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nice blackboard

thx

Can you solve this question f is real valued function. f^3(x)+f(x)=x. Find integration of f(x) from 0 to 2.

Very good video

appriciated

Thankyou sir I am an Indian high school students

Amazing. Is there a non creative solution that one can come up with on the spot?

2nd Method- 🌞☀🌄 1st Method- 👹

Mueller Via

The statement requires that P be prime, so it's necessary to use that P is prime in the proof, but I'm struggling to see where it was used.

I took pre college maths as optional in my tuition and they taught this there

This channel is amazing

Amazing! Thank you.

We have a lemma says that If lim an+1/an =L<1 with an >0 for all n Thus lim an =0 Since lim an+1/an=lim A/n+1=0 Thus lim an =0

How could epsilon be less then 2|a| and 4|a| at the same time Since we said for all epsilon if someone choose epsilon 12 and our a is 2

It's similar of using Cauchy sequence to prove that is convergent

this is the best youtube proof on this

Excellent video Original and good proof We can also prove it using Jensen inequality and AM-GM inequality

Niceee

黑板寫得亂七八糟，講得模模糊糊，難怪觀看數那麼少。這題很簡單啊，答案心算就出來了。

Thankyou

Quite nice solution!

How can tan(x)/2 be a solution if tan is not continuous on pi/2 + pi*k for arbitrary integer values of k and one is asking for solutions continuous on the whole real line?

Thank you for the problem and its solution. I think one could alternatively observe that (1-x) = (1-x^(1/k))*(1+x^(1/k)+x^(2/k)+x^(3/k)+ ... +x^((k-1)/k)) for an arbitrary positive integer k and replace each of the factors in the denominator by such products, which leads to the initial expression being equal to 1/((1+x^(1/2))*(1+x^(1/3)+x^(2/3))* ... (1+x^(1/n)+x^(2/n)+x^(3/n)+ ... +x^((n-1)/n))). After letting x tend to 1 one then obtains 1/(2*3 ... *(n)) = 1/(n!).

Thanks

I like the use of that lemma.

I actually guessed the correct answer using a few memorized logarithms: The base ten logs for 1-10 are approximately 0, 0.3, 0.48, 0.6, 0.7, 0.78, 0.85, 0.9, 0.95, 1 Looking at the big number, it's basically 22x10^24, so its base 10 log is 24 + log(11) + 0.3. The 11 log isn't in the list I memorized, so I used linear interpolation between 12 and 10: log(12) ~ 1.08, so log(11) ~ 1.04. Then log(big number) ~ 25.34. Dividing by 13 to get the 13th root logarithm, I just do 3 digits mentally to get 1.94. This is extremely close to what I would compute for log(90) ~ 1.95, so my first guess would be that the 13th root of big number is 89.

Hello from iran

(x ➖ 3x+2)

Another approach without using modulo and log (this is "no-pen-no-paper" method): 1. the last digit of x must be 9, because only 9 goes in cycles (9,1) with the 13th power being 9. 2. 100^13 is 27-digit number, so x must be 2-digit number ----> 26(digits) : 13 = 2. 3. x can't be 99, because the given 26-digit number is too small (it's obvious). 4. 8 = (2^3)^13 = 2^39; let's calculate 2^40 5. 2^10 = 1024, let's use "1000" 6. 2^40 = approx. 8^13 = approx 1000 ^4 7. 80^13 = approx 1000 ^4 * 10^13 = 10^12 * 10^13 = 10^25 8. It is less than the given 26-digit number 9. therefore x = 89

1. Since 100^23 has 27 digits, x must be a two digit number. 2. Last digit of x^(4*3+1) must be same as last digit of x^1, therefore, last digit of x must be 9. 3. Digital sum of the given number is 8, therefore, the digital sum of x^13 must be 8. Solving which we get x=89.

i think you can stop and show contradiction at M(1/2)^4 < M(1/2)^k, since from the observation its implied that k > 4 but obviously in that case the inequality does not hold.

