A more elegant approach is to expand (1+1/100)^n for n = 200 , using the binomial formula. All the terms are positive and the sum of the first two terms equals 3, which definitely proves that (1+1/100)^200 > 3.
Multiplying any number by 1.01 increases the value by a percent. If instead you simply added .01 repeatedly, 199 times to 1.01, it would give you 3, exactly. But since raising to the 200 power is akin to "compounding interest" in the financial realm, the resulting value is certain to be larger.
This was my intuition as well. I knew 1.01^200 was greater just on the fact that 1.01^100 was simply more than 2 and going another 100 times must be even more.
By the limit definition of e, (1 + 1/100)^100 is approximately e, which is ~2.718, so therefore 1.01^200 is useful to think of as [(1+ 1/100)^100]^2, which is approximately e^2, and 2.7^2 >> 2^2 = 4 >> 3. Edit: should have read other comments first as others already posted the same idea! :)
That is the spirit. The question for Chinese pupil would be 8 compare to (1 + 1/100)^100. And you method is lukaluka the standard anwser. (1 + 1/100)^100 < lim (n->oo) (1+1/n)^n = e , so e^2 < 8
If you know the definition of e, you know f(x)=(1+1/x)^x is increasing when x>0. Since f(1)=2, f(100)>2, i.e. 1.01^100>2. So 1.01^200=(1.01^100)^2>2^2=4>3. Simple!
apply rule 72 take it as 1% per annum so the original value double in 72/1=72 years so for next 72 years it will be 4 times the original no (1.01)^145 >4 so by this way first part is greater
Multiplying any number by 1.01 means increasing it by 1%. Increasing any number by 1% 200 times means increasing it by *at least* 200%. Increasing any number by 200% means multiplying it by three. Thus, multiplying any number by 1.01 200 times will definitely give a result larger than or *at least* equal to its three times. 1.01×3=3.03>3.
I used finance to figure this out in 5 seconds. Rule of 72. Take the compounding interest rate and divide 72 by it. That's a nonexact estimate on how long it would take to double the value. 72 goes into 200 nearly three times so it doubles nearly three times. 1 -> 2 -> 4 -> almost 8 (I'm guessing 7 point something).
This is very straightforward. Just write down the first two terms of the binomial expansion of (1 +0.01)^200. These are 1 , 2 and other positive quantities. Clearly they sum to something greater than 3.
A much easier explanation is this: Exponentials = compound interest $1 at 1% simple interest after 200 years gets you $3. This is because you just add $0.01 per year for 200 years till you reach $3. Since compound interest is greater (eg. At the $2 mark you're getting $0.02 per year instead of $0.01 per year), the end result will be > $3 This 1.01^200 > 3
1.01 multiplier adds at least 1% to number 1. So after 100 multiplications it will be > 2, then after 50 multiplications it will be > 3. So 1.01^150 > 3. Then 1.01^200 > 1.01^150 > 3 That’s it.
A nice way to solve this is to realize that 1.01 * 1.01 is easy enough to do and it’s equal to 1.0201, so we can conclude that every time you multiply the number by itself it gains about 0.01 more so 200 * 0.01 = 2, so the new number that you will get after multiplying 200 times is higher than 3 (because of the decimal) therefore (1.01)^200 is indeed bigger than 3
That solution is incorrect as your comparing a linear growth(200*0.01) with an exponential growth. Also, (1.01)^200 is not slightly higher than 3, its actually considerably higher. If you still think your logic is correct, try doing it for (1.01)^150
@@vinicus508 I did not say 1.01 times 200 you misread my whole comment, I said if you do just 1.01 times itself you get 1.0201, the second number after the decimal goes up by AT LEAST 1 digit, which means it will be AT LEAST 200 * 0.01 + 1, I SAID AT LEAAAAAAAST.
@@josukehigashikata1494 I did not misread anything. I never mentioned 1.01 times 200. All I said your reasoning is flawed cus you compared exponential vs linear growth. Also, saying its AT LEAST 200*0.01 + 1 which is equals to 3 doesnt prove anything. since we dont care if it its equals to 3 or not, we want to know if its bigger or not, saying its AT LEAST doesnt prove that it can be bigger.
An approach that would work is: (1 + 1/n)^n with n being any natural number (positive integer). This will always be greater or equal to 2 (which can easily be shown). And strictly greater with n>1. It reaches ultimately, as a limit, Euler‘s number e. (1 + 1/100)^200 = ((1 + 1/100)^100)^2 > 2^2 = 4 > 3.
I happen to know that 1.01^70 is slightly more than 2 -- which is to say, after 70 repetitions of *1.01, the original value is doubled. At this point it should be obvious that the left side is MUCH greater than the right side (relatively speaking), because 1.01^140 is already 4, and 1.01^200 should be at about 7.30-ish.
I found another non-intuitive approach to this, which leads to a very simple solution. For the moment, assume that 1.01^N = 1 + .01N. For N=200, this guestimate becomes 3.01 which is greater than 3. Next, we can try just a couple cases of N to show that for all N the real answer is greater than the guestimate, therefore 1.01^N > 3. For N = 2: 1.01 * 1.01 = 1.0201, which is greater than 1+.01N (1.02). For N = 3, the actual result is 1.030301, which is again greater than the guestimate (1.03). The inequality persists at each N value. You can prove this by comparing the ratio of 1.01^(N+1)/1.01^N with (1+.01(N+1))/(1+.01N). In the first case, the ratio is always 1.01. In the second case the ratio decreases at each step. For N = 3, that's 1.04/1.03. for N=4, it's 1.05/1.04. 1.01 > 1.04/1.03, or 1.05/1.04, etc. The real result is always greater than 3.01 which is also greater than 3.
