are you tired of the a^b vs b^a questions?
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- Опубліковано 21 вер 2024
- Here's a very common question that asks to compare a^b vs. b^a. We will use the calculus derivative of the function f(x)=x^(1/x) to find its maximum. It's a great practice for calculus 1 students.
This video is inspired by my old video e^pi vs. pi^e: • e^pi vs pi^e (no calcu...
For more calculus tutorials, check out my new channel "just calculus": / justcalculus
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#calculus #blackpenredpen #mathteacher
TWO THINGS!
1. Be sure to watch 3:25 for the derivative of x^(1/x) the superman way! #KickingCalculusInItsHead #CalculusFinisher
2. Please try the problem at 11:21 2.5^3 vs. 3^2.5 I do not know if that is even possible. So, any thoughts will be greatly appreciated!
2.5 is around 0.22 smaller than e. 3 is around 0.28 larger than e. But since < e of y=x^(1/x) increases a lot faster in general that >e, and both values are still fairly far from e for that difference to dominate, 2.5^3 < 3^2.5.
Really fuzzy reasoning there though, and it requires remembering the graph of that function in a fair amount of detail, so I don't think it's useful. Also not 100% confident that it's true.
@@JoshuaHillerup But the graph isn't symmetrical about x=e, so yea...
@@blackpenredpen
2.5^3 = 2.5^2.5 * 2.5^0.5
3^2.5 = 2.5^2.5 * (3/2.5)^2.5
So the question becomes which is bigger 2.5^0.5 or (3/2.5)^2.5
(3/2.5)^2.5 simplifies to (72sqrt(3)/125) * 2.5^0.5
So the question becomes is 72sqrt(3)/125 bigger or less than 1? Square it and the answer is 15552/15625, which is less than 1, hence 3^2.5 is smaller than 2.5^3.
2.5^3 vs 3^2.5
(5/2)^3 vs 3^(5/2)
125/8 vs (√3)^5
125/8 vs 9√3
125/72 vs √3
1.736111... Vs 1.732...
1.736111... > 1.732...
(2.5^3)/9 > (3^2.5)/9
2.5^3 > 3^2.5
Call the first number a and the second number b, then solve for x where a^x = x^a , then if the second number is greater than x, then a^b is greater than b^a and vice versa
“I’m not doing 2019” proceeds to do 2020 instead
poor guy never saw it coming.
😅
But this comment was made in 2021
@@wiellnyan Yeah it's 4 months into 2021, we are still living the misery of 2020
For the challenge question of "bases on opposite sides of the maximum" (a
I found a solution but it still requires a calculator. Let me know what you think.
docs.google.com/document/d/1n-dCsfetjwaku9coaFiyJlzhy7Cx10YrndfRAElLm_I/edit?usp=sharing
Great!
I agree.
Nice video. I'm thinking of starting my own math channel in German once I'm better. Today I started chemo, and so far feeling all normal. Hope it stays this way for the rest of the cycles.
Л.С. Мото glad to hear!! Wish everything the best for you!
@@blackpenredpen Thank you very much bprp. Much appreciated :)
Did you end up starting a German-language math channel?
@@LS-Moto How’s it going dude?
Are you alive my man??
For a while, I was confused what chen lu is, but then it got clear. What made it funny is that you call it that on purpose.
: ))))))
Dr. P started it!
His videos go very swimmingly because he does a lot on porpoise!
lmao it sounds funny
blackpenredpen isn’t it?
@@blackpenredpen 😀😀😀😀😀😀
This is definitely a much more rigorous mathematical proof to this type of problem, and not one I would've thought of until now. The way I solved 9^10 vs 10^9 in my head almost as quickly was this:
1. For two positive integers x and y, such that the question asked is x^y vs. y^x, which is bigger?
2. Take the y-th derivative of x^y, you ultimately end up with y!x.
3. Take the x-th derivative of y^x, you ultimately end up with x!y.
4. If you compare x and y, then whichever one is bigger will have a duplicated factor in its factorial:
> A. If x < y, then y!x will have an x^2 term in it, whereas x!y will merely be x!y. Therefore, x^y > y^x.
> B. But if y < x, then x!y will have a y^2 term in it, whereas y!x will just be y!x. Therefore, y^x > x^y.
The 10-th derivative of 9^10 is 10!x9 whereas the 9-th derivative of 10^9 is 9!x10 which is just 10!. 10!x9 is strictly greater than 10! because it has an extra factor of 9 in it. Therefore, 9^10 > 10^9.
The logic of this method is this: An n-adic function (quadratic, cubic, quartic, etc.) always increases in value faster than a linear function. Therefore, taking the respective n-th and m-th derivatives of the two functions will reveal which one is n-adic and which one is linear by virtue of their coefficients. You can conclude then that one is ultimately greater than the other.
