Maclaurin Polynomial

katchat64

Q: Find the Maclaurin polynomial of order 3 for the function f(x) = e^(-2x).

I know that the Maclaurin polynomial for e^x = 1 + x + x^2/2! + x^3/3! + ...

So my question is if I would have to plug in (-2x) in the Maclaurin polynomial of e^x so I get

e^(-2x) = 1 - 2x + 2x^2 - 4x^3/3 + ...

Is this right or wrong?

yeKciM

to get the Maclaurin use the formula :

$$\displaystyle f_{(x)}= \displaystyle \sum_{n=0} ^{\infty} \displaystyle \frac {f_{(0)}^{(n)}}{n!} x^n$$

where is $$\displaystyle f_{(0)}^{(n)}$$ nth derivative...

so it goes

$$\displaystyle f_{(0)}=e^{-2\cdot0}=1$$

$$\displaystyle f'_{x}=(e^{-2x})'=\displaystyle -\frac {2}{e^{2x}} \mid _{x=0} = -2$$

$$\displaystyle f''_{x}=(e^{-2x})''=\displaystyle \frac {4}{e^{2x}} \mid _{x=0} = 4$$

$$\displaystyle f'''_{x}=(e^{-2x})''' =\displaystyle -\frac {8}{e^{2x}} \mid _{x=0} =-8$$

$$\displaystyle f^{(4)}_{x}=(e^{-2x})^{(4)}=\displaystyle \frac {16}{e^{2x}} \mid _{x=0} = 16$$

and so on .... and u have :

$$\displaystyle f_{(x)}= \displaystyle \sum _{n=0} ^{\infty} \frac {(-1)^{n}}{n!} \cdot ......$$

i think u can do that

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Also sprach Zarathustra

Q: Find the Maclaurin polynomial of order 3 for the function f(x) = e^(-2x).

I know that the Maclaurin polynomial for e^x = 1 + x + x^2/2! + x^3/3! + ...

So my question is if I would have to plug in (-2x) in the Maclaurin polynomial of e^x so I get

e^(-2x) = 1 - 2x + 2x^2 - 4x^3/3 + ...

Is this right or wrong?

It is right! (But why?)

katchat64

I know the formula I was just looking to see some work to see if I did the problem correctly. In another problem that I had in an exam consisted of the Maclaurin polynomial of 1/(1-x) = 1 + x + x^2 + X^3 + ...

and the professor wanted us to find the power series representation of 1/(1-x^3) and to figure that out you needed to put an x^3 wherever there was an x in the original given equation.

So I thought since I know the Maclaurin polynomial of e^x I thought that you were able to plug in (-2^x) wherever there was an x in that original polynomial.

yeKciM

I know the formula I was just looking to see some work to see if I did the problem correctly. In another problem that I had in an exam consisted of the Maclaurin polynomial of 1/(1-x) = 1 + x + x^2 + X^3 + ...

and the professor wanted us to find the power series representation of 1/(1-x^3) and to figure that out you needed to put an x^3 wherever there was an x in the original given equation.

So I thought since I know the Maclaurin polynomial of e^x I thought that you were able to plug in (-2^x) wherever there was an x in that original polynomial.
yes but i think yours 3rd member is wrong should be $$\displaystyle 4x^2$$

just to note that when u mentioned that one $$\displaystyle \displaystyle \frac {1}{1-x}$$ that is from $$\displaystyle \displaystyle \sum _{n=0} ^{\infty} x^n$$ when $$\displaystyle |x|<1$$ we had have a lot of times to find $$\displaystyle \displaystyle \frac {1}{(1-x)^2}$$ or $$\displaystyle \displaystyle \frac {1}{(1-x)^3}$$ which are just a derivate of $$\displaystyle \displaystyle \frac {1}{1-x}$$ and actually using that Maclouren $$\displaystyle \displaystyle \frac {1}{1-x} = \displaystyle \sum _{n=0} ^{\infty} x^n$$ u can have a lot of another to do based on that one, you just have to tune it so it looks like that one

there is like 5 - 10 depending on school to school that u should know by heart and everything goes around them (at least here )

Edit:just remembered something to note

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HallsofIvy

MHF Helper
yes but i think yours 3rd member is wrong should be $$\displaystyle 4x^2$$
No, the third term for $$\displaystyle e^x$$ is $$\displaystyle \frac{1}{2}x^2$$. Replacing x by -2x gives $$\displaystyle \frac{(-2x)^2}{2}= \frac{4x^2}{2}= 2x^2$$ as katchat64 has.

katchat64, yes, your method is completely correct.

yeKciM

yeKciM

No, the third term for $$\displaystyle e^x$$ is $$\displaystyle \frac{1}{2}x^2$$. Replacing x by -2x gives $$\displaystyle \frac{(-2x)^2}{2}= \frac{4x^2}{2}= 2x^2$$ as katchat64 has.

katchat64, yes, your method is completely correct.
thank you for some reason i leave factorials as they are because in my mind i always thinking of it as sum there and then just looking for how does changes... leaving down that factorial thanks again