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April 1

Solving sums of powers

Consider the equation $2^{x}+3^{x}=13$. The solution, of course, is $x=2$. What I am trying to understand is how to approach the general case; given $A^{x}+B^{x}=C$, for any known $A,B,C$, what is the value of $x$? I just can't wrap my head around a reasonable approach. I tried taking the natural log on both sides, but that didn't seem to help. Had it been a single term on the left-hand-side of the equation, I could have extracted the exponent and solved it without much fuss. Any suggestions/pointers on how to proceed? Earl of Arundel (talk) 02:17, 1 April 2023 (UTC)

In the following, I assume that $A,B,$ and $C$ are all positive. In only a few cases can equations with a sum of powers with the unknown $x$ in the exponents be solved easily, by trial and error or analytically. If $C=2,$ the solution is given by $x=0.$ If $B=1,$ we find $x=\log _{A}(C-1).$ But there is no analytical way to solve this in general, so we have to resort to a numerical approach. I think Newton's method will do well, using

$f(x)=A^{x}+B^{x}-C,$

$f'(x)=A^{x}\log A+B^{x}\log B,$

and repeatedly computing

$x_{n+1}=x_{n}-{\frac {f(x_{n})}{f'(x_{n})}}$

until a desired level of convergence. For the initial estimate, we can use, assuming wlog that $|\log A|\leq |\log B|,$

$x_{0}=\log _{A}C$  if  $C<2,$

$x_{0}=\log _{B}C$  otherwise.

Convergence is not guaranteed, though. When $A$ and $B$ are not at opposite sides of $1$, the equation may not have a solution or may have two solutions. The computation may also fail by dividing by $0,$ as when $A=B=1.$ If there is an integer solution, it will be found.  --Lambiam 07:32, 1 April 2023 (UTC)

Convergence to high precision may also fail due to the lack of precision in floating point arithmetic. For $2^{x}+3^{x}=60073,$ which is solved by $x=10$, numerical computation (using exponentiation by $z^{x}=\exp(x\log z)$) results in $x_{4}=10.0$ on the dot, but then computing $f(x_{4})$ does not result in zero but produces $-1.309672370553017\cdot 10^{{-}10}.$  --Lambiam 08:04, 1 April 2023 (UTC)

In general there are several ways for Newton's method to fail; see the section "Failure analysis" in that article. I think the main point though is that the equation is not solvable in closed form and one must resort to numerical methods to get an approximate value. Of course nowadays you can plug an equation in Wolfram Alpha and get a numerical answer. --RDBury (talk) 17:00, 1 April 2023 (UTC)

The issue above is not among the failure modes mentioned in that section, which are mathematical in nature. Mathematically, $2^{10}+3^{10}-60073$ is exactly equal to $0.$ But the actual computed value is not; we get

$3^{10.0}=\exp(10\times \log 3)=\exp(10\times 1.09861228866810956...)=$

$=\exp(10.9861228866810956...)=59048.99999999986...<59049=3^{10}.$

 --Lambiam 20:04, 1 April 2023 (UTC)

Thanks, Lambiam, I had no idea this was such a difficult problem to solve for the general case! I wonder if there is some way to guarantee convergence (for most cases anyway) by dividing through by C? Playing with that idea I even found an interesting approximation for x:

$A^{x}+B^{x}=C$

$A^{x}/C+B^{x}/C=1$

$A^{x}/C=1-B^{x}/C$

$xln(A)-ln(C)=ln(C-B^{x})-ln(C)$

$xln(A)=ln(C-B^{x})$

And since $ln(C-B^{x})\approx ln(C)$ as x approaches infinity, we get a rough estimate:

$x\approx ln(C)/ln(A)$

In the case of $2^{x}+3^{x}=13$, we find that

$x=2\approx ln(13)/ln(2)\approx 3.7$

Ignoring extreme corner cases, setting our initial value x0 using this estimated root should yield a numerical convergence in most cases, no?

Earl of Arundel (talk) 18:58, 1 April 2023 (UTC)

Whoops, that isn't right! Earl of Arundel (talk) 19:14, 1 April 2023 (UTC)

You can get to faster to your fifth line without dividing by $C$:

$A^{x}+B^{x}=C$

$A^{x}=C-B^{x}$

$\ln(A^{x})=\ln(C-B^{x})$

$x\ln A=\ln(C-B^{x}).$

It appears plausible that using $f(x)=x\ln A-\ln(C-B^{x})$ gives faster convergence, or, when $|\log A|\leq |\log B|,$ swapping $A$ and $B$ and using $f(x)=x\ln B-\ln(C-A^{x}).$ I have not investigated this, though. If $A$ and $B$ are similar in size and not at opposite sides of $1,$ we can approximate $A^{x}+B^{x}$ by $2\left({\tfrac {1}{2}}(A+B)\right)^{x}$ and use

$x_{0}={\frac {\ln C-\ln 2}{\ln(A+B)-\ln 2}}.$

For the case of $2^{x}+3^{x}=13$, we then get $x_{0}\approx 2.04280\,.$  --Lambiam 19:38, 1 April 2023 (UTC)

Nice! That does seem to work pretty well. Where does the $\ln {2}$ term come from though? It seems to randomly pop up in a lot of equations, come to think of it. Why is that, I wonder? Earl of Arundel (talk) 20:58, 1 April 2023 (UTC)

