
<BLOCKQUOTE><font size="1" face="Verdana, Arial">quote:</font><HR><font face="Verdana, Arial" size="2">Originally posted by Sarvis:
<B>*mutter* Stupid calculus... has to be dependant on crap I never bothered to learn in high school....
How would you evaluate this:
(4/5)(1/(k^5/4) if k = infinity?
Step by step instructions would be nice... so I don't have to come and ask you guys every time I see one of these... heh.
Sarvis</B></font><HR></BLOCKQUOTE>
Ok, don't quote me, but it looks like you could make that into 4/(5k^5/4). Now the answer really depends on what exactly the question says. Does the question ask as K approaches Infinity or K= Infinity? Anyway I think the answer would be zero since the denominator will be infinitely larger than the numerator.
Any other opinions?
<B>*mutter* Stupid calculus... has to be dependant on crap I never bothered to learn in high school....
How would you evaluate this:
(4/5)(1/(k^5/4) if k = infinity?
Step by step instructions would be nice... so I don't have to come and ask you guys every time I see one of these... heh.
Sarvis</B></font><HR></BLOCKQUOTE>
Ok, don't quote me, but it looks like you could make that into 4/(5k^5/4). Now the answer really depends on what exactly the question says. Does the question ask as K approaches Infinity or K= Infinity? Anyway I think the answer would be zero since the denominator will be infinitely larger than the numerator.
Any other opinions?
Actually, we're trying to determine whether or not the series 1/n^(1/4) converges, using the Integral Test. I integrated and got
lim -(4/5)n^(-5/4) from 1 to k.
k->infinity
(really wish I could write in the actual math symbols here...)
which comes out to be
-(4/5)k^(-5/4) - -4/5 = ?
I thought that the k part would evaluate to 0, and the answer would be 4/5... but then I started thinking you needed a common denominator in order to combine the -4/5 and the 1/k^(5/4)... which would change things. Maybe I was right the first time... I hate math.
Thanks
Sarvis
lim -(4/5)n^(-5/4) from 1 to k.
k->infinity
(really wish I could write in the actual math symbols here...)
which comes out to be
-(4/5)k^(-5/4) - -4/5 = ?
I thought that the k part would evaluate to 0, and the answer would be 4/5... but then I started thinking you needed a common denominator in order to combine the -4/5 and the 1/k^(5/4)... which would change things. Maybe I was right the first time... I hate math.
Thanks
Sarvis
-
Guest
<BLOCKQUOTE><font size="1" face="Verdana, Arial">quote:</font><HR><font face="Verdana, Arial" size="2">Originally posted by Sarvis:
<B>*mutter* Stupid calculus... has to be dependant on crap I never bothered to learn in high school....
How would you evaluate this:
(4/5)(1/(k^5/4) if k = infinity?
Step by step instructions would be nice... so I don't have to come and ask you guys every time I see one of these... heh.
Sarvis</B></font><HR></BLOCKQUOTE>
I'll take a stab at it, it's been a while, but you have to take each part individually and analyze how it reacts as k approaches infinity. A number can never equal infinity, but you have to take the limit of it as it approaches infinity.
4/5 you know will always be 4/5, no matter what k is. 1/(k^5/4) as k approaches infinity will be 0. Don't get bogged down with all the powers and the fractions. lim (1/k^n) k->infinity with k >= 1 and n >= 0 will always be 0. The power on k will just make it approach it more rapidly (affecting the arc, etc). In this instance it's k^5/4.
(4/5)(0) = 0.
So lim k->infinity of the equation will be 0.
Math rules, wait until you start taking surface areas and volumes of curves. Makes it possible to get objects with an finite volume, but an infinite surface area. Meaning, you can fill up the object, but never paint the outside.
Figure that one out.
[This message has been edited by Kiaransalee (edited 10-12-2001).]
<B>*mutter* Stupid calculus... has to be dependant on crap I never bothered to learn in high school....
How would you evaluate this:
(4/5)(1/(k^5/4) if k = infinity?
Step by step instructions would be nice... so I don't have to come and ask you guys every time I see one of these... heh.
Sarvis</B></font><HR></BLOCKQUOTE>
I'll take a stab at it, it's been a while, but you have to take each part individually and analyze how it reacts as k approaches infinity. A number can never equal infinity, but you have to take the limit of it as it approaches infinity.
4/5 you know will always be 4/5, no matter what k is. 1/(k^5/4) as k approaches infinity will be 0. Don't get bogged down with all the powers and the fractions. lim (1/k^n) k->infinity with k >= 1 and n >= 0 will always be 0. The power on k will just make it approach it more rapidly (affecting the arc, etc). In this instance it's k^5/4.