For N=200, your guesstimate would be exactly 3; .01*200 is exactly 2. But still, since the real value is always greater than the guesstimate, your method is sound.
Your initial assumption is False ((1.01)^N cannot be equals to 1 + 0.01N for any natural number N), therefore anything you state based on it is also False. Also, for N = 200 your result ain't 3.01, its exactly 3. If you still think your resonating is valid, try proving that (1.01)^150 is greater than 3 using that method...
@@vinicus508 1.01^150 > 3 can't be proven by my approach even though 1.01^150 is greater. My approach to the original question is valid though. I chose a simple approximation of 1.01^N which is provably LESS than the actual number. If my simple estimate for 1.01^200 = 3 and my estimate is provably low, then the actual 1.01^N is definitely > 3. With 1.01^150 compared to 3, a more complicated method is required.
I thought like this: After every multuplying on 1,01 if other number higher or equals 1, this number will be added minimum 0,01 and this just logic because 1 × 1,01 = 1,01 and we like added 0,01. When we know what 1,01 × 1,01 = 1,0201 we don't make another 198 multiplying but instead add 1,98 because our logic trick and 1,0201 + 1,98 = 3,0001 > 3. And can you say, what this is not elegant?
Another simple approach is to use that (1+a)^n > 1+n*a (a>0) ,with first a=1/100 and n= 100 ,leading to (1+1/100)^100 >2 and then (1+1/100 )^200 > 2^2 =4 .
Theoretical approach without calculating at all and assuming no knowledge of exponential growth: The term 1.01^200 implies that an operation is performed that increases the initial value by 1 % and that this operation is repeated 200 times. An increase of 1 % repeated 200 times will increase the initial value by a total of at least 200 %, so for example 1 would become at least 3. Since the initial value is already over 1, the final value will definitely be greater than 3.
We have (1+1/100)^200={(1+1/100)^100}^2. Now Euler's number e=lim n-> infinity (1+1/n)^n. Therefore, (1+1/100)^100 is nearly Euler's number. We have e^2 > 3. QED
I did it in a different way: If you get 10% interest every year on you bank account you doubled your starting capital after 7 years. Therefore 1*(1.1^7) is let's say 1*2 = 2 This fraction is an "interest" of 1%. So the duration has to be multiplied by 10. So: 1*(1.01^70) is let's say 1*2 = 2 Let's do it again and the new starting capital is 2. So: 2*(1.01^70) = 2 * 2 = 4. These are the second "70" that are used so we are at 1.01^140. As 1.01^140 is still below 1.01^200 but greater that 3 we can see that 1.01^200 has to be greater than 3. 🙃😉
I think we can use euler's number, cause we can write like this: [(1+1/100)¹⁰⁰]² and (1+1/100)¹⁰⁰~e, so the number stays nearly to e², that is bigger than 3
when programmers enter!! var a= math.pow(1.01,200); var b= 3; If(b> a){ Console.log("b is greater "); Else{ Console.log("a is grater and its value is",a); } } Result a is greater and it's value is 7.316
Awesome but better & easier way is to apply the binomial theorem .in any case highly appreciate you considering the problem from junior school level maths knowledge.
Everyone in the comments is doing too much to solve this particular question. To solve this problem, we realize that multiplying by 1.01 refers to a growth of 1%. Knowing this, we can set the absolute minimum bound of growth as an additional 1.01 * 1.01 > 1 * 1.01 = 0.01 per exponentiation. We can easily tell that 1.01^200 > 1 + 0.1*200 = 3.
My Logic: 1.01^2 = 1.01 x 1.01 means it's 0.01 bigger than the original, so ^200 mean do it 200 times: 0.01 x 200 = 2 2+1.01 > 3 It's not a good solution because 1.01^111 also > 3. however it's work for this question.
There is an even easier method 1.01 ^ 200 = ((1 + 1 /100) ^ 100) ^ 2 (1 + 1 /100) ^ 100 is approximately e So 1.01 ^ 200 is approximately e * e = 2.718 * 2.718 > 7 Therefore 1.01 ^ 200 > 3
A abordagem mais simples é fazer (y=1,01^x) e investigar a partir de qual valor de "x" y>3. y = 1,01^x 3 = 1,01^x log 3 = x . log (101/100) x = log 3 / (log 101 - log 100) x = 110,41 A partir de 1,01¹¹⁰ ⁴¹ y > 3, portanto, é claro que 1,01²⁰⁰ > 3
The second option which is WAY simpler is (1+1/100)^200 = ((1+1/100)^100)^2 = e^2 ( very close to e squre ) - e = ~2.7 - when you square it you get more then 6 ... i.e. bigger then 3
1.01^2=1.0201 which increased by .0101 . We can assume it will increase at least .0101 for every consecutive compound of the exponent for a fact 200 in total. .0101 x 200 = 2.02 so we know it will increase by at least 2.02 (far more with compounding interest but never the less) 2.02 plus our initial amount of 1.01 equals 3.03 which is already greater than 3. Took 5 seconds to come to a conclusion.
Obviously first is bigger since by power rule we get 1.01^100^2, and 1.01^100 is approximately Euler's number. (Not exactly but close) And e^2 is of course bigger than 3
For mental calculation, I've memorized the following pairs of numbers: rt%, e^(rt) 10, 11 20, 22 30, 35 40, 49 50, 65 60, 82 70, 100 Because the term (1+) r multiplied t times is close to e^rt. This means that if r times t equals .70, whether it is 7% during 10 years or 3.5% during 20 years or whatever, the growth will be 100%, i.e. 1 will grow to 2, because e^0.7=2 (close enough for mental calculation). 1.01^200 means rt=200% which is 0.7+0.7+0.6 i.e. two doublings to 4 and 82% growth on top of that, so a bit more than 7.2, or 620% growth. To grow 1 to 3, rt needs to be 70%+40%=110% (double to 2 and then 50% on top of that) which is far less than 200%. Convenient for comparing for example different companies historic growth rates in a glance at a table.
