I think statistics would be one of the more important things people should know, just so they know how to interpret stuff in newspapers. Averages are a poor description of a population in a lot of cases.
i wouldn't hate on someone for not having the same type of brain as me and not being able to think mathematically
but i do hate people who pretend to know statistics and then misuse it
Remember when I posted a week ago about having a bit of a hard time with Calculus? It all feels so easy now.
[editline]9th March 2013[/editline]
[QUOTE=Jo The Shmo;39839444]you have to look at it from a new perspective
math is a [i]very[/i] abstract idea and its not at all a natural way of thinking
yes the natural world behaves according to mathematical laws, but the way our brain works is not in the slightest mathematical
some people are lucky and they can get a strong grasp on math which leads them to highly academic careers
but for a large portion of people, math is simply a weird thing that they dont need to think about very much
there is some truth to when people say "everything about math that i need i learned in elementary school!"
its not completely true, but for the large majority of the population, there never comes a moment where they need to use anything but basic arithmetic and algebra[/QUOTE]
The same holds for many other subjects too. Also, doing math is not just about the math itself, but also the logical thinking practice that comes with it.
What I think is critical to understanding maths is looking at it with no expectations or assumptions. You have to learn to take certain rules and just those. Once I realized that, understanding completely new math went a lot easier (I'm still no math wizz though). Some people are able to do this "intuitively", others have to learn it.
[QUOTE=Number-41;39854503]What I think is critical to understanding maths is looking at it with no expectations or assumptions. You have to learn to take certain rules and just those. Once I realized that, understanding completely new math went a lot easier (I'm still no math wizz though). Some people are able to do this "intuitively", others have to learn it.[/QUOTE]
What kinds of assumptions are we talking about here? I don't think I understand exactly what you mean.
[QUOTE=ArgvCompany;39854599]What kinds of assumptions are we talking about here? I don't think I understand exactly what you mean.[/QUOTE]
A lot of people have to understand that it doesn't matter if you "feel" something is right or not in mathematics. If it works, it works. If doesn't work, it doesn't work. No if's and but's about it.
Course 8: Root and logarithmic functions
Expectations after completing the exam: 9 (Scale from 4 to 10)
Result: 7
Words do not describe the disappointment after seeing that result, FML :=
[QUOTE=Glorbo;39856953]A lot of people have to understand that it doesn't matter if you "feel" something is right or not in mathematics. If it works, it works. If doesn't work, it doesn't work. No if's and but's about it.[/QUOTE]
No no no.
Math is supposed to feel right.
If it doesn't feel right, you never know if you learned something correctly or if you got it all wrong.
If you have, for instance, proven something, and you know it's a solid proof but the result doesn't "feel" right, you need to adjust your feelings to suit the mathematics. Not the other way around.
[editline]9th March 2013[/editline]
A computer can verify a proof or a calculation without any feeling. You can know something is right without it feeling right.
Are there any programs that are good for drawing graphs?
Trying to draw an argand diagram, and want some 'digital graph paper' :v:
[QUOTE=JohnnyMo1;39860703]If you have, for instance, proven something, and you know it's a solid proof but the result doesn't "feel" right, you need to adjust your feelings to suit the mathematics. Not the other way around.
[editline]9th March 2013[/editline]
A computer can verify a proof or a calculation without any feeling. You can know something is right without it feeling right.[/QUOTE]
If it is mathematically true then there exists a way to explain it in English that feels right.
For every mathematical theory or proof, no matter how abstract, there is a way to explain it that makes sense to a layman, and thus feels right - you just can't necessarily think of such an explanation right away.
A proof is worth little if you can read it, but don't understand it.
Edit: What I'm saying is, if a proof doesn't feel right, that doesn't make it imperfect. It means your understanding of it is imperfect. Building upon imperfect understanding is fundamentally wrong - you should just stop and meditate on the proof until you can say that everything makes sense about it.
I don't equate "feeling right" with "understanding." I can see why the Banach-Tarski paradox is true, I can understand the proof, but that doesn't mean it doesn't seem strange.
[editline]10th March 2013[/editline]
It's a case of having to adjust our intuitions to meet our mathematical understanding.
I found this website:
[url]http://www.proofwiki.org/wiki/Main_Page[/url]
Pretty much have my ass covered, permitting that I know what to search for.
Proofwiki is fantastic. Hard to resist the temptation to look up proofs on my homeworks but you really learn a lot doing proofs for yourself. Plus it's satisfying. :v:
So I've decided to get an AS degree. Is this the first step to becoming a math nerd? He who knows the most math wins.