Value of X must be odd, since the result is odd, and X is not 1 or 5 (where the result has unit 1 and 5) For K=1,2,3,4,5,..., then ?3^K, the unit result is 3,9,7,1,3,9,7,1,... ?7^K, the unit result is 7,9,3,,1,7,9,3,1,... ?9^K the unit result is 9,1,9,1,9,1,9,.... For K=13, then ?3^K has unit 3, ?7^K has unit 7 and ?9^K has unit 9 So, possible answer is 09,19,29,39,49,59,69,79,89,99 Number of digit 99^13 is about 36 (less than 36) so 99 is rejected, then X=88 (** if there is no typo like this 21,982,145,917,038,330,487,013,369 😊😊😊)

I just thought... Ends with 9 -> The original number must end with 3, 7, or 9 3 -> Final digits of 3 9 7 1 that repeats itself, will be 3 in 13 (Fail) 7 -> Final digits of 7 9 3 1 that repeats itself, will be 7 in 13 (Fail) 9 -> Final digits of 9 1 that repeats itself, will be 9 in 13 (Success) There are 26 digits, close to 10^26 with 27 digits (AKA 100^13), but not too close I'm now left with 89 and 99 as the most likely answers, and thought, "It shouldn't be too close" and picked 89...i was right...

Last digit is 9, it's easy. 80^13 < 10^25, so it can be 99 or 89. If it 99 then the given number mast be divided by 11. Easy to se that it doesn't. So it is only 89

Rather easy to check that by sheer size x it has to be between 80 and 90 and has to end with 3 7 or 9. Of which only 9 can work. So x is 89.

For people who doesn't know, this equation was the first counterexample, to Euler's conjecture: If the sum of n many kth powers of positive integers is itself a kth power, then n≥k This conjecture was proved wrong in 1966, by L. J. Lander and T. L. Parkin. Their example was 27⁵+84⁵+110⁵+133⁵ = 144⁵

0:08 X is a real number 0:18 X is an integer number WTF!? How did you figured out (in just 10 seconds) that solution is an integer number?

Despite your content not being serious, it is very enjoyable to watch you.

Hahahahahahaha.

I counted digits and figured it had to be 2 digit number. Last digit had to be 9. Then I compared to 100 and saw we ended up at 22% of what 100 would have. This means we haved a little over two times so a half life of roughly 6 years. Law of 72 says we are decreasing by 12% per year to have a half life of six years so answer should be close to 88. Since last digit is 9 the answer must be 89.

10<x<100, since 10^13 would be too small and 100^13 is too big. x cannot end with an even, 1, or 5, because then x^13 would end in an even, 1, or 5, respectively. 7^2 and 3^2 end in 9, so 7^4 and 3^4 end in 1, so 7^12 and 3^12 end in 1. So 7^13 ends in 7 and 3^13 ends in 3. So the last digit cannot be 7 or 3. 9^2 ends in 1, so 9^12 ends in 1. So 9^13 ends in 9. The last digit can be 9. So we know that x is in the form 10y+9, with 0<y<10. Looking at the first 13 digits, y^13 <= 2198214591730, or about 2.198 * 10^12 because if it was bigger, the most significant 13 digits would have to be bigger. 7^2 < 50, so 7^12 < 50^6. So 7^13 < 50^6*7, or 1.09375 * 10^11. This is far too small. 8^3 = 2^9 > 500. So 8^12 > 500^4, and 8^13 > 5^4 * 8 * 10^8. So 8^13 > 5*10^11, which is in the right ballpark. 9^2 = 81 > 80, and 9^13 > 80^6 * 9 = 8^6 * 9 * 10^6 > 500^2 * 9 * 10^6, or 25*9 * 10^10, or 2.25*10^12, which is too big. So y=7 is proven too small, y=9 is proven too big, and y=8 looks plausible. So the answer is 89.

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Strangely, x^13 = my phone number multipled by my wife's number plus our door number to the power of the numbers in my post code.

Thank you very much for this proof. There aren't many out there actually proving the existence of the limit, by using lim inf and lim sup

Gonna guess like 91?

Sir, doesn't this depend on the choice of e? We could choose e>1/2, and then a_n <= M(2e)^k -> inf, implying a_n is unbounded. Am I missing something here?

It's obviously 89. Because 9^9 has the last digit of 89 and same as 9^(10n+9). Which means that 9^9^9^9=9^9^(....89)=9^(....89)=.....89.

89

some classic hard anal. great video