@@MathAndScience Dr. Gibson, can you please make a video on bayesian statistics in research? I am having hard time understanding it, and I think you can definitely help me get some sense of it. Thank you in advance. 😊
Hello Mr. Gibson, I just wanted to take this chance to thank you for helping me in many subjects like chemistry. I am good at chemistry, but you were always a second address if I had to know something for school. Thanks to you help I am very successful in school, that's the reason I'm writing this comment. Greetings from Germany
Oh my gosh I probably watched like 10 different videos explaining the same topic and I couldn’t understand anything but just watching your video to the 6 minute mark made everything so clear now. THANK YOU. You have a talent for simplification!
Already I learnt this geometric series from one of your lessons. One point I cleared from this lesson is the range of common ratio. Success is neither a minute work nor a day long work. Rather, it's how much time you spend in a day to reach your goal.
I have been looking at divergent series for a while, and think statements like "-1 = 1 + 2 + 4 + 8 + ... " are just badly stated, but not actually wrong. What's wrong is that in the " + ... " you are not told what is in the tail. If you use recursion, it is concrete, and you can just do the algebra; and never resort to appeals to infinity in any way. T = Sum(T,n) + Tail(T,n). T = 1 + 10 T. (1-10)T = 1. -9 T = 1. T = -1/9. ??? because common ratio is 10... but carry on.... T = (1 + 10 T) = (1 + 10 (1 + 10 T)) = (1 + 10 + 100 T) = (10^0 + 10^1 + 10^2 + ... + 10^n) + 10^{n+1} T = Sum(T,n) + Tail(T,n). T is -1/9, yet it expands like: "1 + 10 + 100 + ... ". But notice that we dropped the recursive tail, because we don't believe in its importance. T is only Sum(T,n) when Tail(T,n) goes to zero. So, T doesn't mean what people think it does. But T is meaningful. It's a number required to calculate the closed form of the sum of n terms. T -Tail(T,n) = Sum(T,n) T(1 - 10^{n+1}) = Sum(T,n) (-1/9)(1 - 10^{n+1}) = Sum(T,n) (10^{n+1}-1)/9 = Sum(T,n) This takes the mystery out of divergent series being associated with a number. The sum is not T. It's Sum(T,n), which is equal to T-Tail(T,n). Recursion might be the only really concrete way to specify infinite processes; because it still uses finite iteration. When you have a series where there's not a common ratio, but it goes up by 1 for a coefficient, you can usually get there by differentiating a series like this one; so you can do term-by-term addition of series. Because of this, you don't run into any of the odd issues of commutativity/associativity that you will get when infinity is explicitly involved. Recursion is a much more sound basis, when you have a way to use it. You will run into this problem pretty early on if you try to automate computer algebra. If you follow the algebraic rules and both expand a sequence to a sequence, and also solve it for a number; a computer has no good way to object to it. All you can do is to change the algebra rules until they give you what you want, without breaking unrelated things. So, you have to face this issue of divergent series. It almost always means that we are interpreting it wrongly The fact that you end up subtracting the tail off of an "infinite" sum to recover a finite sum up to n terms says something deep about infinite processes. It says to me that the concept of an infinite process is kind of hand-wavy, when compared to algebraic rules that you follow without objecting to the results; where you sometimes need to carefully re-interpret something that doesn't make sense. If S is not the sum here when it's true for EVERY value of n, including the largest one... S = Sum(S,n) + Tail(S,n) and S is an actual value, but Sum(S,n) is large-positive and Tail(S,n) is large negative; then that means that we are misinterpreting what S even is. S-Tail(S,n) = Sum(S,n) is unbounded. But S is a constant. And as n->inf, Tail(S,n) doesn't go to zero, then you can't say that S = Sum(S,n) at all. It's just that we can confuse the two when -1 < r < 1, S = Sum(S). Note where you get a square wave of 0,1 values: A = 1 - A. (1+1)A = 1. A = 1/2. // written without a tail, it's nonsense. A = (-1)^0 + (-1)^1 + (-1)^2 + ... = 1 - 1 + 1 - 1 + ... A isn't the sum. A-Tail(A,n) is the sum. Sum(A,n) oscillates between 0 and 1. Yet A has a value, and it's 1/2. A has to have a value, to be true no matter how many times you expand the recursion. And if you expand to a particular pattern through recursion, then you are equal to that infinite sequence; though different sequences can be equal to the same number.
thanks so much. i'm re-learning high school maths, plus want to learn some new stuff, your channel is awesome, i love the pacing of the lessons in addition to the content itself
Sir you are out a wonderful teacher, you are helping me a lot. This question is troubling me, please help me sir, the sum to infinity is equal to 36 and the sum of the first three terms of a Geometric progression is 81/2, find the first three terms.
