|
On September 29 2017 18:38 Acrofales wrote:Show nested quote +On September 29 2017 18:34 Shalashaska_123 wrote:+ Show Spoiler [long] +On September 29 2017 01:39 HKTPZ wrote:
Hey y'all.
I've been messing around with differentiation of exponential functions last few days and there's something that I cant seem to figure out so I thought I'd ask TeamLiquid for advise/help/inspiration.
By the definition of differentiation, the derivative of a^x is a^x times the limit of ((a^h))-1/h) as h goes to 0. Evaluating that limit is trivial by applying L'H BUT that rests upon already knowing the derivative of a^x so assuming that isnt known how does one figure out that limit?
the brute approach obviously is just convincing oneself that the derivate of 2^x is 2^x itself times a constant which appears to be approximately 0.7
then one could do the same for 8^x and come to the realization that the derivative of 8^x is 8^x times a constant which is about 3 times as great as 0.7 - which suggests we're looking at something logarithmic - the base of which can be approximated by a Taylor Series etc etc
this approach seems so bruteish and non-elegant so my question is: is there a cleverer and more elegant way to evaluate the limit which holds the key to differentiation of exponential functions?
(help me know if my wording was too poor/confusing - English is not my first language) Hello, HKTPZ. Honestly, I would do it the way Liebig suggested. On September 29 2017 02:37 Liebig wrote:
a^x = exp(ln(a^x))=exp(x*ln(a))
so the derivative of a^x is ln(a)*a^x If we can't use that approach, though, then we have to go back to the definition of the derivative. ![[image loading]](https://i.imgur.com/zvpRqOv.gif) The function we have is f(x) = a^x. Plug this into the definition. ![[image loading]](https://i.imgur.com/WtftWll.gif) Use one of the properties of exponents. ![[image loading]](https://i.imgur.com/Chb7Y0i.gif) Factor a^x and bring it in front of the limit. This can be done because it doesn't depend on h. ![[image loading]](https://i.imgur.com/mCWiw9t.gif) We have here the indeterminate form 0/0, so we can apply L'hopital's rule. ![[image loading]](https://i.imgur.com/V1YPlxy.gif) The denominator is just 1, so it disappears. In the numerator, the constant disappears. We're just left with what we set out to find from the beginning. ![[image loading]](https://i.imgur.com/75l0wsQ.gif) This is just f'(0). ![[image loading]](https://i.imgur.com/yifwYdn.gif) We stated at the beginning that f(x) = a^x. Substitute it here. ![[image loading]](https://i.imgur.com/QoqivQZ.gif) We thus have a differential equation for f(x). Divide both sides by f(x). ![[image loading]](https://i.imgur.com/T4OZ9kM.gif) The left side is just the derivative of ln f. ![[image loading]](https://i.imgur.com/5M0cFs6.gif) Now integrate both sides with respect to x. ![[image loading]](https://i.imgur.com/fAgrjpK.gif) Exponentiate both sides so that we just have f(x) on the left side. ![[image loading]](https://i.imgur.com/nSdXWNx.gif) Use one of the properties of exponents. ![[image loading]](https://i.imgur.com/8okhpgN.gif) Use a new constant for e^C. ![[image loading]](https://i.imgur.com/Z1MAnFI.gif) Now substitute a^x for f(x). ![[image loading]](https://i.imgur.com/zcZ71qS.gif) We can determine A by setting x = 0. Doing so gives us 1 = A. ![[image loading]](https://i.imgur.com/DGeswe2.gif) All that's left to do is to solve for f'(0). Take the natural log of both sides. ![[image loading]](https://i.imgur.com/i9hbB25.gif) Use the property of logs that allows us to bring the power of the argument to the coefficient. ![[image loading]](https://i.imgur.com/ZZS5wTs.gif) ln e = 1 and x cancels from both sides. We're left with the value of f'(0). ![[image loading]](https://i.imgur.com/h5XesHx.gif) f'(x) = a^xf'(0). Therefore, ![[image loading]](https://i.imgur.com/s2hJezG.gif) I hope this helped you out. Sincerely, Shalashaska_123 Holy wowzerz. You generated a million different images from latex, uploaded them to imgur and included them here. You're my hero. Also, a super elegant solution, but there is one slight hitch: "The left side is just the derivative of ln f." It's easy to prove, but you'd still have to return to first principles. I think you don't escape needing to define e^x in order to be able to define ln x as its inverse. And then you'd need to prove that sentence there. Which means a lot of work.
