Overcome Problems With One Simple Trick
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- Опубликовано: 5 ноя 2023
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Explore the intriguing mathematical showdown between 1.001^1000 and the number 2 ,a must-watch for Math Olympiad aspirants and math enthusiasts alike. We break down this complex comparison with engaging visuals and clear explanations, perfect for those preparing for challenging competitions like the International Mathematical Olympiad (IMO). This video not only sharpens your problem-solving skills but also delves into the kind of exponential expressions and numerical reasoning found in Math Olympiad problems. Whether you're aiming to compete in the World Math Olympiad or simply love to challenge your mathematical prowess, join us for this insightful analysis.
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4 days ago?
What font do you use in your videos?
"... our representation of two looks extremely familiar [sic] to this representation ..." -- you speak as though representations were people; "familiar" is no substitute for "similar".
I mean I was doing all that, but then I realized that 1.001^1000 is an approximation of e
Yeah, as soon as he expanded it to (1 + 1/1000)^1000 I immediately saw e
Yes
For anyone wondering, the reason why 1.001^1000 (approx 2.71692) is so close to e (2.71828...) is because 1.001^1000 is of the form (1+1/x)^x which approaches e as x approaches infinity, now obviously 1000 is much smaller than infinity so we get a value that's close to but not equal to e
Here's another question compare mathematically 1.001^1000 vs. 3. Do it mathematically rigorously. If so, that could potentially prove that e is between 2 & 3.
nice
ok
It is close to e because in the first place, that is the definition of e!!!
oh gosh you're challenging me as the most liked comment on this video
My solution:
By the binomial theorem and the knowledge that "n choose 1" equals n, we get
1.001¹⁰⁰⁰ = (1 + 0.001)¹⁰⁰⁰ = 1¹⁰⁰⁰ + 1000*1⁹⁹⁹*0.001 + (more terms which are all positive) = 1 + 1 + (something positive) > 2.
I also thought this
Same here
Though not very expandable, I just figured: 1.001 on each integer exponentiation will grow by at least .001 (1.001^2=1.002001). Since you're doing this 1000 times, .001*1000=1, which means the number necessarily grows by more than 1, so 1+ >1 = >2.
That's how I got it, too.
1.001 ^ 1000 = (1 + 1/1000)^1000
and (1 + 1/1)^1 = 2
lol obvious answer and tho (1 + 1/x)^x approaches e as x approaches infinity
ofc 1.001^1000 > 2
thats exactly my thought process also
me with my calculator
By Bernouli inequality (1+x)^n >1+nx this is immediately
My method before watching the video:
0.001 is 0.1% and notice that adding 0.1% 1000 times is equal to 1. Using simple interest as an example, that would be the same as getting 0.1% simple interest 1000 times, which would leave you with 2 (or 2x the original amount)
Now compare that to 1.001^1000 which would be the same as getting 0.1% interest 1000 times, except that now its compounded interest.
Now its very clear that compound interest grows "faster" than simple interest, hence 1.001^1000 > 2
Another easy way is to use the binomial theorem.
(1+1/1000)^1000 = 1^1000 + (1000C1)*1^999*(1/1000)^1 + ...
We know that all the other terms in the expansion will be positive, so we can just remove them and the result will be smaller. So,
(1+1/1000)^1000 > 1 + 1000*1*1/1000
=1 + 1
=2
Therefore 1.001^1000 > 2.
Great solution nonetheless.
Yes, exactly, came here to say this. Much quicker and simpler.
that's how i did it
Yeah I did the same! very neat and easy since the first term is obviously 1 and we only need the second term to prove it larger than 2, proving it larger than (sqrt5) would be trickier using this method though.
PS- the third term is enough for sqrt5! just tried and it is 999/2000 which i greater than 0.254 or whatever
I was trying to calculate e recently, so this was easy :>
there actually is another really nice way to solve this equation!
Guessing 🎉🎉
That's a neat solution! Mine wasn't quite as interesting, but it hasn't been said yet, so I'll share it.
In xy = x+z, z gets larger as x does if y stays constant. If we take the product 1.001¹⁰⁰⁰ in steps, we are multiplying larger and larger x (x = partial products) by the same y = 1.001. The first product is 1.001*1.001 = 1.002001, so the first z is 0.001001. For the first x and z, x+1000z = 2.002 > 2, and all z are greater than or equal to the first, so 1.001¹⁰⁰⁰ > 2.
