@@jejojoje9521 I tried to include a link, but it seems to have gotten clobbered. If both terms are less than 'e' then yes you are correct. Largest base wins. Otherwise it is complicated as the crossover point follows a fussy curve involving the Lambert W (product log) function. I'll try to post the link to the graph in the next comment to see if it's just my inclusion of the link causing it to get deleted.
Yeah the comment with the link gets auto-deleted. The critical curve where x^y=y^x is solved by using the Lambert W function. x = -yW(-log(y)/y)/log(y) which gives a curve that looks like it asymptotically approaches (1,∞) and (∞,1) while going through the point (e,e).
It takes seconds to determine that lim(ln(x)/x) as x goes to infinity is zero, and, further, that there's a maximum at e. Since e < 3 < pi, we have everything we need.
I set PI = X + 3 where X is decimal portion of PI. 3^(X+3) vs (X+3)^3. Left side is 3^X*3^3. Since X is < 1, this is (3)*27 which is < 81. The right side expanded is X^33+9X^2+27X+27. Since X
I do not follow your reasoning. In both cases, you managed to establish only upper bounds and you have not actually proved the inequality. In other words, you actually have to show that 3^pi is greater than something and that pi^3 is less than something. In your argument above, you only established that both are less than two numbers. That is not very helpful as 3^pi can still be less than pi^3.
well.. the function f(x) = x^(1/x) is decreasing when x>e. (defferetiate it for proof) hence if e < a < b than a^(1/a) > b^(1/b) hence (a^(1/a))^(ab) > (b^(1/b))^(ab) hence a^b > b^a here 3 > e and pi > e. and 3 < pi. so 3^pi > pi^3 peace to all
@@wostin this is true if they both are not less than e (the base of natural lorarithm 2.72..). example e^pi > pi^e and if the both are not greater than e ***upd. AND they both are not less than 1. (a < b) => (a^b < b^a) example 2^2.5 < 2.5^2 and.. if one less than e and another greater than e example 2.5^3 vs. 3^2.5 huh i dunno ¯\_(ツ)_/¯
Thank you for the consistent posts. I get a little smarter every time. Your explanations always seem to come together, even if it takes me half an hour or so to get some concepts. 😊
Thank you. Clear explanation without going off topic. I am glad I checked out your video. I am looking forward to more postings and going through what you already have made.
You could use an induction proof. If π was equal to 3, then 3^3 = 3^3. if π was equal to 4, then 3^4 > 4^3 (81>64) if π was equal to 5, then 3^5 > 5^3 (243>125) But π = 3.142... and 3 > π > 4 > 5 Hence 3^π > π^3
@@robertveith6383 What do you mean pi is not equal to 3.142... That is actually rounded. It should be 3.14159... But, what I'm really saying is that it is between 3 and 4. If it's not induction, what do you call it?
@@tullfan2560 Pi, of course, is *not* equal to 3.142... The three dots (an elliipsis), means the digits continue after the 2, which they do not. Pi = 3.141... instead. Induction will involve a base case, and using a variable, for instance.
@@robertveith6383 You can see that when the difference between the number and its exponent increases, the value of 3^N - N^3 increases. The difference between the two terms is zero when N=3 and monotonically increases as N gets bigger. As π is greater than 3, 3^π will be greater than π^3.
Or just use induction combined with intuition. Substitute PI with 2. You then have 3 to the power of 2 (which is 9) on the left side. And you have 2 to the power of 3 (which is 8) on the right side. So is 9 bigger than 8? YES! And this is true for all numbers N. If you have PI instead of 2 you know the left hand side will always "win".
I considered this approach, but wasn't able to completely prove it. Suppose we start the sequence with 1^2 and 2^1, the left side is less than. Going to 2^3 and 3^2, the result is greater than as you showed. Working the next few (3^4 4^3. etc) they lead to the left side greater than. Intuitively, if this continues to infinity, the two sides seem to approach equality asymptotically. But how does one prove it? If you're at point N and the left side is greater, how to prove that point N+1 also has left side greater? I couldn't figure a method to prove it that doesn't involve calculus (and likely log as well).