However, this technique fails when either x or y are not positive integers. Although Dr. Peyam has shown how to take the a-th derivative of a functionfor some positive real number a, it's a laborious process involving integration, and doesn't mix and match with the other differentiation rules at all. And afaik it's not possible to take the a-th derivative of a function for ANY negative value of a, integer OR real.
Also, when you have two numbers that are on either side of the critical value e in x^(1/x) (or the x-th root of x), something to look for is whether a or b is less than or equal to 1. If that is the case, then the function with that (a|b) e. Therefore, 1 is another special value of the function x^(1/x). I haven't done the second derivative of it yet (and I plan to after I finish posting this comment, just for the fun of it), but I would wager that x=1 will be a point of inflection for the function because it has that special property of also being the point where the left tail of the function is permanently less than the right tail, by virtue of y=1 being the horizontal asymptote and the right tail therefore always being greater than 1. But I'm curious to see where the point of inflection is for the right tail where x > e.
Oh my God
You just solved one of my greatest doubts in mathematics
I knew it had some sensible method using calculus
But I never tried to venture and solve it
Thank you very much
here's my method:
9^10 [ ] 10^9 {ln both sides}
note: the [ ] is where an equal or inequality sign goes
ln (9^10) [ ] ln (10^9) {by taking the powers out of the ln}
10 ln(9) [ ] 9 ln(10) {dividing both sides by 9 & dividing both sides by ln(9)}
10/9 [ ] ln(10)/ln(9)
now it's obvious the left side is bigger, because the difference between ln 10 and ln 9 is very little compared to 10 and 9, the slope for any log function, including ln, is very small after you pass the base number, which is e for ln
well my method is more about logic than algebra I guess
Log_10 makes this even more obvious.
Log(9^10) = 10 log(9)
Log(9) ~ Log(10) = 1
Log(10^9) = 9
10 > 9
Bro the easiest way is using binomial expansion that make this question really easy to work on
@@speedcode5795 mind explaining??
@@orlandobinungcaliii3175 u can take 9^10 = (10-1)^10 and u can use permutations and combinations style of binomial expansion there will be total 11 terms and just by writting that terms u will get to know which number is biggere 9^10 or 10^9
@@orlandobinungcaliii3175 u can just search a topic called binomial expansion and u can see many vedios on that as it's taught In senior secondary classes
Not the solution that you’re looking for in general,but since the numbers are nice...
(3^2.5)^2=3^5=243
(2.5^3)^2=(2.5^2)(2.5^2)(2.5^2)=6.25^3>244
Can be done by hand if you’re up to it.
So (2.5^3)^2>(3^2.5)^2
Implies 2.5^3>3^2.5
You don't even have to do 6.25^3 by hand all the way...simply expand (6+.25)^3
(6+.25)^3 = 6^3 + 3(6^2).25+3(6)(.25^2)+.25^3
First term = 216, second term = 27, so sum of first two terms is 243. Sum of last two terms is obviously > 0 , so 6.25^3 >243
tz very nice!
Technically (2.5^3)^2 > (3^2.5)^2 implies that |2.5^3| > |3^2.5|
@@jwl_william9276, Therefore you can't use the validity of the proof with the condition
e < a < b.
And just because the condition isn't fulfilled, there is reason why the question 2.5 vs. 3 could be extra interesting for being taken apart.
The same of course with the well known
2^4 vs. 4^2.
by the way 2.5 is closer to e!! 😂
Let's just use a calculator.
Let a < e < b
Is it easy to find the unique c > e such that c^(1/c) = a^(1/a)? Because then we could just use our first theorem.
I wanted to share something I discovered based on your suggestion. Suppose we are comparing 2^(1/2) to 5^(1/5). Using the Lambert W function, I figured out that 2^(1/2) = 4^(1/4), which is easy to see. We know 4^(1/4) is greater than 5^(1/5) so 2^(1/2) is also greater. I think that we may have finally geeralized this problem. What do you think, blackpenredpen?
@@GreenMeansGOF so the idea is to get to the form x^(1/x) vs y^(1/y) using the Lambert W function, with x and y greater than e? I was missing Lambert if this indeed works!
The idea is to get two numbers that are either both greater than e or both less than e. Note that 2
For your follow up question, we know that the function you describe increases first then decreases. If we pick 1 < a < e < b (if a < 1, there are no solution and b^1/b will always be larger than a^1/a), we first pick a and want to find the value x for which x^(1/x) = a^(1/a) with x>e. If we have this, then we know that for b > x, b^(1/b) will be smaller than a^(1/a), and larger if e < b < x.
