Which of the numbers $99^{100}$ & $100^{99}$ is the larger? Solve without using logarithms.
8 Answers
Note that $$\begin{align} 99^{100} > 100^{99} &\iff 99 \cdot 99^{99} > 100^{99} \\ &\iff 99 > (100/99)^{99} \\ &\iff 99 > \left( 1 + \frac{1}{99}\right)^{99} \end{align}$$
Since $(1 + \frac{1}{n})^n < 3$ for all integers $n$, the above inequalities are all true. Thus, $99^{100} > 100^{99}$. In general, you should expect that $x^y > y^x$, whenever $y > x$.
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3
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8$\begingroup$ Mr. Man, sir, can you please cite where $(1 + \frac{1}{n})^n < 3$ comes from, or prove it, or something? I'm Rusty on this sort of thing. $\endgroup$– Ed StaubCommented Jan 29, 2012 at 2:38
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7$\begingroup$ @EdStaub: The inequality comes from the calculation of e. e=2.7(ish), and is defined as the limit of that equasion, as n= infinity. $\endgroup$ Commented Jan 29, 2012 at 2:54
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4$\begingroup$ @EdStaub: Expanding $(1 + \frac{1}{n})^n$ using the binomial theorem, it is enough to show $\frac{1}{2!} + \frac{1}{3!} + \dots + \frac{1}{n!} < 1$. But this follows immediately since $n! \geq 2^{n-1}$ for positive integers $n$ (which you can prove by induction), and hence $$\sum_{k =2}^n \frac{1}{k!} \leq \sum_{k=2}^n \frac{1}{2^{n-1}} < 1$. Adding $2$ to both sides gives the result. There are many places where you can read more (by googling "limit definition of e"), say for example, here: physicsforums.com/showthread.php?t=176076 $\endgroup$– JavaManCommented Jan 29, 2012 at 3:25
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7$\begingroup$ @Pearsonartphoto: no it doesn't. What you get from the calculation of $e=2.71828\dots$ is that there is an $N$ so that for $x>N$, $\left(1+\frac1x\right)^x<3$. However, what happens for $x\le N$? $\endgroup$– robjohn ♦Commented Jan 29, 2012 at 22:39
$99^{100} - 100^{99}$ is:
3560323412732295049306160265725173861897
1207663892369140595737269931704475072474
8187196543510026950400661569100652843274
7182356968017994158571053544917075742738
9035006098270837114978219916760849490001
Since this number is positive, $99^{100}$ is the bigger number.
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34$\begingroup$ Nice reminder that the world has changed since I first did mathematics. $\endgroup$ Commented Jan 29, 2012 at 1:57
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15$\begingroup$ @AndréNicolas I suppose it doesn't use logarithms... $\endgroup$– MattCommented Jan 29, 2012 at 2:06
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21$\begingroup$ @Matt, technically correct is the best kind of correct. $\endgroup$ Commented Jan 29, 2012 at 2:24
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16$\begingroup$ So, Hammerite, is $999999^{1000000}$ larger than $1000000^{999999}$? $\endgroup$– MyselfCommented Jan 29, 2012 at 2:48
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9$\begingroup$ @Myself - Yep, wolframalpha.com/input/?i=999999%5E1000000+%3E+1000000%5E999999. Might have to give it a few seconds ;) $\endgroup$– DManCommented Jan 29, 2012 at 4:18
A purely math solution: Using AM-GM inequality:
$$(x+1)^x\times \frac{x}{2} \times \frac{x}{2} < \left(\frac{x(x+1)+x}{x+2}\right)^{x+2}=x^{x+2}.$$
Therefore
$$(x+1)^x < 4x^x$$
and easily we see that $(x+1)^x< x^{x+1}$ for any $x\ge 4$.
