MHB How to Expand the Quantity (t + P)^(1/2) in Terms of t/P?

[ends]

Expand the quantity (t + P)^(1/2) about 0 in terms of t/P. Give four non-zero terms.

(t + P)^(1/2) ~
=

 

[chisigma]

Expand the quantity (t + P)^(1/2) about 0 in terms of t/P. Give four non-zero terms.

(t + P)^(1/2) ~
=

Applying the series expansion...

$\displaystyle \sqrt{1 + x} = 1 + \frac{1}{2}\ x - \frac{1}{2\ 4}\ x^{2} + \frac{1\ 3}{2\ 4\ 6}\ x^{3} - ...\ (1)$

... You obtain...

$\displaystyle \sqrt{t + P} = \sqrt{P}\ \{1 + \frac{1}{2}\ \frac{t}{P} - \frac{1}{8}\ (\frac{t}{P}^{2}) + \frac{1}{16}\ (\frac{t}{P})^{3} - ...\}\ (2)$

Kind regards

$\chi$ $\sigma$

 

[ends]

Applying the series expansion...

$\displaystyle \sqrt{1 + x} = 1 + \frac{1}{2}\ x - \frac{1}{2\ 4}\ x^{2} + \frac{1\ 3}{2\ 4\ 6}\ x^{3} - ...\ (1)$

... You obtain...

$\displaystyle \sqrt{t + P} = \sqrt{P}\ \{1 + \frac{1}{2}\ \frac{t}{P} - \frac{1}{8}\ (\frac{t}{P}^{2}) + \frac{1}{16}\ (\frac{t}{P})^{3} - ...\}\ (2)$

Kind regards

$\chi$ $\sigma$

Thanks a lot! Makes perfect sense. Could you also help me with this one - I have the integral right, but I'm having trouble approximating it perfectly for the online source to consider it correct -

Assume e^x equals it's Maclaurin series for all x. Use the Maclaurin series e^(-4x^4) to evaluate the integral:

*integral sign* from 0 to 0.2 of (e^(-4x^4))dx

Your answer will be an infinite series. Use the first two non-zero terms to estimate it's value, it must match this value to within 10^-7.

Im not sure if I am supposed to enter the decimal approximation, or the series for e^((-4x)^4)