Thanks, mate. 
The only justification needed for that step is this. We know that the derivative of ln x is 1/x. If we have ln f(x) instead, then we have to apply the chain rule: (1/f)*f', which is f'/f. Having to go any further than that is just splitting hairs in my humble opinion and would take focus away from the original problem.
|
On September 29 2017 18:34 Shalashaska_123 wrote:+ Show Spoiler +On September 29 2017 01:39 HKTPZ wrote:
Hey y'all.
I've been messing around with differentiation of exponential functions last few days and there's something that I cant seem to figure out so I thought I'd ask TeamLiquid for advise/help/inspiration.
By the definition of differentiation, the derivative of a^x is a^x times the limit of ((a^h))-1/h) as h goes to 0. Evaluating that limit is trivial by applying L'H BUT that rests upon already knowing the derivative of a^x so assuming that isnt known how does one figure out that limit?
the brute approach obviously is just convincing oneself that the derivate of 2^x is 2^x itself times a constant which appears to be approximately 0.7
then one could do the same for 8^x and come to the realization that the derivative of 8^x is 8^x times a constant which is about 3 times as great as 0.7 - which suggests we're looking at something logarithmic - the base of which can be approximated by a Taylor Series etc etc
this approach seems so bruteish and non-elegant so my question is: is there a cleverer and more elegant way to evaluate the limit which holds the key to differentiation of exponential functions?
(help me know if my wording was too poor/confusing - English is not my first language) Hello, HKTPZ. Honestly, I would do it the way Liebig suggested. On September 29 2017 02:37 Liebig wrote:
a^x = exp(ln(a^x))=exp(x*ln(a))
so the derivative of a^x is ln(a)*a^x If we can't use that approach, though, then we have to go back to the definition of the derivative. The function we have is f(x) = a^x. Plug this into the definition. Use one of the properties of exponents. Factor a^x and bring it in front of the limit. This can be done because it doesn't depend on h. We have here the indeterminate form 0/0, so we can apply L'hopital's rule. The denominator is just 1, so it disappears. In the numerator, the constant disappears. We're just left with what we set out to find from the beginning. This is just f'(0). We stated at the beginning that f(x) = a^x. Substitute it here. We thus have a differential equation for f(x). Divide both sides by f(x). The left side is just the derivative of ln f. Now integrate both sides with respect to x. Exponentiate both sides so that we just have f(x) on the left side. Use one of the properties of exponents. Use a new constant for e^C. Now substitute a^x for f(x). We can determine A by setting x = 0. Doing so gives us 1 = A. All that's left to do is to solve for f'(0). Take the natural log of both sides. Use the property of logs that allows us to bring the power of the argument to the coefficient. ln e = 1 and x cancels from both sides. We're left with the value of f'(0). f'(x) = a^xf'(0). Therefore, I hope this helped you out. Sincerely, Shalashaska_123
This is exactly what I was looking for. Elegant and simple. Thank you very much for spelling it out to such length.
I feel so ambivalent right now xD happy that it makes sense to me now and sad that I was too stupid to figure it out on my own
anyway thanks everyone for all of your replies
|
On September 29 2017 19:17 Shalashaska_123 wrote:Show nested quote +On September 29 2017 18:38 Acrofales wrote:On September 29 2017 18:34 Shalashaska_123 wrote:+ Show Spoiler [long] +On September 29 2017 01:39 HKTPZ wrote:
Hey y'all.
I've been messing around with differentiation of exponential functions last few days and there's something that I cant seem to figure out so I thought I'd ask TeamLiquid for advise/help/inspiration.
