But if you really want to get a grip over mathematics, you should solve problems by yourself...It's just my two scents as I am myself a mathematics lecturer buddy...
this video is nowhere to be found in the kahn academy website :(((( it interrupts the thread of Series. Anyway thank you very much for having all this available
Hi Cant we take a common in numerator and deduce the equation to a(1--r)^n+1 divided by 1-r) which reduces to a(1-r)^n? may be am wrong but just a thought
ROMEO NANDRAJOG If a is pulled out of the numerator, the numerator becomes "a(1-r^(n+1))" which does not equal "a((1-r)^(n+1))". You can check this with small positive numbers for r and n. Hope this helps.
Excuse me can you help me to solve this problem please Find the positive value of n satisfying ∑(1/2)^k = 31/32 where lower limit is k = 1 and upper limit is k = n
How can it exist? the sum diverges...every term would be greater than its preceding term.. So we can't approximately give an answer..but if r is less than one, the series converges and the last few terms would be nearly equal to zero and we can ignore them and give an approximate value..
You will, assuming that you will take Geometry next, run into it in 9th grade (Algebra 2/Trig). It is covered in Chapter 11 of the Algebra and Trigonometry Book 2 published by Houghton Mifflin. Hope that helped!
I think Khan made mistake when he mentioned 1 as a first term and put it in the numerator ( 4:23 ). Because in numerator we need multiplier "a", not the first number of series and in this case, it seems to be also 1, as 1*(1+1/3+1/3^2+1/3^3+...+). I think that mentioning 1 as a first number in a serie and putting in numerator is a little confusing.
you're basically finding the area under those Sums. so like though the n is from 1 to infinity, the area underneath that could still equal a number because it just gets so so small that all those added up could equal one. think of it like this: draw a square. the area of the square is 1. draw a line that cuts that square in half, 1/2 + 1/2 is still one. Now cut it in half again...etc. you can keep cutting that square in half an infinite amount of time but the square's area will always be one.
It's quite amazing, isn't it! Basically what is happening is that the terms you are adding together are getting smaller and smaller (because for a converging geometric series, the ratio is a proper fraction). This means, that when you go to infinity, you are eventually adding zero to the sum. At this point, you have the answer to the infinite sum of the series.
Jason Parry thanks. yeah i know what you mean by the numbers getting very small, but it will never reach zero; and ure adding an infinite amount of positive non zero numbers, so sum should be infinity, no matter how small the numbers get, as long its positive and non zero
+chechennyboiiy When we add together our terms of the series, we notice that some of these series get really close to a finite value, such as 1, 0.75, 0, and so on. When the sum of these numbers keeps getting closer, to the point that only adding them an infinite amount of times will derive our solution, we say the series converges, or equals a finite number despite being added infinitely. If the series does not converge, then it diverges, or has no solution because it gets larger and larger. Not all infinite series will converge, so you want to check if they do or not. There's a bunch of different ways to do this.
I learned 2 new things from this video.
1. Infinite Geometric Series Formula.
2. "Mildly Amazing Thing"
waw i just saw what does it mean mildly on google translate lol
So beautiful i have a tear in my eye
Awww!!!
Awww!!!
Awww!!!
Awww!!!
Awww!!!
I understood in 4.5 minutes what I couldn't in a 45 minute lecture. Sal rocks
1:46 Of course I didn't give it a go. The reason I watch your videos is because I'm too lazy to figure stuff out for myself.
thats the dumbest reason ever
HAHAHAHAHAHA
@@avoncosmeticsandskincareal3163 aka the most realistic reason ever.
But if you really want to get a grip over mathematics, you should solve problems by yourself...It's just my two scents as I am myself a mathematics lecturer buddy...
I’m confused as to why someone would dislike this. Why?
haters gonna hate
First time ever Sal isnt repeating the sentences again and again. Maybe just a few times. Its just wonderful
This is really more than mildly amazing 💕😍
this video is nowhere to be found in the kahn academy website :(((( it interrupts the thread of Series. Anyway thank you very much for having all this available
Thank you Sal!
The ratio could also be less than 0 but more than -1 right? So the common ratio can be -1
I know this is a bit late but your thinking was correct. That's the reason why he referred to the absolute value of the common ratio.
Yes you are right 🥲
Wonderful! Thank you
Awesome! I needed this to understand a proof that "e" is irrational.
Mind blown.
why is it a r^n+1 rather than ar^n ? this is the only thing i am confused about,
Hi
Cant we take a common in numerator and deduce the equation to a(1--r)^n+1 divided by 1-r) which reduces to a(1-r)^n?
may be am wrong but just a thought
ROMEO NANDRAJOG If a is pulled out of the numerator, the numerator becomes "a(1-r^(n+1))" which does not equal "a((1-r)^(n+1))". You can check this with small positive numbers for r and n. Hope this helps.
