QNA > H > How To Find The Real Root Of The Equation X=E^-X

How to find the real root of the equation x=e^-x

Consider the equation as f(x)=x - e^(-x)

f(0)=0-e^(-0)=0–1=-1

f(1)=1-e^(-1)=1-(1/e) since 2

-(1/3)<-(1/e)<-(1/2)

Adding 1

1-(1/3)< 1 - (1/e)<1-(1/2)

2/3 < 1-(1/e)<1/2

So, f(1)>0 always, so the root of the equation f(x)=0 lies in the interval (0,1).

f(0.5)=0.5 - e^(-0.5)

f(0.5)=(1/2)- 1/√e

f(0.5)=-0.106

f(0.25)=-0.528800

So, the root lies between (0.25+0.5)/2=0.75/2=0.375

So the root lies between (0.375,0.5)

f(0.375)=0.375 - e^(-0.375)

f(0.375)=-0.312289

(0.375,0.5)

Now the root lies between

x=(0.5+0.375)/2=0.875/2=0.4375

(0.4375,0.5)

f(0.4375)=0.4375 - e^(-0.4375)

f(0.4375)=-0.20148

The new root is (0.4375+0.5)/2= 0.9375/2=0.46875

The root lies between (0.46875,0.5)

f(0.46875)=0.46875 - e^(-0.46875)

f(0.46875)=-0.15703400960

x=(0.46875+0.5)/2=0.96875/2=0.484375

f(0.484375)=-0.13170712779

So, the root lies between (0.484375,0.5)

x=(0.484375+0.5)/2=0.984375/2=0.4921875

f(0.4921875)=0.4921875 - e^(-0.4921875)

f(0.4921875)=-0.11910023863

Since it is negative the root is now

x=(0.4921875+0.5)/2

x=0.9921875/2=0.49609375

So the root lies between (0.49609375,0.5)

f(0.49609375)=0.49609375 - e^(-0.49609375)

f(0.49609375)=−0.11281080359

x=(0.49609375+0.5)/2

x=0.498046875

f(0.498046875)=0.498046875 - e^(-0.498046875)

f(0.498046875)=−0.10966957252

So, x=(0.498046875+0.5)/2

x=0.998046875/2

x=0.4990234375

The root is vaery near to the

x≈0.4990234376

Di Bridgette Brentson

Qual è la differenza tra un PC e un Chromebook? :: Come disattivare le notifiche su un Android durante il gioco
Link utili