Hey I just met u and this is crazy ..
Here my problems , Help me maybe ? :))
prove approximation of Mac Laurin Polinom from this function

- anonymous

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- schrodinger

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- anonymous

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- lgbasallote

No.

- anonymous

okay .. :(

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## More answers

- anonymous

Harsh. @lgbasallote

- lgbasallote

haha lol

- anonymous

So help her already. I dont know what that is, lol. Can you atleast try? @lgbasallote

- lgbasallote

i have no idea

- anonymous

yeah yeah , at least try if u can .. :)

- TuringTest

I'm not sure what you mean, you want to derive the three series expansions above?

- TuringTest

@greysica do I understand that you just want to see how the above statements are derived?

- anonymous

I have to find the way how it works as a function based on mac laurin polinom

- TuringTest

I can derive the mcLauren series from the function, but that doesn't sound like what you want.

- anonymous

hmm , I'm not sure actually about question for my homework , but would u like show me how it works based on ur opinion ?

- TuringTest

the Taylor expansion of a function about x=0 is\[f(x)=\sum_{n=0}^\infty{f^{(n)}(0)\over n!}x^n\]where \(f^{(n)}(0)\) is the \(n^{th}\) derivative of the function at x=0

- anonymous

is that for A ?

- anonymous

@greysica what Turing Test has mentioned above is general formula for Taylor series about a=0 also Known as Maclaurin series.

- anonymous

@greysica you need Maclaurin Polynomials of the given functions??

- anonymous

ya :)

- TuringTest

ok @greysica you there?
I'll walk you though it must you must participate
the formula for the series expansion of a function about \(x=0\) is\[f(x)=\sum_{n=0}^\infty{f^{(n)}(0)\over n!}x^n\]where \(f^{(n)}(0)\) is the \(n^{th}\) derivative of \(f(x)\) at the value \(x=0\)
so let's start with the first function, \(f(x)=a^x\)
what is the first term of the sequence according to the formula?
i.e. when \(n=0\)

- anonymous

x=0 ?

- TuringTest

not not x=0

- anonymous

|dw:1344008695732:dw|

- TuringTest

what is \(f^{(0)}(x)\) when \(f(x)=a^x\)
that is, what is the 0th derivative?

- anonymous

devivative of 0 = 1 ?

- anonymous

I dont understand about derivative actually .. is that ax^x-1 ?

- TuringTest

no, the \(0^{th}\) derivative means we don't take the derivative at all; it's just the function
so \(f^{(0)}(x)=f(x)=a^x\)
now what then is \(f^{(0)}(0)\) ?

- TuringTest

you say you don't know how to take the derivative of a function?

- anonymous

ya , I dont know how , I really confuse about this subject

- anonymous

f(0) ?

- TuringTest

yes, and what is f(0) in this case?

- anonymous

so f(0)=a^x ?

- anonymous

f(0)=a^0

- anonymous

f(0)=1 ?

- anonymous

or f(0)= a ?

- TuringTest

f(x)=a^x
f(0)=1
correct
what about the rest of the parts of the formula?\[T_n(a^x)={f^{(n)}(0)\over n!}x^n\] we agree that \(f^{(n)}(0)=1\) right?
now where else in this formula do we need to plug in zero to get the first term?

- anonymous

f(n)(0)=1
---------1 ??
0!

- TuringTest

right
should have written f^(0)(0) but still, that's right
so what does all that simplify to?

- anonymous

1/0 * 1 ?

- TuringTest

\[f^{(0)}(0)=f(0)=a^0=1\] and \[x^0=1\]true, but\[0!\neq0\]

- TuringTest

\[0!=1\]by definition

- TuringTest

so try again, what does it simplify too?

