Math Olympiad Question | Which Is Larger? | You should learn this method
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- Опубликовано: 8 фев 2025
- 1.005^200 and 2, which one is larger? A fantastic math problem. A common method to this kind of questions.
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Most proofs in the comments here are needlessly complicated. The binomial expansion of (1+x)^n is 1+nx + additional positive terms. The first two terms add to 2 in this instance (n=200, x=.005) so the additional terms make the LHS larger than 2.
Note added: in no way am I criticizing the original video. There’s all different ways to solve problems. However I do find some of the methods discussed in the comments to be needlessly complicated. I shared my own thoughts on it and many agreed. I don’t feel I was being nasty about it.
Finally someone with some sense. I was going to comment the same but thank you for saving everyones time
Snap, it is exactly what I thought too.
i agree but the idea will be good
Also IIRC (1+x)^n>1+nx is learned in the first few weeks of calculus, before binomial
EDIT: I did not remember the name of this inequality, but according to another comment it is bernoulli's inequality, which is true for x>-1 (while your derivation naively only holds for positive x).
On the other hand while binomial distribution in learned later it is much easier to intuitively explain
@@감나빗-26 Exactly :)
0.005 is half a percent, and if you increase something by half a percent 200 times it will more than double. Intuitive way I solved the problem.
1.005^200 is kind of close to e.
This is actually the definition of e instead of 200 take Infinity
Is your name in reference to the Jerky Boys?
@@hanskywalker1246 this is not the definition of e. There's no limit and x is only 200, hence my point.
@@efisgpr actually because of Han solo in Star wars, he's family too
@@Gideon_Judges6 yes, but I wanted to point why this number is so close to e and that's nearly the definition of e to correct myself
As it was mentioned, 1.005^200 is (1 + 1/200)^200, so it is a number from the sequence that defines the e number. That sequence is monotone, it increases, and its second number is (1 + 1/2)^2 = 2.25, which is larger than 2. So 200th number is surely even larger than 2.
Good point , provided you are able to prove that the sequence is increasing.
@@ShiMon- that's pretty easy using derivatives.
pretty sure the sequence commonly used to define e is the sum of all 1/n!, which you can then use to prove (1+1/n)^n -> e
@@ratzou2 no, this sum is not the definition.
@@alkinooskontopodias5919 the derivation of derivatives is based on the fact, that this sequence has a limit.
I think it can just use Bernoulli's Inequality (1+x)^n>=1+nx for x>=-1 is real number and n>=1 is integer.
When x isn't equal to 0 and n isn't equal to 1, it can be (1+x)^n>1+nx,so x=0.005 n=200,(1+0.005 )^200>1+200*0.005=2 Q.E.D
You are a certified ASIAN
@@sciencefanboy8239 hahahaha... In the past my math ancestors would say HE WOULD BE GERMAN!!! LoL
(Maybe he is an engineer of any kind!!)
LoL
Anyway he is excellent. He has vast math culture and good memory.
Brasil
@@KRYPTOS_K5 🤣
Actually I'm a junior in college, and major in statistics in Taiwan XD.
This inequality was taught to let us learn mathematical induction in tenth grade.
By the way, this can be derived from Binomial Theorem and is useful to set significant level α when we do multiple tests.
For example, (1-α/n)^n >= 1-n*(α/n) = 1-α, so the type I error
Be careful, the part at 6:35 is not always true ! It only works with positive numbers. Take a= -1, b=3, c=2, here b>c (3>2) yet a/b > a/c (-1/3 > -1/2)
Excellent.
She shoulda used absolute values to be less confusing
@@andrewberdahl9922 or just specify that
a ∈ N (N is the set of natural numbers, which doesn't contain negatives.)
There is a faster and “stupider” way, but which I think was meant to be the preferred solution: 1.005^2 is 1.010025, which means that at every multiplication we add 0.005 and a “tiny” bit. Since we are doing this 200 times we can just do 200 x 0.005 and we add 1 so we are already at 2. But we have a lot of “tiny” bits we didn’t add, so we can be sure the answer is above 2. I think this could be the preferred solution as it would justify why the question involves precisely 1.005 and 200 :)
if you find a better quantified for "tiny bit" then the proof would be accepted
Cool that's pretty much I what did too
Lemma: 1.005^x - 1 > x*0.005 for all x in N; x >= 2
Base case x=2 proven by hand calculation.
Suppose it's true for some n in N; n >= 2
Then
1.005^(n+1) - 1
= 1.005 * 1.005^n - 1 (exponentiation axioms)
= (1 + 0.005) * 1.005^n - 1 (because 1+0.005 = 1.005)
= (1*1.005^n + 0.005*1.005^n) - 1 (distributivity axiom)
= (0.005*1.005^n + 1*1.005^n) - 1 (commutativity axiom of addition)
= (0.005*1.005^n + 1.005^n) - 1 (because 1 is a multiplicative identiy)
= 0.005*1.005^n + (1.005^n - 1) (associative axiom of addition)
> 0.005*1.005^n + n*0.005 (using the assumption)
> 0.005 + n*0.005 (because 1.005^n > 1)
= (n+1) * 0.005 (distributivity axiom again (in reverse))
Which proves the lemma (by induction)
Theorem: 1.005^200 > 2
Proof: According to the lemma we have
1.005^200 - 1 > 200*0.005
=> 1.005^200 > 1+1 = 2
QED
Simunator, happy?
