Sequences and series (part 2)
Summary
TLDRThis educational video script explores the concept of geometric series, demonstrating how to calculate their sum using the formula S = (a^{n+1} - 1) / (a - 1). It illustrates the process with examples, including the intriguing scenario where the sum of an infinite geometric series with a base of 1/2 converges to 2. The script effectively conveys the beauty of mathematics by showing how an infinite series can result in a finite, calculable sum.
Takeaways
- 🧮 The video continues from where the previous one left off, discussing the geometric series.
- 🔢 A geometric series involves summing terms of the form 'a^k' where 'a' is a constant and 'k' is the exponent.
- 🧠 The speaker defines the geometric sum (S) and introduces another sum (a * S) to help simplify the calculation.
- ➖ Subtracting the second sum (a * S) from the first sum (S) leads to the cancellation of many terms, simplifying the expression.
- ✏️ The resulting formula for the sum of a geometric series is S = (a^(n+1) - 1) / (a - 1).
- 🔍 The formula is useful for calculating the sum of finite geometric series, such as summing powers of 3 up to 3^10.
- ♾️ The speaker transitions to discussing infinite geometric series, which converge if the base 'a' is less than 1.
- ⚖️ When the base 'a' is a fraction (like 1/2), the terms get smaller and smaller as the series progresses.
- 🔗 The formula can also be applied to infinite series, and as n approaches infinity, the sum converges to a finite number.
- 🎯 A specific example is given: the sum of 1/2, 1/4, 1/8, etc., converges to the value 2.
Q & A
What is a geometric series?
-A geometric series is a sequence of numbers where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio.
What is the sum of a geometric series with the first term 'a' and the last term 'a^n'?
-The sum S of a geometric series from a^0 to a^n can be calculated using the formula S = a^(n+1) - 1 / (a - 1), provided that the common ratio 'a' is not equal to 1.
Why did the presenter correct 'a^(n-2)' to 'a^n' in the script?
-The presenter corrected 'a^(n-2)' to 'a^n' because the original notation was incorrect. The series should include terms up to 'a' raised to the power of 'n', not 'n-2'.
How does the presenter define the sum 's' in the script?
-The presenter defines 's' as the sum of a geometric series, which is the sum of terms from 'a^0' to 'a^n'.
What is the purpose of defining another sum 'a*s' in the script?
-Defining 'a*s' allows the presenter to create an equation that, when simplified, helps derive the formula for the sum of a geometric series.
What happens when the presenter subtracts 's' from 'a*s' in the script?
-When 's' is subtracted from 'a*s', all terms except the first ('a^0') and the last ('a^(n+1)') cancel out, leaving 'a^(n+1) - 1'.
Why does the presenter multiply the top and bottom of the fraction by -1 in the script?
-The presenter multiplies the top and bottom by -1 to simplify the fraction and make it easier to understand that the sum of an infinite geometric series with a ratio of 1/2 is 2.
What is the significance of the presenter's question about subtracting 's' from 'a*s'?
-This question is significant because it leads to the discovery that the sum of a geometric series can be represented by a simple formula, which is crucial for understanding the properties of geometric series.
How does the presenter demonstrate the concept of an infinite geometric series?
-The presenter demonstrates an infinite geometric series by considering a series with a common ratio of 1/2 and showing that the sum converges to a finite number (2) as the number of terms approaches infinity.
What does the presenter suggest as a project to further understand geometric series?
-The presenter suggests drawing out an infinite geometric series as a pie chart to visualize how adding smaller and smaller pieces results in a finite sum.
Why is the sum of an infinite geometric series with a ratio of 1/2 equal to 2?
-The sum equals 2 because as the number of terms approaches infinity, the terms become infinitesimally small and approach zero, leaving only the first term 'a^0' which is 1, and the last term 'a^(n+1)' which approaches 1, resulting in the sum being 1/(1 - 1/2) = 2.
Outlines
📐 Understanding Geometric Series
The paragraph introduces the concept of a geometric series, where a base number 'a' is raised to increasing powers and summed up. The presenter corrects a previous mistake in notation, clarifying that the series should sum up to 'a^N'. They define 's' as the sum of this series and then create another sum 'a*s', which is essentially the original sum multiplied by 'a'. By distributing 'a' across the sum, they show how each term's exponent increases by one. The presenter then explores what happens when 'a*s' is subtracted from 's', resulting in a simplified expression of '-a^0 + a^(N+1)'. They conclude by deriving the formula for the sum of a geometric series as 'a^(N+1) - 1' divided by 'a - 1', emphasizing its utility.