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11$\begingroup$ "purely math solution" as opposed to? $\endgroup$ Commented Jan 29, 2012 at 7:42
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2$\begingroup$ In this case, it's no computation. In general, it's just personal sense. Since the question is pretty easy with calculus, I expected that an "elementary" solution (secondary-school) is a best fit. $\endgroup$– hiroCommented Jan 29, 2012 at 13:48
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1$\begingroup$ @GeorgesElencwajg I don't see how using a hand calculator, as say I did in my answer, doesn't come as something check-able. Or for that matter if we calculate $99^{100}$ by multiplying 99s out on paper, it only involves 100 multiplications. Sure, that's several sheets of paper and probably takes a few hours to do, but we could all do that before we die if we had the persistance and desire to do so. $\endgroup$ Commented Jan 29, 2012 at 17:24
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3$\begingroup$ @Doug: But why would we want to do such a thing, since (a) this is most boring, (b) using our brain a few seconds would solve the question and would furthermore suggest easy generalizations? $\endgroup$– DidCommented Jan 29, 2012 at 17:40
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1$\begingroup$ @Doug: If I didn't trust my calculating machines, and I wanted to know how much greater one of the numbers is than the other, I would much prefer to know that $(x+1)^x=\varrho(x)x^x$, where the function $x\mapsto \varrho(x)$ is increasing on $x\geqslant1$ from $\varrho(1)=2$ to $\varrho(+\infty)=\mathrm e$. The proof is easier to check and the result is more informative. $\endgroup$– DidCommented Jan 30, 2012 at 6:13
$x^{x+1}=x x^x$ while for large $x$, $(x+1)^x\sim e x^x$. Since $99>e$, I would say that $99^{100}>100^{99}$.
More Detail:
To show that $(x+1)^x=\left(1+\frac1x\right)^xx^x<ex^x$, without just saying so and without using logarithms, consider the binomial expansion $$ \left(1+\frac1x\right)^x=1+1+\frac12\frac{x-1}{x}+\frac16\frac{(x-1)(x-2)}{x^2}+\frac{1}{24}\frac{(x-1)(x-2)(x-3)}{x^3}+\dots $$ and note that, at least for $x\in\mathbb{N}$, each term is monotonically increasing. Thus, $\left(1+\frac1x\right)^x$ monotonically increases to $e=\sum\limits_{k=0}^\infty\frac{1}{k!}$.
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1$\begingroup$ Is your approximation here necessarily an over or under approximation? If not, then how do you have anything more than a good guess? $\endgroup$ Commented Jan 29, 2012 at 3:48
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2$\begingroup$ $$\frac{(x+1)^x}{x^x}=\left(1+\frac{1}{x}\right)^x<e<99$$ for all $x>0$. $\endgroup$ Commented Jan 29, 2012 at 3:59
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1$\begingroup$ @Doug: At least for natural $x$, we can show monotonicity simply with the binomial theorem. I have added the details. $\endgroup$– robjohn ♦Commented Jan 29, 2012 at 14:37
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1$\begingroup$ @robjohn I've now upvoted your answer, since the details can help here, while the older version didn't help much. $\endgroup$ Commented Jan 29, 2012 at 16:44
Proof that $x^y > y^x$ for all $y > x > e$: Raising both sides to the ${1 \over xy}$ power, this is equivalent to $x^{1 \over x} > y^{1 \over y}$. The derivative of $x^{1 \over x}$ with respect to $x$ is ${\displaystyle {1 - \ln(x) \over x^2} x^{1 \over x}}$, which is negative whenever $\ln(x) > 1$ i.e. when $x > e$. Thus $x^{1 \over x}$ is a decreasing function of $x$ for $x > e$.