I believe this is the best approach because it does not require one to realise any sort of "representation" as being the needed representation to solve the problem, or the analysis of series.
It also makes clear that the statement isn't just true, it's obviously true. The reason it's "obvious" is because, as the OP says, you don't even have to care about the magnitude of the extra "little bits". This would be true even if it was just 1.005^2 > 1.01. There is no "n" that we have to reach before it those little bits add up to get us over the threshold. We are over the threshold immediately, from n=2 onwards.
It illustrates the difference between an "explicit" and "synthetic" proof.
To give another example, if we wanted to prove that the factorisation of x^2 + 2x + 1 is (x+1)^2, then there are two ways we could do it:
a) (explicit) apply the quadratic formula to the trinomial
b) (synthetic) multiply out (x+1)^2 and using distributivity and show that it's x^2 + 2x + 1
In this example the synthetic method is easier, but it leaves on wondering, how did you know in the first place what the factorisation was? The proof is correct but how did you know that in the first place? On the other hand, the proof using the quadratic equation does not require fore-knowledge of what the factorised form will be.
Many of Archimedes's proofs are presented synthetically. He will say "construct this line" and you will think "how the hell did he know to do that?"
Archimedes's "The Method" has the answer. He first solved the problem explicitly (or, sometimes, mechanically), and once he knew the answer, only then did he create his proof, which was usually synthetic, leaving out his initial calculations, leaving one wondering how he came up with that idea.
Likewise, the method shown in the video is correct, but leaves on wondering "how did they know to do it like that?"
On the other hand the method in the OP comment and which I have elaborated on here is obvious. You aren't constructing anything or changing any representations. It's just a calculation.
The OP's only error is to refer to this method as "stupider". If anything it's smarter. Though "stupid" and "smart" aren't technical terms so let's just stick with "different" and "more explicit"
Love your comment. I agree, stupider is wrong, but “stupider” maybe ok? XD
@@neildutoit5177 dude who are you pythagoras or someone?
If it didn't need to be proven then one can easily use the rule of 72 that can be used to approximate how many times an interest rate has to applied to double the initial value. This is very often used to quickly check how many years it would take to double your initial capital. Since the "interest" here is 0.5% 72/0.5 = 144. So 1.005^144 is approximately equal to 2, which would mean the 200th power will be a lot larger (somewhere around e).
that is very cool and probably the intended solution for the question based on how quick it can be done
This is how I solved it as well. You can think of the problem as, "If I put my money in a bank account paying 0.5% per year, will my money have doubled in 200 years?" and know right away that your money will double after less than 140 years if compounded continuously.
Yep, this is how I solved it as well. 72/0.5 which is just double 72 or 144. No paper or calculator needed!
This was precisely my response too. Instant intuition that 1.005^144 is about 2.0, and it’s a pretty good approximation. The actual equivalent exponent is closer to 1.005^139 = 2, but even with the slight error the answer to the original question is very obvious.
I don't know where did You find that rule of 72, but it seems that a rule of 69 is a little better: 69/.5 = 138, wich is exactly the last one before it gets to 2. It is more accurate when the interest
Here's what I did:
Using binomial expansion: (a+b) ^ n = a^n + n ( a^n-1)(b) + ...etc (there are n+1 terms but in this case we only need to look at the first 2)
So, (1 + .005) ^ 200 = 1^200 + 200(1^199)(.005) + ... (the other 199 terms are much smaller but all are positive)
which simplifies to 1 + 200(1)(.005) + ....(199 smaller positive numbers).
which further simplifies to 1 + 1 + ....(199 smaller positive numbers) = 2 + .... (199 smaller positive numbers).
Thus 1.005^200 must be larger than 2.
Think of (1.005) as getting 1/2% interest on your money. Because it is compound interest, you get some interest ON your interest. Must be greater than 2.
That's what I did
@ Yes of course. Pencil on paper. But a true math guy just looks at it and says "200 time half percent already equals two, before any compounding." Talking about insight and intuition here. Thanks.
@@JonRobert Well, actually, I solved it while I was in bed falling asleep. I am used to count rows in Pascal's triangle to fight insomnia.
This is what i also thought. Binomial theoram
I just thought of it as 1.005 represents a 0.5% increase. If that increase was additive, then after 200 increases, the value would be 200% (2). But here it is compounded, which would always be greater.
same here.
Exactly. As simple as that, a few seconds' thinking and no need to write anything.
wow good idea
That's the intuitive idea behind my method, which is more formal: creating a linear approximation of the function f(x) = x^200 at x=1
My thoughts exactly. This is a nice proof, but the question didn't require a proof.