For every power it increases by .01 so by the time it gets to 100 it becomes 2. something and when it gets to 200 it becomes 3. something which is greater than 3.
Soooo 1.01 * 1.01 = 1.0201. The number goes up more than .01 for each power. 3 - 1.01 = 1.99. To get an additional 1.99, at most we will need a power of 199. And because 200>199, 1.01^200 > 3
Too much for such an easy task. :) see, when you multiply N by 1.01 you get N*1.01 = N + N/100. This means that multiplying by 1.01 is the same as adding 100-th part of the number N. And now we have 1.01^200 that is 1.01*1.01*... It means that the first multiplication adds more that 0.01 to 1.01 and each next adds even more. So, the left part has to be bigger than 1.01 + 0.01*199 = 1.01 + 1.99 = 3. Solved :) 1.01^200 > 3
This is too easy, we know the first two term (1+x)^n expansion, it give (1+nx+...) > 1+200x0.01 = 3. If it was 1.01^120, then it may involve more tricks.
Some comments use e to estimate 1.01^200. The way to use e in an actual proof is to consider the function (1+1/x)^x where x>0. You can use calculus to show this function is increasing, this is the hardest part. Note that e is the limit at positive infinity. Since the function is increasing, (1+1/n)^n < (1+1/100)^100 =1.01^100, for 0 < n < 100. We can simply choose n=1 to show that 1.01^100 > 2. Then 1.01^200 > 4 > 3. If you choose n=2, you can get a lower bound of 5. After this, calculations become tedious.
@@toto-yf8tc It makes sense. My approach is formalizing the instinct to use the limit for e. It has the power to bound 1.01^200 from above. So it is more useful to solve the general problem where 3 is replaced by something else.
2:20 / 9:56 Which Is Larger? | Nice Maths Question｜Use this method and solve it quickly! Math Window 51,3 N người đăng ký Đã đăng ký 56 Chia sẻ Cảm ơn 2 N lượt xem 13 giờ trước Comparison, which one is larger? A fantastic math problem. A common method to this kind of questions. 10 bình luận Nguyễn Phùng Gia Hung`s channel Viết bình luận... Allan Flippin Allan Flippin 6 giờ trước I found another non-intuitive approach to this, which leads to a very simple solution. For the moment, assume that 1.01^N = 1 + .01N. For N=200, this guestimate becomes 3.01 which is greater than 3. Next, we can try just a couple cases of N to show that for all N the real answer is greater than the guestimate, therefore 1.01^N > 3. For N = 2: 1.01 * 1.01 = 1.0201, which is greater than 1+.01N (1.02). For N = 3, the actual result is 1.030301, which is again greater than the guestimate (1.03). The inequality persists at each N value. You can prove this by comparing the ratio of 1.01^(N+1)/1.01^N with (1+.01(N+1))/(1+.01N). In the first case, the ratio is always 1.01. In the second case the ratio decreases at each step. For N = 3, that's 1.04/1.03. for N=4, it's 1.05/1.04. 1.01 > 1.04/1.03, or 1.05/1.04, etc. The real result is always greater than 3.01 which is also greater than 3. Phản hồi Sasuntidictous Rhoireiphapos Sasuntidictous Rhoireiphapos 12 giờ trước 1.01^200 = (1.01^(100))^2 is approximately e^2, so it is larger. 2 Phản hồi Cath Cath 12 giờ trước Never underestimate the power of powers! 3 Phản hồi Маркиз Карабас Маркиз Карабас 12 giờ trước Красиво! Цікаво було б порівнять 1.01^70 та 2. Це була б взагалі 💣 1 Phản hồi Pão Pão 3 giờ trước yeah, quite larger... 1.01^111 is already larger than 3 Phản hồi Robbie Katanga Robbie Katanga 7 giờ trước Nostalgic with Addma aka additional mathematics. 1 Phản hồi Alexey Evpalov Alexey Evpalov 13 giờ trước Спасибо. 1 Phản hồi D B D B 6 giờ trước شكرا على مجهوداتكم يمكن استعمال حدانية نيوتن نجد بسرعة ان 3 اصغر بكثير..... 1 Phản hồi D B D B 5 giờ trước العدد اكبر من 3 بل اكبر من 7 1 Phản hồi faith(٠٠٩٦٧٧١٣٨٦٢٨٥٧) faith(٠٠٩٦٧٧١٣٨٦٢٨٥٧) 13 giờ trước Hرسالتي إلي كل مسلم "يااخواني قال الله٠ ٠ ٠ عزوجل: يَٰٓأَيُّهَا ٱلَّذِينَ ءَامَنُواْ لَا تُبۡطِلُواْ صَدَقَٰتِكُم بِٱلۡمَنِّ وَٱلۡأَذَىٰ كَٱلَّذِي يُنفِقُ مَالَهُۥ رِئَآءَ ٱلنَّاسِ وَلَا يُؤۡمِنُ بِٱللَّهِ وَٱلۡيَوۡمِ ٱلۡأٓخِرِۖ فَمَثَلُهُۥ كَمَثَلِ صَفۡوَانٍ عَلَيۡهِ تُرَابٞ فَأَصَابَهُۥ وَابِلٞ فَتَرَكَهُۥ صَلۡدٗاۖ لَّا يَقۡدِرُونَ عَلَىٰ شَيۡءٖ مِّمَّا كَسَبُواْۗ وَٱللَّهُ لَا يَهۡدِي ٱلۡقَوۡمَ ٱلۡكَٰفِرِينَ}}}ياناس ياامة محمد اني اقسم بالله علئ كتاب الله اني لااكذب عليك ولاانصب ولااحتال اني بنت يمنيه نازحين بسبب الحرب اناواسرتي عايشين اناوامي واخوتي سغار والدنا متوفي الله يرحمه ومامعنااحدفي هذاالدنيا يقف بجنبنا في هذا الضروف القاسيه ومامعي