@@allanflippin2453 ..did graph this a^x=x^a , at e thers one intersection point, under and over thers two under ae, a^x is always greater after the second intersection point for very lage values of a the first intersection point goes toward 1 and the second is x=a^a tested a between 0 < a < 1 000 000 ..and its always true... edit: actually 0's not tested as 0^0 is undefined, just close to zero values tip: use a log axis scale, makes life easier ;D (a^x becomes a traight line)
@@allanflippin2453I think his point is, we inuitively know the answer without proof, so if the question is just "which is larger?" we'd get the points. But I agree, we need proof since intuition can be wrong, and an answer (even a correct one) without proof is dissatisfying.
@@allanflippin2453 so, you are trying to prove N^{N+1} > (N+1)^N for N≥3? First, we can prove 2^k < (k+1)! for k≥2. Indeed, for k=2, 2² = 4 < 6 = 3! = (2+1)! ✓ If it is valid for k≥2, then 2 2×2^k < (k+2)×(k+1)! => 2^{k+1} < (k+2)! Done. We proved 2^k < (k+1)! for k≥2, which implies 1/(k+1)! < 1/2^k (*1*) for k≥2. Now we prove sum 1/k! < 3 (*2*) for any k. Indeed, 1/0! = 1 1/1! = 1 1/2! = 1/2 1/3! = 1/6 < 1/2², by (*1*) 1/4! = 1/24 < 1/2³, by (*1*) ... Adding everything, sum 1/k! < 1+1+1/2+1/2²+... < 1+2 = 3 So we proved (*2*). Now we prove (1+1/N)^N ≤ sum_k 1/k! (*3*) for any N, with k = 0,...,N. Indeed, expanding (1+1/N)^N using the binomial formula (1+1/N)^N = sum_k C(N,k)(1/N)^k with k=0,...,N and C(N,k) = N!/(k!(N-k)!) So, fixing k, 1≤k≤N, we have the term C(N,k)(1/N)^k = (N!/(k!(N-k)!))(1/N)^k = (N(N-1)...(N-k+1)/k!)(1/N)^k = (N/N)((N-1)/N)...((N-k+1)/N)(1/k!) = 1(1-1/N)...(1-(k-1)/N)(1/k!) ≤ 1/k! because for each factor 1-j/N, j=0,...,k-1, 1-j/N ≤ 1 So we proved (*3*). This means we have, for any N, (1+1/N)^N ≤ sum_k 1/k! , by (*3*) < 3 , by (*2*) proving (1+1/N)^N < 3 for any N, which implies (N+1)^N < 3N^N (!!) Finally ... for N≥3, we obtain the inequality, (N+1)^N < 3N^N , by (!!) ≤ N×N^N , by 3≤N = N^{N+1} Done. That's a way.
Pi is between 3 and 4 3 cubed is the same as 3 cubed 3 to the 4th is greater than 4 cubed Therefore 3 raised to the pi is greater than pi cubed. This is a more simple approach, but it not necessarily proves it*😅
Simple to solve really. 3^x = x^3 when x = 3. 3^4 = 81, and 4^3 = 64 therefore 3^4 > 4^3. Therefore 3^pi > pi^3. The only hand wavy thing was 3^x = x^3 when x = 3. You should technically prove this is the only real solution. But since cube root's alternative solutions and logarithms alternative solutions only matter in the complex I do believe it is easy to see there is only one.
Sorry, but your logic appears to be flawed. According to your logic, we can also say: 3^x = x^3 when x = 3 . 3^2 = 9 , and 2^3 = 8 , therefore 3^2 > 2^3 . Therefore (you'd conclude that) 3^e > e^3 . Which is obviously incorrect, because 3^e < e^3 .
Since this, and similar integer problems, are so common on math channels, 5 years ago I chose to remember this rule of thumb: if the exponents are larger than e, the larger exponent wins over the larger base. I think it holds true, at least for the problems I have come across
Really enjoyed the analysis! Interesting that you take the derivative with the product rule, not the quotient rule. I used to do that, but then decided it was better to just memorize the quotient rule.
Yes, I feel it's the logical first step to take. It makes the result easy to see from the (well-known) fact that log(x) increases slower than x if log>1, and faster if log
@@bkkboy-cm3eb ok here we go: 3^π vs π³, take the cube root of both, 3^(π/3) vs π = 3(π/3) let x = π/3 3^x vs 3x. for x > 1, 3^x > 3x, and since π > 3, 3^(π/3) > 3(π/3) = π ie 3^(π/3) > π, cube both sides 3^π > π^3 Cedit to you, I just took your solution and did it reverse.