I did it this way on a slide rule. ln(1.005) is about 0.00495 (LL1 to D). Then the product of that and 200 is 0.99 (standard CD multiply). Get rid of the ln by exponentiation, e raised to 0.99 is 2.68 (D to LL2) just a bit under e. e>2.
Using a taylor's expansion of ln(x) around x=1, it is easy to see that ln(1.005) is very close to 0.005, so ln(1.005^200)=200*ln(1.005) is approximately 1. From this we can conclude that 1.005^200 is very close to euler's number "e" (since ln(e)=1), which is around 2.71, hence 1.005^200 is approximately 2.71.
Cool!
Yes, but 'very close' is a vague notion and your proof isn't actually proving anything
@@fahrenheit2101 While it's true that using the Taylor series isn't the best idea in this case, since 1.005^200 is far from incalculable, the margin of error is still small enough, and the difference between the estimate and the number 2 large enough, that the results are fairly conclusive
Square and multiply with force to the lower bound of 1.005^2 being 2.771. Since 200 has a low amount of ones in it's binary form (11001000), it makes it even easier. Start with 1 in your result and 1.005 as your working number. If the last digit in binary of the exponent is 1, multiply result by your working number. Square the working number and round it DOWN to 3-4 decimals, the more you do, the closer the lower bound will be. Delete the rightmost digit in the binary exponent. Repeat until you delete everything from your exponent. You just did a multiplication of 2 4-digit numbers exactly 10 times and proven, that 1.005^2 > 2.771 and that's certainly greater than 2. That's what I thought of when I saw this first time, nonetheless, the video shows a much more elegant solution :D
Binomial theorem - (1 + 0.005)^200 = 1 + 200*1/200 + something > 0 = 2 + a with a> 0 so > 2. That was easy.
This is how I would do it.
f(x) = 1.005^x is strictly increasing. Thus, we need to find the value of x when f(x) is 2. If x is less than 200 then 1.005^200 is greater than 2 otherwise it is not.
1. 1.005^x = 2
2. log(1.005^x) = log(2)
3. xlog(1.005) = log(2)
4. x = log(2)/log(1.005)
5. x =138.976
Since the value for x that causes f(x) equal to 2 is less than 200 and f(x) is strictly increasing then 1.005^200 > 2
I like this solution best. Efficient, clean, straightforward. Good work.
There's a problem here. How would u get log(1.005) without calculator?
I u could use one, why not just multiply 1.005 200 times, or use power function?
This is the way I was thought math in school (about 23 years ago). Still the most efficient manner in my opinion.
This is useless. You require a calculator for your method, but a calculator answers this question for you instantly anyways...
if you divide logs you may as well just use a calculator and compute 1.005^200.
The idea is to get a proof without doing any calculations. And that was the solution in the video. We want to apply mathematical induction not computation to get the answer.
It can be easily solved using binomial expansion,
(1+x)^n~1+nx when x2.
Therefore,
1.005²⁰⁰>2
If you are familiar with Euler's number (e), you instantly know, without any calculations, that the first number must be greater than 2 since it is fairly close to e.
1.005²⁰⁰= (1+0.005)²⁰⁰
The *Bernoulli Aproximation:*
*_(1+a)^x ≈ 1+ax, for a ≪ 1_*
Using it:
(1+0.005)²⁰⁰ ≈ 1+0.005*200 = 2
Now, let's see if the *aproximation* is *greater* or *lower* than the real one:
(1+0.005)² - 1+0.005*2 = 1,010025 - 1,01 = 0,000025.
As the *difference is positive: Real > Aproximation*
Thus, *Real > 2* and so *1.005²⁰⁰ > 2*
I give the proof that *Bernoulli's Aproximation* is always smaller than the real one for you 😉
I should know "The Bernoulli Aproximation" before this huh...
Can you explain why 1.005^2 - 1.01 ?
@@kneticnrg he just did the math but using a 2 instead of 200 'cause the math with 2 is easier to do, and since in both cases, x would be the same on both sides of the equation, the experience remains, so if using 2 you get a result on which side of the equation is greater, then the same would apply to 200.
Benoulli also did some good shit with fluid mechanics. I met him once way back when I was a young man. Oh that was only about 244 years ago. Good man that Bernoulli. He was even a hippie. ha ha
It's simple doing homework with Bernoulli! Try alone man! lol
Very cool method. I guess using the binomial theorem also works.
Yes and it is faster. But it is not the motif of my little friend's video.
Brasil
I don’t know, I thought of using mathematical induction to do it because it’s straightforward way.
@@1mol831 Like by proving that 1.005^n is larger than or equal to 1+0.005n? I guess that also works.
@@SmartWorkingSmartWorker in general it could be (1+x)^n > 1 + xn, provided x > 0 and n > 1
Multiply both terms for 1000^200, on the left you get 1005^200 while on the right 2*(1000^200) and written like this it is pretty clear that the left one is bigger, even though i have proved nothing like this
Just multiply fist 0.005 by 200 itself gives 1. so, LHS is greater than 2 ( by binomial expansion)
The solution shown is very elegant and uses only basic algebra. Enjoyed the presentation.