By the definition of differentiation, the derivative of a^x is a^x times the limit of ((a^h))-1/h) as h goes to 0. Evaluating that limit is trivial by applying L'H BUT that rests upon already knowing the derivative of a^x so assuming that isnt known how does one figure out that limit?
the brute approach obviously is just convincing oneself that the derivate of 2^x is 2^x itself times a constant which appears to be approximately 0.7
then one could do the same for 8^x and come to the realization that the derivative of 8^x is 8^x times a constant which is about 3 times as great as 0.7 - which suggests we're looking at something logarithmic - the base of which can be approximated by a Taylor Series etc etc
this approach seems so bruteish and non-elegant so my question is: is there a cleverer and more elegant way to evaluate the limit which holds the key to differentiation of exponential functions?
(help me know if my wording was too poor/confusing - English is not my first language) Hello, HKTPZ. Honestly, I would do it the way Liebig suggested. On September 29 2017 02:37 Liebig wrote:
a^x = exp(ln(a^x))=exp(x*ln(a))
so the derivative of a^x is ln(a)*a^x If we can't use that approach, though, then we have to go back to the definition of the derivative. The function we have is f(x) = a^x. Plug this into the definition. Use one of the properties of exponents. Factor a^x and bring it in front of the limit. This can be done because it doesn't depend on h. We have here the indeterminate form 0/0, so we can apply L'hopital's rule. The denominator is just 1, so it disappears. In the numerator, the constant disappears. We're just left with what we set out to find from the beginning. This is just f'(0). We stated at the beginning that f(x) = a^x. Substitute it here. We thus have a differential equation for f(x). Divide both sides by f(x). The left side is just the derivative of ln f. Now integrate both sides with respect to x. Exponentiate both sides so that we just have f(x) on the left side. Use one of the properties of exponents. Use a new constant for e^C. Now substitute a^x for f(x). We can determine A by setting x = 0. Doing so gives us 1 = A. All that's left to do is to solve for f'(0). Take the natural log of both sides. Use the property of logs that allows us to bring the power of the argument to the coefficient. ln e = 1 and x cancels from both sides. We're left with the value of f'(0). f'(x) = a^xf'(0). Therefore, I hope this helped you out. Sincerely, Shalashaska_123 Holy wowzerz. You generated a million different images from latex, uploaded them to imgur and included them here. You're my hero. Also, a super elegant solution, but there is one slight hitch: "The left side is just the derivative of ln f." It's easy to prove, but you'd still have to return to first principles. I think you don't escape needing to define e^x in order to be able to define ln x as its inverse. And then you'd need to prove that sentence there. Which means a lot of work. Thanks, mate. The only justification needed for that step is this. We know that the derivative of ln x is 1/x. If we have ln f(x) instead, then we have to apply the chain rule: (1/f)*f', which is f'/f. Having to go any further than that is just splitting hairs in my humble opinion and would take focus away from the original problem.
Clearly it's academic, because HKTPZ is happy with it, but the way I interpreted the question was to remove all of the plumbing, which also means you can't just take as known that the derivative of ln x is 1/x, but would have to prove that, which would need some basic knowledge of what the natural logarithm is.. and HKTPZ kinda forbade us from using that knowledge.
It's a bit strange to ask to prove that the derivative of f(x) = a^x is ln(a) * a^(x), WITHOUT using the knowledge that a^x = e^(x ln a), but WITH using the general knowledge of what ln is. One follows quite naturally from the other. But I guess it's homework  Placing arbitrary limitations is a great way of figuring out what students actually know 
|
On September 29 2017 19:59 Acrofales wrote:Show nested quote +On September 29 2017 19:17 Shalashaska_123 wrote:On September 29 2017 18:38 Acrofales wrote:On September 29 2017 18:34 Shalashaska_123 wrote:+ Show Spoiler [long] +On September 29 2017 01:39 HKTPZ wrote:
Hey y'all.
