e^pi vs pi^e (no calculator)
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- เผยแพร่เมื่อ 29 มิ.ย. 2016
- e^pi vs pi^e, which one is larger? We will use calculus, NOT calculators, to find out. This is a very classic math-for-fun a^b vs b^a exponential question. It is very suitable for all calculus 1 and AP calculus students. My approach is to find the maximum of the function f(x)=x^(1/x) by doing logarithmic differentiation, finding the critical numbers, and then doing the first derivative test. Watch to find out the final answer!
中文版(2021) 👉 • e^pi vs pi^e (外加兩個小故事)
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By the fundamental theorem of engineering, e = pi = 3. Both of those numbers are 27, which is approximately 30.
@Another Random Cuber Depends if you're making a legitimate calculation or just back of the envelope. For sure back of the envelope engineers do exactly this, because really you're just looking for order of magnitude. On a real calculation the engineer will use the exact value + 20% for safety, plus an extra 20% for safety on the final result of the calculation!
😂👌
Well, that looks like an approximation that would cause a lot of errors
@@hatasesasd9631 , its joke
... and 30 is "close enough" to 42.
There's actually another way
Since e^x = 1 + x + x^2/2 +… by Macluarin series, e^x > 1 + x
Then put x = pi/e - 1
e^(pi/e-1) > 1 + pi/e - 1
e^(pi/e) / e > pi/e
e^(pi/e) > pi
(e^(pi/e))^(e) > pi^(e)
e^pi > pi^e
Beautiful!!!
Jonathan Ma loved it. I must congratulate you on this beautiful proof
Amazing friend
I have a doubt how x^(1÷x) can't be zero.
If we put x=0 we get 0^infinity and we know 0^(any number) is zero.
😕😕
sonu shaw x^infinity doesn't have an answer.
sonu shaw Watch this video. It doesn't explain your question exactly, but the thinking process is in the correct place:
th-cam.com/video/BRRolKTlF6Q/w-d-xo.html
Dude... this is dank. Mindblowing! But why are you holding a thermal detonator?
LOL. I have heard so many different nicknames for my mic and "thermal detonator" is so funny!
In case the proof doesn't work.
Because 50 thousand, no less
OOOOH its a mike
Exactly Pi hi exactly Tau /2
It makes me feel very strange to find myself at 2.00 in the morning really enjoying videos like this. Its not even that I really understand it.... I think I just enjoy watching someone being so enthusiastically clever. Thank you. Now I'm going to go try and sleep.
japeking1 mysteries of life....it’s enjoyable to see a thought a human mind solve a non trivial question.....logically....its bueatiful actually ..like the Mona Lisa painting...
Is the blank space... is ok I guess...
Lies again? EPA EPI
I'm seeing it at 1, four years later
You solved that with e's (ease). It was easy as pi.
Goddamnit
It was a piece of cake for sure!
r/iamverrysmart
SpeedyMicherGD r/whooosh
@@castroploiin r/ihavereddit
Simplify the calculus immensely: x^y > y^x, log both sides and bring down the exponents to get ylogx > xlogy, then solve to get y/logy > x/logx, so the function to analyze is x/logx. We know x is bigger than 1 in both cases so we can ignore the asymptote. Derivative is 1/logx - 1/(logx)^2, and setting equal to zero gives logx(1-logx) = 0, so logx = 0 or logx = 1. Since x is bigger than 1, logx is bigger than 0, so logx=1 which means x=e. It's obvious that x grows faster than logx for large x, so the limit to infinity is infinity. Similarly, as x goes towards 1, the denominator goes to 0 while the numerator goes to 1, so the function goes to infinity. Thus, x=e is a minimum of the function, so e/loge = e is less than x/logx for any x>1, in particular, pi. So, pi/logpi > e/loge, so from the algebra in the beginning, we get e^pi>pi^e.
Very nice!!
Essence of Abstract analysis.
I was going to write same thing bro
Nice proof without a calculator!
e=3=pi， which means 3^pi =pi^e😂😂
Engineers be like
Funny thing: in astronomy this is actually correct, you're never concered about what's behind the decimal symbol but rather about the magnitude of the result. So 3*3=pi*pi=10
@@alexbarac 10?