Sure - I am working on a comprehensive grade level series through 8th grade math in full detail. I was terrible in math at those grade levels and some important skills are formed so it is very important to me. I'll also be re-doing calculus 1,2,3 physics and chemistry in much greater detail. I have a lot planned!
You nailed it. Much greater detail in calculus, physics, and chemistry, 4k resolution, more problems, and more topics. I consider it my life's work, seriously. I have a lot planned. Thank you! Jason
You are such an awesome teacher Jason Gibson. If there was a Nobel Prize for teaching, you would have been winning that one every year
Thank you so much!
@@MathAndScience Dr. Gibson, can you please make a video on bayesian statistics in research? I am having hard time understanding it, and I think you can definitely help me get some sense of it. Thank you in advance. 😊
Hello Mr. Gibson,
I just wanted to take this chance to thank you for helping me in many subjects like chemistry.
I am good at chemistry, but you were always a second address if I had to know something for school.
Thanks to you help I am very successful in school, that's the reason I'm writing this comment.
Greetings from Germany
That is so awesome! Good luck to you! Jason
Oh my gosh I probably watched like 10 different videos explaining the same topic and I couldn’t understand anything but just watching your video to the 6 minute mark made everything so clear now. THANK YOU. You have a talent for simplification!
Already I learnt this geometric series from one of your lessons. One point I cleared from this lesson is the range of common ratio.
Success is neither a minute work nor a day long work. Rather, it's how much time you spend in a day to reach your goal.
I have been looking at divergent series for a while, and think statements like "-1 = 1 + 2 + 4 + 8 + ... " are just badly stated, but not actually wrong. What's wrong is that in the " + ... " you are not told what is in the tail. If you use recursion, it is concrete, and you can just do the algebra; and never resort to appeals to infinity in any way.
T = Sum(T,n) + Tail(T,n).
T = 1 + 10 T.
(1-10)T = 1.
-9 T = 1.
T = -1/9. ??? because common ratio is 10... but carry on....
T = (1 + 10 T)
= (1 + 10 (1 + 10 T))
= (1 + 10 + 100 T)
= (10^0 + 10^1 + 10^2 + ... + 10^n) + 10^{n+1} T
= Sum(T,n) + Tail(T,n).
T is -1/9, yet it expands like: "1 + 10 + 100 + ... ".
But notice that we dropped the recursive tail, because we don't believe in its importance. T is only Sum(T,n) when Tail(T,n) goes to zero. So, T doesn't mean what people think it does. But T is meaningful. It's a number required to calculate the closed form of the sum of n terms.
T -Tail(T,n) = Sum(T,n)
T(1 - 10^{n+1}) = Sum(T,n)
(-1/9)(1 - 10^{n+1}) = Sum(T,n)
(10^{n+1}-1)/9 = Sum(T,n)
This takes the mystery out of divergent series being associated with a number. The sum is not T. It's Sum(T,n), which is equal to T-Tail(T,n). Recursion might be the only really concrete way to specify infinite processes; because it still uses finite iteration.
When you have a series where there's not a common ratio, but it goes up by 1 for a coefficient, you can usually get there by differentiating a series like this one; so you can do term-by-term addition of series. Because of this, you don't run into any of the odd issues of commutativity/associativity that you will get when infinity is explicitly involved. Recursion is a much more sound basis, when you have a way to use it.
You will run into this problem pretty early on if you try to automate computer algebra. If you follow the algebraic rules and both expand a sequence to a sequence, and also solve it for a number; a computer has no good way to object to it. All you can do is to change the algebra rules until they give you what you want, without breaking unrelated things. So, you have to face this issue of divergent series. It almost always means that we are interpreting it wrongly
The fact that you end up subtracting the tail off of an "infinite" sum to recover a finite sum up to n terms says something deep about infinite processes. It says to me that the concept of an infinite process is kind of hand-wavy, when compared to algebraic rules that you follow without objecting to the results; where you sometimes need to carefully re-interpret something that doesn't make sense.