I've been messing around with differentiation of exponential functions last few days and there's something that I cant seem to figure out so I thought I'd ask TeamLiquid for advise/help/inspiration.
By the definition of differentiation, the derivative of a^x is a^x times the limit of ((a^h))-1/h) as h goes to 0. Evaluating that limit is trivial by applying L'H BUT that rests upon already knowing the derivative of a^x so assuming that isnt known how does one figure out that limit?
the brute approach obviously is just convincing oneself that the derivate of 2^x is 2^x itself times a constant which appears to be approximately 0.7
then one could do the same for 8^x and come to the realization that the derivative of 8^x is 8^x times a constant which is about 3 times as great as 0.7 - which suggests we're looking at something logarithmic - the base of which can be approximated by a Taylor Series etc etc
this approach seems so bruteish and non-elegant so my question is: is there a cleverer and more elegant way to evaluate the limit which holds the key to differentiation of exponential functions?
(help me know if my wording was too poor/confusing - English is not my first language) Hello, HKTPZ. Honestly, I would do it the way Liebig suggested. On September 29 2017 02:37 Liebig wrote:
a^x = exp(ln(a^x))=exp(x*ln(a))
so the derivative of a^x is ln(a)*a^x If we can't use that approach, though, then we have to go back to the definition of the derivative. The function we have is f(x) = a^x. Plug this into the definition. Use one of the properties of exponents. Factor a^x and bring it in front of the limit. This can be done because it doesn't depend on h. We have here the indeterminate form 0/0, so we can apply L'hopital's rule. The denominator is just 1, so it disappears. In the numerator, the constant disappears. We're just left with what we set out to find from the beginning. This is just f'(0). We stated at the beginning that f(x) = a^x. Substitute it here. We thus have a differential equation for f(x). Divide both sides by f(x). The left side is just the derivative of ln f. Now integrate both sides with respect to x. Exponentiate both sides so that we just have f(x) on the left side. Use one of the properties of exponents. Use a new constant for e^C. Now substitute a^x for f(x). We can determine A by setting x = 0. Doing so gives us 1 = A. All that's left to do is to solve for f'(0). Take the natural log of both sides. Use the property of logs that allows us to bring the power of the argument to the coefficient. ln e = 1 and x cancels from both sides. We're left with the value of f'(0). f'(x) = a^xf'(0). Therefore, I hope this helped you out. Sincerely, Shalashaska_123 Holy wowzerz. You generated a million different images from latex, uploaded them to imgur and included them here. You're my hero. Also, a super elegant solution, but there is one slight hitch: "The left side is just the derivative of ln f." It's easy to prove, but you'd still have to return to first principles. I think you don't escape needing to define e^x in order to be able to define ln x as its inverse. And then you'd need to prove that sentence there. Which means a lot of work. Thanks, mate. The only justification needed for that step is this. We know that the derivative of ln x is 1/x. If we have ln f(x) instead, then we have to apply the chain rule: (1/f)*f', which is f'/f. Having to go any further than that is just splitting hairs in my humble opinion and would take focus away from the original problem. Clearly it's academic, because HKTPZ is happy with it, but the way I interpreted the question was to remove all of the plumbing, which also means you can't just take as known that the derivative of ln x is 1/x, but would have to prove that, which would need some basic knowledge of what the natural logarithm is.. and HKTPZ kinda forbade us from using that knowledge. It's a bit strange to ask to prove that the derivative of f(x) = a^x is ln(a) * a^(x), WITHOUT using the knowledge that a^x = e^(x ln a), but WITH using the general knowledge of what ln is. One follows quite naturally from the other. But I guess it's homework Placing arbitrary limitations is a great way of figuring out what students actually know
well fortunately for my odd request, the derivative of ln x does not rely on knowing what the derivative of a^x is, so all's good ^^
I dont think its a completely arbitrary limitation - in fact I think it's very reasonable. But to each their own :D Also it isnt even for homework - it was just something i was wondering about
|
On September 29 2017 20:13 HKTPZ wrote:Show nested quote +On September 29 2017 19:59 Acrofales wrote:On September 29 2017 19:17 Shalashaska_123 wrote:On September 29 2017 18:38 Acrofales wrote:On September 29 2017 18:34 Shalashaska_123 wrote:+ Show Spoiler [long] +On September 29 2017 01:39 HKTPZ wrote:
Hey y'all.