@@alexbarac also, its 3^3, not 3*3. 3*3=9, 3^3=27
@@methatis3013 Yeah, it's an accepted approximation. In many cases you tend to get out of the equations pi*pi and to simplify the calculations, you just aproximate it to 10.
wow.what an amazing application of maxima minima concept.
Yay!!
I find your math videos quite relaxing to follow :) It's therapeutic.
I love your channel. It's been awhile since I took calculus and I enjoy how you use it to solve understandable problems. Keeps me refreshed.
I am glad to hear! thank you
e=pi=3
both equal 27
e = 2 = 3 = pi
You can also demonstrate the inequality by using the monotony of the 1/x * ln x function
In a video of yours, you claimed and demonstrated that if a>b>e, then b^a > a^b. Both e^pi and pi^e are greater than e, so we can deduce instantly that e^pi > pi ^e because pi>e
I used e*ln(x)/x which is 1 at x=e and decreasing (take derivative) for x>e. So e*ln(Pi)/Pi
Excellent video! You explain things VERY well!! Fortunately, I was able to figure it out after your statement at 1:46; that insight you gave was genuis! What I have found and love about math is that, often, the beauty comes, not in the answer or even the problem itself, but in the method used to find the answer and the creativity that such problems often require. Great presentation!
alkankondo89 thank you for finding my video enjoyable. I also have another playlist called the "math for fun" you can check that out too
@@blackpenredpen Sir how would you even have known that the maximum value of x^(1/x) is e^(1/e)? It is clear when we take the derivative but honestly I don’t see how anyone would even have thought of doing this in the first place to solve this problem.
@@captainhd9741 we have 2 statements as e^(1/e) and π^(1/π) and we want to find which one is bigger. İn those statements, different values are e and π so we just write x for them. Than we can make the graph of the x^(1/x) function and see the results. Or try to prove as he did in the video.
We wrote x for them mainly because we don't want to calculate the result, rather we want to prove that one is bigger than the other.
Also if we have a proper function, it's sometimes easier to find the extremum points rather then calculating.
Nice pfp
Idk why but in this video the way you talk is relaxing af XD
It's pretty relaxing in every video, but in this one it's at it's maximum
TheMichus
Wow!! Ok!!!!! Maybe Bc of the music??
@@blackpenredpen
I think this might be the case
TheMichus : )
beautiful! this proof is soooo satisfying!
What an absolutely awesome explanation! Just love all your videos.
This is absolutely amazing! I had no idea you could use calculus to solve problems like this and I am mind-blown. Thank you for your videos and your channel 💙
2.7 < e < 2.8
3.1 < pi < 3.2
e^pi > 2.7^3.1 > 3.2^2.8 > pi^e
Therefore, e^pi > pi^e.
No need for exact values. Just the first two decimal places of each (2.7... vs 3.1...)
That doesn't make any sense in this world , how can u say that 2.7^3.1 >3.2^2.8 without calculator
@@theimprudentman7272 just simply calculate it in the head !!!
e to the pi is greater. I cheated..I used a computer.
You were a computer? How did you become a human?
Joel c.l what if he is a cat, not a hooman?
Mark Zuckerberg was an android, so you are in good company.
r/Not_cheating
Now are you human?
When I first saw this problem I used a similar approach but with y=e^x-x^e and using that you can prove that y>=0 for all x>0, with y=0 only when x=e. Thus you can say that for any x (obviously other than e) e^x is bigger than x^e and thus must also be true for x=pi
I like how great amd simple your explanations are and also how nice your accent sounds
I have a better question.
Which is bigger: key lime pi^e or apple pi^e?
Project Overturn aka RareBeeph apple pie^e is bigger
apple
Dude, that's brilliant! I've been a math professor for 8 years and never seen such innovation. Where did you find such a question? Let alone the inspiration to solve it.
Sad Prof.
It's a JEE ADVANCED Question.