اخوان كبار اناالكبيره في اخوتي ولكن انابنت لااستطيع مثلك ان اروح اشتغل بين الرجال واصرف علئ اسرتي والله ثم والله يااخي انناقدلنايومين محرومين من لقمت العيش ومعي اخوان سغار انظركيف حالتهم اقسم بالله يااخي انهم خرجومن البيت للشارع وشافو الجيران ياكلو راحووقفوعندبابهم لجل يعطوهم ولوخبزه يابسه يسدوبها جوعهم والله الذي له ملك السموات والارض انهم غلقو الباب وطردوهم ورجعويبكوايموتومن الجوع مااحدرحمهم وعطانهم لقمت عيش والان لومااحد ساعدنا بحق كيلو دقيق اقسم بالله اننا انموت من الجوع فيااخي انادخيله علئ الله ثم عليك واريدمنك المساعده لوجه الله انشدك بالله وبمحمد رسول الله يامن تحب الخير واتساعدني ولو ب500ريال يمني ان تراسلي واتساب علئ هذا الرقم٠٠٩٦٧٧١٣٨٦٢٨٥٧ وتطلب اسم بطاقتي وترسلي ولاتتاخر وايعوضك الله بكل خير فيااخي انت رجال إذاشفت اسرتك جاوعين تعمل المستحيل من اجل تامن لهم ألاكب ولكن انابنت عيني بصيره ويدي قصيره ليس لي اب مثلك واخواني سغار شوف كيف حالتهم وساعدناوانقذناقبل ان يطردونا في الشارع نتبهدل او نموت من الجوع انااقسم بالله الذي رفع سبع سموات بلاعمدوبسط ألأرض ومهداني لاأكذب عليك بحرف من هذا ألرساله واني ماطلبتك إلئ من ضيق ومن قسوت الضروف والحال الذي احنافيه أناوسرتي نسالك بالله لولك مقدره علئ مساعدتنا لاتتاخر عليناوجزاك آلّلّهً آلّفُ خيِر/////٠ ٠ ٠ ٠ ،،،ن
A more elegant approach is to expand (1+1/100)^n for n = 200 , using the binomial formula. All the terms are positive and the sum of the first two terms equals 3, which definitely proves that (1+1/100)^200 > 3.
👍🏻👍🏻👍🏻
@To infinity And beyond Why not ?
@To infinity And beyond the lowest 2 terms (in term of power) are 1+200*x. Not what you wrote. And x=0.01
thanks for spoiling it
No. You're meant to prove from first principles, not from a memorised formula sheet!
Multiplying any number by 1.01 increases the value by a percent. If instead you simply added .01 repeatedly, 199 times to 1.01, it would give you 3, exactly. But since raising to the 200 power is akin to "compounding interest" in the financial realm, the resulting value is certain to be larger.
i thinked same way
Thank you, I knew there had to be a simple way to figure it out.
This was my intuition as well. I knew 1.01^200 was greater just on the fact that 1.01^100 was simply more than 2 and going another 100 times must be even more.
By the limit definition of e, (1 + 1/100)^100 is approximately e, which is ~2.718, so therefore 1.01^200 is useful to think of as [(1+ 1/100)^100]^2, which is approximately e^2, and 2.7^2 >> 2^2 = 4 >> 3.
Edit: should have read other comments first as others already posted the same idea! :)
That is the spirit. The question for Chinese pupil would be 8 compare to (1 + 1/100)^100.
And you method is lukaluka the standard anwser. (1 + 1/100)^100 < lim (n->oo) (1+1/n)^n = e , so e^2 < 8
If you know the definition of e, you know f(x)=(1+1/x)^x is increasing when x>0. Since f(1)=2, f(100)>2, i.e. 1.01^100>2. So 1.01^200=(1.01^100)^2>2^2=4>3. Simple!
apply rule 72
take it as 1% per annum
so the original value double in 72/1=72 years
so for next 72 years it will be 4 times the original no
(1.01)^145 >4
so by this way
first part is greater
1.01^100 is close to e, so 1.01^200 is roughly e^2c which is much greater than 3
In order for that proof to be valid you need to prove (1.01)^100 is close to e, lol.
Multiplying any number by 1.01 means increasing it by 1%. Increasing any number by 1% 200 times means increasing it by *at least* 200%. Increasing any number by 200% means multiplying it by three. Thus, multiplying any number by 1.01 200 times will definitely give a result larger than or *at least* equal to its three times. 1.01×3=3.03>3.
Any *positive* number.
Fred
nice.
@ffggddss yes, thanks for the correction
I used finance to figure this out in 5 seconds. Rule of 72. Take the compounding interest rate and divide 72 by it. That's a nonexact estimate on how long it would take to double the value. 72 goes into 200 nearly three times so it doubles nearly three times. 1 -> 2 -> 4 -> almost 8 (I'm guessing 7 point something).
I saw the same thing, 1% interest for 200 time cycles. It should double almost 3 times, so way over 3.
This is very straightforward. Just write down the first two terms of the binomial expansion of (1 +0.01)^200. These are 1 , 2 and other positive quantities. Clearly they sum to something greater than 3.
A much easier explanation is this:
Exponentials = compound interest
$1 at 1% simple interest after 200 years gets you $3. This is because you just add $0.01 per year for 200 years till you reach $3.