The problem is to solve the general equation x^(1/x) = k. If we take the log, we get to ln(x) = h.x (with h = ln(k)) which can be solved using the Lambert W function:
x = e^(hx) => x.e^(-hx) = 1 => -hx.e^(-hx) = -h, we apply W on both sides to get -hx = W(-h), thus x = -W(-h)/h.
In our problem, we probably need to take the other branch of the W function, since we'd get x = a otherwise.
If we for example plug in a = 2, this gives us x = -2 * W(-1, -ln(2)/2) / ln(2), but since W(-1, ln(2)/2) is -2ln(2) according to Wolfram Alfa, we get to the expected value x = 4.
If you plug in a = 2.5, there's no shortcut, we get x = -2.5 / ln (2.5) * W(-1,-ln(2.5)/2.5), Wolfram Alpha gives and approximate value of 2.97029.
You can verify that 2.5 ^ (2.5) is approximately the same as 2.97029 ^ (2.97029), within 6 decimal places.
Poor calculus :( why did you kick him ?
Canadada yaşiyourmusun? Galiba babani biliyorum...
In my case, calculus kick me instead
Calculus gave me chock slam!!
Eye for an eye!
You mean Evil calculus right?
With *a < e < b* there is no simple rule, since all cases may occur. Counterexamples:
*2^3 < 3^2 2^4 = 4^2 2^5 > 5^2*
Concerning the challenge: Define the function *f(x) = x^(1/2)* increasing for *x > 0* and notice
*3^(2.5) / (2.5)^3 = f( 3^5 / (2.5)^6 ) = f( 2 * 6^5 / 5^6 ) = f( 15552 / 15625 ) =: X*
Since *f* maps *(0; 1) -> (0; 1),* the result *X* lies in *(0; 1),* leading to *3^(2.5) < (2.5)^3*
Consider explaining it !
@@samueljehanno Which part would you say needs further explanation?
@@carstenmeyer7786 this
Cool trick
"Thank you for this cool t-shirt that I'm hiding behind my giant ball mic so you can't see it!"
Even without checking the comments my first thought was to take logarithms of both sides. It is then usually trivial to demonstrate the comparative values. This works on all positive values. Maybe negative ones too. (No idea about complex numbers and quaternions.) And you don't have to take logs using base 10 or base e. Still, it was nice to be shown the general proof.
@@pichisnoweasel7977 If f(x) = x^(1/x), then let g(x) == ln(f(x)) = ln(x)/x. g'(x) = 1/x^2 - ln(x)/x^2 = (1-ln(x))/x^2.
This is negative for ln(x) > 1 or x>e. The same must be true for the original f(x) because the logarithm is a monotonic function.
@@pichisnoweasel7977 taking logarithm base 10, we need to compare:
10log9 and 9
it is sufficient to show log9>0.9 to see that 9^10 is greater
log9=log(10-1)=1+log(1-1/10)>=1-1/10=0.9 (taylor's expansion) from where it is clear.
similar ineq of ln(a)/a > ln(b)/b, which is "obvious" because 1/x gets smaller much faster than ln(x) gets bigger for x>e (compare derivatives)
(5/2)^3 vs 3^(5/2) is equivalent to 5^6 vs 2^6 * 3^5. You could mentally multiply this out, but it's easier to note that 3^5 = 243 < 244 = 122 * 2, and then note that 128 * 122 = 5^6 - 3^2 from the difference of squares identity. So 5^6 > 2^6 * 3^5, and thus (5/2)^3 > 3^(5/2).
Of course, this solution isn't general, but I suspect there isn't one for the case a < e < b.
Me in homeworks: "please dont make me do the second derivative" :v
Derivative is easy but please I don't want those integrals in my notebooks 😭
You can use the shape of the function and the equality 2^4 = 4^2 to split the problem in 4 intervals 0 < w < 2 < x < e < y < 4 < z. You get some insight for some values across e.
Then w^y < y^w and x^z > z^x
However, if 0
Right. The issue is if 1 < a < e < b.
please do the second derivative
@Erik Awwad he he
Me? I want to see how many derivatives it takes to reach 0.
@@thebloxxer22 infinite
@@thebloxxer22 infinte
@@thebloxxer22 it will never reach the function f(x)=0, it will just keep getting more complicated cause it's not as simple as a polynomial
This reminds me about some math a did a couple years ago with tetration! I found that when you infinitely tetrate x, you get a function of x=y^(1/y). But only for 0
I actually have a lot more on this but it won't fit in a youtube comment.
@@MCredstoningnstuff : )))))))
@@MCredstoningnstuff A little inaccurate. Rather, infinite tetration is defined everywhere by analytic continuation, but the sequence of partial tetrations only converges if 1/(e^e) < x < e^(1/e).
