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

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GOODMAN
 2 years ago
Best ResponseYou've already chosen the best response.0So help her already. I dont know what that is, lol. Can you atleast try? @lgbasallote

greysica
 2 years ago
Best ResponseYou've already chosen the best response.1yeah yeah , at least try if u can .. :)

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.3I'm not sure what you mean, you want to derive the three series expansions above?

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.3@greysica do I understand that you just want to see how the above statements are derived?

greysica
 2 years ago
Best ResponseYou've already chosen the best response.1I have to find the way how it works as a function based on mac laurin polinom

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.3I can derive the mcLauren series from the function, but that doesn't sound like what you want.

greysica
 2 years ago
Best ResponseYou've already chosen the best response.1hmm , I'm not sure actually about question for my homework , but would u like show me how it works based on ur opinion ?

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.3the 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

sami21
 2 years ago
Best ResponseYou've already chosen the best response.0@greysica what Turing Test has mentioned above is general formula for Taylor series about a=0 also Known as Maclaurin series.

sami21
 2 years ago
Best ResponseYou've already chosen the best response.0@greysica you need Maclaurin Polynomials of the given functions??

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.3ok @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 \(n=0\)

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.3what is \(f^{(0)}(x)\) when \(f(x)=a^x\) that is, what is the 0th derivative?

greysica
 2 years ago
Best ResponseYou've already chosen the best response.1I dont understand about derivative actually .. is that ax^x1 ?

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.3no, 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
 2 years ago
Best ResponseYou've already chosen the best response.3you say you don't know how to take the derivative of a function?

greysica
 2 years ago
Best ResponseYou've already chosen the best response.1ya , I dont know how , I really confuse about this subject

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.3yes, and what is f(0) in this case?

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.3f(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?

greysica
 2 years ago
Best ResponseYou've already chosen the best response.1f(n)(0)=1 1 ?? 0!

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.3right should have written f^(0)(0) but still, that's right so what does all that simplify to?

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.3\[f^{(0)}(0)=f(0)=a^0=1\] and \[x^0=1\]true, but\[0!\neq0\]

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.3\[0!=1\]by definition

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.3so try again, what does it simplify too?

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.3\[T_0(a^x)={f^{(0)}(0)\over 0!}x^0={a^0\over0!}x^0=?\]

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.3correct, that is the first term :)

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.3so 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
 2 years ago
Best ResponseYou've already chosen the best response.3in 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

greysica
 2 years ago
Best ResponseYou've already chosen the best response.1(f^1)(0) x^1 1!

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.3and what is \(f^{(1)}(0)\) for \(f(x)=a^x\) ?

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.3I 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
 2 years ago
Best ResponseYou've already chosen the best response.3*I think you have...

greysica
 2 years ago
Best ResponseYou've already chosen the best response.1n^th ? what does mean ?

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.3it means the nth number derivative n=1 is the first derivative n=2 is the second derivative n=3 is the third derivative etc.

greysica
 2 years ago
Best ResponseYou've already chosen the best response.1so 1^1 1 ?? 1!

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.3first 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 \(f(x)=a^x\) ?

greysica
 2 years ago
Best ResponseYou've already chosen the best response.11! = 2 right ? 1! means 1+1*1 = 2 ? like that ?

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.3no, 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

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.3ignore every other part of the formula for now, we are just trying to figure out the numerator part at the moment

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.3look 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

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.3no 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

greysica
 2 years ago
Best ResponseYou've already chosen the best response.1ughh .. okay .. thanks ,,,

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.3the derivative formula you want is under the part of the list I gave you that says "common derivatives" top row, third column

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.3what is \[\frac d{dx}(a^x)\]?

greysica
 2 years ago
Best ResponseYou've already chosen the best response.1what is the meaning of In ?

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.3natural 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)

greysica
 2 years ago
Best ResponseYou've already chosen the best response.1@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 :(

greysica
 2 years ago
Best ResponseYou've already chosen the best response.1I'm so desperate and shame about this .. thank u for trying help me

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.3Taylor 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

greysica
 2 years ago
Best ResponseYou've already chosen the best response.1yeah !! 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
 2 years ago
Best ResponseYou've already chosen the best response.3...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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