Taking double logarithm of pi^e: Ln(Ln(pi^e))=Ln(e.ln(pi))=1+ln(ln(pi)). Since 1+x < exp(x), we get.
Ln(ln(pi^e)) < exp(ln(ln(pi)) = ln(pi).
Taking exponent on both sides: ln(pi^e)
No need to take double log, take single log and one no. Will be negative other will be positive , the positive one will be greater
Very nicely done! You have a very pleasant presentation style.
*starts using calculus on every question*
Me: .-.
r u crazy?
e^x>1+x
replace x by (pi/e)-1
answer obtained.
very impressive!...
but this is just devine inspiration. I find his systematic approach a lot more useful.
thats true
Equation isn't correct for x=0 ?
≥
Hi sitangshu sekhar bhattacharjee
Managed to do it by considering a^b and b^a where a>b. You can consider both scenarios and get a contradiction in a^b > b^a. Was very satisfying indeed
I had this problem on a calculus exam 14 years ago. It was interesting to see a youtube video that tackled it (and did so in a very different way from how I did it back then).
Thank you for helping me to remember first derivate test.
There's one final observation you could add here --- because x=e was the maximum, the other number didn't matter at all. e^x > x^e for ALL non-negative x except x=e. So e^2 > 2^e, e^420000 > (420000)^e, etc, and, of course, e^pi > pi^e.
EDIT: inserted the word "non-negative" -- thanks all who pointed that out
as long as x > 0
To expand on zyrohnmng's comment:
Need to avoid using negative numbers in such an example, because a negative number raised to a non-integer power is highly problematic, and almost always, non-real & multi-valued. And complex numbers do not have ordering.
That, and during his proof, he made the assumption that x was positive on multiple occasions for the very reason you just gave.
True -- I should have said ALL *non-negative* x except x=e, and not used -420000 as an example! (Note to +zyrohnmng: the relation holds ok for x=0)
The proof doesn't work for x=0 since the function and its derivative is undefined for x=0. You can check the case separately.
I solved this with one simple observation i.e. a^b > b^a (when a π^e
I've always wondered how to do this one! Thank you!
My pleasure!!
I love how you solved it but you could just plug in the values to a calculator :P
now this is e^pic
Where c=1, mind you.
Just consider the monotonicity of the function f(x) = ln(x)/x, since x^y > y^x i.e. ln(x)/x > ln(y)/y
That was a fun one! Thanks!!
SUNDAY MORNING RAIN IS FALLING
Thank you. I was confident that pi^e would be bigger, and I'm glad to be humbled.
Yah but it's hard to predict between (1.23)^e and e^1.23? By intuition
..
Calculus is crazy....it was created with intuition and it give solutions to problems which are not at all intuitive....
This is why you can't assume anything in math! Appearances are often deceiving
Brilliant. Never saw an answer to this question before. And taking the derivative of x^(1/x) the way you do is very smart
as a corollary to x^(1/x) having a single maximum at e, we can generalize the result:
for ANY x,y: x^y > y^x when either the triples (e,x,y) or (y,x,e) are in order, i.e.:
(1) e
we notice that x^(1/x) falls faster from its maximum for x < e than for x > e, i.e., x^(1/x) is flatter right of e
in case (3a) above, i.e., when x < e < y, can we show x^y < y^x when the x distance to the max (e) is bigger than the y distance to e, i.e., e-x > y-e, i.e.:
3a: if x
How about 2.001^1.999 vs 1.999^2.001
I graphed them on my mac grapher app:
y=e^x
y=pi^x
x=e
x=pi
and it turns out they are closer than I would've thought!
e^pi > pi^e. Because both of them are >= e. This is true for ANY 2 numbers that are >= e; if e
Superb...
Out of all your lectures this is my favorite😍😍😍
That's one big ass microphone!
Sad about being on high school, i stopped understanding when he used calculus :(
@Sulayman Hussain the reason why implicit differentiation works is not that hard. It basically works because of the chain rule, keep in the back of your mind that y is a function of x ;)
Dy/dx (y) = d/dy(y) x (dy/dx) = 1 dy/dx = dy/dx
You have always been implicit differentiating and you never realised. Every time you differentiate a y=___ equation. The LHS goes to dy/dx because of this reason.