Hi @delta3244, i really liked your solution, even though you said it was not that nice. But i couldn’t understand why you said that two was equal to x + 1000z. I thought that ln xy = x + z and two wasn’t in this form. Could you explain why, please? Or/and recommend where to find it’s deduction and explanation. I would really appreciate
@@victorsilva9779 Thank you. In retrospect, that sentence is deceptive thanks to l and I being near-indistinguishable in YT's font. In that sentence, "In" = "in." The setence begins with "in [xy = x+z], ..." (brackets added around the mathematical expression for clarity). Sorry for the confusion.
Hey delta, I didnt understand why You multiplied z by 1000 so that the expression is equal to 2.002, could You please explain it?
@@guilhermenoronha1978 multiplying 1.001 by 1.001 is equivalent to adding 0.001001 ("z") to 1.001. Multiplying a larger number than 1.001 by 1.001 is equivalent to adding more than z to the number. When 1.001 is raised to the 1000th power, it is equivalent to adding ≥z to it 1000 times, so the final sum is ≥[1.001 + 1000z]
(1 + 1/1000)^1000 is just a rough approximation of e
Nice solution. We can use the binomial expansion too. The first and the second term are equal to 1, the rest of the 999 terms are all positives, so the sum it will be larger them 2
Really underated comment
The binomial expansion of (1+x)^n gives
(1+x)^n = 1 + n*x + [...]
For x = 0.001 and n = 1000:
1.001¹⁰⁰⁰ = 2 + [...]
So it is greater than 2 because it is 2 plus a positive quantity.
You can actually kind of tell just from looking at it which is larger. 1.001^1000 is (1 + 1/1000)^1000 which is in the form of the limit definition of e and 1000 is the term for n which is a pretty large number so you know it must be at least 2.7 something or something around there.
Sure, but that’s not a guarantee it’s larger than 2. And there are easier ways to “kind of tell” without using the limit definition of e.
@@Alex-02 What are easier ways to tell just from looking at it?
@@maxhagenauer24 Let’s say you have a number a larger than 1. We all know intuitively that a*1.001 > a + 0.001.
If you were to add 1 + (0.001 a thousand times) you’d get exactly 2. When you multiply by 1.001 a thousand times this number will then be larger than 2.
In my head I’d go like this:
1 * (1.001^1000) > 1 + (0.001 * 1000) = 2
Let me know if this was understandable, it’s always difficult to explain math in a yt comment. Great thing about this is a high schooler could do it. Also, you could do a more rigorous proof based on this as I’ve seen one person do in the comments.
@@Alex-02@Alex-02 You took a=1, which is not >1 and for a=1, a+0.001 is = a*1.001,
not a+0.001 < a*0.001.....
How then is, acc. to you, 1.001¹⁰⁰⁰>2
@@Alex-02And yeah, one more thing.... e limit definition method IS A GUARANTEE for (1+1/1000)¹⁰⁰⁰>2 because you know: in f(x)=(1+1/x)^x............
f(x) = 2 if x=1 (calculable)
It's = 2.036 if x=1.1
It's = 2.25 if x=2 (yet calculable)
It's = 2.593... if x=10
It implies that if x>1, then f(x)>2,
because x=1 implies f(x)=2
All the stuff above thereby implies:
(1+1/1000)^1000 > 2
100% ✅✅
One more thing..in ur other comment you say that a*1.001>a+0.001, is *INTUITIVELY* correct for a>1, here's mathematical tiny proof:
a*(1.001)=a*(1+0.001)
=a+a*0.001>a+0.001
NO MORE INTUITION needed lol
Have a good day
The brief, if good, is twice good! Binommial expansion is the best method.
I used the factor 72 to estimate that 1.001 ^ 1000 was larger... 1.001 can be written as 1 + 0.1%; You can estimate the number of compoundings to just about double the initial number by using 72 / 0.1 = 720. This is less than 1000, so 1.001 ^ 1000 must be somewhat larger than 2.
Since 1.001 ^ 1000 can be expressed as ( 1 + 1/1000 ) ^ 1000 and we know it converges rather quickly to Euler's constant, it is bigger than 2
I love this collapsing telescoping technique....
I remember when I saw a question like this once I plugged the fraction into the “ratio to cents” formular for intervals in microtonal music (1200*log_2(ratio))
we can directly use binomial approximation (1+x)^n = (1+nx) + n(n-1)x²/n when x
lim n->inf (1+1/n)^n = e, and e > 2, so for large n values we can reasonably guess the expression will be close to e and therefore greater that 2
I thoght the problem in this way:
First i took the natural logarithm:
ln(1.001 ¹⁰⁰⁰) = 1000 ln(1 + 10-³)
Using a Taylor series aproximation:
ln(1+x) ≈ x if x->0
So
ln(1.001 ¹⁰⁰⁰) ≈ 1000•10-³ = 1
Then
1.001 ¹⁰⁰⁰ ≈ e > 2
It is not a perfect demonstration but It shows why the numer Is so near to the Euler number.