@@angelmendez-rivera351 you're right! It does follow this function on the lower part of the bifurcation after 1/(e^e) but the whole thing isn't defined as a function. I actually don't know what function the upper half follows. Thanks for the correction
bro that video made me so excited. like, it is kind of an art and magic! thank you for those awesome videos. I have learned a lot from your videos and always kept my math passion alive.
Or use e^ln(9^(10)) and then solve
Ad you did for the other video for e^π and π^e
Yes, this is the general version.
Hmm. The equation x^(1/x)=a has two solutions for 1
2.5^3 vs 3^2.5
(squaring)
2.5^6 vs 3^5
(multyplying by 2^6)
5^6 vs 2^6*3^5
5^6 vs 2*6^5
Clearly the left hand side is greater. My issue with this is that this technique likely will not work in general.
GreenMeansGO
Yea... that’s my main question too. For example (sqrt3)^pi vs pi^sqrt(3)
@ GreenMeansGO: Clearly?
5⁶ = 15625
6⁵ = 7776; 2·6⁵ = 15552
. . a pretty close shave; about ½% difference.
PS: Kudos for getting down to integers! I stopped short of that in my own answer.
Fred
@@blackpenredpen I was writing a comment in which I developed a general theorem to compare a^b and b^a, but my phone died in the process :( I will rewrite it again, so comparing pi^sqrt(3) and sqrt(3)^pi will not be so much of a problem any longer.
Hi Fred. You know, I honestly do not know how I made the final conclusion. My analysis is incomplete. I lack a justification as to why the left hand side is greater. Regardless, we need a more general technique. Also, thank you for your acknowledgement of the integers. This worked because our numbers were positive and rational. Perhaps similar results can be derived for algebraic numbers. I want to think about this more.
@@GreenMeansGOF Well, you were actually correct in the end :-)
As far as a more general technique, I'm not sure there is one.
Sounds like maybe Angel M-R may be on the trail to something, though, so let's stay tuned....
Fred
If numbers are on the opposite side in some cases there is a way.
1. We have numbers a and b, that 0 < a
@CogitoErgoCogitoSum Then they are on the same side of the graph, so the one closer to e is our winner, I'm talking when they aren't on the same side "If numbers are on the opposite side"
Cool! For the case in the question, I just noticed that the first two terms of the binomial expansion of (10-1)^10 cancel out and the third (dominating) term 90/2*10^8 is big enough to easily beat both the fourth term -720/6 * 10^7 and the other side of the equation, 10^9.
Regarding the derivative y'=x^(1/x), there's a faster way: take y=exp{ln(x^(1/x))}. Then y'=[exp{ln(x^(1/x))}][ln(x^(1/x)]'=x^(1/x)[(1/x^2)-(1/x^2)lnx].
Nice explanation, as usual!
Woah that's some real superman way to do the derivative of x^1/x
I always wished if I could do that someway without taking "ln" both sides...
you can if you have had multivariable calculus
if you call f(u,v)=u^(1/v) and then say that u=x and v=x then there exists a "chain rule" that says df/dx=(df/du)*(du/dx)+(df/dv)*(dv/dx), du/dx and dv/dx are obviously equal to 1 and the rest you can easily work out, this rule applies for any multivariable function even with more then 2 variables
take a = x^1/x and apply log on both sides bruhh.
There is one case where you do know the answer of the comparison when a is below e and b above e: if 0 < a 1 for any b > e will yield that b^(1/b) > 1 >= a^(1/a). Which means a^b < b^a.
2.5 is going to be tough, but since most people will probably just use 2 as one of the numbers, it's useful to know that the graph has the same value at 4 allowing you to still use this method. 3^2 is "closer" to e than 2^3 because 2 is as "far" (vertically) as 4.
Note: I have no idea what e means, thus here is my idea 2³
The second derivative yields slope on each side of e. This could be used to compare rate of change on those sides and give a criterion for the a and b on opposite sides though the formula will be more complicated than whichever is nearer to e.
Case 1 and Case 2 - when both a and b are greater than e, then the exponent dominates regardless.
Case 3 and Case 4 - when both a and b are less than e, then the base dominates regardless.
Case 5 - When b
*Cries in bluepen*
if abs(e-a) almost = abs(e-b) then and 0
Very nice! Wish I had known this trick when my Calc 2 prof gave us a similar problem for homework
: )))))
I think that 2.5 is closer to e than 3 is, so therefore I say 2.5^3 is bigger than 3^2.5. it's not a bullet proof way to argue, since obviously you could take 0.9^10 vs 10^0.9, and you know that 0.9^10 is strictly less than 1, whilst 10^0.9 has to be more than 1.