Just wrote my first 'shitty' proof for a maths assignment:
"Prove the following inequalities for all real numbers x and y."
[IMG]http://latex.codecogs.com/gif.latex?\left%20|%20x+y%20\right%20|\geq%20\left%20|%20x%20\right%20|%20-%20\left%20|%20y%20\right%20|[/IMG]
**Can I get an extremely honest critical analysis on how this is presented, or suggestions on how you would present it. I'd rather my feeling be hurt, than my GPA :v:
Proof (Not quite sure which, but I'm guessing Induction):
Lets assume that this is true:
[IMG]http://latex.codecogs.com/gif.latex?\left%20|%20x+y%20\right%20|\geq%20\left%20|%20x%20\right%20|%20-%20\left%20|%20y%20\right%20|[/IMG]
[IMG]http://latex.codecogs.com/gif.latex?\left%20|%20x+y%20\right%20|^{2}\geq%20\left%20(%20\left%20|%20x%20\right%20|%20-%20\left%20|%20y%20\right%20|\right%20)^{2}[/IMG]
[IMG]http://latex.codecogs.com/gif.latex?x^{2}+2xy+y^{2}\geq%20\left%20x^{2}-2\left%20|%20x%20\right%20|\left%20|%20y%20\right%20|\right+y^{2}[/IMG]
[IMG]http://latex.codecogs.com/gif.latex?2xy\geq%20-2\left%20|%20x%20\right%20|\left%20|%20y%20\right%20|[/IMG]
Since,
[IMG]http://latex.codecogs.com/gif.latex?\left%20|%20x%20\right%20|\left%20|%20y%20\right%20|\geq%200[/IMG]
It must follow from the trichotomy of real numbers that if x or y is less than zero, that we would have an equality.
It must also follow that if x and y are less than zero, that the LHS is greater than the RHS.
[IMG]http://latex.codecogs.com/gif.latex?\left%20|%20x+y%20\right%20|\geq%20\left%20|%20x%20\right%20|%20-%20\left%20|%20y%20\right%20|[/IMG]
Holds true [IMG]http://latex.codecogs.com/gif.latex?\forall%20(x,%20y)\in%20\mathbb{R}[/IMG]
I haven't been doing math in a while, but are you starting a proof by assuming that what you want to prove is true? This seems... Wrong. Maybe I just didn't understand.
[editline]12th March 2013[/editline]
I'm getting my white board out to see how I would try and prove this. I miss maths
[editline]12th March 2013[/editline]
Yeah ok you're just doing your proof completely backwards, at least to me :v:
Do not begin with what you are trying to prove.
[editline]12th March 2013[/editline]
It's not really a valid argument if you do it backwards because if you assumption is false, you can still derive true statements from it.
[QUOTE=JohnnyMo1;39887064]Do not begin with what you are trying to prove.[/QUOTE]
How should I start it.
As, I know how and why it is correct... I just don't know how to set it up. :/
Can I do it by contradiction, assume that it isn't true, do the same working, saying that it shouldn't work out, and if it does work out, then it must have been true...
[editline]Try something new[/editline]
Could I work backwards, starting out with this:
"Prove the following inequalities for all real numbers x and y."
[IMG]http://latex.codecogs.com/gif.latex?\left | x+y \right |\geq \left | x \right | - \left | y \right |[/IMG]
[B]Proof by induction:[/B]
Lemma 1:
Proposition:
[IMG]http://latex.codecogs.com/gif.latex?ab \geq -|a||b|[/IMG]
Proof:
Where (a, b) ∈ ℝ
If [I]a[/I] and [I]b[/I] are less than zero, then the LHS will be [I]greater than the [/I]RHS, as (ab ≥ 0) and (0 ≥ -|a||b| ).
If [I]a[/I] or [I]b[/I] are less than zero, then the LHS will be [I]equal to[/I] the RHS, as (ab ≤ 0) and (-|a||b|≤ ab ≤ 0).
Given the lemma, it then follows:
[IMG]http://latex.codecogs.com/gif.latex?2ab \geq -2|a||b|[/IMG]
This gives us a perfect square:
[IMG]http://latex.codecogs.com/gif.latex?a^{2} + 2ab +b^{2} \geq a^{2} -2|a||b| + b^{2}[/IMG]
When this perfect square is factored:
[IMG]http://latex.codecogs.com/gif.latex?(a+b)^{2} \geq (|a| - |b|)^2[/IMG]
Using the following as an axiom:
[IMG]http://latex.codecogs.com/gif.latex?|x| = \sqrt{x^{2}}[/IMG]
Square root both sides:
[IMG]http://latex.codecogs.com/gif.latex?LHS = \sqrt{(a+b)^{2}} =|a+b|[/IMG]
[IMG]http://latex.codecogs.com/gif.latex?RHS = \sqrt{(|a|-|b|)^{2}} =|a|-|b|[/IMG]
Therefore,
[IMG]http://latex.codecogs.com/gif.latex?|a+b| \geq |a|-|b|[/IMG]
[QUOTE=Bradyns;39887102]How should I start it.