You can also note that in general, for {ai} with each ai>=1, then a1*...*an>=1+(a1-1)+...+(an-1)=a1+...+an-(n-1) (for a,b>=1, with a>1, then ab=(1+(a-1)(1+(b-1))=1+(a-1)+(b-1)+(a-1)(b-1)>=1+(a-1)+(b-1), and inductively it holds) (essentially multiplication is more powerful than addition in a particular way if everything is greater than or equal to 1)
Nice!
The boring was to find this out would be to take 2^(1/1000).
The more fun way would be to use the graph (1+1/x)^x, and see that it’s not only always increasing, but also that at x=1, y=2, and x=1000 would net a larger number!
This is fun 🙃
I wound up taking the derivative of (1 + 1/x)^x to make sure it's strictly increasing everywhere it matters
And yes, for x > 0 it is increasing
But… you used a calculator to solve the problem… At that point just put (1.001)^1000 in the calculator and get the answer directly
Simpler way. Use the fact that ln(1+x) is close to x if x is small. Done. Left side is close to e > 2
Hey briii! Can you please do i^infinity?
The way I did it isn't as elegant... I can write 2 = (1 + 1/1)^1 to put it in the same for of f(x) = (1 + 1/x)^x
Then I proved the function is strictly increasing by taking the derivative and making sure it's positive everywhere it should (which happens to be when x > 0)
since f(1000) > f(1), 1.001^1000 > 2
I could have also used binomial theorem to make it quick:
(1 + 1/1000) ^ 1000 = 1 + 1000(1/1000) + positive stuff
= 2 + extra positive stuff
> 2
I did it slightly diffrent
(1+1/x)^x = 1 + (x choose 1)*1/x + ... + (1/x)^x where x is a natural number and x > 1
(1+1/x)^x = 1 + x*1/x + ... + (1/x)^x
(1+1/x)^x = 1 + 1 + ... + (1/x)^x
(1+1/x)^x = 2 + ... + (1/x)^x
since every term is positive the equation has to be:
2 + ... + (1/x)^x > 2
Just transform 1.001^1000 into 2^(log(1.001)/log(2)*10000), then you will get that the exponent is about 1.44, which is bigger than the exponent for 2, which is 1.
limit of (1+1/n)^n as n goes to infinity is e, if n = 1, the expression is equal to 2, since 2 is smaller than e that means for any value n larger than 1 the expression must be larger than 2.
It's been a long time since l had any math classes (50+ years) aside from applied math related to work. I worked as a machinist, tool maker and machine rebuilder. But just at a glance l was thinking that obviously 2 > 1.001 to the 1000th power (my device won't do powers). At a glance my first thought was the portion after the decimal is going to keep getting pushed farther out. But even working it by hand to the cube it becomes blatantly clear that 2
A slight bit of nitpicking here, but at 3:15 you say that the first equation must be greater than the second one. I fully understand what you meant. But equations are statements of equality and are either true or false. They don’t really have size. The left and right-hand sides of each do. Of course that’s what you meant to compare. Again, sorry if I’m being overly stringent. Good video!
An easier way of figuring out which is bigger, in this example, would be by taking 0.001x1000 as an approximation, which is 1, and add the leading 1. Together thats 2, excluding the compounding effect, and hence it must be bigger than 2
Use the binomial equation which tells us that (a+b)^n = sum over k from 0 to n of a ^k *b^(n-k)*binomial (n,k). Now take a=1 ,b=1/n .
Then we have 1+1+higher order terms,all of them positive.Note that binomial (n,0) =1 ,binomial (n,1)=n ,
which gives the first two terms with 1.This shows that (1+1/n) ^n >2 for n>1.
Well I use log and its much easier that way
We use log rules on 1.001^1000 and by using logx^y=ylogx and log(m/n)=logm-logn
And log of 1.001^1000 comes out around 0.4 while log2 is 0.301
I don't know if we can use this or not but math is fun
Hmmm. I was expecting the compounding power to give a much larger number than 2.7. Nice exposition
There is a simpler way to prove it.
Just take the first 2 terms of the expansion
(1+1/1000)^1000 equivalent to 1.001 ^ 1000:
(1+1/1000)^1000 = 1^1000 + Combin(1000, 1)*(1/1000)^1 + Combin(1000, 2)*(1/1000)^2 + ....
