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.
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. 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.
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. 😊
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 .
@@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).
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
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*😅
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)
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.
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
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.
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.
@@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”.
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.
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.
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
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
@@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.
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
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
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.
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
Wait, so for every two numbers when one is greater than the other (a b^a ??? Or am I just confused?
@@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. 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.
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. 😊
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.
I love your substitution of x for pi. This makes the problem obvious. You can graph 3^x and x^3.
I love your substitution of x for pi. If you graph 3^x and x^3, it becomes clear that for any value of x greater than 3, 3^x is greater than x^3.
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 .
Really simple rule. If a and b are both greater than e (2.71828...) then the expression with the bigger exponent always wins.
And otherwise, the bigger base always wins?
@@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).
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 graphed y=3^x and y=x^3 by hand. They are equal at x=3. And y=3^x is greater after that.
7:12 sounds like you're threatening me on the street
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*😅
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)
i mixed up NAZ and ZAN again and got confused q3q
3^x > 3x (x>1)
∴3^(π/3) > 3(π/3) = π
∴3^π > π³
Excellent explanation really. Instantly subscribed, thank you for this video
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
That is not a correct proof. Pi is not equal to 3.142...
That is nor an induction proof.
That is not an induction proof.
@@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.
Nice, but unnecessarily detailed.
Essentially ln(3) ~ ln(pi), and then pi*ln(3) > 3*ln(pi)
Why didn’t you take a calculator for 5€…?
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
Why cant we use log to simply solve it
Excellent. Very well explained... with a simple yet concise flow...
Great!❤
I haven’t watched the video yet. My guess is 3 raised to pi is larger because the exponent is larger.
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.
@@allasar Yup, I am dissatisfied that I can't prove it :D
@@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.
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.
Compare these two: sqrt(6)^pi and pi^(sqrt(6)).
Коварство запредельное: √6 и π лежат по разные стороны от e. Хотя можно и усугубить: √7
@@-wx-78- Неа, лучше (е-1/е)^π и π^(е-1/е)
@@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”.
I guessed and got the right answer
3 is less than pie...
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.
What are you on? This is a property of logarithms.
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.
Pi times three
pi^3 = 3^3*(1 + (pi-3)/3)^3 = 3^3 * (1 + (pi-3)/3)^(3/(pi - 3) * (pi - 3)/3 * 3) = 3^3 * ((1 + (pi - 3)/3)^(3/(pi - 3))^(pi - 3) < 3^3 * e^(pi - 3) < 3^3 * 3^(pi - 3) = 3^pi
yawn 3^pi is large as it is closer to e. this is already forever answered. in any a^b versus b^a if a,b > e and given a
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.
3^3,14 is greater
dude...lol too much. This is good practice for all you know....
Advice: Don’t rattle off the chain rule. Show your work. I gave you a thumbs up expecting you’ll improve.
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
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
this is only by heuristic, though.
3^2 > 2^3
Your post fails.
This person keeps writing smaller and smaller. In the limit thereof this analysis equals zero. Poof!
Uff so tired of this type of videos... a^b > b^a if b>a. END.
@@juaneliasmillasvera not quite right 2^3 < 3^2 and 3>2
@@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.
@@joshualee1595 Actually the video already solves most of that by finding the maximum value at e. Assume ab^a, if both a and b are smaller than e, a^b
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
For positive bases & positive exponents, I use the rule of abba or a^b () b^a
If a > b then a^b > b^a
If a < b then a^b < b^a
If a = b then a^b = b^a
This rule just doesn't work like if a = 10 and b = 2, a^b < b^a and not the opposite
@@octalbert7280 It didn't work for the problem in this very video, either, since 3 < pi and 3^pi > pi^3
This is pure trash
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
@@prasoon7916-- Write a sentence.