Have a look at the essence of calculus series on the channel 3blue1brown
@Sulayman Hussain what you are saying is true. The fact that we can simplify differentials is not well defined in cal1 and cal2. With teachers telling us to pretend differentials as fractions. We don't get any concrete proof why until real analysis.
that is a very elegant and easy proof, thank you so much 😊
Very good logic! Excellent presentation.
Sir, you teach in an excellent way, you cover those numerical which are neither present in any of the textbooks, nor they are taught by teachers in Colleges and in Universities.
Sir can you please compare Phi (The Golden Ratio) with e and pi? I mean "phi to the e", "e to the phi", "phi to the pi", "pi to the phi", "e to the pi" and "pi to the e". Sir can you please compare them all together in a single video?
Thank you! The work will be very similar. The ones with base e will win. I think I can work out e^phi vs. phi^e for fun : )
blackpenredpen thank you for you reply Sir, but why are you so late in reply? I've commented on you "sine of 18 degree" video about a week earlier, I asked a question in that comment, but you havn't replied me,
مشكور جدا حل ممتاز
Muhammad Hussain Sarhandi lol
@@divyoroy9056 what happened?
You are awesome.
Indranath Mukherjee thanks!!!
Excellent demo.
Very clever demonstration. Congratulation.
Wow, by choosing to do implicit derivation instead of the chain rule you made that derivative waaaaay harder than it had to be.
Brianna Baker waaaay harder? Please make it waaaaay easier for us. Apple sauce.
Okay I'll be honest, I forgot that using the exponential rule doesn't work for functions of base x, which is where I thought the majority of the optimization would come from. Still, it can be done without implicit derivation though apart from the first step this is fundamentally the same process you took. I guess I just hate doing implicit derivation unless it is absolutely necessary. Anyways here's the steps for those interested:
imgur.com/a/sghux
-_-......
I was expecting something waaaaaay easier.......
easier solution:
e^pi>pi^e ln(e^pi)>ln(pi^e) pi>e*ln(pi) pi-e*ln(pi)>0
let f(x)=x-e*ln(x). we want to prove f(pi)>0:
f'(x) = 1-e/x
so for (x > e) we have f'(x)>0 so we have f(x) increasing and f(e)=0 which means f(pi)>0.
which means e^pi>pi^e
blackpenredpen... since you were expecting a derivative waaaaay easier, here it is:
derivatives are just rules you memorise... like the product rule, similarly there is a power-exponent rule.
For the product u*v the rule says "derivate u*v keeping u constant, plus derivate u*v keeping v constant".
To derivate u^v "derivate u^v keeping u constant (exponential) plus derivate u^v keeping v constant (power)".
Therefore
d(u^v)/dx= u^v*Ln(u)*dv/dx + v*u^(v-1)*du/dx
Using this rule (which is analogous to the multiplication rule) you can write the derivative in just one step. :-)
Great proof, nice job. But I think you should write the inequality : e^(1/e) > pi^(1/pi) > 1 . Only then you can raise both sides to the same exponent, which is pi*e, and be right with the inequality you finally get. In fact, this stems from the fact that the exponential function a^x is monotonically increasing for a>1, and monotonically decreasing for 0
well I think it's obvious since 1/e and 1/pi are positive they are both greater then 1
Destroctive Blade neither 1/pi nor 1/e are greater than 1.
of course they aren't but 1/pi and 1/e are positive so e^(1/e)and pi^(1/pi) are greater then 1
this is what you want to prove, right?
My initial comment was aiming to the fact that the proof was incomplete, that's all.
yeah sure I just said that it is obvious
Dude. I love your videos. I'm still learning algebra but I can't wait to be able to do these maths.