Now, the first term of this expansion is 1, and the second term is also 1 because Combin(1000, 1) = 1000.
Like all other terms in the expansion, although small they are positive, it can be concluded that 1.001^1000 is greater than 2.
Bem mais fácil aliás
A good way to solve this is to recognise that it is (1+1/1000)^1000, where 1000 is a big number and that is how e is defined, so 1.001^1000 would be slightly less than e, which is greater than 2.
That's a good argument to convince yourself, but it doesn't actually prove it.
You know that (1+1/x)^x must be bigger than 2 eventually, but you don't know how big x needs to be to get there.
@@anticorncob6 oh true
Dang I would not have thought of that
I took ln then divided =ln(2-1.001) which is close to 1 so close to 0 which mean 1000> almost 0
I think the binomial theorem is also useful since in the expansion of (1+1/1000)^1000 we get the first two terms as '1' so we have 2 + non-zero terms = greater than 2
(1±x)ⁿ when x tends to zero is approximately equal to 1±nx
Anyways great video 😁!
i'd just approach it with the binomial expansion method:
1.001^1000 = (1 + 0.001)^1000 = 1 * 1^1000 * 0.001^0 + 1000 * 1^999 * 0.001^1 + (loads of very small, yet positive terms)
< 1 + 1000 * 0.001 = 2
This fight will be legendary
„Hey learn to overcome math problems“
„Just rewrite 2“
Just looking at (1+1/a)^2 = 1+2/a+a^2, we get that it is larger than 1+2/a, and from this we can conclude that (1+1/a)^n > 1+n/a. This shows us (1+1/1000)^1000>2
I like his proof but it's easier to work with 2 than 1.001^1000. All you need to do is express 2 in the form { 2^(x) }^1000 and check if {2^(x)} is greater or less than 1.001. This can be done: {2^(1/1000)}^1000 since 2^(1/1000) < 1.001 therefore, 1.001^1000 > 2
What font do you use
« overcome this problem with one simple trick »
pretty sure using a calculator doesn’t take 5 minutes good sir
Just open the binomial expansion of 1+x, and you get 1+nx + .... And you already have 2 by these first two terms + some other positive small numbers. So obviously it's bigger than 2.
easier way, if the exponent for this similar thing is larger than 1, it is greater.
1.5^2,1.01^100, 1.0000000001^10000000000, etc
The formula for e: (1 + 1/n) ^ n already gives 2 when n=1. The larger n is, the larger the result is. The problem has n=1000 and since 1000 is larger than 1, the result must be larger than 2
i used this way
1.001^1000
=(1+1/1000)^1000
=1+1000(1/1000)+1000C2(1/1000)(1/1000)+...
=2+...
>2
judging by the looks of it, being definition of e the left part should approach it~2.72 therefore it's bigger
Dude i think that we should use the binomial theorem that can also be done
Legal, mas dá para usar o binômio de newton que vai ser bem mais rápido
1.001^1000 is fairly obviously, to me at least, larger. Multiplying by 1.001 adds at least .001 to the original 1, and a bit more to the remainder. Doing that 1000 times will give you 1.001 + .001*999 + a bit more, which is 2 + a bit more, which is more than 2
Or you can use an approximation of the logarithm around 0;
Exp(1000*ln(1+0.001))=exp(1000*(0.001+o(0.001))=exp(1+o(1)) ~ e
And e>2
Here's how I figured it out:
Start with (1+1/n)^n.
The first product is n=1000, and the second n=1. As the limit is bigger than 2 as n blows up, the product gets bigger as n does.
Since I fairly routinely use the fact that ln(1 + x) ≈ x for small x, it seems much easier to me to compare ln(1 + 0.001)¹⁰⁰⁰ ≈ 1000(0.001) ≈ 1 vs. ln2 ≈ 0.693.
that's neat!
This is pretty cool but the easy way is to just put a ln on both sides then bring the 1000 in front.
1000*ln(1.001) > ln(2)
When there are exponents in a problem always think ln because 90% of the time that is the quickest way to solve it
My first impression is that 1.001^1000 is same as (1+1/1000)^1000, so definitely it is bigger than 2 because (1+1/1)^1=2 and it goes to e...
The moment you expanded the 1.001^1000 I let out a “dammit that’s just e” 😭
I saw 2.71 and nearly got a heart attack, euler has plagued my life
Very interesting but could we get an equation that can find what number in the first form is equal to a given quantity in integer form?
the binomial is one, but very tedious for large exponents like these
easily solvable using (1+x)^y>1+(x*y) when x>0 and y>1
3:20 Minor nitpick: Expression or term, not equation.