I've been messing around with differentiation of exponential functions last few days and there's something that I cant seem to figure out so I thought I'd ask TeamLiquid for advise/help/inspiration.
By the definition of differentiation, the derivative of a^x is a^x times the limit of ((a^h))-1/h) as h goes to 0. Evaluating that limit is trivial by applying L'H BUT that rests upon already knowing the derivative of a^x so assuming that isnt known how does one figure out that limit?
the brute approach obviously is just convincing oneself that the derivate of 2^x is 2^x itself times a constant which appears to be approximately 0.7
then one could do the same for 8^x and come to the realization that the derivative of 8^x is 8^x times a constant which is about 3 times as great as 0.7 - which suggests we're looking at something logarithmic - the base of which can be approximated by a Taylor Series etc etc
this approach seems so bruteish and non-elegant so my question is: is there a cleverer and more elegant way to evaluate the limit which holds the key to differentiation of exponential functions?
(help me know if my wording was too poor/confusing - English is not my first language) Hello, HKTPZ. Honestly, I would do it the way Liebig suggested. On September 29 2017 02:37 Liebig wrote:
a^x = exp(ln(a^x))=exp(x*ln(a))
so the derivative of a^x is ln(a)*a^x If we can't use that approach, though, then we have to go back to the definition of the derivative. The function we have is f(x) = a^x. Plug this into the definition. Use one of the properties of exponents. Factor a^x and bring it in front of the limit. This can be done because it doesn't depend on h. We have here the indeterminate form 0/0, so we can apply L'hopital's rule. The denominator is just 1, so it disappears. In the numerator, the constant disappears. We're just left with what we set out to find from the beginning. This is just f'(0). We stated at the beginning that f(x) = a^x. Substitute it here. We thus have a differential equation for f(x). Divide both sides by f(x). The left side is just the derivative of ln f. Now integrate both sides with respect to x. Exponentiate both sides so that we just have f(x) on the left side. Use one of the properties of exponents. Use a new constant for e^C. Now substitute a^x for f(x). We can determine A by setting x = 0. Doing so gives us 1 = A. All that's left to do is to solve for f'(0). Take the natural log of both sides. Use the property of logs that allows us to bring the power of the argument to the coefficient. ln e = 1 and x cancels from both sides. We're left with the value of f'(0). f'(x) = a^xf'(0). Therefore, I hope this helped you out. Sincerely, Shalashaska_123 Holy wowzerz. You generated a million different images from latex, uploaded them to imgur and included them here. You're my hero. Also, a super elegant solution, but there is one slight hitch: "The left side is just the derivative of ln f." It's easy to prove, but you'd still have to return to first principles. I think you don't escape needing to define e^x in order to be able to define ln x as its inverse. And then you'd need to prove that sentence there. Which means a lot of work. Thanks, mate. The only justification needed for that step is this. We know that the derivative of ln x is 1/x. If we have ln f(x) instead, then we have to apply the chain rule: (1/f)*f', which is f'/f. Having to go any further than that is just splitting hairs in my humble opinion and would take focus away from the original problem. Clearly it's academic, because HKTPZ is happy with it, but the way I interpreted the question was to remove all of the plumbing, which also means you can't just take as known that the derivative of ln x is 1/x, but would have to prove that, which would need some basic knowledge of what the natural logarithm is.. and HKTPZ kinda forbade us from using that knowledge. It's a bit strange to ask to prove that the derivative of f(x) = a^x is ln(a) * a^(x), WITHOUT using the knowledge that a^x = e^(x ln a), but WITH using the general knowledge of what ln is. One follows quite naturally from the other. But I guess it's homework Placing arbitrary limitations is a great way of figuring out what students actually know well fortunately for my odd request, the derivative of ln x does not rely on knowing what the derivative of a^x is, so all's good ^^ I dont think its a completely arbitrary limitation - in fact I think it's very reasonable. But to each their own :D Also it isnt even for homework - it was just something i was wondering about