(4/5)(0) = 0.
So lim k->infinity of the equation will be 0.
Math rules, wait until you start taking surface areas and volumes of curves. Makes it possible to get objects with an finite volume, but an infinite surface area. Meaning, you can fill up the object, but never paint the outside.
Figure that one out.
[This message has been edited by Kiaransalee (edited 10-12-2001).]
actually the answer is zero. As K->infinity, the 2nd term->0 which means you have something along the lines of (4/5) * 0.00000000001, which is approaching zero. I could actually do the entire limit for you, but im too lazy. If you are really desperate, email me at mike@zazzle.com.
Also, what level calculus is this. . .meaning can you just take the derivative or do they want you to do it manually?
Out,
Tanras
Also, what level calculus is this. . .meaning can you just take the derivative or do they want you to do it manually?
Out,
Tanras
Ok, everyone is right in saying that the limit of that expression is 0 as k -> inf, but since you're trying to determine convergence that's not the actual question at hand.
Now, just before you do anything, from the p-series you know that this will diverge because p = 1/4 < 1. For the actual work however:
The integral of this expression will either diverge or converge. If it diverges, you know that the series diverges.
So you take:
INT(1, inf, n^(-1/4))
= 4n^(3/4) / 3, evaluated over 1 to inf.
= lim(b -> inf) 4n^(3/4) / 3, evaluated over 1 to b.
= lim(b->inf) 4b^(3/4) / 3 - 4 / 3
Taking the limit as b -> inf clearly gives you inf, so this integral diverges, and thus the series diverges.
Now, just before you do anything, from the p-series you know that this will diverge because p = 1/4 < 1. For the actual work however:
The integral of this expression will either diverge or converge. If it diverges, you know that the series diverges.
So you take:
INT(1, inf, n^(-1/4))
= 4n^(3/4) / 3, evaluated over 1 to inf.
= lim(b -> inf) 4n^(3/4) / 3, evaluated over 1 to b.
= lim(b->inf) 4b^(3/4) / 3 - 4 / 3
Taking the limit as b -> inf clearly gives you inf, so this integral diverges, and thus the series diverges.
-
Gindipple
- Sojourner
- Posts: 676
- Joined: Fri Jan 26, 2001 6:01 am
- Location: O' Fallon, MO. USA
- Contact:
Sarvis, I used to help out at our school in what was called the mathlab.
I learned a lot there, more than what was taught to me.
I suggest you see if your school has such a thing, it's usually students helping students.
If you're going to be doing anything that seriously requires calculas, like Engineering, you might consider re-taking calc II if you feel in over your head already.
PS. The worst thing you can probably do for your situation is MUD!
[This message has been edited by Gindipple (edited 10-12-2001).]
I learned a lot there, more than what was taught to me.
I suggest you see if your school has such a thing, it's usually students helping students.
If you're going to be doing anything that seriously requires calculas, like Engineering, you might consider re-taking calc II if you feel in over your head already.
PS. The worst thing you can probably do for your situation is MUD!
[This message has been edited by Gindipple (edited 10-12-2001).]
Bleh. I don't have time to retake calc2... I seriously am sick of school and need to graduate. Besides, taking calc2 3 times was far more than enough. *sigh* I'm a computer science major... have never encountered any math in any programming I've done, except computer graphics... and that wasw enough to make me give up any hope of becoming a game programmer. Now I just want to get paid loads of cash to code print drivers and shit, but gotta finish calc3 first.
Not sure if this will help but...
I too took Cal3. The weekly homework usually took me the whole sunday to do and left me completely clueless. When I took the final exam it was in the form of 8 questions with none of them look even half remotely familiar to me. Anyhow, I applied the Cal3 methods to the best of my knowledge and got an A for the class. Although, I never did have the courage to ask my professor what my actual grade was on the final =P
Yssilk
I too took Cal3. The weekly homework usually took me the whole sunday to do and left me completely clueless. When I took the final exam it was in the form of 8 questions with none of them look even half remotely familiar to me. Anyhow, I applied the Cal3 methods to the best of my knowledge and got an A for the class. Although, I never did have the courage to ask my professor what my actual grade was on the final =P
Yssilk
<BLOCKQUOTE><font size="1" face="Verdana, Arial">quote:</font><HR><font face="Verdana, Arial" size="2">Originally posted by Sarvis:
Bleh. I don't have time to retake calc2... I seriously am sick of school and need to graduate. Besides, taking calc2 3 times was far more than enough. *sigh* I'm a computer science major... have never encountered any math in any programming I've done, except computer graphics...</font><HR></BLOCKQUOTE>
WOAH dude! Reading that hurt. Sorry, this is an old thread but I have to comment on it :-P. Computer Science is 100% based on mathematics. You have to know your shit before you realize it though. (Programming is not computer "science" btw. Its a basic tool). Heheh so for example:
-Computers in general are an abstract concept of a Turing Machine, a mathematical engine to build a computer from the 1930's (10 years before they built a computer)
-A programming language was initially viewed as a way to program a Turing machine. The Turing machine would read a tape, that had instructions that would make the tape go Left or Right. This would program the simple Turing machine by reading the tape and causing the Turing machine to go into various states.