I’d just use the binomial theorem. See immediately that (1+1/1000)^1000 = 1 + 1 + lots of positive numbers > 2
a cool solution but if you just wrote out 1.001^2 and 1.001^3 you would also quickly notice a pattern that will get slightly bigger than 2 after a thousand iterations, and I feel like most people could do that a lot faster
I had a simpler solution:
1*1.001 adds 1/1000 to the original number. As you do this with bigger and bigger numbers, you're gonna be adding more and more.
2 is just adding 1/1000 to 1, 1000 times. That's exactly like the first one, except it's not increasing
Yo todo un Alberto Instantáneo: * Lo pone en la calculadora *
A lot of people are saying that the limit ot (1+1/x)^x is e which is 2.718... so plugging in 1000 means it's almost e and thus bigger than 2.
That's good to convince yourself but it doesn't actually prove it.
There's a simple approximation: (1*ε)^n, where ε is a small value, is about 1 +n*ε, (follws from the binomial formula), and this approximation is always a bit low (also from the binomial formula). No more proof necessary.
U could also use logarithms and determine that 1.001^693.5 is just over 2
Why is the value so close to the constant e? I tried doing it 1.0001^1000 or 1.001^10000 and other variations first under the assumption that it would approach e but it’s only specifically 1.001^1000 which is so close to e.
read the top comments, this form IS e
log base 1.001 of 2 is less than 1000 therefore 1.001^1000 > 2
i read it on the thumbnail as 1001^1000 vs 2 and i cackled thinking it's a joke video, then i realized my mistake and cackled at myself XD
(1 + 1/x)^x is always >= 2 unless x => 1
2
At around x=700, the 1.001 to the x beats the 2.
3:30 e Jumpscare 🤭
I finished my last maths exam5 months ago Yet here i an
if you know binominal theorem you'd know that (1+1/1000)^1000 = 1+1000*(1/1000)*1+...
First the sum of first 2 numbers is already equal two, and other numbers in this sum are larger than 0, so this sum is greater than 2. That's a faster solution
(1+x)^n>=nx and because 2^(1/1000) is irrational its not going to be equal here: (1+1/1000)^1000>=1+1000/1000=2 qed.
If you know that lim(x→∞) (1 + x⁻¹)ˣ = e, then the answer is pretty easy.
For x = 1, it is already equal to 2.
For x = 2, it is 2,25 > 2.
So, for x = 1000, it is also bigger than 2, of course.
You also have to show that (1+1/x)^x is increasing.
Solved it in my mind getting that the first two terms of the binomial expansion of (1+0.001)¹⁰⁰⁰ which are 1¹⁰⁰⁰+1000•1⁹⁹⁹•0.001 = 2, all the other ones are strictly greater than 0, so 1.001¹⁰⁰⁰ > 2. (one could also get that by knowing that e ≈ 2.718 >> 2)
That's what I did too. But where does e come from? Thank you in advance.
@@Ninja20704 (1+1/n)^n approaches e as n approaches infinity, so n = 1000 would be pretty close to it (it's not a proof, just intuition)
@@lukandrate9866 ah i see, I forgot about that definition of e. Thank you
@@Ninja20704 No problem! Could I share a different problem with you? (i've got noone to share my "discovery" with)
@@lukandrate9866 Share with me! (If you will) I will be glad to hear it! Be aware though that I won't be able to reply to you in a timely manner, maybe a day late or two. But I can assure you that I will reply when I have the time.
it is approximately e because (1+1/x)^x when x is large is approximately e!！
1.001^1000 = e^(1000*ln(1.001)) = e^(1000)^ln(1.001)≈ e^1000^0 = e ≈ 2,72.
(1+1/x)^x > 2, {x >= 1}
Looks like eulers number
(1+x)^n = 1+nx+...
(1+0.001)^1000 = 1+(0.001x1000)+... = 2+... > 2
1.001*1.001 = 1.002 allready give you answer what will be bigger and bcoz its *1.001 result get bigger every time
so it sure over 2
thanks my mind is flashed
POV: you have heard of e at least one time in your life:
"BRO THE ANSWER TO THAT IS OBVIOUS."
easy. first take the 1000th root of 2, if its smaller than 1.001 than 1.001^1000 is bigger, otherwise its smaller.
I just guessed and got it correct :D
When I saw the (1+1/1000)^1000, I exactly knew that it'll be greater than 2, as e = 2.718...