Yeah I know, I used logarithms. But someone needed to say this ;)
I cheat and use a basic fact about $e$.
$${99^{100}\over 100^{99}} = 99\left({99\over 100}\right)^{99}\approx {99\over e} > 1.$$
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$\begingroup$ Nice cheat, what is the general formula for this fact about $e$ please? Thanks. $\endgroup$– NoChanceCommented Jan 29, 2012 at 15:09
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1$\begingroup$ For large $n$, $(1 + \lambda/n)^n \sim e.$$ $\endgroup$ Commented Jan 29, 2012 at 15:18
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2$\begingroup$ @ncmathsadist Your comment has a small typo. Surely, you meant to write either $(1 + \lambda/n)^n \sim e^{\lambda}$ or $(1 + 1/n)^n \sim e$. $\endgroup$ Commented Jan 29, 2012 at 17:27
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6$\begingroup$ Somebody should probably mention that such asymptotics, per se, can give NO information about the $n$th term of a sequence, even for $n=99$, hence the (true) basic fact used here CANNOT prove the desired inequality. $\endgroup$– DidCommented Jan 29, 2012 at 19:19
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3$\begingroup$ Starting from $2\le1+1/n\le3$, I doubt one can go far. And if you wish to use $(1+1/n)^n\le3$ for every $n$, then mention it... $\endgroup$– DidCommented Jan 30, 2012 at 16:29
$100^{99}$=$(10*10)^{99}$=$(10^{99})(10^{99})$=$10^{198}$ exactly.
$99^{100}$=$(9*11)^{100}$=$(9^{100})(11^{100})$ exactly. My "hand" calculator approximates $9^{100}$ as about $(2.656)(10^{95})$.
11=(2)(2)(2.75).
$2^{100}$ equals about (1.267)($10^{30}$), $2.75^{100}$ equals about (8.575)($10^{43}$). Dropping the coefficients here we can thus approximate ($11^{100}$) by a lower bound of ($10^{30}$)($10^{30}$)($10^{43}$)=$10^{103}$.
Keeping the coefficients on the approximation of $9^{100}$ we have a lower bound for $99^{100}$ as $(2.656)((10^{95}$)($10^{103}$))=(2.656)($10^{198}$) which comes as greater than $10^{198}$.
So, $99^{100}$>$100^{99}$.
Note that if we kept the coefficients in here, we would also have more of an idea as to how much greater $99^{100}$ is than $100^{99}$. Some of the other answers do this, some don't. This doesn't necessarily make this answer better though, since such information might come as extraneous to the problem.
From experimenting with small numbers:
scala> (0 to 5).map (x=> (math.pow (x, x+1) - math.pow (x+1, x))).mkString ("; ")
res18: String = -1.0; -1.0; -1.0; 17.0; 399.0; 7849.0
scala> (0 to 5).map (x=> (math.pow (x, x+1), math.pow (x+1, x))).mkString ("; ")
res19: String = (0.0,1.0); (1.0,2.0); (8.0,9.0); (81.0,64.0); (1024.0,625.0); (15625.0,7776.0)
you can conclude, that the first one is growing faster than the second. Of course this is only an indication.
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$\begingroup$ Do you have computer code in this answer? I simply don't know how to interpret "scala> (0 to 5).map" and the like. $\endgroup$ Commented Jan 30, 2012 at 4:40
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$\begingroup$ @DougSpoonwood: Yes, Scala code. (0 to 5) creates a Range, a collection of the numbers (0, 1, ..., 5). The
map (x =>
takes each of them, and puts them, namedx
, into a function, math.pow (x, x+1) - ... so for the first element it is math.pow (0, 0+1) or 0^(1), then 1^2, 2^3 and so on minus 1^0, then 2^1, 3^2 and so on. mkString is only used for formatting the output a bit. (3^4-4^2) = 81.0-64.0 = 16 $\endgroup$ Commented Jan 30, 2012 at 14:50
99**100 > 100*99 == True
$\endgroup$1 <= x < 9
), so it was interpreting99**100 > 100**99 == True
as one of these. True has an integer comparison value of 1, so this is really99**100 > 10**99 == 1
, which is false.99**100 > 100**99
and(99**100 > 10**99) == True
both return True as you'd expect. $\endgroup$