-Data basing is 100% based on Set Theory. You won't neccisarily realize it if you do data basing, but its foundation is from hard core mathematics. Set Theory is brutal, avoid at all costs.
-You could make strong arguements on the similarity of an object and a group. Both encapsulate properties. Group theory is WAY cool.
-If you ever want to compare or contrast two algorithms, whip out the math boy. Ouch. This stuff is painfull.
-I'm a cryptographer, so hehehe there is a tonne of math in the stuff I do. Lots of algebra.
Hehhe. Yea, when I was in 2nd/3rd year CS, I use to wonder why they taught the type of math they did. If you ever study the high end CS (especially the theory), it all starts to gel. But you are correct, you can do quite well in the industry by just being a programmer.
Trogar, Mathematician and Computer Scientist extraordinare!
Bleh. I don't have time to retake calc2... I seriously am sick of school and need to graduate. Besides, taking calc2 3 times was far more than enough. *sigh* I'm a computer science major... have never encountered any math in any programming I've done, except computer graphics...</font><HR></BLOCKQUOTE>
WOAH dude! Reading that hurt. Sorry, this is an old thread but I have to comment on it :-P. Computer Science is 100% based on mathematics. You have to know your shit before you realize it though. (Programming is not computer "science" btw. Its a basic tool). Heheh so for example:
-Computers in general are an abstract concept of a Turing Machine, a mathematical engine to build a computer from the 1930's (10 years before they built a computer)
-A programming language was initially viewed as a way to program a Turing machine. The Turing machine would read a tape, that had instructions that would make the tape go Left or Right. This would program the simple Turing machine by reading the tape and causing the Turing machine to go into various states.
-Data basing is 100% based on Set Theory. You won't neccisarily realize it if you do data basing, but its foundation is from hard core mathematics. Set Theory is brutal, avoid at all costs.
-You could make strong arguements on the similarity of an object and a group. Both encapsulate properties. Group theory is WAY cool.
-If you ever want to compare or contrast two algorithms, whip out the math boy. Ouch. This stuff is painfull.
-I'm a cryptographer, so hehehe there is a tonne of math in the stuff I do. Lots of algebra.
Hehhe. Yea, when I was in 2nd/3rd year CS, I use to wonder why they taught the type of math they did. If you ever study the high end CS (especially the theory), it all starts to gel. But you are correct, you can do quite well in the industry by just being a programmer.
Trogar, Mathematician and Computer Scientist extraordinare!
Yes, yes... I'm aware of how much computer science is based on mathematics. Then again so are cars, but you don't need advanced calculus to get a job at your local Chevy plant.
But it might not be so bad if they taught it differently. I've always hated the way math was taught... everything is a proof, or a theorem and has no obvious practical applications. Even when they give word problems they are so based in fiction that my first thought is why the hell would anyone ever want to calculate this!?! If they taught it in terms of usage I could probably deal with it much better. Also if they taught it from computer science perspective. We take the same courses as engineers, so even when they do give some concrete usage of a theorem it's somethng all the engineers can nod at, and I don't even know what they are talking about! Bleh... enough ranting for now...'
Sarvis
But it might not be so bad if they taught it differently. I've always hated the way math was taught... everything is a proof, or a theorem and has no obvious practical applications. Even when they give word problems they are so based in fiction that my first thought is why the hell would anyone ever want to calculate this!?! If they taught it in terms of usage I could probably deal with it much better. Also if they taught it from computer science perspective. We take the same courses as engineers, so even when they do give some concrete usage of a theorem it's somethng all the engineers can nod at, and I don't even know what they are talking about! Bleh... enough ranting for now...'