He and calculus are like BFF.....who personally beats him and publicly promotes his friend.
LOL!🤣
9^10 vs 10^9
If log a base x> log b base x
Then a>b. if x>1
Taking log base 10
Log(9^10) base 10= 20 log(3) base 10
= 20*0.48
= 9.6
Log(10^9) base 10= 9
9.6>9
Then log(9^10) base 10> log(10^9) base 10
Then 9^10 > 10^9
I think this way is more easier...
Use binomial theorem.. write 9^10 as (10-1)^10 then compare the two
I remember Doing it using the variations of the function lnx/x.
Edit: this is a decreasing function for all x>e.. (in fact (e,1/e) is an absolute maximum)
then for e lnb/b then blna > alnb, or ln(a^b)>ln(b^a), which finally gives a^b > b^a.
I love Chan luuu!
*Gives a flying kiss*
Suppose a>b, let n=a/b (n>1), after doing some algebra( it's kinda long so I won't show but you can do it by isolating b ) now we compare n^(1/(n-1)) vs b with the same inequality as a^b vs b^a. And we can approximate: n^(1/(n-1))~1+4/n(n>=30)~1.1+2/n(3=
I did the lazy estimate method.
I used 10^10 to divide both.
10^10 / 10^9 = 10
10^10 / 9^10 = (10/9)^10
(10/9)^10 = (10/9)^9 * 10/9
(10/9)^10 = [(1000/729)^3 * 10/9] < [2^3 * 1.25] (which is 10)
Hence 9^10 is larger.
We can create another bound where it becomes easy again if we use the horizontal asymptote y=1. Choose any value a such that f(a) < 1 and we know it can never be greater than f(b) if b >= e
Hey Steve, i have to say that i just saw your channel and i already love it. You really do a great job and i love the way you are fascinated by math. Just gained another fan :)
Joao Pedro Leal Maran thank you Joao!!! I am very glad to hear it!!
4^8=65536>4096=8⁴. For 2,5³ and 3^(2,5), we have simply 2,5³=15,625 and 3^(2,5) is irrational because 243 is not a perfect square so we'll have something like 15,582 - ok I can stop there, the next number will be 0 - we can see that 2,5³>3^(2,5). It's funny when you say that you don't know, even by considering the same function you have your answer because when 0
I think 2.5^3 is is bigger coz 2.5 is closer to e than 3
0^1000 or 1000^0. Which one's bigger?
@@SlipperyTeeth 1000^0 bigger
I don't think it works like that? You're assuming symmetry of the graph of log(x)/x around x=e which is obviously not correct.
But considering the second derivative maybe such an argument can be made to work, but not the other way round.
@@SlipperyTeeth 0^1000 is 0, 1000^0 is 1
You can also use the tangent line to y=ln(x) (which is concave down) at x=9 to get ln(10)
Wouldn't it be easier to take the ln on both sides of y=x^(1/x), bring the 1/x in front, and then take the derivative? Although I believe it is essentially the same thing, it would make more sense to me.
When function like x^(1/x) is spotted, I think the Lambert W function can came to rescue again.
The 'other' solution to x^a=a^x is (see wikipedia page for derivation)
-a/log(a) W[0, -log(a)/a ], if a>e &
-a/log(a) W[-1, -log(a)/a ], if a
I was working on this problem and came to the same conclusion as you did. Lambert W function seems to give us a methods of solving any problem of this form.
Also, just as a small note, -x/ln(x)*W(-ln(x)/x) can be written as
e^-W(-ln(x)/x) which looks nicer to me. LOL
At first I forgot that they have to be gte e, thought "hey, 8 < 9", and then guessed wrong.
From now on I'll remember that 81> 64
@@JoshuaHillerup LOL : )
@@ertcet7679 Nope, 2 4^3
(2.5)^3 = 125/8 = 15.625
3^(2.5) = 9 sqr(3) = 15.588
for small numbers it is easy to evaluate the quantities ... your analysis is very useful to answer the question when big numbers are considered
(2.5)^3 and 3^(2.5)
((2.5)^3)^2 and (3^(2.5))^2
(2.5)^(3*2) and 3^(2.5*2)
2.5^6 and 3^5
Next calculations easy
(Not must need to know square of 3)
Me: but what if ...
1 minute later - 11:02 bprp: *_I KNOW_*
There is a much easier way out-
Take 9^10 = a
Taking log base 10 both sides
Log 9^10 = log a
10* log 9 = log a ) ----eq 1
Now let 10^ 9 = p
Log 10^ 9 = log p (log base 10 )
9 log 10 = log p
9 = log p (log 10 = 1)----ii
We get from i and ii p>a hence 9^10 is greater .
Such problems used to haunt me like anything. Thanks for this brilliant approach!!!!