Can I do this that way?
f(x)=(1+1/x)^x
by definition:
lim_(x->oo) f(x) = e
and
lim_(x->1) f(x) = 2
so
lim_(x->1) f(x) = 2 < lim_(x->200) f(x) < lim_(x->oo) f(x) = e
I suppose that i must to prove that f(x) is increasing function in (0,+oo)
nice :)
I was thinking about it as interest without compounding interest, as in if you only get interest on the original 1 and skip the .005, adding 0.005 200 times gives you 2, so if we count compound interest we are definitely above 2 :).
Another solution is, solve for x in the equation 1.005^x = 2. And we know 1.005^x is exponentially increasing.
1.005 is 0.5% per period. 200 is number of periods. This is compound interest. Quick an dirty estimation when compound interest makes x2 is - 70/0.5 = 140 times. Means 1.005^140 ~= 2.
200 > 140, so 1.005^200 > 2
other than the application of the e limit that you can use, you can also just calculate log2(1.005) if you're given a calculator, or you can also think about how 1.005 x 1.005 behaves. 1.005*1.005 equals to something higher than 1.01, which i don't care about (easy to calculate, i just care about it being a higher number). So we can apply a minoration and multiply 1.01x1.01 and do the same thing for 3 times. Same reasoning for multiplication by 5 done 2 times (200=2^3*5^2). With the minoration we should get to 2. This means that we know for sure that the 200th root of 2 is a smaller number than 1.005, elevate both members to the 200th power and we get the disequation we were trying to figure out earlier.
Nice. Framing this as a partial progression of the limit-definition of e makes a lot of sense.
Here is what I did:
If you do 1.005^n+1 and n=1, than 1.005^n+1 = 1.005. Each time you add 1 to n, it will increase by more than .005, and if you do that 200 times that means that it will increase bymore than (.005 * 200), and .005*200 = 1, so because you started at 1, and increased by more than 1, you will end up with a number greater than two.
Very nice video, remembered how much I liked mathematics, then and now in my 40s. Awesome handwriting too.
No need to solve..only 1 rule and 1 step
Rule 1.. (1+1/n)^n = e when n is infinity
Step 1.. (1+1/n)^n = 2 When n=1
Answer ... here n=200 so answer is between 2 and e, so 1.005^200 greater than 2
💟👍🏻👍🏻
Now I am seeing that the power in the left is greater than 2. This is because we have (1.005)^200>2.5 because this number is almost equal to e=2.71828.... Hence (1.005)^200>2. Thank you very much!
You know that it converges to e but you should have find any n such that (1+1/n)^n>2 and show that f(n) increases strictly.
ln(x) is a convex function. 200 x ln ( 1005/1000) > 200 x 0.005 x ln(2) = ln(2).
I used the rule of 72 and found an estimate of doubling at 1.005^144 (took less than 3 seconds). From there it was safe to say that it'd be ">". Made a couple of videos explaining this.
ruclips.net/video/c9aRHid0kfU/видео.html
f(n) = (1+1/n)^n
Want f(200) vs 2
Know: f'(n) > 0 and f(1)=2
Then f(200) > 2. QED
yes, I was going to suggest binomial theorem is much less clunky than video method, but I noticed many others already said this :)
That will not give clear proof..it is an approximation
@@karthikcind1981 The first term of the binomial expansion is 1. The second term is also equal to 1, so adding them gives 2. All of the remaining terms are positive, so the overall sum is greater than 2.
This is quite easy to answer, because it deals with interest calculation we know from daily life. We distinguish between linear and exponential return. Usually, banks do linear return when the time span is less than a year, but exponential (compound interest) over several years. The factor 1.005 means that the capital is multiplied by 1.005 after one period, that means that 0.5 percent interest rate is applied. 0.5 percent is 1/200.
Now if linear return is applied, it takes 200 periods to double the initial capital (e.g. 1000 dollars initial capital, interest is 5 dollars per period, after 200 periods you have gained 200 times 5 dollars = 1000 dollars, resulting in 2000 dollars overall capital).
However, if exponential return is applied, the capital of course is growing faster (due to the interest of interest effect), so you will have more than 2000 dollars after the same 200 time periods. And this is exactly the result of 1.005^200, using the exponential function with base 1.005 and exponent 200. So you will more than double your capital within these 200 time periods.
In summary, 1.005^200 > 2.
Great approach! Just recognizing this as compounded vs simple interest gets you an instant answer.
@@j10001 Many thanks! I'm a math teacher and explain exponential functions exactly this way in my lessons. I always use the general formula
y(t) = a * b^t
with a = Initial value (Anfangswert in German), b = Basis (base), t = time.
or its analogon from interest calculation, the capital function
K(n) = K0 * (1 + p/100)^n
with K0 = Initial capital, p = interest rate in %, n = number of years
Using thos formula, we can solve Almosen any exrcise at school, including textbook exercises from pactical Situationskomik like interest calculation or radioactive decreasing etc.
Almost any, practical situations. Typos everwhere.
Tremendous Approach. Especially useful when the challenge is to use basic Mathematics and Arithmetic.
