While the Lagrange error bound method here gives 4, I think the actual correct answer is 3. The Maclaurin polynomial of sin(x)=x-x^3/3!+..., so sin(0.4)=0.4-0.4^3/3!+...≈0.389418. Since the degree of a polynomial is the highest degree of its terms when the polynomial is expressed in its canonical form consisting of a linear combination of monomials, we can see that the polynomial x-x^3/3! has a degree of 3. Notice that 0.4-0.4^3/3!≈0.389333 and the difference sin(0.4)-(0.4-0.4^3/3!)≈0.000085
Thanks for this, confirms to me that the way these exercises are worded is problematic / difficult to determine without making an exhaustive table. Although I think using the alternating series error analysis method would yield better results.
Spencer, you are correct...When you find the degree "n" needed for the Taylor polynomial, using the method shown in this video, the answer you get isn't necessarily the minimum required degree for the Taylor polynomial. The reason: When you find an upper bound for the (n+1)st derivative, the upper bound isn't unique. A larger upper bound may give a larger answer for "n". (Note: When you're asked to find "an upper bound" it does not imply you're required to find the "smallest upper bound"). In the video his upper bound for the absolute value of the (n+1)st derivative of sin(x) was 1, which led to the answer n = 4 for the required degree of the polynomial. As you pointed out, you can do some calculations to show that the minimum n needed is actually n = 3. This illustrates that the "n" you find, using the Taylor Remainder Theorem (the Lagrange error bound), is not unique, but the good news is: It is guaranteed to give an "n" which assures an error smaller than that specified. To get a little more technical about the example in the video, we can look at getting a better error estimate. For n = 3, we'll need an upper bound for the absolute value of the 4th derivative of sin(x). But that 4th derivative is just sin(x), so between x = 0 and x = 0.4, an upper bound on that 4th derivative is 0.4, using the fact that sin(x) is an increasing function, and sin(x) < x for all x > 0. So using that "better" upper bound of 0.4, for n = 3 the absolute error is less than (0.4)(0.4)^4/(4!) which is 0.000426...That's less than the specified allowable error, so that proves a MacLaurin polynomial of degree 3 will do the job. To summarize: The answer n = 4 in the video is not "the LEAST degree of the polynomial that assures an error smaller than 0.001". But it is a correct answer in the sense that a MacLaurin polynomial of degree 4 will assure an error smaller than 0.001 when estimating sin(0.4).
Yes his solution only works under the assumption that the n + 1 th derivative is a cosine function thus the range of the n + 1 th derivative is from cos(0) to cos(0.4) so cos(0) = 1 is the Max value, M. Under this assumption n MUST be even otherwise M may take on a different (smaller) value, so his solution finds the smallest value of n that is even. If he had done the same thing for odd n he would have found that the n + 1 th derivative is bounded by sin(0.4) so M = sin(0.4)
I just wanted to comment that since abs(Rn(x)) equals to m*(this is important)(x to the power of n + 1) / (n + 1) factorial, the remainder evaluated at 0.4 will be smaller than 0.001 at the "third" degree because it is lower than 0.001 when Rn(x) is n = 3.
looks like n = 4 but since fourth degree polynomial for sin x will end up being a third degree polynomial; you will have to take a 5th degree polynomial to get the desired accuracy according to video.
Pretty sure you are incorrect here. The n=4 polynomial is exactly equal to the n=3 polynomial, so its upper bound (as determined here) also works for n=3, in this instance.
Moe 590 you could take this in highschool if ur good. this is calc 3, next is multivariable calc which completes the basic requirements for getting into a math bachelors program.
ACTUALLY it has really bad been bounded taking M as 1 is bad bound knowing that n=1 is the correct answer and taking this approximation give an n at 4 not do this technique
Can anyone explain this? Since M = f^(n+1)(x), whereas x belongs to [0, 0.4]. The maximum value for f(x) = sin(0.4) which isn’t 1. Please help me if anyone have any idea about this.
The entire purpose of this calculation is to calculate functions such as trig functions without a calculator. So it wouldn’t make much sense if you would need to calculate sin(0.4) so we just use the best estimate of the maximum of the function which is always 1 for sin.😊
I love how he repeats the words while he writes
While the Lagrange error bound method here gives 4, I think the actual correct answer is 3. The Maclaurin polynomial of sin(x)=x-x^3/3!+..., so sin(0.4)=0.4-0.4^3/3!+...≈0.389418. Since the degree of a polynomial is the highest degree of its terms when the polynomial is expressed in its canonical form consisting of a linear combination of monomials, we can see that the polynomial x-x^3/3! has a degree of 3. Notice that 0.4-0.4^3/3!≈0.389333 and the difference sin(0.4)-(0.4-0.4^3/3!)≈0.000085
Thanks for this, confirms to me that the way these exercises are worded is problematic / difficult to determine without making an exhaustive table. Although I think using the alternating series error analysis method would yield better results.