As, I know how and why it is correct... I just don't know how to set it up. :/
Can I do it by contradiction, assume that it isn't true, do the same working, saying that it shouldn't work out, and if it does work out, then it must have been true...[/QUOTE]
The most obvious (to me) way of doing it just trivially follows from the triangle inequality:
The triangle inequality says |a+b|<= |a|+|b|
Substituting a = x+y and b = -y gives |x| <= |x+y| + |y|
Then just rearrange to get the result.
Proof by contradiction is totally legitimate, and would start with you supposing the opposite - i.e. that there exist [i]some[/i] x and y with |x+y| < |x| - |y|, but in this case it's easier to prove it directly.
As JohnnyMo says - never begin by assuming what you want to prove! That's circular logic...
However, your method of proof does work when reversed, so long as you're careful when taking square roots (given that you can take the positive or negative root):
xy >= -|x||y|
x^2 + 2xy + y^2 >= x^2 - 2|x||y| + y^2
|x+y|^2 >= (|x|-|y|)^2
|x+y| >= |(|x|-|y|)| (taking positive square root on both sides!)
|x+y| >= |x|-|y| (since being negative only makes it smaller)
[QUOTE=Joey90;39887158]The most obvious (to me) way of doing it just trivially follows from the triangle inequality:
The triangle inequality says |a+b|<= |a|+|b|
Substituting a = x+y and b = -y gives |x| <= |x+y| + |y|
Then just rearrange to get the result.
Proof by contradiction is totally legitimate, and would start with you supposing the opposite - i.e. that there exist [I]some[/I] x and y with |x+y| < |x| - |y|, but in this case it's easier to prove it directly.
As JohnnyMo says - never begin by assuming what you want to prove! That's circular logic...
However, your method of proof does work when reversed, so long as you're careful when taking square roots (given that you can take the positive or negative root):
xy >= -|x||y|
x^2 + 2xy + y^2 >= x^2 - 2|x||y| + y^2
|x+y|^2 >= (|x|-|y|)^2
|x+y| >= |(|x|-|y|)| (taking positive square root on both sides!)
|x+y| >= |x|-|y| (since being negative only makes it smaller)[/QUOTE]
Didn't even notice the third page; glad we came to the same conclusion though.
I quite like your short work around though..
Wouldn't you still have to show with the substitutions it holds true for all reals?
[editline]OffTheHook[/editline]
So, the method yourself and I used was a 'Direct proof'?
Essentially, thrown in the deep end at the moment; trying to learn the lingo, and acquire good habits.
[QUOTE=Bradyns;39887102]Could I work backwards, starting out with this:
"Prove the following inequalities for all real numbers x and y."
[IMG]http://latex.codecogs.com/gif.latex?\left | x+y \right |\geq \left | x \right | - \left | y \right |[/IMG]
[B]Proof by induction:[/B]
Lemma 1:
Proposition:
[IMG]http://latex.codecogs.com/gif.latex?ab \geq -|a||b|[/IMG]
Proof:
Where (a, b) ∈ ℝ
If [I]a[/I] and [I]b[/I] are less than zero, then the LHS will be [I]greater than the [/I]RHS, as (ab ≥ 0) and (0 ≥ -|a||b| ).
If [I]a[/I] or [I]b[/I] are less than zero, then the LHS will be [I]equal to[/I] the RHS, as (ab ≤ 0) and (-|a||b|≤ ab ≤ 0).
Given the lemma, it then follows:
[IMG]http://latex.codecogs.com/gif.latex?2ab \geq -2|a||b|[/IMG]
This gives us a perfect square:
[IMG]http://latex.codecogs.com/gif.latex?a^{2} + 2ab +b^{2} \geq a^{2} -2|a||b| + b^{2}[/IMG]
When this perfect square is factored:
[IMG]http://latex.codecogs.com/gif.latex?(a+b)^{2} \geq (|a| - |b|)^2[/IMG]
Using the following as an axiom:
[IMG]http://latex.codecogs.com/gif.latex?|x| = \sqrt{x^{2}}[/IMG]
Square root both sides:
[IMG]http://latex.codecogs.com/gif.latex?LHS = \sqrt{(a+b)^{2}} =|a+b|[/IMG]
[IMG]http://latex.codecogs.com/gif.latex?RHS = \sqrt{(|a|-|b|)^{2}} =|a|-|b|[/IMG]
Therefore,
[IMG]http://latex.codecogs.com/gif.latex?|a+b| \geq |a|-|b|[/IMG][/QUOTE]
This works
It's basically the first proof you tried but not backwards :v:
That being said, finding the proof backwards (starting from what you want to prove) is a good way to find out how to do it. Then write it in the correct order and bam, solved.