The derivative of lnx kind of depends on having a good idea what lnx is, though. Which requires a good idea of what e^x is, which means you probably know the derivative of e^x, at which point you can use a^x = e^(xlna)
|
On September 29 2017 07:23 Joni_ wrote:Show nested quote +On September 29 2017 06:46 AbouSV wrote:On September 29 2017 06:16 jodljodl wrote:
exp is not defined by exp'(x) = exp(x). Its defined by a power series (exp(x) = sum x^n/n! , n= 0 to infinity) or as the limit of the sequence (1 + x/n)^n , n -> infty. Exponential can also be defined as the only solution of y' = y, with y(0) = 1, implying exp'(x) = exp(x). You need a lot more mathematical background to prove that this differential equation has a unique solution than you need to define the exponential function by it's series representation. Also the series representation can be generalized more easily (say to the case of bounded operators acting on possibly infinite-dimensional banach spaces) and it allows you to verify properties of the (real) exponential function like its limit behaviour when concatenated with polynomial functions in a very simple way.. I dont really see the advantage of using your proposed definition, as long as it's not part of a class for say engineers that are primarily interested in solving differential equations. Mind you I dont want to say that either definition is really more "correct" (at least in the case of real exponential functions), I just think that using the series representation fits more naturally into the way an analysis class is structured. At least where I'm from, maybe it's different if you do several classes of "calculus" before developing an axiomatic theory.. Edit: The more I think about it the less I like your way of defining the exponential function. It really only works properly in the case of functions defined on at least some open neighbourhood of 0 on the real axis. And at the very least as soon as you try to generalize to the complex plane you will inevitably need the series representation to prove uniqueness of the solution anyway, bc you will need some form of an identity theorem for holomorphic functions... Just feels like it really only is useful in a setting where the people being taught are interested only in the theory of ordinary differential equations on the set of real functions and their strong solutions....
The background knowledge you need is different indeed. But I learnt both the same year (how to properly solve a differential equation, or at least know we can, and the power series and what goes with them), which is why I tend to put both those definitions in the same bag.
Although, I agree with you that the series expansion definition is more convenient anyway, especially when it comes to complex numbers, but also matrices and other objects.
|
random statistics question I want to verify. I missed a point for this on my exam.
We had to draw a box plot based on some data. They gave us 15 values.
15
20
22
22
24
24
24
28
30
30
34
36
37
so I calculated the quartiles in this way.
q2 = median = 24
then remove the median from the data.
then find the median of the lower half of the data, and the median of the upper half of the data.
lower half:
15
20
22
22
24
24
median = 22
upper half:
28
30
30
34
36
37
median = 32 (the mean of 34 and 30)
so for q3 I used 32. I got docked a point, they crossed it out and wrote 30.
Did I do something incorrect, or were they wrong to dock me a point?
Thanks.
|
I see 13 , not 15 values. Assuming that there are 15 values, using the simpliest method, then then surely the q3 would be 4th from the highest value, which is 30. If there are 13 values, then it should be 3rd from the highest which is 34. Using another simple method, with 15 values, then q3 will be the mean from the 4th and 5th from the highest and assuming there are 13 values then q3 will be 4th from the highest which is 30. The other methods are probably too complicated for you and you can't seem to count either 13/15 properly anyways.
Statistics is strange. It isn't a universal maths, so you have to follow whatever methodology is taught. So tell us first the method you are supposed to use.
|
sorry i said 15 when I meant 13, that doesn't mean I can't count properly lol. what does "then then" mean? clearly you don't understand english, since you made a mistake in what you typed.
anyways, they never told us a method we are supposed to use.
I used method 1 here
https://en.wikipedia.org/wiki/Quartile
|
What kind of exam tests something that you have not been taught?
