Is \(\displaystyle{x}^{{{2}}}-{a}{x}+{a}^{{{x}}}-{1}\) real and less than 2?

Heather Dickson

Heather Dickson

Answered question

2022-04-15

Is x2ax+ax1 real and less than 2?

Answer & Explanation

xcentricccrhb1

xcentricccrhb1

Beginner2022-04-16Added 12 answers

Step 1
Consider the function
f(x)=x2ax+ax3
Since there is no restriction on x, we must consider a>0
For this function
f(x)=2xa+axlog(a) and f(x)=2+axlog2(a)
The first derivative cancels at
x=a2W(12aa2log2(a))log(a)>0a>0
where W() is Lambert funtion.
f(x)=W(12aa2log2(a))(W(12aa2log2(a))+2)log2(a)a243<0a>0
f(x)=2(W(12aa2log2(a))+1)>0a>0
So, x corresponds to a minimum of the function
So f(x)>0 as soon as x is greater than the solution of f(x)=0. The problem is that this solution does not show analytical expression even using special functions. This would require numerical methods.
Step 2
For the solution of f(x)=0, we can try approximations using Taylor expansion around x=x This gives
f(x)=f(x)+12f(x)(xx)2+O((xx)3)
and then the estimate of the two roots
x±=x±2f(x)f(x)
which is not too bad as shown in the table below
(axestxsolx+estx+sol0.50.9039130.8644091.895481.922961.01.0000001.0000002.000002.000001.50.9361680.9474951.939261.924772.00.8122080.8568351.826931.769122.50.6871170.7667241.717811.622783.00.5788480.6861961.626031.509373.50.4892480.6165581.551841.424444.00.4157570.5569421.492231.360214.50.3552120.5059571.444071.310655.00.3048660.4622211.404771.27158)
Using these estimates as x0, Newton method should converge quite fast using, as usual,
xn+1=xnf(xn)f(xn)
In the present case
xn+1=axn(xnlog(a)1)+xn2+3axnlog(a)a+2xn
Using it for a bad case a=5, the iterates would be
(nxn00.3048660910.4714835220.4622471430.46222053)
(nxn01.4047725111.2884279521.2718864231.2715799441.27157984)

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