Calculating f(x) at x=0.01 using Transcedental Equation Cos(x)-e^(-x^2/2)

[mite]

f(x)=Cos(x)-e^(-x^2/2)
Value of f(x) at x=0.01

I tried linearising at x=0 but higher derivatives are zero at x=0 .Please help me proceed

 

[chiro]

f(x)=Cos(x)-e^(-x^2/2)
Value of f(x) at x=0.01

I tried linearising at x=0 but higher derivatives are zero at x=0 .Please help me proceed

Lets for the moment forget the exponential term.

f'(x) = -sin(x)
f''(x) = -cos(x)
f'''(x) = sin(x)
f''''(x) = cos(x) and so on

f'(0) = 0 but f''(0) = -1 so you're argument doesn't stand.

As for e^(x^2/2) we get

g'(x) = x * e^(x^2/2)
g''(x) = e^(x^2/2) + x^2 * e^(x^2/2)
g'''(x) = x * e^(x^2/2) + 2x * e^(x^2/2) + x^3 * e(x^2/2)

Not of all of these terms will be zero.

 

[mite]

It is e^(-(x^2)/2) .when both are considered second derivative is zero

 

[chiro]

It is e^(-(x^2)/2) .when both are considered second derivative is zero

Use a few more terms in the Maclaurin series

 

[mite]

i tried doing that one time when there is a sine function other terms have x terms so at x=0 it is zero another time when there is cos function there is one term such as exp(-(x^2/2)) which cancels it so at x=0 it is zero

 

[HallsofIvy]

Yes, up to second derivative the difference is 0. So go past the second derivative.

cos(x)= 1- \frac{x^2}{2!}+ \frac{x^4}{4!}- \frac{x^6}{6!}+ \cdot\cdot\cdot
= 1- \frac{x^2}{2}+ \frac{x^4}{24}- \frac{x^6}{720}+ \cdot\cdot\cdot

e^{-x^2/2}= 1- \frac{x^2}{2}+ \frac{x^4}{(4)(2!)}- \frac{x^6}{(8)(6!)}+ \cdot\cdot\cdot
= 1- \frac{x^2}{2}+ \frac{x^4}{8}- \frac{x^6}{5760}

Subtract to get
-\frac{x^4}{12}+ \frac{7x^6}{5760}+ \cdot\cdot\cdot

 

[mite]

Yes I got it thank you

 

[Dickfore]

http://www.wolframalpha.com/input/?i=Series[Cos[x]-Exp[-x^2/2],{x,0,8}

 

[Calrik]

Might be worth noting that the second term is basically an error function, might not but it does tell you that this is transcendental and needs either a trick, or a series that only converges at infinity. Of course you already knew that but...

Might help in future. To solve by some sort of inspection by realising that this essentially is another function you can take the series from and use after the -?