Since compound interest is greater (eg. At the $2 mark you're getting $0.02 per year instead of $0.01 per year), the end result will be > $3
This 1.01^200 > 3
1.01 multiplier adds at least 1% to number 1. So after 100 multiplications it will be > 2, then after 50 multiplications it will be > 3. So 1.01^150 > 3. Then 1.01^200 > 1.01^150 > 3
That’s it.
A nice way to solve this is to realize that 1.01 * 1.01 is easy enough to do and it’s equal to 1.0201, so we can conclude that every time you multiply the number by itself it gains about 0.01 more so 200 * 0.01 = 2, so the new number that you will get after multiplying 200 times is higher than 3 (because of the decimal) therefore (1.01)^200 is indeed bigger than 3
That solution is incorrect as your comparing a linear growth(200*0.01) with an exponential growth. Also, (1.01)^200 is not slightly higher than 3, its actually considerably higher. If you still think your logic is correct, try doing it for (1.01)^150
@@vinicus508 I did not say 1.01 times 200 you misread my whole comment, I said if you do just 1.01 times itself you get 1.0201, the second number after the decimal goes up by AT LEAST 1 digit, which means it will be AT LEAST 200 * 0.01 + 1, I SAID AT LEAAAAAAAST.
@@josukehigashikata1494 I did not misread anything. I never mentioned 1.01 times 200. All I said your reasoning is flawed cus you compared exponential vs linear growth. Also, saying its AT LEAST 200*0.01 + 1 which is equals to 3 doesnt prove anything. since we dont care if it its equals to 3 or not, we want to know if its bigger or not, saying its AT LEAST doesnt prove that it can be bigger.
@@vinicus508 saying it’s at least, there is also many decimals that follow the 0.01 but you’re blind.
The method is ok. But you should delete the word "slightly".
An approach that would work is: (1 + 1/n)^n with n being any natural number (positive integer). This will always be greater or equal to 2 (which can easily be shown). And strictly greater with n>1. It reaches ultimately, as a limit, Euler‘s number e. (1 + 1/100)^200 = ((1 + 1/100)^100)^2 > 2^2 = 4 > 3.
I happen to know that 1.01^70 is slightly more than 2 -- which is to say, after 70 repetitions of *1.01, the original value is doubled. At this point it should be obvious that the left side is MUCH greater than the right side (relatively speaking), because 1.01^140 is already 4, and 1.01^200 should be at about 7.30-ish.
Using the Rule of 72, .01 should double every 72 cycles, so 1.01^200 would be nearly 8x and be greater than 3.
I found another non-intuitive approach to this, which leads to a very simple solution. For the moment, assume that 1.01^N = 1 + .01N. For N=200, this guestimate becomes 3.01 which is greater than 3. Next, we can try just a couple cases of N to show that for all N the real answer is greater than the guestimate, therefore 1.01^N > 3.
For N = 2: 1.01 * 1.01 = 1.0201, which is greater than 1+.01N (1.02). For N = 3, the actual result is 1.030301, which is again greater than the guestimate (1.03). The inequality persists at each N value. You can prove this by comparing the ratio of 1.01^(N+1)/1.01^N with (1+.01(N+1))/(1+.01N). In the first case, the ratio is always 1.01. In the second case the ratio decreases at each step. For N = 3, that's 1.04/1.03. for N=4, it's 1.05/1.04. 1.01 > 1.04/1.03, or 1.05/1.04, etc. The real result is always greater than 3.01 which is also greater than 3.
Yeah used your approach
You are basically using the binomial expansion and approximating the result
For N=200, your guesstimate would be exactly 3; .01*200 is exactly 2. But still, since the real value is always greater than the guesstimate, your method is sound.
Your initial assumption is False ((1.01)^N cannot be equals to 1 + 0.01N for any natural number N), therefore anything you state based on it is also False. Also, for N = 200 your result ain't 3.01, its exactly 3. If you still think your resonating is valid, try proving that (1.01)^150 is greater than 3 using that method...
@@vinicus508 1.01^150 > 3 can't be proven by my approach even though 1.01^150 is greater. My approach to the original question is valid though. I chose a simple approximation of 1.01^N which is provably LESS than the actual number. If my simple estimate for 1.01^200 = 3 and my estimate is provably low, then the actual 1.01^N is definitely > 3. With 1.01^150 compared to 3, a more complicated method is required.
Ln on both sides; as ln(1+X)=X per X close to 0 we have 200 by 0.01 or 2 that is larger than ln(3), i.e. the lhs is larger than the rhs
Never underestimate the power of powers!
Vergil was a math nerd all along
1.01^200 is larger the composting effect increases it super quickly so it turned to 7.31
I thought like this:
After every multuplying on 1,01 if other number higher or equals 1, this number will be added minimum 0,01 and this just logic because 1 × 1,01 = 1,01 and we like added 0,01.
When we know what 1,01 × 1,01 = 1,0201 we don't make another 198 multiplying but instead add 1,98 because our logic trick and 1,0201 + 1,98 = 3,0001 > 3.
And can you say, what this is not elegant?
Another simple approach is to use that (1+a)^n > 1+n*a (a>0) ,with first a=1/100 and n= 100 ,leading to (1+1/100)^100 >2 and then
(1+1/100 )^200 > 2^2 =4 .
use binomial thm and the result follows immediately.
Absolutely right. The solution is immediate with no messing around.
Greetings from the BIG SKY. Back in 1968 I'd say left side.
1.01^200 would approach e^2 (fundamental limit), but taking the log may also help.
The fundamental theorem can't be used for n=200 lol. It does give a hint that it could be true, but it's doesn't prove it.