How’re u doing now?
much easier: e^pi > pi^e amounts to pi/ln(pi) > e = e/ln(e). take the function f(x)= x/ln(x); then is f'(x) = (ln(x)-1)/ln^2(x) so f'(e) = 0 with f'(e+h) > 0 > f'(e - h) for 0 < h < 1, which implies e = f(e) < f(pi) = pi/ln(pi)
for a^b > b^a, could you describe a function to give you the the upper limit of a for any value of b?
in other words could the problem of e^pi vs pi^e be generalized to ANY real numbers ?
yes, and it was generalized in the video using the function x^(1/x). a^b > b^a iff e = a > b
I got an ad asking why I need to watch math videos from 2006 to do my homework...
The Epic Gamer I hate that ad too...
You can also try this method
Take ln both sides
Pi vs e ln(pi)
Consider a function x-elnx
Differentiate
We get, from e to infinity, the function is monotonically increasing
Therefore minimum value 0(at x=e)
Therefore x-e lnx> 0 for x>e
Put x= pi (since pi>e)
We get pi>e ln pi
Take e power both sides
Hence proved
Very good! I did it like this: consider f(x) = e*ln(x) - x; take a derivative; find the maximum value is 0 at x = e. So, f(pi) < 0. So e*ln(pi) < pi. So, ln(pi^e) < pi. So, pi^e < e^pi.
It's a better proof
I sort of knew intuitively what the answer is but the beauty in math is seeing the proof :)
wp.
e^π and the π^e you like good p and I like the trees
More elegantly put, use e^x >= x + 1; substitute x as x = pi / e - 1, getting; e^(pi / e - 1) > pi / e, hence e^(pi / e) > pi and finally e^pi > pi^e
good proof, may you solve problems with conic sections please?
Which one is bigger?
I was told, size doesn't matter.
The smaller one to the power of the bigger one.
Simon Ruszczak The problem is, the sizes do matter. If you, for example, have 3^2 vs 2^3, 3^2 is greater. This is the simplest example without using zeroes and ones (which would naturally mess things up), and it goes against the rule.
wouldn't 3^2 being bigger show that π^e be bigger ( pi~3 and e~2.6), which was proved otherwise?
WOW!! I WORSHIP YOU IN THE SAME WAY AS I WORSHIP KHAN ACADEMY AND PATRICKJMT!
Thank you~! = )
Same this guy is amazing. Does mathematical hocus pocus to make hard problems easy.
+blackpenredpen was there a context like this in Stewart's Calculus?
Hmm, I am not sure..
+blackpenredpen it's okay! Anyway I hope you could do Calc III videos. Just a suggestion though.
Beautiful solution
wow, this was an incredible solution to watch. very inspiring, very cool :)
Evil MrMuffinz thank you!!
Aren't they both equal to 9?!
As all engineer knows, e = π = 3, so
3³ = 3³ = √81 = 9
Man, 3^3 is 27.😅
E is approximately 2.71828 whereas pi is approximately 3.14159
S.H. Kaifee and if you’re an engineer you round to the closest whole number and both e and π is 3 to their closest while numbers
Ohhh
Getting smarter every day
I didn't know that fact about engineers
Thanks @@lucifersdevilishdetails.
If you just take natural logs, you get pi vs. e.ln(pi). pi is between 3 and 4 and e is between 2 and 3, so ln(pi) must be greater than zero but less than 1, and thus e.ln(pi) must be smaller than both e and pi. Converting back to exponents takes you back to the original form of e^pi > pi^e.
ln(pi) > 1
The natural log of pi is greater than 1 tho.
Let lp(a) = log(pi) (a)
x = e^pi
lp(x) = pi
y = pi^e
lp(y) = pi lp(e)
e < pi then lp(e) y (e^pi > pi^e)
That was such a wonderful explanation...thankyou so much ...
Let's say e=2 and pi=3
2^3=8
3^2=9
gotta love physicists
quik maffs
Mr. American man that’s a bad estimate 😂
but e^pi is greater
Wow same method bro, but wrong estimation, it supposed to be
e = 2.5
π = 3
I think it’s obvious that e^pi is greater than pi^e lol. Still, great video
I loved this!! you are awesome!
Take natural log for both sides and it's way more easier to differentiate if you take fx as x/lnx
That was e’s (easy) as pi.
That was e'asy pieasy
you derivate ln(y) by dx ...
yea, you can do that so long as you leave the dy/dx at the end.