[editline]12th March 2013[/editline]
[QUOTE=Joey90;39887158]The most obvious (to me) way of doing it just trivially follows from the triangle inequality:
The triangle inequality says |a+b|<= |a|+|b|
Substituting a = x+y and b = -y gives |x| <= |x+y| + |y|
Then just rearrange to get the result.[/QUOTE]
I'm pretty sure that this would be considered cheating, you're bypassing the entire point of the proof :v:
Unless that first inequality has been proven earlier in the same problem of course
[QUOTE=Zanpa;39888141]I'm pretty sure that this would be considered cheating, you're bypassing the entire point of the proof :v:
Unless that first inequality has been proven earlier in the same problem of course[/QUOTE]
The triangle inequality is an axiom for the metric in metric spaces and of the norm in normed vector spaces, so I don't see why he'd have to prove it.
[QUOTE=JohnnyMo1;39888246]The triangle inequality is an axiom for the metric in metric spaces and of the norm in normed vector spaces, so I don't see why he'd have to prove it.[/QUOTE]
Because if he used it then there would basically be no point to this proof. I just wouldn't use it. Or I'd prove it beforehand.
[QUOTE=Zanpa;39888425]Because if he used it then there would basically be no point to this proof. I just wouldn't use it. Or I'd prove it beforehand.[/QUOTE]
Then the proof is pointless. If you can't use the axioms, what are you supposed to use? :v:
[QUOTE=Bradyns;39887852]
So, the method yourself and I used was a 'Direct proof'?
Essentially, thrown in the deep end at the moment; trying to learn the lingo, and acquire good habits.[/QUOTE]
A proof by contradiction works by assuming that the statement is false and arriving at a contradiction, thus you deduce that the original assumption can't have been correct (i.e. the statement must be true).
I would call anything that doesn't do this a 'Direct proof' - since you show 'directly' that the statement is true, rather than showing the contrapositive. (If you are a [url=http://en.wikipedia.org/wiki/Constructivism_(mathematics)]Constructivist[/url] this is in fact the [i]only[/i] type of proof allowed)
Proof by induction is a type of direct proof for when you have a statement involving a natural number n=1,2,... You prove it for n=1, and you prove that if it's true for n=1,...,k then it is true for n=k+1. Thus you deduce that it is true for all n=1,2... (This can also be generalised and stated in a number of ways)
[QUOTE=Zanpa;39888141]I'm pretty sure that this would be considered cheating, you're bypassing the entire point of the proof :v:
Unless that first inequality has been proven earlier in the same problem of course[/QUOTE]
This is a fair point, it rather depends what context the question is in... Though I'd hope that one of the first things you do when looking at the modulus/absolute value/norm is prove the triangle inequality.
Hey guys, while we're discussing proof by induction, could someone help me with a homework problem I have tonight? It asks me to evaluate the summation (k to n) of k/(k+1)! for n=1 to 5. So far I've computed like my teacher has asked me to do (1/2+1/3+1/8+1/30+1/144=719/720), but I'm struggling to make a conjecture for a new formula to prove. Any pointers, please?
[QUOTE=mastoner20;39889146]Hey guys, while we're discussing proof by induction, could someone help me with a homework problem I have tonight? It asks me to evaluate the summation (k to n) of k/(k+1)! for n=1 to 5. So far I've computed like my teacher has asked me to do (1/2+1/3+1/8+1/30+1/144=719/720), but I'm struggling to make a conjecture for a new formula to prove. Any pointers, please?[/QUOTE]
If you've worked out the sum for n=1 to 5 I would have thought the conjecture to make would be fairly clear...
The values you should get are 1/2, 5/6, 23/24, 119/120, 719/720
If you really need another hint, write out n! for n=2,...,6
is it possible to find a point of tangency without the tangent line equation. i have a problem that shows a tangent line that passes through a point at (3pi/2, 0) for a function of sin(x), and that's all that it gives.
Could you post the problem? It seems vague.
If the function is just sin(x), it doesn't really make sense because that point isn't on sin(x).
With the actual function and assuming the point is actually on the curve, you can find a tangent line with a little calculus.
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