10.Find k if f f(k)=5 where f(k)=2k-1
Answers
Answer:
We have the equation f(t/2)4=f(t) . Expanding both sides in a Taylor series expansion for small t , we find
f(t/2)4=f(0)4+2f(0)3f′(0)t +12f(0)2[3f′(0)2+f(0)f′′(0)]t2+⋯
and
f(t)=f(0)+f′(0)t+12f′′(0)t2+⋯
Comparing the coefficients at each order in t , we find the equations
At order 0
f(0)[f(0)3−1]=1⟹f(0)=0 or 1
At order 1
f′(0)[2f(0)3−1]=0⟹f′(0)=0 . This is true because 2f(0)3−1 is either -1 or 1, but not zero.
At order 2
f′′(0)[f(0)3−1]=0 . This implies that f′′(0)=0 if f(0)=0 and undetermined if f(0)=1 .
Following this procedure to higher and higher orders, we get the following result, in the two cases
If f(0)=0 , then we get f(n)(0)=0 for all n≥1 . This implies that the function identically vanishes. This solution is not of interest.
If f(0)=1 , then we get f(2n+1)(0)=0 for all n≥0 . For the even derivatives, we find f(2n)(0)=3⋅5⋯(2n−1)f′′(0)n and f′′(0) is not fixed. Let us write f′′(0)=2a . The Taylor expansion for f(t) in this case is then
f(t)=∑∞n=0f(2n)(0)t2n(2n)! =∑∞n=0(2at2)n2⋅4⋯(2n) =∑∞n=0(at2)nn!=eat2
In the final step we have summed the series. This converges only in the range t2∈[0,1/|a|) . However, once the series has been summed, we can analytically continue it uniquely to all t . Finally, we write ea=c . Note that c>0
Thus, the only non-trivial solution to the equation that you had is f(t)=ct2 for any real c>0
Answer:
3
Step-by-step explanation:
5=2k-1
5+1=2k
2k=6
k=6/2
k=3