Spencer, you are correct...When you find the degree "n" needed for the Taylor polynomial, using the method shown in this video, the answer you get isn't necessarily the minimum required degree for the Taylor polynomial.
The reason: When you find an upper bound for the (n+1)st derivative, the upper bound isn't unique. A larger upper bound may give a larger answer for "n". (Note: When you're asked to find "an upper bound" it does not imply you're required to find the "smallest upper bound").
In the video his upper bound for the absolute value of the (n+1)st derivative of sin(x) was 1, which led to the answer n = 4 for the required degree of the polynomial. As you pointed out, you can do some calculations to show that the minimum n needed is actually n = 3.
This illustrates that the "n" you find, using the Taylor Remainder Theorem (the Lagrange error bound), is not unique, but the good news is: It is guaranteed to give an "n" which assures an error smaller than that specified.
To get a little more technical about the example in the video, we can look at getting a better error estimate. For n = 3, we'll need an upper bound for the absolute value of the 4th derivative of sin(x). But that 4th derivative is just sin(x), so between x = 0 and x = 0.4, an upper bound on that 4th derivative is 0.4, using the fact that sin(x) is an increasing function, and sin(x) < x for all x > 0.
So using that "better" upper bound of 0.4, for n = 3 the absolute error is less than (0.4)(0.4)^4/(4!) which is 0.000426...That's less than the specified allowable error, so that proves a MacLaurin polynomial of degree 3 will do the job.
To summarize: The answer n = 4 in the video is not "the LEAST degree of the polynomial that assures an error smaller than 0.001". But it is a correct answer in the sense that a MacLaurin polynomial of degree 4 will assure an error smaller than 0.001 when estimating sin(0.4).
Yes his solution only works under the assumption that the n + 1 th derivative is a cosine function thus the range of the n + 1 th derivative is from cos(0) to cos(0.4) so cos(0) = 1 is the Max value, M. Under this assumption n MUST be even otherwise M may take on a different (smaller) value, so his solution finds the smallest value of n that is even. If he had done the same thing for odd n he would have found that the n + 1 th derivative is bounded by sin(0.4) so M = sin(0.4)
@@ameerq8767 I think you're right. Thank you for the explanation!
I did another method and it gave me 3 as well and i was confused why it was not matching....thankfully i am not the only one
Is there any way to solve this without brute forcing n?
I just wanted to comment that since abs(Rn(x)) equals to m*(this is important)(x to the power of n + 1) / (n + 1) factorial, the remainder evaluated at 0.4 will be smaller than 0.001 at the "third" degree because it is lower than 0.001 when Rn(x) is n = 3.
Lagrange
Keep up the great work😀👍
Ben Vanderzwan no
Merry Christmas mr. khan!!!!
looks like n = 4 but since fourth degree polynomial for sin x will end up being a third degree polynomial; you will have to take a 5th degree polynomial to get the desired accuracy according to video.
Pretty sure you are incorrect here. The n=4 polynomial is exactly equal to the n=3 polynomial, so its upper bound (as determined here) also works for n=3, in this instance.
Isn't M the maximum value of the n+1 derivative on the interval [a,x]? How come we use the absolute maximum of the derivatives?
What grade is this for because I'm not look forward to it.
vault boy If you are good, this is for 11th Grade, or 12th Grade
vault boy 12 for most good people
This is for whenever you take Calculus BC.
AP calculus BC its not that bad it just takes a bit more studying to really understand it
Did this in college Calculus. Revisiting it for numerical analysis 🙃
Is this Professor degree mathematics
Barr
Moe 590 you could take this in highschool if ur good. this is calc 3, next is multivariable calc which completes the basic requirements for getting into a math bachelors program.
Cory Ziegler I'm no expert here I'm just ok at maths (I'm from uk btw)
I showed this to my pre-algebra class and I swear only half the kids got it.
@@chessandmathguy I guarantee you no one in a pre algebra class understood this
ACTUALLY it has really bad been bounded taking M as 1 is bad bound knowing that n=1 is the correct answer and taking this approximation give an n at 4 not do this technique
It’s way more convenient to solve it by alternating series error bound. Isn’t it?
Not sure . . . does that work here? What alternating series can you think of? It seems like it tho . . .
@@calebcoudriet7275 You know sin's Taylor series is literally an alternating series right?
Can anyone explain this? Since M = f^(n+1)(x), whereas x belongs to [0, 0.4]. The maximum value for f(x) = sin(0.4) which isn’t 1. Please help me if anyone have any idea about this.
The entire purpose of this calculation is to calculate functions such as trig functions without a calculator. So it wouldn’t make much sense if you would need to calculate sin(0.4) so we just use the best estimate of the maximum of the function which is always 1 for sin.😊
@@junghyunkim2247 Thank you so much, I spent around half hour trying to figure out two weeks ago
lagrunge
❤