Taking natural log on both sides:
200ln(1.01) ? ln(3)
Since ln(x) ~= x-1 when x aproaches 1 then
200ln(1.01) ~= 2
And ln(3) ln(3) -> 1.01^200 > 3
Theoretical approach without calculating at all and assuming no knowledge of exponential growth: The term 1.01^200 implies that an operation is performed that increases the initial value by 1 % and that this operation is repeated 200 times. An increase of 1 % repeated 200 times will increase the initial value by a total of at least 200 %, so for example 1 would become at least 3. Since the initial value is already over 1, the final value will definitely be greater than 3.
We have (1+1/100)^200={(1+1/100)^100}^2. Now Euler's number e=lim n-> infinity (1+1/n)^n. Therefore, (1+1/100)^100 is nearly Euler's number. We have e^2 > 3. QED
Easy to see that if you’re earning 1% interest, your principal will double in less than 100 years, so 1.01^200 has to be greater than 4
Very interesting approach.
I did it in a different way:
If you get 10% interest every year on you bank account you doubled your starting capital after 7 years.
Therefore 1*(1.1^7) is let's say 1*2 = 2
This fraction is an "interest" of 1%. So the duration has to be multiplied by 10.
So: 1*(1.01^70) is let's say 1*2 = 2
Let's do it again and the new starting capital is 2. So: 2*(1.01^70) = 2 * 2 = 4.
These are the second "70" that are used so we are at 1.01^140.
As 1.01^140 is still below 1.01^200 but greater that 3 we can see that 1.01^200 has to be greater than 3. 🙃😉
I think we can use euler's number, cause we can write like this: [(1+1/100)¹⁰⁰]²
and (1+1/100)¹⁰⁰~e, so the number stays nearly to e², that is bigger than 3
when programmers enter!!
var a= math.pow(1.01,200);
var b= 3;
If(b> a){
Console.log("b is greater ");
Else{
Console.log("a is grater and its value is",a);
}
}
Result a is greater and it's value is 7.316
I didn’t know 1.01^200 and 3 could make a huge equation by just trying to find out who’s bigger
Awesome but better & easier way is to apply the binomial theorem .in any case highly appreciate you considering the problem from junior school level maths knowledge.
1.005^200 is just a little bit less than e, which is only a stone's throw away from 3. 1.01^200 would have to be much, much larger than 3.
1.01^200 is around e^2, not so much much larger.
for a max value of n (1 + 1/n)^n is formula to e which is approximately 2.718 therefore it can't be greater than 3.
In general: (1+a)^n > 1 + n*a, if a and n are positive numbers.
And can you compare it to an arbitrary number, that isn't 3? Like 7 or 8 for example?
Everyone in the comments is doing too much to solve this particular question. To solve this problem, we realize that multiplying by 1.01 refers to a growth of 1%. Knowing this, we can set the absolute minimum bound of growth as an additional 1.01 * 1.01 > 1 * 1.01 = 0.01 per exponentiation. We can easily tell that 1.01^200 > 1 + 0.1*200 = 3.
Yeah the one with the bigger exponent is usually way bigger.
Got nothing that was darn mindconfusing!😵
One term of binomial expansion and you're done (and the knowledge that there are more terms). Proof by inspection. Don't even need a pencil.
My Logic: 1.01^2 = 1.01 x 1.01 means it's 0.01 bigger than the original,
so ^200 mean do it 200 times: 0.01 x 200 = 2
2+1.01 > 3
It's not a good solution because 1.01^111 also > 3.
however it's work for this question.
There is an even easier method
1.01 ^ 200 = ((1 + 1 /100) ^ 100) ^ 2
(1 + 1 /100) ^ 100 is approximately e
So 1.01 ^ 200 is approximately e * e = 2.718 * 2.718 > 7
Therefore 1.01 ^ 200 > 3
A abordagem mais simples é fazer (y=1,01^x) e investigar a partir de qual valor de "x" y>3.
y = 1,01^x
3 = 1,01^x
log 3 = x . log (101/100)
x = log 3 / (log 101 - log 100)
x = 110,41
A partir de 1,01¹¹⁰ ⁴¹ y > 3, portanto, é claro que
1,01²⁰⁰ > 3
The second option which is WAY simpler is (1+1/100)^200 = ((1+1/100)^100)^2 = e^2 ( very close to e squre ) - e = ~2.7 - when you square it you get more then 6 ... i.e. bigger then 3
If one knows that lim (1+1/n)^n ,n-> inf is e ,then (1+1/100)^200 is close to e^2 > 7 .
1.01^2=1.0201 which increased by .0101 . We can assume it will increase at least .0101 for every consecutive compound of the exponent for a fact 200 in total. .0101 x 200 = 2.02 so we know it will increase by at least 2.02 (far more with compounding interest but never the less) 2.02 plus our initial amount of 1.01 equals 3.03 which is already greater than 3. Took 5 seconds to come to a conclusion.
Obviously first is bigger since by power rule we get 1.01^100^2, and 1.01^100 is approximately Euler's number. (Not exactly but close)
And e^2 is of course bigger than 3
For mental calculation, I've memorized the following pairs of numbers:
rt%, e^(rt)
10, 11
20, 22
30, 35
40, 49
50, 65
60, 82
70, 100
Because the term (1+) r multiplied t times is close to e^rt. This means that if r times t equals .70, whether it is 7% during 10 years or 3.5% during 20 years or whatever, the growth will be 100%, i.e. 1 will grow to 2, because e^0.7=2 (close enough for mental calculation). 1.01^200 means rt=200% which is 0.7+0.7+0.6 i.e. two doublings to 4 and 82% growth on top of that, so a bit more than 7.2, or 620% growth. To grow 1 to 3, rt needs to be 70%+40%=110% (double to 2 and then 50% on top of that) which is far less than 200%.
Convenient for comparing for example different companies historic growth rates in a glance at a table.
what would do if 200 replaced with 111 or 110?
Or you could simply say (1+x)^200 being convex with a derivative of 200 in 1 (1+x)^200>1+x*200. Put x=1/100 and voilà.