Personally I used the properties of the natural logarithm on the sequence 1.005^k to find from which integer 1.005^k>2. We then find k=139 as 1.005^k is an increasing sequence it follows that since 200>139 and indeed 1.005^200>2
This lady's hand works like a color plotter. Great talent.
Easier way could be to use binomial approx
(1+x)^n tends to 1+nx when x is small
Which means (1+0.005)^200 is 2+(some smaller powers) i.e. left term is greater
Yes but not the aim of this little friend in this video.
Brasil
In this specific case, there is NOTHING to think about. 1.005^200 is OBVIOUSLY larger than 2 because 0.005 * 200 = 1 and 1.005 * 1.005 is more than 1.01 while the extra bit gets progressively bigger. Therefore, there is no chance 2 could be larger or equal to 1.005^200.
Same, it was apparent after about two seconds of thought 😁 I guess having a formal proof is the thing at these sorts of events though.
@@PeterCooperUK Well, at school, yes. In practical life, no.
Rule of 72 is good. It’s popular in finance. 72 / Interest rate = Doubling period.
For example, 10% interest rate has doubling period of 7.2 years.
In this case, .5% has doubling period of 144 years. So 1 * 1.005^144 is approximately 2.
For lower interest rates use Rule of 69 because ln(2) ~ 0.69. 1.005^139 ~ 2.
1.005 = 1 + 1/200
1.005^2 = (1 + 1/200)(1 + 1/200)
1 + 2/200 + x where x is a small positive integer (we can ignore x because everything related to x will be positive)
1.005^3 = (1 + 1/200)(1 + 1/200)(1 + 1/200)
(1 + 1/200)(1 + 2/200)
1 + 3/200 + some other positive number x
Thus (1 + 1/200)^y = 1 + y/200 + some positive x
y is 200 in this case so
1 + 200/200 plus some positive x
2 + x
Again, x is positive and thus
1.005^200 > 2
Edit: Fixed typo and wording
Excellent👍🏻👍🏻👍🏻
Excellent👍🏻👍🏻👍🏻
Wow, beautifully simple solution.
It is not an integer tho, more like a 200 orders of magnitude smaller than 1/200 real number. Nevertheless, this argument still holds👏🏻
@@Trip_Fontaine Yeah but I think you mean 200 times smaller, not 200 orders of magnitude smaller.
I think a very quick solution can be obtained if you use an estimate with the logarithm of 2. The idea is that 1.005^200 is the same as 2^(log2(1.005^200)). Now all you have to do is to look how the exponent log2(1.005^200) compares to 1. log2(1.005^200) = 200 * log2(1.005). That is < 1 when log2(1.005) < 0.005. Since log2(1) = 0 and the deviation of log2 < 1 for every argument > 1, log2(1.005) must be smaller than 0.005. Therefore log2(1.005^200) < 1 and hence 2^log2(1.005^200) = 1.005^200 < 2.
I thought of similar way . I also took Logarithm but I made 2 cases then got 1 false and one true
Expand 1.005^200 : The first two terms sum to 2, and all the rest are positive. QED.
Exactly what I dii.
Thank you for nice solution, that doesn't use other theorems and remarks.
e is approximately equal to 2.718... and is the limit of (1+1/n)^n
in the case of 1.005^200, n=200
as (1+1/2)^2=2.25 and (1+1/n)^n is strictly increasing
surely, (1+1/200)^200 is relatively close to 2.718 and greater than 2
therefore, 1.005^200 is greater than 2
1.005^200 is larger than 2.
Interest calculated by compounding is always greater than simple interest for same rate of interest. Principle=1,rate =0.005,time =200. SI= 1,amount =2
Therefor CI >1, P+CI >2.
What about this, though?
1.005^138 < 2
Simple binomial expansion for any real index (1+x)^n leads to the conclusion
(1+0.005)^200 = 1+1....some positive terms which clearly states that it is greater than 2
There are two methods to figure out which value is larger without using a calculator.
I would go with the calculus method as an easy way. (1.005)^200 can be written as (1+0.005)^200 or (1+1/200)^200. We can take the limit on both values as x is approaching to infinity. We know that limit of 2 as x tends to go to ∞ is 2. The limit of (1+1/x)^x as x tends to go to ∞ is e by limit definition. You can also compute this limit using l'Hopital's rule since 1^∞ is one of the 7 indeterminate forms. The limit (1+1/x)^x as x goes to infinity can be rewritten using base e as e^(xln(1+1/x)). The exponent here is an indeterminate form of 0(∞). Turn this as e^(ln(1+1/x)/(1/x)). The last indeterminate form in this step is now 0/0. We have to take the derivative of the top and derivative of the bottom (not a quotient rule derivative), which is e^((-1/(x(x+1)))/(-1/x^2))). Using algebra, it simplifies to e^(x/(x+1)). We can also use l'Hopital's rule since it's again an indeterminate form of ∞/∞, which is e^1=e. The limit as x tends to go to ∞ of e is e. You can also divide by the highest power of x, which is e^(1/(1+1/x)). We know that the limit of 1/x as x tends to go to ∞ is 0 from what we know from the graph of y=1/x, so the limit of e^(1/(1+0))=e^1=e, which is the limit as x goes to ∞.