🔍 Applying the Geometric Series Formula
This section applies the geometric series formula to a specific example using the base '3' raised to powers up to '3^10'. The presenter substitutes 'a' with '3' and 'n' with '10' in the formula to find the sum. They then transition to discuss the concept of infinite series and their convergence. The focus is on how an infinite geometric series with a fraction as the base, such as '1/2', can converge to a finite sum. By taking the limit as 'n' approaches infinity, the presenter demonstrates that the sum of an infinite geometric series with a base of '1/2' converges to '2'. The presenter expresses amazement at the result, highlighting the counterintuitive nature of summing an infinite number of terms to get a finite number.
Mindmap
Keywords
💡Geometric Series
💡Base
💡Exponents
💡Sum
💡Convergence
💡Infinite Series
💡Limit
💡Fraction
💡Formula
💡Powers of a Number
💡Distribute
Highlights
Introduction to the concept of a geometric series
Definition of a geometric series with base 'a'
Explanation of increasing exponents in a geometric series
Summation of a geometric series up to a^n
Correction of a mistake from the previous video
Definition of sum 's' as the geometric series sum
Introduction of another sum 'a times s'
Distributive property applied to the sum 'a times s'
Result of distributing 'a' across the sum
Derivation of the formula for the sum of a geometric series
Subtraction of the original sum 's' from 'a times s'
Result of the subtraction leaving only 'a^(n+1) - 1'
Final formula for the sum of a geometric series derived
Practical application of the formula with base 'a' as 3
Exploration of the sum of powers of 3 up to 3^10
Introduction to the concept of infinite series
Condition for an infinite series to converge
Example of a geometric series with base 1/2
Summation of an infinite geometric series with base 1/2
Limit of the sum as 'n' approaches infinity
Result of the infinite sum converging to 2
Amazement at the finite result from an infinite sum
Suggestion for a project to visualize the summation
Conclusion and anticipation for the next video
Transcripts
Welcome back.
So where we left off in the last video, I'd shown you
this thing called the geometric series.
And, you know, we could have some base a.
It could be any number.
It could be 1/2, it could be 10.
But that's just-- but some number.
And we keep taking it to increasing exponents, and we
sum them up, and this is called a geometric series.
And so I want to figure out the sum of a geometric series of,
you know, when I have some base a, and I go up to some
number a to the n.
What-- is this a to the-- why did I write
a to n minus 2 there?
That should be a to the big N.
My brain must have been malfunctioning in
the previous video.
That always happens when I start running out of time.
But anyway.
Let's go back to this.
So I defined s as this geometric sum.
Now I'm going to define another sum.
And that sum I'm going to define as a times s.
And that equals-- well, that's just going to be a times
this exact sum, right?
And that's the same a as this a, right?
That a is the same as this a.
So what's a times this whole thing?
Well, it's the a times a to the zero is-- let me
write it down for you.
So this'll be a because I just distribute the a, right? a
times a to the zero, plus a times a to the 1, plus a times
a squared, plus all the way a times a to the n minus one,
plus a times a to the n.
I just took an a and I distributed it along
this whole sum.
But what is this equal to?
Well, this is equal to a times a to the zero.
That's a one-- a to the first power-- plus a squared, plus a
cubed, plus a to the n, right?
Because you just add the exponents, a to the n.
Plus a to the n plus 1.
So this is as.
And we saw before that s is just our original sum.
That is just a to the zero, plus a to the 1, plus a
squared, plus up, up, up, up.
All the way to plus a to the n, right?
So let me ask you a question.
What happens if I subtract this from that?
What happens?
If I say, as minus s.
Well, I subtracted this from here, on the left hand side.
What happens on the right hand side?
Well, all of these become negative, right?
Let me do it in a bold color.
This becomes-- because I'm subtracting-- negative,
negative, these are all negatives.
Negative.
Negative.
Well, a to the first, minus a to the first.
That crosses out. a squared minus a squared crosses
out. a to the third, it'll all cross out.
All the way up to a to the n, right?
So what are we left with?
We're just left with minus a to the zero, right?
We're just left with that term.
And we're just left with that term.
Plus a to the n plus 1.
And of course, what's a to the zero?
That's just 1.
So we have a times s minus s is equal to a to
the n plus 1 minus 1.
And now let's distribute the s out.
So we get s times a minus 1 is equal to a to the n
plus 1 minus 1, right?