Instead of solving $A^{x}+B^{x}=C,$ we set off by solving

$2\left({\tfrac {1}{2}}(A+B)\right)^{x_{0}}=C$

$\left({\tfrac {1}{2}}(A+B)\right)^{x_{0}}={\tfrac {1}{2}}C$

$x_{0}\ln \left({\tfrac {1}{2}}(A+B)\right)=\ln({\tfrac {1}{2}}C)$

$x_{0}(\ln(A+B)-\ln 2)=\ln C-\ln 2$

--Lambiam 21:21, 1 April 2023 (UTC)

Ah, right. Subtracting $\ln {2}$ has the same effect as dividing by $2$ in that context. Thanks again! Earl of Arundel (talk) 21:39, 1 April 2023 (UTC)

April 6

30th row of Pascal's triangle

The first row is simply 1. The second row is 1-1. The third row is 1-2-1. The fourth row is 1-3-3-1. The fifth row is 1-4-6-4-1. And so on. Do you know the 30th row?? (It's easy to guess correctly that the first 2 points in it are 1-29 and the last 2 are 29-1, but do you know all 30 members of the 30th row of Pascal's triangle?? Georgia guy (talk) 14:33, 6 April 2023 (UTC)

As described at Pascal's triangle, it's based on a simple, easily repeatable algorithm, and there are several options to find it. You can do it by brute force by simply repeating the algorithm yourself, you can apply the process described in that article in the section titled "Calculating a row or diagonal by itself". Or you can use an online calculator such as this one, which will generate any row you wish. --Jayron32 14:43, 6 April 2023 (UTC)

(ec) Sure, it's $30 \choose 0$, $30 \choose 1$, ..., $30 \choose 30$, cf. binomial coefficient (or simply go through the triangle until you reach row 30). As I do not have more time than you have, I won't do the numbers for you. --Wrongfilter (talk) 14:49, 6 April 2023 (UTC)

The 30th row has big numbers; the largest is more than 70 million. (The 15th row is 1-14-91-364-1001-2002-3003-3432-3003-2002-1001-364-91-14-1.) Georgia guy (talk) 14:52, 6 April 2023 (UTC)

Georgia guy: If you're already aware of how to find the 30th row, and are aware of it enough to know the scale of the numbers on it, what kind of answer were you seeking? --Jayron32 14:56, 6 April 2023 (UTC)

User:Jayron32, the main thing I want to know is what row of Pascal's triangle has "2n, 3n, 4n" as 3 consecutive terms. The 15th row has n, 2n, and 3n. (To find what row of the triangle you're on, just add one to the second number in the row; the 30th row starts with 1,29... Georgia guy (talk) 15:38, 6 April 2023 (UTC)

Well then you should have asked exactly that. If you just wanted to know the 30th row, that's academic. If you want to know which rows have some specific property that's a different question. The 30th row doesn't have that property. It has an "n 2n" sequence (10015005, 20030010) but the pattern breaks on either side of it. In researching your question, I also found that, in Commons, the image File:Pascals triangle 30 lines.png exists. You can explore it yourself. --Jayron32 16:15, 6 April 2023 (UTC)

I see that my guess of the 30th row is wrong. The rows where 3n and 4n occur as consecutive members of the triangle are the rows divisible by 7, which 30 is not. So the row that would have 2n,3n,4n is not the 30th, but an even lower row, the 35th. Its terms are 1,34,561...561,34,1. Georgia guy (talk) 16:20, 6 April 2023 (UTC)

OK. --Jayron32 16:36, 6 April 2023 (UTC)

Yes, it is the 35th row (the row that is designated as "34" by people who use cardinal numbers to designate rows of Pascal's triangle.) The value of n where 2n, 3n, and 4n occur on this row is 463,991,880. Georgia guy (talk) 16:53, 6 April 2023 (UTC)

OK. --Jayron32 17:43, 6 April 2023 (UTC)

Enter for(n=0,30,print1(binomial(30,n)", ")) at https://pari.math.u-bordeaux.fr/gp.html. vector(31,n,binomial(30,n-1)) is shorter but may be less instructive if you want to do other problems. You can also download PARI/GP and run it on your own computer. PrimeHunter (talk) 15:05, 6 April 2023 (UTC)

The 3435th row is the one you're looking for, elements 13, 14 and 15, with n = 463,991,880 (if I'm not mistaken). It's fairly easy to find, the conditions are ${N \choose k+1}={\frac {3}{2}}{N \choose k}$ and ${N \choose k+2}={\frac {4}{3}}{N \choose k+1}$. A lot of terms cancel and these conditions lead to two linear equations with solution N=34 and k=13. (correction: the triangle starts with N=0, so N'=34 is the 35th row as you would have it). --Wrongfilter (talk) 16:35, 6 April 2023 (UTC)

In general, the system of equations

${\binom {N}{k}}:{\binom {N}{k{+}1}}:{\binom {N}{k{+}2}}=p:p{+}1:p{+}2$

is solved by

$N=4p^{2}+8p+2,k=2p^{2}+3p-1.$

--Lambiam 21:20, 6 April 2023 (UTC); corrected 06:48, 7 April 2023 (UTC)

You already said that the first number in the row is binomial(29, 0) = 1. You can compute each entry in the row from the previous one by binomial(29, k) = ((29 + 1 - k) / k) · binomial(k - 1). You get:

1, 29, 406, 3654, 23751, 118755, 475020, 1560780, 4292145, 10015005, 20030010, 34597290, 51895935, 67863915, 77558760, 77558760, 67863915, 51895935, 34597290, 20030010, 10015005, 4292145, 1560780, 475020, 118755, 23751, 3654, 406, 29, 1

– b_jonas 15:33, 7 April 2023 (UTC)