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ganeshie8
 one year ago
[Weekend Challenge  Solved by @adxpoi and @oldrin.bataku ] Suppose \(P(x)\) is a polynomial of degree \(10\) such that \(P(k) = 2^k\) for all \(k=0,1,2,\ldots,10\).
Find the value of \(P(11)\)
ganeshie8
 one year ago
[Weekend Challenge  Solved by @adxpoi and @oldrin.bataku ] Suppose \(P(x)\) is a polynomial of degree \(10\) such that \(P(k) = 2^k\) for all \(k=0,1,2,\ldots,10\). Find the value of \(P(11)\)

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freckles
 one year ago
Best ResponseYou've already chosen the best response.1\[P(x)=a_{10}x^{10}+a_9x^9+a_8x^8+a_7x^7+a_6x^6+a_5x^5+a_4x^4+a_3x^3 + \\ +a_2x^2+a_1x^+a_0 \\ P(0)=1 \\ a_0=1 \\ \] well I know there is a really long way... pluggin in all those critters to find the constant coefficients of the polynomial P

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.111 equations and 11 unknowns is solvable for sure :)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1@ospreytriple nope, but it seems you got it! the answer is \(2047\)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1please post your method, im pretty sure it is just a typo somewhere..

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Darn. Added an extra factor of 2 somewhere. :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Not sure if I understood the question but the condition that \(P(k)=2^k\) got me thinking that the answer was simply \(P(11)=2^{11}+2^{10}+...+2^1+2^0

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1Because we only know \(11\) values of function \(P(x)\), namely, \(P(0), P(1), P(2),\ldots,P(10)\) we don't know the value of\(P(x)\) for other values of \(x\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Consider the polynomial \[Q(x) = \sum_{j=0}^{10} (1)^{10j}\frac{2^j}{j!(10j)!}l_j(x)\] where \(l_j(x) = \frac{x(x1)...(x10)}{(xj)}\) [\(Q(x)\) is basically the Lagrange Interpolation polynomial with data points \((0,1),(1,2),(2,2^2),...,(10,2^{10})\)] Note that \(l_j(j) = (1)^{10j}j!(10j)!\) Therefore, \(Q(j) = 2^j\) for \(j = 0\) to \(10\) \(Q(x)\) is a polynomial of degree at most \(10\) so, \(Q(x)P(x)\) is also a polynomail of degree at most \(10\) Also, \(Q(x)P(x) = 0\) for \(11\) values of \(x\). Therefore, \(Q(x) P(x)\) ia identically zero. That is, \(Q(x) = P(x)\) Therefore, \(P(11)\) \( = Q(11)\) \( = \sum_{j=0}^{10} (1)^{10j}\frac{2^j}{j!(10j)!}l_j(11)\) \( = \sum_{j=0}^{10} (1)^{10j}\frac{2^j}{j!(10j)!}\frac{11!}{(11j)}\) \( = \sum_{j=0}^{10} (1)^j2^j\left(\begin{matrix}11 \\ j\end{matrix}\right)\) \( = 2^{11}  (21)^{11}\) \( = 2^{11}  1\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@adxpoi did it correctly, although I used Newton polynomials

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0consider the finite differences: 1 1 2 1 2 4 2 4 8 4 8 16 8 16 32 which is a consequence of the fact that \(2^k=2\cdot2^{k1}=2^{k1}+2^{k1}\). now consider $$ P(n)=\sum_{k=0}^{10}\binom{n}{k}\Delta^k[P](0)\\P(11)=\sum_{k=0}^{10}\binom{11}k\cdot 1=\sum_{k=0}^{10}\binom{11}k $$ now using the binomial theorem, we have that: $$2^{11}=\left(1+1\right)^{11}=\sum_{k=0}^{11}\binom{11}k\\2^{11}1=\sum_{k=0}^{10}\binom{11}k$$ giving us \(P(11)=2^{11}1=2047\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0alternatively, if you understand that \(2^n\) has identical finite differences, i.e. \(\Delta^k[2^n]=2^n\), then we should expect that for our \(10\)th degree polynomial once we reach our \(10\)th difference we'll hit all constants, which must be \(1,1,\dots\). however, for \(2^n\) it would have been changing again, so \(1,2,\dots\).

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so for the \(11\)th term, in which we'll make use of our eleventh difference at \(0\), we expect \(\Delta^{11}[P]=0\), while for \(2^n\) we'd have \(\Delta^{11}[2^n]=1\). hence our result will be \(1\) less than\(2^11\)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1Excellent! both methods look great! In summary, the polynomial that satisfies given data points is \[\large P(x) = \sum\limits_{i=0}^{10} \binom{x}{i}\] plugging in \(x=11\) gives \(P(11) = 2^{11}1\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I tried the brute force method. With\[P(x) = a_{10}x^{10} + a_9x^9+a_8x^8+a_7x^7+a_6x^6+a_5x^5+a_4x^4+a_3x^3+a_2x^2+a_1x+a_0\]and knowing that\[P(0)=1\]\[P(1)=2\]\[P(2)=4\]\[P(3)=8\]\[P(4)=16\]\[P(5)=32\]\[P(6)=64\]\[P(7)=128\]\[P(8)=256\]\[P(9)=512\]\[P(10)=1024\]I created an augmented matrix and after a lot of GaussJordan elimination determined that\[a_{10}=2.75576\times10^{7}\]\[a_9=9.64519\times10^{6}\]\[a_8=1.65347\times10^{4}\]\[a_7=1.59560\times10^{3}\]\[a_6=1.01449\times10^{2}\]\[a_5=3.87451\times10^{2}\]\[a_4=1.12450\times10^{1}\]\[a_3=1.05290\times10^{1}\]\[a_2=3.77246\times10^{1}\]\[a_1=6.45634\times10^{1}\]\[a_0=1\]I kept about double these number of decimal places and confirmed the results using Excel. After confirming the value of the coefficients, I was able to calculate\[P(11)=2047\]I'm tired now :)
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