- TuringTest

\[T_0(a^x)={f^{(0)}(0)\over 0!}x^0={a^0\over0!}x^0=?\]

- anonymous

1/1 * 1= 1

- TuringTest

correct, that is the first term :)

- TuringTest

so let's write that down somewhere because we will need it later:\[T_0(a^x)=1\]now we try to find \(T_1(a^x)\)
so now we use the formula\[T_n(a^x)={f^{(n)}(0)\over n!}x^n\]with \(n=1\)
what do you get? (this should take you a moment to figure out, i'll brb.

- TuringTest

in case you have forgotten your derivatives use this list
http://tutorial.math.lamar.edu/pdf/Calculus_Cheat_Sheet_Derivatives_Reduced.pdf
and tell me what you get for the next term
like I said, I'm going to the store, brb

- anonymous

(f^1)(0)
----------x^1
1!

- TuringTest

and what is \(f^{(1)}(0)\) for \(f(x)=a^x\) ?

- anonymous

0/1 * 1 ?

- TuringTest

I don't think you have forgotten what I said that \(f^{(n)}(x)\) means
it means the \(n^{th}\) derivative of the function \(f(x)\)
so what is \(f^{(1)}(x)\) considering that \(f(x)=a^x\)

- TuringTest

*I think you have...

- anonymous

n^th ? what does mean ?

- TuringTest

it means the n-th number derivative
n=1 is the first derivative
n=2 is the second derivative
n=3 is the third derivative
etc.

- anonymous

so 1^1
--------1 ??
1!

- TuringTest

first you need to find the correct derivative of the function
since n=1 that means we need the first derivative of the function
what is the first derivative of \(f(x)=a^x\) ?

- anonymous

1! = 2 right ? 1! means 1+1*1 = 2 ? like that ?

- anonymous

or 1 ! = 1*1 = 1 ?

- TuringTest

no, the question we need to answer is:
if\[f(x)=a^x\]what is\[f^{(1)}(x)=f'(x)=?\]there should be no number in the answer to the question I am asking you

- anonymous

oh ..

- TuringTest

ignore every other part of the formula for now, we are just trying to figure out the numerator part at the moment

- anonymous

i dunno :(

- anonymous

ax^x-1 ?

- TuringTest

look on the list I gave you
http://tutorial.math.lamar.edu/pdf/Calculus_Cheat_Sheet_Derivatives_Reduced.pdf
find your function
though if you have problems taking the derivative of functions this is going to be nearly impossible for you

- anonymous

f (x) + f '(x) ?

- TuringTest

no
sorry I can't really help you if you can't take the derivative
you may need to retake calculus 1 if this is a problem

- anonymous

f^0 (x) ?

- anonymous

ughh .. okay .. thanks ,,,

- TuringTest

the derivative formula you want is under the part of the list I gave you that says "common derivatives"
top row, third column

- TuringTest

what is \[\frac d{dx}(a^x)\]?

- anonymous

a^x ln(a)

- TuringTest

yes

- anonymous

what is the meaning of In ?

- TuringTest

natural logarithm (base e)\[\huge\log_e(a)=\ln a\]so what you just found above from the chart is the first derivative of the function is\[f^{(1)}(x)\]so what is\[f^{(1)}(0)\]?
(plug in zero into what you found for the derivative)

- anonymous

@TuringTest I think I will ask about this to my senior .. I really really lost about this subject .. I dontknow why I'm so stupid :(

- anonymous

I'm so desperate and shame about this .. thank u for trying help me

- TuringTest

Taylor series are tough for many people, so don't feel bad.
I was watching the MIT multivariable calc lecture, and when the teacher said they needed Taylor series to solve a problem the whole class was like "NO!" hahaa
so you're not alone, but you won't get there unless you make sure you understand derivatives very well, so work on that in the meantime first.
Don't feel ashamed, just try to identify your problem areas and don't give up!
good luck :D

- anonymous

yeah !! Thank u very much :)
I'll try my best ! I have calculus 2 test in the first week of september , wish can do it properly !!
I hope I can get at least B :)

- TuringTest

...and you are NOT stupid, so don't say that :P
just review your derivatives
you're welcome, and good luck :)

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