If we use the algebra method, we can try those values if we take the log on both values, which is 200log(1.005) (or 200log(1+1/200)=200log(201/200)) and log(2). In the first value, we can use properties of logarithms, which is 200log(201)-200log(200). We need to compare something with the second value for log(2). Here, we can break 200log(200) again using log properties as 400+200log(2). The first new value with the second value comparison becomes 200log(201)+200log(2)+400 and log(2). We can now subtract 200log(2) on those two values, which is 200log(201)+400 and -199log(2). We don't need to expand more steps since it's obvious that 200log(201)+400 is a positive value and -199log(2) is a negative value meaning that 200log(201)+400>-199log(2). Therefore, (1.005)^200>2.
1st one doesn't work because you are approximating arbitrarily a partial sum to the complete sum to infinity. It's not complete.
It's much easier than that. You just need to prove (1+1/(n+1))^(n+1) is greater than (1+1/n)^n for all positive integers. Which is trivial using algebra. That proves the sequence is monotally increasing. And at n=1 it's 2. So done. No need to assume anything, use approximations. And the eaiet and fastest method there is. Without a calculator. And it works for any weird n too.
@@twakilon that is correct it's not. It's actually that we are taking the limit on both values as you get closer to the value of 200. However, if we take something that is getting larger values without any bounds, it will tend to get something that is close to e.
Are you generalizing? Because it appears to me the simple approach is realizing 1.005^200 must be larger because 0.005•200 = 1 ...
@@user-zu1ix3yq2w That has next to no relevance. Power of 200 and multiplying by 200 behave completely differently.
@@dav356 It is relevant. We don't need to abstract it out or generalize a solution.
There is an issue with your statement. You said "Power of 200" but what we're actually looking at is 1.005 raised to the power of 200. My solution works because (1.005 - 1) * 2 is smaller than 1.005^2 - 1. It follows that 1.005^200 will be *larger* than 0.005*200 + 1.
0.005 * 200 + 1 = 2. I hope this explanation was good enough to show you the relevancy.
I used the logic of 1.005^200 would be more than 1+0.005*200. And since 0.005 times 200 is 1 it can be identified that 1.005^200 is greater than 2
I saw the thumbnail and immediately realized that 1.005^200 should be very close to e, which is bigger than 2. Also, adding 0.005 200 times already gets you to 2 without the compounding effect so yeah definitely greater than 2
At a glance I was also certain it had to be greater but I couldn't show much elegant work for it
Adding 0.005 200 times equals 1
@FUG Slayer Nominee yes but you're adding all that to one so 2
@@wissamkadamani Ahh got it, thanks a lot.
While not a proof, just an intuition, it’s interesting that the right side is in the form of the limit for e, (1+1/n)^n; given it’s a large n, we can assume it is reasonably close to 2.71828… which is clearly greater than 2
This challenge is for you!🔥
Solve:
a³ + b² = 1 ;
a² + b³ = -1
Note: The solutions of these equations are real integers.
a=0
b=-1
??
@@shubhamaggarwal3469 ya correct! I just assumed that values in my mind and created this question! But.. the fun fact is I am get tired to solve it mathematically! 😅 And I want to know the trick to solve it! So, I challenged him to solve! So, please Math window! Put a video for solving this question!
@@nishanthproyt it’s too easy I think
you should able to solve this.
Q. 7x + 12y = 220 ; where x,y ∈ ℤ(+)
then find:- max [|x - y|]/min [|x - y|] = ??
Video link:- ruclips.net/video/t0c6EFHExn8/видео.html
To start, we observe that since a^2 must be non-negative, then b^3 (and b, as well) must be non-positive for a^2+b^3=-1 to be true.
Now let’s consider three cases.
Case 1: b-1
Since we’ve established b must be non-positive with our first observation, and we’re only dealing in integers, this would mean b=0. This leaves no possible value of a such that-
a^2+b^3=a^2+0=-1.
In conclusion-
If we consider all real numbers, I suspect there may be a solution with both a and b between 0 and -1. But with just integers a=0 and b=-1 is the only solution.
Start the expansion. The first term is 1, the second 200* 1^199*(1/200)=1. Thus we have the total of 2 already.
very well explained, thanks for sharing
If you need to guesstimate it under time pressure, the rule of 72s tells you the doubling occurs at somewhere near an exponent of 144 so the value of the first expression is more than 2.
This was a fairly nice question 👍
two more simple methods are binomial expansion as 1+nx+n(n-1)x²+... only three term will exceed 2......
second one is approximation using differentiation...both works just fine
Isn't this one of the basic questions on binomial theorm
Why so complicated, just apply binomial expansion on (1+1/200)²⁰⁰. Sum of first two terms of expansion is 2, hence adding all terms will make it greater than 2. Simple!
Yeah
1.005^200 к числу e стремится (при n=200), и точно больше, чем 2.25 (при n=2), следовательно, 1.005^200 > 2
An interesting solution, however, it can be much simpler if we use the second sonderful limit. lim (n->inf) (1+1/n)^n = e.