Maybe someone has already made this point, but isn't it sufficient to observe that the left side must be greater since a log varies more slowly than its argument? π/3 vs ln(π)/ln(3) It may not be quite as explicit as Math Window's demonstration, but if you're aware of that fact (and maybe need to come up with a fast answer) I think it's all you need to know.
This one I can get behind. I'd say for a more rigorous proof, we'd use the property that (d ln(x)/dx) < 1 for x > 1 and ln(1)=0 which means that ln(x) grows strictly slower in that interval and is guaranteed to be a smaller value than the constant function f(x)=x
Write sentences! What you have written are groups of words that do not have capitalized first words or needed punctuation. Do not expect readers to look at your sloppy, lazy, ignorant post.
Take powers of 1/(3π) 3^(1/3) ~ π^(1/π) if cam be shown that the function x^(1/x) has a maximum at x= e and is monotonic decreasing away from ths maximum. π >3 thus 3^(1/3) >π^(1/π)
Might be overkill, but I just threw calculus at this problem immediately... f(x) = 3^x / x^3 f'(x) = (1/x^6)( (ln3)(x^3)(3^x) - (3^x)(3x^2) ) What are the roots of f'(x) ? x clearly can't be cannot be zero, so... (ln3)(x^3)(3^x) - (3^x)(3x^2) = 0 (3^x)( (ln3)x^3 - 3x^2 ) = 0 (ln3)x^3 - 3x^2 = 0 (x^2)( (ln3)x - 3 ) = 0 (ln3)x - 3 = 0 x = 3/ln3 Since 3^4 > 4^3 ( 81 > 64 ) And 3^3 = 3^3 ( 27 = 27 ) And 3/ln3 < 3 We can conclude that for all x in (3, +infinity) 3^x > x^3 Therefore, 3^pi > pi^3
cube root of both sides and dividing by 3 gives 3^0.1415.... compared to pi/3 3^0.1415 approx: but greater than 3x0.38=1.16 and pi/3 is approx but def less than 1.05, thus 3^pi > pi^3
(3 + 1)^3 = 64 and 3^(3+1) = 81. Now let the 1 be reduced by small amounts to zero - would the inequality ever change sign? So 3^pi is greater. Why make it so complicated? And you don't have to remember any tricks!
@@romank.6813 Месье знает толк в извращениях. 😉 P.S. В который раз вижу под русским текстом “Translate to Russian”, сподобился нажать - и вместо объяснения про константу Эйлера получил замену «Неа» на какой-то “her”.
@@santer70 well how do you think your calculator calculates math for you? You think it's magic? NO, PEOPLE HAVE TO USE THIS MATH TO DERIVE A FORMULA FOR CALCULATORS TO CALCULATE THESE EXPRESSIONS BRO
Sorry but ln(π^3) is not equal of 3×ln(π). ln is the natural logarithm so the base number is e. π based logarithm is logπ(x)where π is the base number of the logarithm function.
For any real number a>0 and any real number b, a^b = (e^ln(a))^b = e^(ln(a) * b) = e^(b*ln(a)) and hence ln(a^b) = b*ln(a) This also holds true when a = π and b = 3 , and therefore ln(π^3) = 3*ln(π) There is no mention or implication of a "π based logarithm" in the video.
@SFefy @yurenchu Here's a complete proof using fundamentals for the expression: ln(π^3) = 3 × ln(π) We will define the natural logarithm of π^3 or ln(π^3) as such: I. For any logarithmic function, we define as: f(b, x, y): b^x = y OR: log b (y) = x we generally read this expression as: "the logarithm base b of y is equal to x" so the natural logarithm is unique, the base is by default Euler's number e ~ 2.718 now we define a function for a natural logarithm by plugging in base b = e: f(e, x, y): e^x = y OR: log e (y) = x OR: ln(y) = x so if you plug in y = π^3 f(e, x, π^3): e^x = π^3 OR: ln(π^3) = x (1) Again, ln(π^3) = x is exactly the same as: log e (π^3) = x which reads as either: "The natural logarithm of π^3 is equal to x" OR, "The logarithm base e ~ 2.718 of π^3 is equal to x" II. We should know: the square root of a number *n* can be expressed as: √n OR ²√n OR n^(1/2) similarly, the cube root of n is: ³√n OR n^(1/3) III. Now, if we cube root both sides of the expression e^x = π^3 It will become: (e^x)^(1/3) = (π^3)^(1/3) we know that (a^b)^c = a^b^c = a^(b×c), therefore: e^(x × 1/3) = π^(3 × 1/3) e^(x/3) = π So now if we plug y = π into f(e, x, y): f(e, x, π): e^(x/3) = π OR: ln(π) = x/3 (2) Notice that: ln(π^3) = x (1) ln(π) = x/3 (2) we can rewrite expression (2) as: 3 × ln(π) = x (3) Combining (1) and (3), hence: ln(π^3) = 3 × ln(π)
Great. I would have appreciate to have the justification of why the greater than symbol does not flip at any point (common mistake with inequalities) (multiplication by positive numbers, ln function continuously increasing). But really great overall.