If u saw the pattern of (1+n%+…%) you could see that the result will more than 1+(n+…)% such as
100 + 20% + 30% = 156 = 150
Nostalgic with Addma aka additional mathematics.
Not only larger than 3, MUCH larger: approx 7.3
(1+(1÷100))^100 is approaching to e. Thus (1+(1÷100))^200 is about e^2, which is larger than 3
For every power it increases by .01 so by the time it gets to 100 it becomes 2. something and when it gets to 200 it becomes 3. something which is greater than 3.
You got the correct answer but your statement is incorrect. It doesn't increase by .01, it MULTIPLIES by .01.
So it is actually equal to ~7.31
ln(1 + x) ≈ x for small x, so compare the natural logarithms. The lhs will be ≈ 200(x) = 200(0.01) = 2 and ln3 is barely larger than 1.
The best and easy solution
Красиво! Цікаво було б порівнять 1.01^70 та 2. Це була б взагалі 💣
(1+0.01)^200 = 1+ 0.01.200 = 3
A 1° approximation gave us 3, and an N number of terms are remaining in this so (1.01)^200 > 3
{ 45 SECOND APPROACH }
Take the log of both sides
은행 원금 불리는 개념으로 보면, 원금 1에 월1% 이자를 앞은 복리, 뒤는 단리로 200월동안 불린거네요
직관적으로 앞이 더 큰듯
Soooo 1.01 * 1.01 = 1.0201.
The number goes up more than .01 for each power. 3 - 1.01 = 1.99. To get an additional 1.99, at most we will need a power of 199. And because 200>199, 1.01^200 > 3
1.01^100 approaches e, 1.01^200 approaches e^2, oviously greater than 3
Binomial expansion will give you the answer in 30 seconds.
1.01^200 is easily over 3.
If its not then I'm stupid.
(1+1/100)^100*2 -> (~e)^2 > 3
Or just type it in a calculator and see that it’s more than 7. Took just a few seconds.
Another way would be the take e as 1+1/n)ñ is approx e
Too much for such an easy task. :) see, when you multiply N by 1.01 you get N*1.01 = N + N/100. This means that multiplying by 1.01 is the same as adding 100-th part of the number N. And now we have 1.01^200 that is 1.01*1.01*... It means that the first multiplication adds more that 0.01 to 1.01 and each next adds even more. So, the left part has to be bigger than 1.01 + 0.01*199 = 1.01 + 1.99 = 3. Solved :) 1.01^200 > 3
This is too easy, we know the first two term (1+x)^n expansion, it give (1+nx+...) > 1+200x0.01 = 3. If it was 1.01^120, then it may involve more tricks.
I looked at 1.01^200 and said okay 1.01^100 is slightly less than e, so 1.01^200 is slightly less than e^2 which is significantly larger than 3.
Спасибо.
1.01^x ~ 1 + 0.01x
Simply put, it's like that
شكرا على مجهوداتكم
يمكن استعمال حدانية نيوتن نجد بسرعة ان 3 اصغر بكثير.....
By bernulli's inequality (1.01¹⁰⁰)²≥2²>3
1.01¹⁰⁰ is very close to e, so 1.01²⁰⁰ is roughly e² which is bigger than 3. e²=7.3895....
1+1/100^200 = 7.34 repeating.
Hmm.
Ofc the left equation will be bigger than 3.
Math magic lol
THis is easy. It is e^2 approximately.
答え 3
ヒント 1^200＝1 1.01^200
Some comments use e to estimate 1.01^200. The way to use e in an actual proof is to consider the function (1+1/x)^x where x>0.
You can use calculus to show this function is increasing, this is the hardest part. Note that e is the limit at positive infinity. Since the function is increasing, (1+1/n)^n < (1+1/100)^100 =1.01^100, for 0 < n < 100.
We can simply choose n=1 to show that 1.01^100 > 2. Then 1.01^200 > 4 > 3. If you choose n=2, you can get a lower bound of 5. After this, calculations become tedious.
Makes no sense at all. Good function is (1+x)^200 and use convexity. Which is easy to prove
@@toto-yf8tc It makes sense. My approach is formalizing the instinct to use the limit for e. It has the power to bound 1.01^200 from above. So it is more useful to solve the general problem where 3 is replaced by something else.
1.01^100 ~ e, so 1.01^200 is definitely bigger than 3
Whenever you feel useless, come and watch this dude doing unnecessarily lengthened calculation so as to find an obviously seemable answer
(1+x)^n>1+nx
>7.3
yeah, quite larger... 1.01^111 is already larger than 3
yes😊
could you please show how you did it .
i mean of course without the calculator ..
Yeah. 1.01^200 is a little more than 7.
using 10 minutes to explain such a simple problem, what a shame
Do 1.01^111
Half a second 😛
1.01
1.01^200
あなたの口から出た言葉ではなく、我々の共通言語で証明して欲しかった
Well, let's see:
$ bc
>>> scale=6
>>> 1.01^200
7.316017
>>> quit
$
200 log1.01>log3
Logarithm
2:20 / 9:56
Which Is Larger? | Nice Maths Question｜Use this method and solve it quickly!
Math Window
51,3 N người đăng ký
Đã đăng ký
56
Chia sẻ
Cảm ơn
2 N lượt xem 13 giờ trước
Comparison, which one is larger? A fantastic math problem. A common method to this kind of questions.
10 bình luận
Nguyễn Phùng Gia Hung`s channel
Viết bình luận...
Allan Flippin
Allan Flippin
6 giờ trước
I found another non-intuitive approach to this, which leads to a very simple solution. For the moment, assume that 1.01^N = 1 + .01N. For N=200, this guestimate becomes 3.01 which is greater than 3. Next, we can try just a couple cases of N to show that for all N the real answer is greater than the guestimate, therefore 1.01^N > 3.