In case of non negative integers the least n is 1, so the value of expression will be equal to (1+1)^1=2. But we have 200. As it's always increasing, the more is the value of n, the more we get value which is nearer to e. Therefore, this stuff is more than 2.
Also, it it can be proved by math induction. But I believe, it's a bit harder :)
Excelent my little friend! Congrats for this video.
Brasil
Use the *finite geometric series formula* with r = 1.005, n = 200:
1.005^0 + ... + 1.005^199 = (1.005^200 - 1)/(1.005 - 1)
Rearranging the equation, we get 1.005^200 = 1 + (1.005^0 + ... + 1.005^199)/200. In other words, 1.005^200 is just one more than the average of the 200 numbers 1.005^0, ..., 1.005^199, and since that average is greater than 1 (since the first number equals 1 and all the following numbers are > 1) that must mean 1.005^200 > 1 + 1, in other words 1.005^200 > 2.
I believe you took the long way to solve it. An Easier way is to solve this equation for X : (1.005)^x = 2 ---> x = ln(2)/ln(1.005) = 138.97 < 200 ---> (1.005)^200 is great than 2 😉
you used a calculator which by that point might as well plug the original values in
In general, I would simply argue that with the binomial expansion, the first two terms alone are equal to two, and every term after that is decreasing but positive, so it would have to be a sum greater than two.
Like a lot of people, I had my own method for figuring out the answer: I just pictured the differential of the first term, and that it was always increasing. Which is a formal way of saying each successive term gets a little bit larger, whereas if you simply multiplied the first term without the "interest", so to speak, you'd get 2.
If this appeared in a math olympiad, it likely required a formal proof, so obviously a rough method like this wouldn't be sufficient. I suppose I could argue that the linear approximation of 1.005^200 is 2, and if the differential is increasing and monotone, a linear approximation of a lower value is always an underestimation. Kind of a shoddy proof though.
btw, my invitation to the Math Olympiad (I think I was 12?) was a necessary wakeup call for me. Like a lot of people, I was the smartest person I knew my age. Then along comes this entrance exam filled with question that are so far beyond my understanding that most of them I can't even read. And who attends this event? Other kids.
It was a timely and propitious corrective to my bloated ego. If that entrance exam doesn't appear in the mail, all roads to any state of "coolness" in high school are shut and bolted.
Take the log of the left side (can approximate it as .005), multiply by 200 to get appr. 1, then exponentiate to get a number close to e which is greater than 2. Much easier.
here's what I did I ignored the small number
(1+1/200)^2 = 1 + 1/200 + 1/200 + 1/40000
ignoring 1/40000. We can say
(1+1/200)^2 is at least 1 + 1/200 + 1/200 = 1 + 1/100
(1+1/200)^4 is at least 1 + 1/100 + 1/100 = 1 + 1/50
follow the pattern
(1+1/200)^8 is at least 1 + 1/25
(1+1/200)^16 is at least 1 + 2/25
(1+1/200)^32 is at least 1 + 4/25
(1+1/200)^64 is at least 1 + 8/25
(1+1/200)^128 is at least 1 + 16/25
(1+1/200)^192 is at least (1 + 16/25) * (1 + 8/25) which is bigger than 2
so if 1.005^192 > 2 then 1.005^200 must be bigger too. Yes, this method only works if we get a result bigger than 2 before or when we reach a power of 200. If we don't, we can't conclude it's smaller. It could be that the result is very close to 2 and the very small values we ignored would've made the final result over 2.
I used the binomial theorem. If you do your calculations using this binomial, then the first two terms will add up to 2, while the remaining 199 terms will add up to some positive real number, say, k. But we don't need to know the value of k (which is unfeasible to calculate it anyway), because we know it is positive, so obviously 2+k > 2. But 2+k = (1.005)²⁰⁰, and thus, (1.005)²⁰⁰ > 2.
f(x)=(x+1)^200 and f'(x)=200(x+1)^199. Plugging in f(0)=1 and f'(0)=200 yields the straight-line approximation, with f(0.005) approximately equal to 1+0.005*200=2. However, f''(0)=200*199>0, so the straight line is an underestimate and f(0.005)>2
By using Class 11 binomial theorem propert of ( 1 + 0.005)²⁰⁰ which is grtr than 2 on equating
(1 + 1/n)^n starts at 2 for n=1 and approaches e as n goes towards infinity. Since e is greater than 2, every (1 + 1/n)^n is for n>1.
I saw that it was of the form (1+1/n)^n which is e.
- As you increase n this always increases
- At n = 1, this is equal to 2
- At n = 2, this is equal to 2.25
- It's already larger than 2 and always increases so it must be larger than 2
Another approach is analysis of function (1+1/x)^x, laying between 1 and e. If we take 2 as x, the result will be 3/2*3/2=9/4=2.25, which is already higher than 2. If we take 3 as x, the result will be 64/27, which is approx. 2.37, which is greater than 2.25, therefore function ascends with rise of x, so as 200>2, with x=200, the result inevitably will be >2.
not rigorous but: I used the "rule of 72" for doubling time to approximate.
