Taylor Polynomial

[stinajeana]

Find P5(x), the fifth degree Taylor Polynomial of f(x)=lnx around a=1, and use it to approximate ln(1.1)

...I don't want the answer, I just want to know how to begin! So if you are able to help can you also explain how you went about doing it as well? Thanks

 

[stinajeana]

So I found the derivative of lnx 5 times and plugged 1 into what I got

1st derivative= 1/x= 1/(1)= 1
2nd derivative= -1/x^2= -1/(1)^2= -1
3rd derivative= 2/x^3= 2/(1)^3= 2
4th derivative= -6/x^4= -6/(1)^4= -6
5th derivative= 25/x^5= 25/(1)^5= 25

... then I substituted what I got into the n-th degree polynomial formula and got:

lnx is approximately equal to 1+(-1/2!)(x-(-1))^4+(2/3!)(x-2)^3+(-6/4!)(x-(-6))^4+(5/5!)(x-5)^5

...So what i'm wondering now is what does the "use it to approximate ln(1.1)" mean?

 

[Bob Brown MSEE]

...So what i'm wondering now is what does the "use it to approximate ln(1.1)" mean?

P(x) = 1(x-1)/1! -1(x-1)²/2! +2(x-1)³/3! -6(x-1)⁴/4! +25(x-1)⁵/5!
F(x) = ln(x)

P(1.1) = 0.09531875...
F(1.1) = 0.09531017...

so F(1.1) ~ P(1.1) = 0.09531875...
or ln(1.1) ~ P(1.1) = 0.09531875...

 

[stinajeana]

P(x) = 1(x-1)/1! -1(x-1)²/2! +2(x-1)³/3! -6(x-1)⁴/4! +25(x-1)⁵/5!
F(x) = ln(x)

P(1.1) = 0.09531875...
F(1.1) = 0.09531017...

so F(1.1) ~ P(1.1) = 0.09531875...
or ln(1.1) ~ P(1.1) = 0.09531875...

I got P(1.1) is equal to ln(1.1) is equal to .0953103333

....I understand the whole point of it i'm just wondering what I did wrong to get the .0953103333

 

[stinajeana]

P(x) = 1(x-1)/1! -1(x-1)²/2! +2(x-1)³/3! -6(x-1)⁴/4! +25(x-1)⁵/5!
F(x) = ln(x)

P(1.1) = 0.09531875...
F(1.1) = 0.09531017...

so F(1.1) ~ P(1.1) = 0.09531875...
or ln(1.1) ~ P(1.1) = 0.09531875...

Ok, never mind. I just realized that it's accurate up to 3 decimals.