ganeshie8
  • ganeshie8
[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 \(P(11)\)
Mathematics
schrodinger
  • schrodinger
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

freckles
  • freckles
\[P(x)=a_{10}x^{10}+a_9x^9+a_8x^8+a_7x^7+a_6x^6+a_5x^5+a_4x^4+a_3x^3 + \\ +a_2x^2+a_1x^+a_0 \\ P(0)=1 \\ a_0=1 \\ \] well I know there is a really long way... pluggin in all those critters to find the constant coefficients of the polynomial P
anonymous
  • anonymous
4095?
ganeshie8
  • ganeshie8
11 equations and 11 unknowns is solvable for sure :)

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

ganeshie8
  • ganeshie8
@ospreytriple nope, but it seems you got it! the answer is \(2047\)
ganeshie8
  • ganeshie8
please post your method, im pretty sure it is just a typo somewhere..
anonymous
  • anonymous
Darn. Added an extra factor of 2 somewhere. :)
anonymous
  • anonymous
Not 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
triciaal
  • triciaal
why not 2048?
triciaal
  • triciaal
never mind
ganeshie8
  • ganeshie8
Because 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
  • anonymous
Consider the polynomial \[Q(x) = \sum_{j=0}^{10} (-1)^{10-j}\frac{2^j}{j!(10-j)!}l_j(x)\] where \(l_j(x) = \frac{x(x-1)...(x-10)}{(x-j)}\) [\(Q(x)\) is basically the Lagrange Interpolation polynomial with data points \((0,1),(1,2),(2,2^2),...,(10,2^{10})\)] Note that \(l_j(j) = (-1)^{10-j}j!(10-j)!\) 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, \(P(11)\) \( = Q(11)\) \( = \sum_{j=0}^{10} (-1)^{10-j}\frac{2^j}{j!(10-j)!}l_j(11)\) \( = \sum_{j=0}^{10} (-1)^{10-j}\frac{2^j}{j!(10-j)!}\frac{11!}{(11-j)}\) \( = \sum_{j=0}^{10} (-1)^j2^j\left(\begin{matrix}11 \\ j\end{matrix}\right)\) \( = 2^{11} - (2-1)^{11}\) \( = 2^{11} - 1\)
anonymous
  • anonymous
@adxpoi did it correctly, although I used Newton polynomials
anonymous
  • anonymous
consider 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 \(2^k=2\cdot2^{k-1}=2^{k-1}+2^{k-1}\). 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
  • anonymous
alternatively, 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
  • anonymous
so 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
  • ganeshie8
Excellent! 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
  • anonymous
I 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 Gauss-Jordan 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 :)

Looking for something else?

Not the answer you are looking for? Search for more explanations.