- #1

deuteron

- 57

- 13

- Homework Statement
- Expand the given functions for very small and very big arguments ##x## giving ##3## terms

- Relevant Equations
- ##T_nf[x;a]= \displaystyle\sum_k \frac 1 {k!}\ f^{(k)}|_a\ (x-a)^k##

The function is

$$ f(x)=\sqrt{1-x}$$

and we are expected to expand it using Taylor expansion for very small ##(x<<1)## and very big ##(x>>1)## arguments

My thought process was the following:

$$T_2f[x;x_0]=\sqrt{1-x_0} -\frac 12 \frac 1{\sqrt{1-x_0}}(x-x_0) -\frac 14 \frac 1 {\sqrt{1-x_0}^3}(x-x_0)^2$$

and then assuming small arguments, namely assuming ##x_0<<1##, I did the following approximation:

$$ T_2f[x;x_0]\approx \sqrt{1} -\frac 12 \frac 1 {\sqrt{1}} (x) -\frac 1 4 \frac 1 {\sqrt{1}^3} (x)^2 = 1-\frac 12 x -\frac 14 x^2$$

For big arguments, I assumed ##x_0>>1## and made the following approximation:

$$T_2f[x;x_0]\approx i\sqrt{a} +\frac 12 \frac 1 {i\sqrt{a}} (a) +\frac 14 \frac 1 {i\sqrt{a}^3} a^2$$

However, the real answers are, for small arguments:

$$f(x<<1)= 1-\frac 12 x - \frac 18 x^2$$

and for big arguments:

$$f(x>>1)= i\sqrt{a} \sqrt{1-\frac 1x} = i\sqrt{x} \ (1-\frac 1 {2x} -\frac 1{8x^2})$$

I see that my mistake is to assume the expansion point to be very big instead of assuming big ##x##, but I don't know how to do that, does anyone understand the in between steps for the solution?

$$ f(x)=\sqrt{1-x}$$

and we are expected to expand it using Taylor expansion for very small ##(x<<1)## and very big ##(x>>1)## arguments

My thought process was the following:

$$T_2f[x;x_0]=\sqrt{1-x_0} -\frac 12 \frac 1{\sqrt{1-x_0}}(x-x_0) -\frac 14 \frac 1 {\sqrt{1-x_0}^3}(x-x_0)^2$$

and then assuming small arguments, namely assuming ##x_0<<1##, I did the following approximation:

$$ T_2f[x;x_0]\approx \sqrt{1} -\frac 12 \frac 1 {\sqrt{1}} (x) -\frac 1 4 \frac 1 {\sqrt{1}^3} (x)^2 = 1-\frac 12 x -\frac 14 x^2$$

For big arguments, I assumed ##x_0>>1## and made the following approximation:

$$T_2f[x;x_0]\approx i\sqrt{a} +\frac 12 \frac 1 {i\sqrt{a}} (a) +\frac 14 \frac 1 {i\sqrt{a}^3} a^2$$

However, the real answers are, for small arguments:

$$f(x<<1)= 1-\frac 12 x - \frac 18 x^2$$

and for big arguments:

$$f(x>>1)= i\sqrt{a} \sqrt{1-\frac 1x} = i\sqrt{x} \ (1-\frac 1 {2x} -\frac 1{8x^2})$$

I see that my mistake is to assume the expansion point to be very big instead of assuming big ##x##, but I don't know how to do that, does anyone understand the in between steps for the solution?