Onkar Gangane : ))))
If a a^a*(a^c - e^c)
which is positive if a > e.
I keep forgetting the order so it helps me to put in outrageous numbers, like 3^1000 vs 1000^3. Obviously the former is greater.
@Irwan Gunardi no, 2 is smaller than e, and 8 vs 9 literally gives you the opposite of what this video is saying.
@@gabrieljohnson6304 point is you instantly solve 2³ or 3², and can know the answer
@@aaykat6078 helpful for solving that specific question, not helpful for remembering which would be bigger when both bases are larger than e
@@aaykat6078 lol
Thank you for the general case. For the case involving 10 and 9, considering 'decimal log' and the fact that 'log' is an increasing function::
It's known from HS log(3)=0.4771... --> 20*log(3) > 20*0.477 --> log(9^10) > 9 = log(10^9). --> 9^19 > 10^9.
he said: don't use a calculator,
me: uses a calculator.
lolol
When considering functions like y = x^f(x) you may also take the natural logarithm of both sides and differentiate implicitly.
thats the easiest one
03:20 WOW! Taking the derivative of x^(1/x) using "power rule plus exponential rule".
I am wondering if it works for all such functions, for example for sin(x)^ln(x).
Maybe with chain rule. Not sure tho
You can also use a property: a^b = e^(b*ln(a)) and then it's a bit easier to get the derjvative if you ever have doubts. My HS math teacher showed us this way when demonstrating this derivative
Yes, the method works for all functions of the form f(x)^g(x) if you apply the chain rule to the función which you are considering variable in each steñ of the derivation. You can check ir by deriving the general expression y=f(x) ^ g(x) using the traditional method of applying ln to both sides, or using the "power rule plus exponential rule". You will arrive to two expressions that can be easily shown to be equal.
There is also another way to determine which number is larger. Here if we let a=9^10 and b=10^9. Take the log of both sides using two equations, and we get log(a)=10log(9) and log(b)=9. Without looking at log(a) and log(b), we know that log(9) is somewhat a bit less than 1, which is log(10). Multiply this by 10, and the value is somewhat greater than 10. This implies that this value is greater than 9. Therefore, if you go back to the original equation and compare the numbers here, 9^10>10^9.
can binomial theorem be used to solve this? (Can write 9^10 as (10-1)^10 and then expand it..?)
AS Wrestling Isnt that just more complicated?
You can solve it using the binomial theorem, but as someone else said, that is noteably more complicated.
@@angelmendez-rivera351 Tbh, at times I find calculations easy for binomial theorem (Like till the power 7-8) I haven't even tried 10 yet.. but i guess it's manageable..
Yes it can!
@@aswrestling9920 i mean.. if you're going that way wouldn't it be easier to just multiply 9 by itself 10 times?
Okay, here goes an attempt at the general form of the problem at 11:21
Start with the equation y=x^(1/x). The lim as x goes to infinity of this is 1. This means that for every value of a between 1 and e there exists a value of b >= e such that a^(1/a)=b^(1/b).
If you rotate this function pi/2 radians counter clockwise and then solve for y in the range x= -(e^(1/e)) to -1 (because this would be rotated as well) you should have two solutions.
To rotate any function, you can use the equation y*cos(theta) - x*sin(theta) = f(y*sin(theta) + x*cos(x)).
Since our theta is pi/2 and our f(x)=x^(1/x) we get:
y*0 - x*1 = (y*1 + x*0)^(1/[y*1+x*0])
which simplifies to:
-x=y^(1/y)
Now, solving this for y is completely beyond me.
According to wolframalpha(sorry, but I just don't know this), for -(e^(1/e))
I don't think you gain anything from rotating, since with the original equation you could just solve for x immediately (it's the same equation as the one you derived up to a negative with x and y reversed) and not have to worry about rotated coordinates
Interesting. I arrived to the same conclusion instinctively, because 2⁴ = 4² and exponential functions rise faster than polynomial functions, so for numbers greater than 2/4 it was the most likely answer.
PS: what’s going on with the first part of your derivative? Isn’t it just chain rule?
Other than a = 1, 2³ < 3² and 2⁴ = 4² are the only integer counterexamples for a < e.
2.5^3 is bigger than 3^2.5. You can think of 2.5 as just 25. 25^2=625 and 625*25=15625. Move decimal point over 3, so 2.5^3=15.625
3^2.5 was a bit more interesting. 2.5 can be reexpressed as 5/2, so 3^2.5=3^(5/2)=sqrt(3^5). We can rewrite 3^5 as (3^2)(3^2)*3 so sqrt((3^2)(3^2)(3))=9sqrt(3). If we approximate sqrt(3) we can say 1.7. So 9*1.7=15.3
To conclude 15.625 > 15.3, therefore 2.5^3>3^2.5
but 2.5 is less than e. Wtf! e=2.732....