And then what do we get?
Well, we can just divide both sides by a minus 1.
Let me erase some of this stuff on top.
I think I can safely erase all of this, really.
Well, I don't want to erase that much.
I want to erase this stuff.
That's good enough.
OK.
So I have just-- dividing both sides of this equation by a
minus 1, I get s is equal to a to the n plus 1 minus
1 over a minus 1.
So where did that get us?
We defined the geometric series as equal to the sum.
From k is equal to 0, to n of a to the k.
And now we've just derived a formula for what that
sum ends up being.
Equals a to the n plus 1 minus 1 over a minus 1.
And why is this useful?
We now know, if I were to say, well, what is-- let me clean
up all of this, as well.
Let me clean up all of this and we can-- OK.
So if I said, you figure out the sum of, I don't know, the
powers of 3 up to 3 to the, I don't know, 3 to
the tenth power.
So, you know, 3.
So 3 to the zero, plus 3 to the one, plus 3 squared, plus all
the way to 3 to the tenth.
So this is the same thing as the sum of k equals zero
to 10, of 3 to the k.
Right?
So this formula we just figured out, a is 3 and n is 10.
So this sum is just going to be equal to 3 to the eleventh
power minus 1 over 3 minus 1.
Which equals-- well, I don't know what 3 to
the eleventh power is.
Minus 1 over 2.
So that's kind of useful.
That is a number.
Although you'd have to memorize your exponent tables to the
eleventh power to do that.
But I think you get the idea.
This is especially useful if we were dealing with-- well, if
the base was a power of ten, it would be very, very easy.
But what I actually want to do now is I want to take this and
say, well, what happens if n goes to infinity?
Let me show you.
So what happens?
So there's two types of series that we can take-- that's
not what I wanted to do.
There are two types of series that we can take that we
can find the sums of.
There's finite series, and infinite series.
And in order for an infinite series to come up to a sum
that's not infinity, they need to-- what we say--
they need to converge.
And if you think about what has to happen for them to converge,
every next digit has to essentially get smaller and
smaller and smaller, as we go towards infinity.
So let's say that a is a fraction.
a is 1/2.
So how does a geometric series look like if we have 1/2 there?
So let's say that we're taking the geometric series from k
is equal to 0 to infinity.
So this is neat.
We're going to take an infinite sum, an infinite number of
terms, and let's see if we can actually get an actual number.
You know, we take an infinite thing, add it up, and it
actually adds up to a finite thing.
This has always amazed me.
And the base now is going to be 1/2.
It's 1/2 and it's going to be 1/2 to the k power.
So this is going to be what?
1/2 to the zero, plus 1/2, plus-- what's 1/2 squared?
Plus 1/4, plus 1/8, plus 1/16.
So as you see, each term is getting a lot, lot smaller.
It's getting half of the previous term.
Well, let's say, what happens if this wasn't infinity?
What happens if this was n?
Well, then we'd get plus 1 over 2 to the n, right?
1/2 to the n is the same thing as 1 over 2 to the n.
And if we look at the formula we figured out, we would say,
well, that is just equal to 1/2 to the n plus 1, minus
1, over 1/2 minus one.
And that would be our answer.
We'd have to know what n is.
But now we want to know what happens if we go to infinity.
So this is essentially a limit problem.
What happens-- what's the limit, as n goes to infinity,
of 1/2 to the n plus one minus 1 over 1/2 minus 1?
Well, all of these are constant terms, so nothing happens.
So what happens as this term, right here, goes to infinity?
What's 1/2 to the infinity power?
Well, that's zero.
That's an unbelievably small number.
Take 1/2 to arbitrarily large exponents, this just goes to 0.
And so what are we left with?
We're just left with this equals minus 1 over 1/2 minus
1, or we could multiply the top and the bottom by negative 1.
And we get 1 over 1 minus 1/2.
Which equals 1 over 1/2, which is equal to 2.
I find that amazing.
If I add 0 plus 1/2 plus 1/4 plus 1/8 plus 1/16 and I never
stop-- I go to infinity-- and not infinity, but I go to 1
over essentially 2 to the infinity-- I end up with
this neat and clean number.
2.
And this might be a little project for you, to actually
draw it out into like maybe a pie and see what happens as
you keep adding smaller and smaller pieces to the pie.
But it never ceases to amaze me, that I added an infinite
number of terms, right?
This was infinity.
And I got a finite number.
I got a finite number.
Anyway, we ran out of time.
See you soon.
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