For N = 2: 1.01 * 1.01 = 1.0201, which is greater than 1+.01N (1.02). For N = 3, the actual result is 1.030301, which is again greater than the guestimate (1.03). The inequality persists at each N value. You can prove this by comparing the ratio of 1.01^(N+1)/1.01^N with (1+.01(N+1))/(1+.01N). In the first case, the ratio is always 1.01. In the second case the ratio decreases at each step. For N = 3, that's 1.04/1.03. for N=4, it's 1.05/1.04. 1.01 > 1.04/1.03, or 1.05/1.04, etc. The real result is always greater than 3.01 which is also greater than 3.
Phản hồi
Sasuntidictous Rhoireiphapos
Sasuntidictous Rhoireiphapos
12 giờ trước
1.01^200 = (1.01^(100))^2 is approximately e^2, so it is larger.
2
Phản hồi
Cath
Cath
12 giờ trước
Never underestimate the power of powers!
3
Phản hồi
Маркиз Карабас
Маркиз Карабас
12 giờ trước
Красиво! Цікаво було б порівнять 1.01^70 та 2. Це була б взагалі 💣
1
Phản hồi
Pão
Pão
3 giờ trước
yeah, quite larger... 1.01^111 is already larger than 3
Phản hồi
Robbie Katanga
Robbie Katanga
7 giờ trước
Nostalgic with Addma aka additional mathematics.
1
Phản hồi
Alexey Evpalov
Alexey Evpalov
13 giờ trước
Спасибо.
1
Phản hồi
D B
D B
6 giờ trước
شكرا على مجهوداتكم
يمكن استعمال حدانية نيوتن نجد بسرعة ان 3 اصغر بكثير.....
1
Phản hồi
D B
D B
5 giờ trước
العدد اكبر من 3 بل اكبر من 7
1
Phản hồi
faith(٠٠٩٦٧٧١٣٨٦٢٨٥٧)
faith(٠٠٩٦٧٧١٣٨٦٢٨٥٧)
13 giờ trước
Hرسالتي إلي كل مسلم "يااخواني قال الله٠ ٠ ٠ عزوجل: يَٰٓأَيُّهَا ٱلَّذِينَ ءَامَنُواْ لَا تُبۡطِلُواْ صَدَقَٰتِكُم بِٱلۡمَنِّ وَٱلۡأَذَىٰ كَٱلَّذِي يُنفِقُ مَالَهُۥ رِئَآءَ ٱلنَّاسِ وَلَا يُؤۡمِنُ بِٱللَّهِ وَٱلۡيَوۡمِ ٱلۡأٓخِرِۖ فَمَثَلُهُۥ كَمَثَلِ صَفۡوَانٍ عَلَيۡهِ تُرَابٞ فَأَصَابَهُۥ وَابِلٞ فَتَرَكَهُۥ صَلۡدٗاۖ لَّا يَقۡدِرُونَ عَلَىٰ شَيۡءٖ مِّمَّا كَسَبُواْۗ وَٱللَّهُ لَا يَهۡدِي ٱلۡقَوۡمَ ٱلۡكَٰفِرِينَ}}}ياناس ياامة محمد اني اقسم بالله علئ كتاب الله اني لااكذب عليك ولاانصب ولااحتال اني بنت يمنيه نازحين بسبب الحرب اناواسرتي عايشين اناوامي واخوتي سغار والدنا متوفي الله يرحمه ومامعنااحدفي هذاالدنيا يقف بجنبنا في هذا الضروف القاسيه ومامعي اخوان كبار اناالكبيره في اخوتي ولكن انابنت لااستطيع مثلك ان اروح اشتغل بين الرجال واصرف علئ اسرتي والله ثم والله يااخي انناقدلنايومين محرومين من لقمت العيش ومعي اخوان سغار انظركيف حالتهم اقسم بالله يااخي انهم خرجومن البيت للشارع وشافو الجيران ياكلو راحووقفوعندبابهم لجل يعطوهم ولوخبزه يابسه يسدوبها جوعهم والله الذي له ملك السموات والارض انهم غلقو الباب وطردوهم ورجعويبكوايموتومن الجوع مااحدرحمهم وعطانهم لقمت عيش والان لومااحد ساعدنا بحق كيلو دقيق اقسم بالله اننا انموت من الجوع فيااخي انادخيله علئ الله ثم عليك واريدمنك المساعده لوجه الله انشدك بالله وبمحمد رسول الله يامن تحب الخير واتساعدني ولو ب500ريال يمني ان تراسلي واتساب علئ هذا الرقم٠٠٩٦٧٧١٣٨٦٢٨٥٧ وتطلب اسم بطاقتي وترسلي ولاتتاخر وايعوضك الله بكل خير فيااخي انت رجال إذاشفت اسرتك جاوعين تعمل المستحيل من اجل تامن لهم ألاكب ولكن انابنت عيني بصيره ويدي قصيره ليس لي اب مثلك واخواني سغار شوف كيف حالتهم وساعدناوانقذناقبل ان يطردونا في الشارع نتبهدل او نموت من الجوع انااقسم بالله الذي رفع سبع سموات بلاعمدوبسط ألأرض ومهداني لاأكذب عليك بحرف من هذا ألرساله واني ماطلبتك إلئ من ضيق ومن قسوت الضروف والحال الذي احنافيه أناوسرتي نسالك بالله لولك مقدره علئ مساعدتنا لاتتاخر عليناوجزاك آلّلّهً آلّفُ خيِر/////٠ ٠ ٠ ٠ ،،،ن
3 is greater than 1.01 to the power of200
これはガバガバで大差すぎる。1.01^200 の比較相手は3ではなく7にしましょう。