Why did you add the two parts of the derivative? It's not a product rule situation is it?
It actually is: y=f(x)^g(x) = e^(g(x)*ln(f(x)))
Do you see that product in the exponent?
2.5³ = 15⅝ = 5⁶/1000 = 1000/2⁶ = 15.625
3²·⁵ = 9√3 = 9·1.73205... = 15.588...
Or if you square both, to make it slightly easier (given memorization of some powers of 3 and 5):
2.5⁶ = 244.140625
3⁵ = 243
showing that it's a close contest in this case.
But for these numbers,
2.5³ > 3²·⁵
Fred
Hey bprp can you solve
x^(x+1)=(x+1)^x ?
I'm stuck with this
hey, i've tried to solve it by hand, but i couldn't. It's quite obvious, that x shouldn't be greater(or equal) than e, otherwise we have a=x and b=x+1(e
@@МаксимИванов-я5п thanks man:)
The second Foias constant is, in a sense, *defined* to be the real solution to x^(x + 1) = (x + 1)^x. There is no closed-form expression for it.
You can take log on both the sides.
Ps: you will have to remember the values of log 1,2 etc
(2.5)³
= 3log(2.5)
=3log(25/10)
=3(log25-log10)
=3(1.3975-1)
=1.1925
On the other side,
3^2.5
=2.5 ( log3)
= 0.7525
Therefore, 2.5³ >3^2.5
log is too close. youd have to know the log up to 3 decimals or more
0.7525 is 2.5*log(2). Took me a hot minute to figure out the mistake.
The actual value is 0.1928.
but 3*log(2.5) is 0.1938, so you came to the right answer accidentally.
At this point it's easier to just keep it within rational numbers (ratio squared is 15625/15552)
If you really want to use log I'd go with base 2 and remember the values from music theory.
log(1.5) = 7/12+1/600-x; // Perfect fifth
log(1.25) = 1/3-7/600+y; // Major third
6*log(2.5)-5*log(3) = 6*log(1.25)+1-5*log(1.5) = 2-42/600+6y+1-35/12-5/600+5x = 3-35/12-47/600+5x+6y
= 1/12-47/600+5x+6y = 3/600+5x+6y
x and y are there to show that the error is in the opposite direction from our comparison.
can you explain why the derivative in the superman way works??? i tried it for a general y = f(x)^g(x) and it works for it! how?!
I don't know what the "superman method" is, but, maybe you can figure that out by using implicit differentiation and look out for a pattern
@@philosandsofost8642 watch the video... he explains it at 3:25
f(x)^g(x) = exp{g(x)·log[f(x)]}, hence the derivative of f(x)^g(x) can be evaluated via the chain rule to be exp{g(x)·log[f(x)]} multiplied by the derivative of g(x)·log[f(x)], which is given by g'(x)·log[f(x)] + g(x)·f'(x)/f(x). Therefore, the derivative of f(x)^g(x) is equal to exp{g(x)·log[f(x)]}·{g'(x)·log[f(x)] + g(x)·f'(x)/f(x)}. Since exp{g(x)·log[f(x)]} = f(x)^g(x), this implies that the derivative of f(x)^g(x) is equal to f(x)^g(x)·{g'(x)·log[f(x)] + g(x)·f'(x)/f(x)} = f(x)^g(x)·log[f(x)]·g'(x) + g(x)·f(x)^[g(x) - 1]·f'(x).
2.5^3 > 3^2.5 ?
First make 2.5 into an improper fraction so it becomes (5/2)^3 and 3^(5/2)
Then go from there, obviously 125/8 is going to be greater than the 5th root of 9 (or 3 sqrd)
In conclusion, 2.5^3 > 3^2.5
There you go! Love your videos man, I’m currently in algebra 2 but I have been learning the basics of calculus thanks to you ❤️
3^(5/2) is not 5th root of 9. It's square root of 3^5. Which is actually 9x√3.
125/8 and 9*√3 are pretty close. 125/8 is greater though 👍
Great you always bring something unique :)
Himanshu gupta thank you!!!
3^2.5 vs. 2.5^3
This can be done trivially by arithmetic, no calculator needed.
3^2.5 = 3^2 * 3^0.5 | 2.5^3 = 2.5 * 2.5 * 2.5
3^2 = 9; 3^0.5 = sqrt 3 | 2.5 * 2.5 = 6.25
Since sqrt 3 is about 1.73, 9*sqrt 3 is app. 15.57 | 6.25 * 2.5 = 15.625
Which is close, but 2.5^3 is bigger.