If S is not the sum here when it's true for EVERY value of n, including the largest one...
S = Sum(S,n) + Tail(S,n)
and S is an actual value, but Sum(S,n) is large-positive and Tail(S,n) is large negative; then that means that we are misinterpreting what S even is. S-Tail(S,n) = Sum(S,n) is unbounded. But S is a constant. And as n->inf, Tail(S,n) doesn't go to zero, then you can't say that S = Sum(S,n) at all. It's just that we can confuse the two when -1 < r < 1, S = Sum(S).
Note where you get a square wave of 0,1 values:
A = 1 - A.
(1+1)A = 1.
A = 1/2.
// written without a tail, it's nonsense.
A = (-1)^0 + (-1)^1 + (-1)^2 + ...
= 1 - 1 + 1 - 1 + ...
A isn't the sum. A-Tail(A,n) is the sum. Sum(A,n) oscillates between 0 and 1. Yet A has a value, and it's 1/2. A has to have a value, to be true no matter how many times you expand the recursion. And if you expand to a particular pattern through recursion, then you are equal to that infinite sequence; though different sequences can be equal to the same number.
the day. I get this good in maths ❤.
thanks so much. i'm re-learning high school maths, plus want to learn some new stuff, your channel is awesome, i love the pacing of the lessons in addition to the content itself
I like your video on infinite geometric series because it has lots of applications to calculus, limits and beyond.
Sir you are out a wonderful teacher, you are helping me a lot. This question is troubling me, please help me sir, the sum to infinity is equal to 36 and the sum of the first three terms of a Geometric progression is 81/2, find the first three terms.
I love you sm rn
Thank you so much for your help.
I’m really proud of you sir
Does the sum of the geometric series is still equals to 1 if the common ratio is a negative fraction?
sn=1/2(1-1/2)n/1-1/2
thanks a lot mr. jason
can you inform us your courses plan for this year ?
Sure - I am working on a comprehensive grade level series through 8th grade math in full detail. I was terrible in math at those grade levels and some important skills are formed so it is very important to me. I'll also be re-doing calculus 1,2,3 physics and chemistry in much greater detail. I have a lot planned!
You nailed it. Much greater detail in calculus, physics, and chemistry, 4k resolution, more problems, and more topics. I consider it my life's work, seriously. I have a lot planned. Thank you! Jason
s10=1/2+1/4+1/8+1/16+1/1024
Thanks my knowledge teacher!!
s2=1/2x2/2=2/4+1/4=3/4
Thank you
s4=1/2+1/4+1/8+1/16
Thank you for the video
Do you have calculus course on UA-cam ?
1/2+1/4+1/8+1/16
Good video
Can I buy this in India?
sn=1-(1/2)n
s=1/2/1-1/2
Best regards
Thank you!
Thanks
I think this is the topic on which Ramanujan worked extensively
s3=1/2+1/4+1/8
sn=t1(1-rn)/1-r
s1=1/2=0.5
r=1/2 r < 1
s=1/2/1/2
s10=961/1024=.938477
s10=512+256+128+64+1/1024
s10=961/1024
Great
1=1/1 > 1/2
4/8=1/2=.5
3940/1024=3
Zabrdast video
1/2+1/2=2/2=1
s10=1023/1024=0000.999022
1 whole number 1/2 fraction
1< 1.5
1/2 = 0.5
1.5> 1
greater > G
6/8+1/8=7/8
1/16=1024/16=64
s=t1/1-r
s10=1/2x 512/512
4/4=1
1>1/2
1=1/1
zero less number then decimal point then 1
0.875
6/8=3/4=.75
8x1024=8192
1.000-.0625
infinite geometry
4880/1024=
1/16=.0625
512/1024
1024/9610
> g greater
240/1024=0
r=12 r < 1
0x1024=0
0 .5 1
3/4=0.75
sn=1
1.0000-.0625
8/8=1
9
s=1
.9375
4x1024=4096
4
16/16=1
1+10+100+1000
1/2
2x1024=2048
formula
.0375
2048-2048=0
2
3x1024=3072
1/32
r=10
less < l
< l less
hahahaha
Please you speak in hindi
s2=1/2+1/4
1/512x2/2=2/1024
1/1x1/2=1/2
s10=961/1024
1/4=1024/4=256