In general, smaller number raised to bigger power is greater than the other way around. 2^8 > 8^2 I hope based on this logic, we can conclude that 3^π > π^3
Let f(x) = x^x . Then f'(x) = (x^x)*[1 + ln(x)] , and f'(x) = 0 ==> x = 1/e . From there, we can show that f(x) is continuously decreasing on the interval (0, 1/e) . Since 1/π and 1/3 are on that interval and since 1/π < 1/3 (because 3 < π), it means that f(1/π) > f(1/3) . ==> (1/π)^(1/π) > (1/3)^(1/3) ... take the reciprocal of both sides, which flips the "greater-than" sign (since 1/x is a continuous and monotonously _decreasing_ function for positive real values of x) ... (π)^(1/π) < (3)^(1/3) ... raise both sides to the power of 3π ; since x^(3π) is a continuous and monotonously _increasing_ function for positive real values of x, the "less-than" sign is unaffected ... (π)^(3π/π) < (3)^(3π/3) π^3 < 3^π
Disagree. I would say most viewers of this channel know how to derive a fraction. Would be tiring to go over each step everytime, already knowing the result of the step. Basically, we skip steps all the time (for instance, we do not need the step by step derivation of lnx to get to 1/x). The question is: which steps do we skip? Personally I would not draw that line at the derivation of a fraction. Most of us have done it so many times, we do not need to see (f'g - fg')/g^2. Nor do we need (1/x × x - lnx × 1)/x^2 as a step, since we already calculated that in our heads before it is written down. There is no clear line, but writing out each step would get old really fast. We are solving an Olympiad question, not filling in a math exam.
@@allasar We’ll agree to disagree, but not even Michael Penn skips that much math in 1 step. Though I admit he has, but in those cases his videos were quite long. Also, I like that ur willing to fight youtube’s algorithm on length. Content creators are making their videos for too long these days. They constantly stretch one or two minutes of information into 15. So, I probably shouldn’t have complained. 😀
I'd rather she explained the chain rule than write out ln (a^b) = b ln(a). If you need to spell that out for someone then they aren't gonna understand chain rule off the top of their head either
for positive numbers, to compare a^b and b^a, then if a>b>e, where e is the euler's constant (around 2.71), then a^b < b^a, and if e>b>a, then b^a > a^b, if a and b are on the other sides of e, then it can go both ways and we need to find something else
3^Pi=Pi^3=12 ACADEMIC MARCELIUS Martirosianas 12 may AcademiC universita della Florida et Semi-protectet edit filter on 9 may 2o22 AcademiC Hcm Bon 11 augusre 2o2o Felds Medalist ACADEMIC Universita di Humboldey General Doctor Expert 2o17 -----2o23 17 moksliu atradejas. Nobel prize study in China -2o22.
@@Shyguy5104 Mmmm, by bad, so there is not a rule for every case of exponents changing variables? or there is a minimun diference between the two variables since what we can approximate the problem in a generallizate way? =)
@@juaneliasmillasvera well I’m not sure if there is complete generalization for whether a^b or b^a is greater, intuition should tell you that the larger the base number, the more exponent size dictates the overall value of the number. Think 2^3 and 3^2 versus 49^50 and 50^49. Intuition tells me that 49^50 is the larger of the two, but 3^2 is the larger of its pair. The question is where this relationship flips, I’m sure someone has already calculated that limit, and it could be a cool exercise to try out for yourself.