And Congo for exhate 3600000 subs
What will be the sum of infinite terms if | r | > 1
Amazing
One way to think about it!
Mindly mind blowing
I think you meant to say.
Mindly mild blowing.
stupid question but what is a? is it just 1 the whole time? does it change?
Hi, where can I find the proof for this method?
Nice one
Excuse me can you help me to solve this problem please
Find the positive value of n satisfying
∑(1/2)^k = 31/32
where
lower limit is k = 1 and upper limit is k = n
Restart Life I'm a year late, but just use Symbol lab, or Wolfram Alpha.
does the sum exists if common ratio is greater than one?
Eunice No it doesn't, for it to have a sum -1 < r < 1
How can it exist? the sum diverges...every term would be greater than its preceding term.. So we can't approximately give an answer..but if r is less than one, the series converges and the last few terms would be nearly equal to zero and we can ignore them and give an approximate value..
@@keshavkrishna2722 Thank you.
why is r an absolute value
Hmm... don't the values in the harmonic series approach zero and that is still divergent?
Still didn't get it paying attention tomorrow.
So it’s literally just a / (1 - r) ??
Yes
T(infinity) = FirstTerm/(1-r)
What does the “E”-ish symbol mean? Edit: sigma notation
Samuel L Jackson @ 0:14 and 0:24
Lol kind of
So true
Colors are disconcerting.
I'm in middle school (7th grade) and I'm learning algebra 1, am I gonna learn this in high school or university or college?
you're learning it now, but you will probably encounter it in college. I took high school calculus and statistics and didn't run into it.
You will, assuming that you will take Geometry next, run into it in 9th grade (Algebra 2/Trig). It is covered in Chapter 11 of the Algebra and Trigonometry Book 2 published by Houghton Mifflin. Hope that helped!
KSGguy I ran across it in high school, math III. It depends on the school's curriculum.
Edit: just realized this is a year old .-.
I think Khan made mistake when he mentioned 1 as a first term and put it in the numerator ( 4:23 ). Because in numerator we need multiplier "a", not the first number of series and in this case, it seems to be also 1, as 1*(1+1/3+1/3^2+1/3^3+...+). I think that mentioning 1 as a first number in a serie and putting in numerator is a little confusing.
magic;)
"sensical"? 2:23
Lol
what if r is greater than 1
then you can't apply the infinite theorem to it since -1
Shubham Malhotra If r is greater than 1 the geometric series doesn't have a sum
Not visible clearly sir
infinity is finite, Illuminati confirmed
We need to rek soom skroobs m90
無限等比数列を英語でどう表現するのか初めて知った。
Is it expressed differently in Japanese?
@@lomouche Yes. it's Japanese.
to the power of n instead ?
Bang ding ow
i dont get it, ur are adding an infinite amount of numbers together yet u can get a finite solution, can some1 explain it
you're basically finding the area under those Sums.
so like though the n is from 1 to infinity, the area underneath that could still equal a number because it just gets so so small that all those added up could equal one.
think of it like this: draw a square. the area of the square is 1. draw a line that cuts that square in half, 1/2 + 1/2 is still one. Now cut it in half again...etc. you can keep cutting that square in half an infinite amount of time but the square's area will always be one.
It's quite amazing, isn't it! Basically what is happening is that the terms you are adding together are getting smaller and smaller (because for a converging geometric series, the ratio is a proper fraction). This means, that when you go to infinity, you are eventually adding zero to the sum. At this point, you have the answer to the infinite sum of the series.
Jason Parry thanks. yeah i know what you mean by the numbers getting very small, but it will never reach zero; and ure adding an infinite amount of positive non zero numbers, so sum should be infinity, no matter how small the numbers get, as long its positive and non zero
+chechennyboiiy When we add together our terms of the series, we notice that some of these series get really close to a finite value, such as 1, 0.75, 0, and so on.
When the sum of these numbers keeps getting closer, to the point that only adding them an infinite amount of times will derive our solution, we say the series converges, or equals a finite number despite being added infinitely.
If the series does not converge, then it diverges, or has no solution because it gets larger and larger. Not all infinite series will converge, so you want to check if they do or not. There's a bunch of different ways to do this.
Can we sum this 2+4+8+16+....∞
That series diverges so the answer is infinity. You can only get a finite value if the series is convergent ( 0 < | ratio | < 1 )
why the hell is Samuel L Jackson in the video
......Iam not knowing this 😅😂😂🙂🤣
thanks, very helpful