72/0.5% = 144 cycles to double. 144
I did it in 30 seconds, just take logs base 10 on both sides. One can approximate which is greater if log table isnt provided
I just did the following, which is way faster:
if you have a^b where a is greater than one, then a^b must be bigger than (a - 1)*b. since (a - 1)*b = 2, a^b must be bigger than two. of course this doesn't always work. If it's 1.005 ^ 199 vs 2 for example, that is still larger than two, but you can't be sure that that's the case, because (a - 1) * b is smaller than 2, so you would need to know how much bigger a ^ b is compared to (a - 1) * b, which this method will not tell you
Can be seen at the red numbers without going any further that with those 200 terms, the first one is 1.005 and the rest get smaller from there, therefore their product is smaller.
Binomial expansion:
(1+x)^n = 1+ n*x + positive terms
(1+0.005)^200 = 1+200*0.005 + positive terms = 2 + positive terms
If b>c. And you compare a/b with a/c... You said a/b < a/c. If a=1, b=2 c=3, you get 1/2 and 1/3 but 1/2 is bigger than 1/3. So a/b > a/c
Excellent basic solution. The question will thus be solvable at the elementary level.
In my country we use comma for decimal separator and dot for thousands separator, and I was trying to understand what was the point of comparing 1005 to 2.
For me, that would be easier and and it requires less ideas to try a proof by induction.
We show (by induction) that
(1 + 1/200)^n > 1 + n/200 for all n > 1
n = 2
(1 + 1/200) ^2 = 1 + 2/200 + 1/40000 > 1 + 2/200
Let's now assume that this is true for an integer m and show that it is true for m+1.
So we know that (1 + 1/200)^m > 1 + m/200 :
(1 + 1/200)^(m+1) = (1 + 1/200)^m * (1 + 1/200)
> (1 + m/200) * (1+1/200) = 1 + m/200 + 1/200 + m/40000
> 1 + (m+1)/200
QED
Now, we just have to take n = 200 and we have
(1,005)^200 = (1+1/200)^200 > 1+200/200 = 2
Tell me if I'm wrong, I may have made mistakes, I'm a little rusty :')
use binomial approximation. this will make it little bit easy
(1+n)^x ≈ 1+nx if x
I did it this way:
You can write 1.005^200 as exp[ ln(1.005) ]^200 = exp[ 200 * ln(1.005) ]
and 2 = exp[ ln(2) ]
Because of the properties of exp[x] you can go on and compare 200 * ln(1.005) to ln(2)
Using ln(x) = (x-1) - (x-1)^2 / 2 + (x-1)^3 / 3 + ... for x in ]0,2]
for ln(1.005) gives 0.005 - something negligible small
So 200 * ln(1.005) = 1 (approximately)
while ln(2) = 1 - 1/2 + 1/3 - 1/4 + ... is significantly smaller than 1
So: 1.005^200 > 2
Just note (1+1/n)^n always increases with n and is already 2 with n=1, so it's >2 for n greater than or equal to 2. The expression always increases because it's a constant 100% interest rate but you're compounding the interest rate more often.
(1+1/n)^n more and more closely approximates e as n gets large and is monotonically increasing. since for n=2 it is already 1.5^2=2.25, each value of n>2 must have a result greater than 2
And here I was, about to try and find a formula for the length of 5^n and see if it would ever beat 1000^n by when n=200!
(1+1/n)^n > 2 for any natural number n >= 2. Proof using logarithms: log2(n) + log2(1 + 1/n) > log2(2), since log2(1 + 1/n) > 0.
if allow to use binomial thm,
(1+1/200)^200 = 1 + 200×(1/200) + other positive terms
> 1 + 1 = 2
thus proved
The answer was easy to find: transpose the expression to the nagtive second dimension. Do an inverse integration of both sides in the equation. You now have the exponential expressions on both sides and the same base numbers and you can compare the exponets directly. 1 < 200 means 2 is less than 1.005^200.
I'm probably doing this in a way that isn't accepted, but
2 = 1+0.005*200
1.005^n > 1+0.005*n
For n > 2, 1.005^2 > 1+(0.005*2)
Since LHS has an exponential increase rather than an additive one, the LHS will increase at a faster rate than the RHS, .: 1.005^200 > 2
(1+x)^n= (1+nx+ lots of small positive values)
Therfore (1+.005)^200 = 1+200×.005 + lots of small positive values = 1+ 1 + lots of small positive values > 2
Therefore (1 +
Very very useful video mem .blessings from India............. 🙏🙏🙏🙏🙏🙏🙏
It's called Bernoulli inequality.
(1+x)^n >= 1+nx for all n and x > -1
Given that x=1/200 and n=200
(1+1/200)^200 > 1+200*1/200 = 2
In general, (1+1/n)^n ≥ 2 is true for all natural numbers n. And the limit as n approaches infinity is e.
1:06-1:38 I think it would be easier visually if we wrote it as (201/200)^200=201^200/200^200.