Of course there is supposed to be some transfer, but still, exams are for testing whether you learned the stuff they taught you. If they didn't teach you how to do a box plot, why would they test that? It is not like it is something that you can derive from other knowledge either. You either know how to do it or not.
|
Yeah, at this point I am just wondering what kind of exam is this?
They don't teach you methodology to a statistical analysis, and from the sounds of it you can't ask any fellow colleagues or the examiner and there is no teacher.
Also this is a maths thread, not an English thread. Which is why I don't mention your lack of capitalisation.
|
On October 05 2017 02:34 travis wrote:
random statistics question I want to verify. I missed a point for this on my exam.
We had to draw a box plot based on some data. They gave us 15 values.
15
20
22
22
24
24
24
28
30
30
34
36
37
so I calculated the quartiles in this way.
q2 = median = 24
then remove the median from the data.
then find the median of the lower half of the data, and the median of the upper half of the data.
lower half:
15
20
22
22
24
24
median = 22
upper half:
28
30
30
34
36
37
median = 32 (the mean of 34 and 30)
so for q3 I used 32. I got docked a point, they crossed it out and wrote 30.
Did I do something incorrect, or were they wrong to dock me a point?
Thanks.
You computed it correctly. The way you solved it is the conventional way that the quartiles are solved, both in AP Statistics in high school as well as in college statistics.
Source: I'm a high school math/ statistics teacher and a college math/ statistics professor. lol.
Also other sources:
This educational page supports your process: http://web.mnstate.edu/peil/MDEV102/U4/S36/S363.html
This math .com site does too: http://www.mathwords.com/t/third_quartile.htm
This math .com site also does... once you put in some data (e.g., typing in 1,2,3,4,5 gives you 4.5 as Q3): http://www.alcula.com/calculators/statistics/quartiles/
Same with this site (again, once you type in the data): https://www.hackmath.net/en/calculator/quartile-q1-q3
From the hackmath.net site:
![[image loading]](https://i.imgur.com/ggGATf1.png)
Furthermore, that 10th data point (the second 30) is only greater than 9/13 of the data, which is 69% of the data and not 75% so... no, it can't be Q3. Your averaged value of 32, on the other hand, is greater than 10/13 of the data, which is 77% (and that's the best average you can get to ensure that you're above 75% of the data, based on there being 13 data points). Your value of 32 does a significantly better job at splitting the data into a lower 75% and a higher 25% than the 30 does.
Q3 is 32.
|
Thanks plasma! I guess I'll go try to get that point back then.
As for them teaching us certain methodology, they taught us how to make box plot and what quartiles are. The professor then mentioned that there are multiple ways to calculate quartiles. He briefly went over 3 methods, I don't remember which. It seemed like he wasn't too concerned about which method we used.
I think this was an error by the grading TA.
|
Hello, travis.
Your "mistake" was in removing the median from the data. If you include 24 in the upper and lower halves and take the median of both, you'll get 22 and 30. This is basically method 2 from that wikipedia article you posted. The way you did the problem (method 1) is legitimate, but I suppose method 2 is how they want you to compute quartiles in this class.
Sincerely,
Shalashaska_123
|
Yeah, that's probably what travis's grader was thinking ^^
|
This is probably below everyone's pay grade, but is there a proof that two negatives multiplying make a positive? It seems obvious to me that the opposite of an opposite is the original, yet my uncle insists that the entire world operates on negatives multiplying out to positives because some guy a while ago made it that way. Any help in convincing a crack pot would be appreciated.
|
your Country52797 Posts
On October 05 2017 12:04 bo1b wrote:
This is probably below everyone's pay grade, but is there a proof that two negatives multiplying make a positive? It seems obvious to me that the opposite of an opposite is the original, yet my uncle insists that the entire world operates on negatives multiplying out to positives because some guy a while ago made it that way. Any help in convincing a crack pot would be appreciated.
I'm bad at proofs but I'll give it a shot.