Yes since you know that y is a function of x you evaluate it as such. By using dy/dx
Yes. It's called *implicit differentiation* because the derivative of y is derived implicitly.
Peter Bočan by using chain rule, we have:
(d/dx)(ln y)
= (d/dy)(dy/dx)(ln y)
= (d(ln y)/dy)(dy/dx)
= (1/y)(dy/dx)
isn't it?
Excellent! I enjoyed that very much. Thank you.
Great job , much more difficult than the video i watched before (e^pi=1/e) . However , our math teacher in school always made us solve the min-max problems by setting df/dx>=0 which is quite hard to solve , especially if the df/dx function is big
Or you could just estimate with 2^3 vs 3^2
ENGINEER DETECTED
that wouldn't have worked, since 2^3>(pi-3)
La Tortue PGM LOL That's exactly how I guessed my answer (I got it right using a slightly better estimate). And I am an engineer.
well, I did it for 2.7^3.1 and 3.1^2.7 and it worked
btw mathematical mathematics memes is the best fb group ever
That was going around when I was in math grad school in the early '70's. My solution was *_a lot_* simpler than yours. :-)
JiveDadson but I have more TH-cam views tho! ：）
JiveDadson Fucking rekt
Very clever and great explanation.
Loved it. You are outstanding!!!
Take e^pi vs pi^e. Rewrite that as: ln(e^pi) vs ln(pi^e) and then pi*lne vs e*ln(pi). Rewrite that as lne/e vs ln(pi)/pi. Now consider the function f(x)=lnx/x. This function is monotonous decreasing, hence lne/e>ln(pi)/pi. It follows pi*lne>e*ln(pi) so lne^pi>ln(pi^e) and in the end e^pi>pi^e
You are very good at explaining this stuff nice!
Is someone calculated at what point we can say that a^b>b^a ?. Seems like it work for every a>e and b>e .
Another way:
take the ln of two sides:
ln(pi^e) vs ln(e^pi)
then take out exponent:
e*ln(pi) vs pi*ln(e)
it equals to a:
ln(pi)/ln(e) vs e/pi
rephrase this:
ln(pi-e) vs e/pi
since e/pi has an order of (0.5,1), and ln(pi-e) has an order about (-1, -0.5) we can outstand the winner:
ln(pi-e) < e/pi
-->>>
pi^e < e^pi. done.
ln(pi)/ln(e) does not equal ln(pi-e), this way is wrong
look at the function log(x)/x (any logarithm base will work but I'll use ln for simplicity of diffirentation) . The first derivative is (1/x^2) * (1-ln(x) ) (just looking at the (1-ln(x)) part ) there is one and only one turning point at x = e. After this the derivative is always negative . Therefore beyond e log(x)/x is a strictly decreasing function . If a>b>e ln(b)/b > ln(a)/a. -> aln(b) > bln(a) -> ln(b^a) > ln(a ^b) -> b^a > a^b . (This also shows e^x >x^e)
A beautiful, simple to understand and elegant proof. I'm a grade 10 student. But I understand calculus because of you! I literally jumped when you proved the maxima value :D
With the function f(x)=lnx/x π>e => f(π)π^e
if you know x^1/x gets smaller for bigger x value. You can apply this
e^1/e > pi^1/pi (pi > e)
(Pi)e Exponent
e^pi > pi^e
God dam bro!!! You seriously destroyed that problem!!!wow!!! As a fellow mathematician I can say you are casual when you speak....but you are really one of the smartest people on earth!!!even to ask such a question is a 1 in a million people....
Let e < x. In this interval, we have e/x < 1. On the interval e < x, both sides are positive. This means that the area under e/x in that interval is less than the area under 1 for the same interval. So the integral from e to pi of e/x dx < integral from e to pi of 1 dx. So e*ln(pi) - e*ln(e) < pi - e. So e*ln(pi) < pi. So e^pi < pi^e. QED
Nice work, bro! Well explained, you could be an excellent teacher.
Matheus Cardoso Thanks. I have been a teacher for many years already.
bro that was so cool kee it up