What is the curvature of the graph of y=x^(1/x)? (:
i do at least know that 2^4 = 4^2, and that may be the only integer solution.
there was a time I was interested in figuring out all spots where a^b = b^a where a < b. with this video I now know how to solve that equation. I need to find where a^(1/a) = b^(1/b), and a < b
so this _must_ mean that 2^(1/2) = 4^(1/4), (which it does)
so that means numbers too far away from e will never give an eqality
i always thought this was only because 2 is unique in the fact that 2 is also the only (nontrivial integer) solution for a + a = a * a (2a = a^2, or a^2 - 2a = 0, so a(a-2) = 0. a is 2 or 0) and as it turns out, both result in the answer being 4
In a math exam would you have to demonstrate why you pick one or the other, or would you just pick one and say "that one is bigger" ?
Also you bring me a lot of joy through the day thanks for you everlasting happiness
11:44 actually we have an horizontal asymptote at 1, so if a is on the "left side" (meaning ae) but a^(1/a)
True but if a1 then it's pretty clear that a^b1 so you can see that directly
Wouldn't differentiation using implicit differentiation also be easy?
This is a somewhat longwinded proof. Provided that both a and b are positive, then:
if a^b>b^a, then it also follows that bln(a)>aln(b), and similarly that ln(a)/a>ln(b)/b.
From this, you can show that, for a general form of y=ln(x)/x that dy/dx is negative for any value of x greater than x=e. This means that ln(8)/8>ln(9)/9, and also that 8^9>9^8
What kind of whiteboard is that? I don't see any marker residue or ghosting at all. How do you keep it so clean?
Peter Kim well... I spent a lot of time erasing before filming....
@@blackpenredpen I see...any tips for erasing so well?
wipe the board with a cloth soaked in ethyl alcohol (ethanol). Works everytime.
@@HK-cq6yf, And use the right marker pens. And - quite cheap tip - don't enjoy all too long but be quick with erasing what you've just filmed. Never let it overnight…
*See replies for proofs and one further comment*
With the case where ae, there are two cases:
i) If a
I mean come on bprp you're basically begging for us to ask you to do the second derivative!
Moreover it's not that difficult, [x^(1/x)]/(x^2) = x^(1/x-2), which makes it "easy" given we already have [x^(1/x)]' 😜
I love this kind of questions, because I have watched e^pi vs pi^e video. Your videos are great man 😍
Loved how you worked with the differentiation part! Is there a proof for this?
he treated it as a multivariable function where the base and power are both variables; the total derivative is equal to the sum of the partial derivatives
comparing a^b to b^a is the same as comparing b.ln(a) to a.ln(b) that makes things so much easier.
Damn i wish they created some equation style for comparison results. Something like
a^b b^a === b.ln(a) a.ln(b)
I somehow remember the values of log3 and log5 (base 10) (crammed for an exam) and that gives me
3log2.5 > 2.5log3
Edit: upto 5 decimal places
EXACTLY WHAT I DID.
Why not just do 20log3 and 9log10?
Use logs
It's easier ig
Like in base10 system
Comparing
9^10 and 10^9
Since log in base 10 is increasing,
10log9 and 9log10
20log3 and 9
20(0.47) vs 9
==> 10log 9 is greater
And hence 9^10 is greater
If a is less than 1 and b is greater than e then it's pretty easy.
I did it in my head but not in 0.4 sec, my logic went:
9^2 is a 2 digit number, 9^3 is a 3 digit number, so 9^10 will be a 10 digit number
10^9 is the lowest possible 10 digit number, so in that case 9^10 is obviously larger because its not equal to exactly 1,000,000,000
Please give lectures on real analysis
Actually when a a^b supposed a
10:48 "I'm not doing 2019".
We were so innocent back then.
It's easier to do it by log, as log is a positively increasing function the log of one number will be higher than the other respectively which boils down to comparing 20*ln2 ans 9*ln10 clearly 1st one is bigger.
I liked the inequality and its maxima
just do this
x= 9¹⁰
applying log(base 10) both sides:-
logx =log9¹⁰
logx=10log9
Similarly,
y = 10⁹
logy=log10⁹
logy=9log10
logy=9
Clearly logx>logy
so x>y
Hence , 9¹⁰>10⁹
Why can we just take log on both sides
10(log9) > 9(log10)
It is not clear as to why 10(log9) is actually bigger than 9(log10).
Andrei Miga Check out change of base formula. If u divide the logarithm to one side then u can use formula to get log10(9)
That is for both number greater than e
If both numbers are smaller than e. Than u need to flip sign when doing division in inequality
@@andreimiga8101 10log9 means 10×(x
You making it look harder binomial distribution is best way to approach these type of questions