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Mat 540 Unit 4 Essay

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Firstly, when n=0, we have F(0)≤4*F(0)+0 ⟹F(0)≥0.
Secondly, when n=2^(k+1), where k is a nonnegative Integer, we have F(2^(k+1) )≤4*F(2^k )+2^(k+1). We can prove that the function F(n) satisfies the propertyP: ∀k∈N,F(2^k )≤4^(k+1)*F(0)+S_k, where S_k is the k^th term of a sequence of integers 〖(S_m)〗_(m=0)^(+∞) that we will calculate later: The case of k=0 is correct, because we have F(2^0 )≤4^1*F(0)+1, which means S_0=1. We suppose that the case of k=m is correct, where m∈N, which means F(2^m )≤4^(m+1)*F(0)+S_m. From this, we have F(2^(m+1) )≤4*F(2^m )+2^(m+1)≤4*(4^(m+1)*F(0)+S_m )+2^(m+1)=4^(m+2)*F(0)+〖4*S〗_m+2^(m+1). As a result, the case of k=m+1 is correct and S_(m+1)=〖4*S〗_m+2^(m+1).

Consequently, we have proved the property P by induction and meanwhile we obtain two …show more content…

We suppose that 〖∀m∈N,S〗_m=4^m*X_m, where X_m is the m^th term of a sequence of integers 〖(X_m)〗_(m=0)^(+∞). We replace S_m and S_(m+1) with X_m and X_(m+1) in the equation (1), we obtain ∀m∈N,4^(m+1)*X_(m+1)=4*(4^m*X_m )+2^(m+1)⟹ X_(m+1)-X_m=〖(1/2)〗^(m+1). Since〖 S〗_0=X_0, we have X_0=1. We can easily deduce that ∀m∈N,X_(m+1)=X_m+(1/2)^(m+1)=(X_(m-1)+(1/2)^m )+(1/2)^(m+1)=⋯=1+(1/2)^2+⋯+(1/2)^(m+1)=2*(1-1/2^(m+2) ). As a result, ∀m∈N,X_m=2*(1-1/2^(m+1) ) and 〖∀m∈N,S〗_m=4^m*2*(1-1/2^(m+1) ).

Finally, we can say that ∀k∈N,F(2^k )≤4^(k+1)*F(0)+4^k*2*(1-1/2^(k+1) ). We replace 2^k with n, the inequality becomes ∀n=2^k,k∈N,F(n)≤(4*F(0)+2)*n^2-n. So far, we have found a function G(n)=(4*F(0)+2)*n^2-n that we can safely say G(n)≥F(n) at some points (when n=2^k,k∈N) instead of at all points. To illustrate G(n)≥F(n) on all points which means G(n) really is an upper bound function of F(n), we can use a software like Matlab to calculate the closest upper bound (it is marked as “real bound” in the figure 1) we can obtain from the initial inequality and compare it to the function

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