Sarvis
-
Gindipple
- Sojourner
- Posts: 676
- Joined: Fri Jan 26, 2001 6:01 am
- Location: O' Fallon, MO. USA
- Contact:
<BLOCKQUOTE><font size="1" face="Verdana, Arial">code:</font><HR><pre>
#include <iostream>
#define INF 1000
using namespace std;
void main(void)
{
double n, total;
total = 0;
for(n = 1; n < INF; ++n)
{
total += 1/(n*(n+2));
if(!((int)n%10))
cout << n << " " << total << endl;
}
}
</pre><HR></BLOCKQUOTE>
results
10 0.662879
20 0.703463
30 0.718246
40 0.7259
50 0.730581
60 0.733739
70 0.736013
80 0.73773
90 0.739071
100 0.740148
110 0.741031
120 0.741769
130 0.742395
140 0.742933
150 0.743399
160 0.743808
170 0.744169
180 0.74449
190 0.744778
200 0.745037
210 0.745272
220 0.745485
230 0.74568
240 0.745859
250 0.746024
260 0.746176
270 0.746317
280 0.746448
290 0.746569
300 0.746683
310 0.74679
320 0.74689
330 0.746983
340 0.747072
350 0.747155
360 0.747234
370 0.747308
380 0.747379
390 0.747446
400 0.747509
410 0.74757
420 0.747628
430 0.747682
440 0.747735
450 0.747785
460 0.747833
470 0.747879
480 0.747923
490 0.747965
500 0.748006
510 0.748045
520 0.748082
530 0.748119
540 0.748153
550 0.748187
560 0.748219
570 0.74825
580 0.74828
590 0.748309
600 0.748337
610 0.748365
620 0.748391
630 0.748416
640 0.748441
650 0.748465
660 0.748488
670 0.748511
680 0.748533
690 0.748554
700 0.748574
710 0.748595
720 0.748614
730 0.748633
740 0.748651
750 0.748669
760 0.748687
770 0.748704
780 0.74872
790 0.748737
800 0.748752
810 0.748768
820 0.748783
830 0.748797
840 0.748812
850 0.748826
860 0.748839
870 0.748853
880 0.748866
890 0.748878
900 0.748891
910 0.748903
920 0.748915
930 0.748926
940 0.748938
950 0.748949
960 0.74896
970 0.748971
980 0.748981
990 0.748991
So yes it's heading to 3/4
PS had to put code in code block.
[This message has been edited by Gindipple (edited 10-24-2001).]
#include <iostream>
#define INF 1000
using namespace std;
void main(void)
{
double n, total;
total = 0;
for(n = 1; n < INF; ++n)
{
total += 1/(n*(n+2));
if(!((int)n%10))
cout << n << " " << total << endl;
}
}
</pre><HR></BLOCKQUOTE>
results
10 0.662879
20 0.703463
30 0.718246
40 0.7259
50 0.730581
60 0.733739
70 0.736013
80 0.73773
90 0.739071
100 0.740148
110 0.741031
120 0.741769
130 0.742395
140 0.742933
150 0.743399
160 0.743808
170 0.744169
180 0.74449
190 0.744778
200 0.745037
210 0.745272
220 0.745485
230 0.74568
240 0.745859
250 0.746024
260 0.746176
270 0.746317
280 0.746448
290 0.746569
300 0.746683
310 0.74679
320 0.74689
330 0.746983
340 0.747072
350 0.747155
360 0.747234
370 0.747308
380 0.747379
390 0.747446
400 0.747509
410 0.74757
420 0.747628
430 0.747682
440 0.747735
450 0.747785
460 0.747833
470 0.747879
480 0.747923
490 0.747965
500 0.748006
510 0.748045
520 0.748082
530 0.748119
540 0.748153
550 0.748187
560 0.748219
570 0.74825
580 0.74828
590 0.748309
600 0.748337
610 0.748365
620 0.748391
630 0.748416
640 0.748441
650 0.748465
660 0.748488
670 0.748511
680 0.748533
690 0.748554
700 0.748574
710 0.748595
720 0.748614
730 0.748633
740 0.748651
750 0.748669
760 0.748687
770 0.748704
780 0.74872
790 0.748737
800 0.748752
810 0.748768
820 0.748783
830 0.748797
840 0.748812
850 0.748826
860 0.748839
870 0.748853
880 0.748866
890 0.748878
900 0.748891
910 0.748903
920 0.748915
930 0.748926
940 0.748938
950 0.748949
960 0.74896
970 0.748971
980 0.748981
990 0.748991
So yes it's heading to 3/4
PS had to put code in code block.
[This message has been edited by Gindipple (edited 10-24-2001).]
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