Math is consistent. Let's say we had positive integers x and y, and -x * -y was negative. I assume your uncle understands that a positive number times a negative number (say, x * -y) is negative. Then (x + (-x)) * -y = x * -y + -x * -y, which is the sum of two negative numbers and therefore negative. However, x + (-x) = 0, and 0 times any number is 0 (which is neither negative nor positive). Therefore we have a contradiction, meaning -x * -y can not be negative.
|
On October 05 2017 12:04 bo1b wrote:
This is probably below everyone's pay grade, but is there a proof that two negatives multiplying make a positive? It seems obvious to me that the opposite of an opposite is the original, yet my uncle insists that the entire world operates on negatives multiplying out to positives because some guy a while ago made it that way. Any help in convincing a crack pot would be appreciated.
I'm not a mathematician, but this 'debt' example made sense to me:
http://mathforum.org/dr.math/faq/faq.negxneg.html
|
On October 05 2017 12:04 bo1b wrote:
This is probably below everyone's pay grade, but is there a proof that two negatives multiplying make a positive? It seems obvious to me that the opposite of an opposite is the original, yet my uncle insists that the entire world operates on negatives multiplying out to positives because some guy a while ago made it that way. Any help in convincing a crack pot would be appreciated.
Your uncle's point of view has some philosophical apparel... After all are there really negative numbers? Can they be observed in the real world? If they arent something we can observe in the real world, how can we be sure what properties they really "have"?
The mathematical answer to the problem of existence is constructing them. This is done the following way (warning: This might not be helpful at all, I included it for sake of completeness):
<math>
The cleanest way to do so is by taking ordered pairs of natural numbers (n,m) and introducing the equivalence relation (n,m) ~ (n',m') iff there ex. a natural number k such that either (n+k,m+k)=(n',m') or (n,m)=(n'+k,m'+k). Then the set of integers can be defined as the set of all equivalence classes and equipped with an addition and multiplication.
Identifying (n,m) with the integer solution x to the equation n+x=m is how one usually imagines these integers.
This construction solves the issue of "existence" because once we have established knowledge about natural numbers, this allows us to construct integers using nothing but equivalence classes of natural numbers.
When reading an integer nobody thinks "wow, I guess that's a nice equivalence class of pairs of natural numbers", so after establishing this, one proves that this construct indeed has all the properties one wants from integers and moves on with life.
</math>
A more "natural" way to think about it is by thinking of integers as a model (that does not necessarily correspond to anything maginary in the real world) that helps us better formulate and formalize certain problems like debt. Then the point of your uncle reduces to "okay but someone had to come up with this model!", which is true, and the question why the product of two negative numbers is a positive number is answered by "because we want the model to behave that way".
Why do we want that? Well, we want to have the laws of distributivity, associativity and commutativity of integers as well as -1*x=-x... So for two positive integers x,y their corresponding additive inverses -x,-y have to allow -x*-y=(-1)*x*(-1)*y=(-1)*(-1)*x*y=x*y.
So while there is no real universal "truth" about why this is the case, we want to have a set of integers that acknowledges these laws, so we made them in a way that they do. Arguing that negative times negative equals positive does not necessarily have to be the case boils down to arguing that at least one of these laws should be omitted. It seems reasonable to assume these laws for what we are trying to model though.
|
Another good argument from y fifth grade textbook:
3*3=?
3*2=?
3*1=?
3*0=?
3*(-1)=?
3*(-2)=?
usw...
And then
(-3)*3=?
(-3)*2=?
(-3)*1=?
(-3)*0=?
(-3)*(-1)=?
(-3)*(-2)=?
Using the consistency of the results in the lines as an argument that the negative * positive is positive, and negative * negative is negative. This is obviously not a complete proof, but a good argument nonetheless.
Also, i would like to mention that this thread is not only for maths students or professors. Anyone that has any problem with math is very welcome here!
And according to this, your uncle is actually kind of right. (-1)*(-1) = 1 is a convention that has been decided by people, but any other possible convention breaks some part of maths that is important to us, because we like maths to be consistent.
|
|
|
|
|
|
EPT
