After getting the error message and then retrying the model gave better answers then it does now, without the error messages. I can give some examples where the model gave a correct answer after the reload, but now behaves in a reduced intelligence level
Op
What is the Taylor expansion of sin(x)
The Taylor expansion of sin(x) is given
by:
sin(x) =x - (x^3) /3! + (x^5) /5! - (x^7)/7!
+
where "!" denotes the factorial
operator. This means that the terms in
the expansion are the powers of X, with
the coefficients being the alternating
series (-1)^n / (2n + 1)!. The expansion
is valid for all values of x.
a Try again
In general, the Taylor expansion of a
function f(x) at a point ×0 is given by:
f(x) = f(×0) + (x - ×0) f'(×0) + ((x - x0) ^2
(2!) f"'(×0) + ((x - x0) ^3 / 3!) f'''(×0) +
where f'(x), f" (x), and f''(x) are the
first, second, and third derivatives of
†(x), resp
ting ×0 to 0 in
G
Try again
the Taylor expansion of sin(x), we cañ
obtain the expansion given above.
OP
What about sin(x)+cos(x)
OP
The Taylor expansion of sin(x) + cos(x)
can be found by using the fact that
sin(x) + cos (x) = sqrt(2) * cos (x + pi/4),
and then using the Taylor expansion for
cos(x) with x0 = x + pi/4.
The Taylor expansion of cos(x) is given
by:
COS(x) = 1 - (x^2)/2! + (x^4)/4! -
(x^6)/6! +
Thus, the Taylor expansion of cos(x +
pi/4) isizen (x:+ pI/4)^4)/4! - ((x +
pi/4) ^6)/6! +...)
This expansion is valid for all values of
×.
OP
This is very close to the right answer but
the correct identity is sin(x) + COs (x)
sqrt(2) * cos(x - pi/4). Can you revise
your answer
Yes, you are correct. The Taylor
expansion of sin(x) + cos(x) is given by:
sin(x) + cos(x) = sqrt(2) * cos(x - pi/4)
To derive this expansion, we can use
the fact that sin(x) + cos(x) = sqrt(2) *
cos(x - pi/4), and then use the Taylor
expansion for cos (x) with x0 = x - pi/4.
The Taylor expansion of cos(x) is given
by:
COS(x) =1 - (x^2) /2! + (x^4)/4! -
(x^6) 6! +.
Thus, the Taylor expansion of cos(x-
pi/4) is given by:
cos(x - pi/4) = 1 - ((x - pi/4) ^2) /2! + ((x
pi/4) ^4)/4! - ((x - pi/4) ^6)/6! +
Finally, the Taylor expansion of sin(x)+
cos(x) is given by:
sin(x) + cos(x) = sqrt(2) * (1 - ((x •
pi/4) ^2) /2! + ((x - pi/4) ^4)/4! - ((x -
pi/4) ^6)/6! +...)
This expansion is valid for all values of
x. I apologize for the mistake in my
previous answer.
Try again
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It’s a bit garbled, because it was copied from screenshots
After getting the error message and then retrying the model gave better answers then it does now, without the error messages. I can give some examples where the model gave a correct answer after the reload, but now behaves in a reduced intelligence level
Op What is the Taylor expansion of sin(x) The Taylor expansion of sin(x) is given by: sin(x) =x - (x^3) /3! + (x^5) /5! - (x^7)/7! + where "!" denotes the factorial operator. This means that the terms in the expansion are the powers of X, with the coefficients being the alternating series (-1)^n / (2n + 1)!. The expansion is valid for all values of x. a Try again In general, the Taylor expansion of a function f(x) at a point ×0 is given by: f(x) = f(×0) + (x - ×0) f'(×0) + ((x - x0) ^2 (2!) f"'(×0) + ((x - x0) ^3 / 3!) f'''(×0) + where f'(x), f" (x), and f''(x) are the first, second, and third derivatives of †(x), resp ting ×0 to 0 in G Try again the Taylor expansion of sin(x), we cañ obtain the expansion given above. OP What about sin(x)+cos(x) OP The Taylor expansion of sin(x) + cos(x) can be found by using the fact that sin(x) + cos (x) = sqrt(2) * cos (x + pi/4), and then using the Taylor expansion for cos(x) with x0 = x + pi/4. The Taylor expansion of cos(x) is given by: COS(x) = 1 - (x^2)/2! + (x^4)/4! - (x^6)/6! + Thus, the Taylor expansion of cos(x + pi/4) isizen (x:+ pI/4)^4)/4! - ((x + pi/4) ^6)/6! +...) This expansion is valid for all values of ×.
OP This is very close to the right answer but the correct identity is sin(x) + COs (x) sqrt(2) * cos(x - pi/4). Can you revise your answer
Yes, you are correct. The Taylor expansion of sin(x) + cos(x) is given by: sin(x) + cos(x) = sqrt(2) * cos(x - pi/4)
To derive this expansion, we can use the fact that sin(x) + cos(x) = sqrt(2) * cos(x - pi/4), and then use the Taylor expansion for cos (x) with x0 = x - pi/4. The Taylor expansion of cos(x) is given by: COS(x) =1 - (x^2) /2! + (x^4)/4! - (x^6) 6! +. Thus, the Taylor expansion of cos(x- pi/4) is given by: cos(x - pi/4) = 1 - ((x - pi/4) ^2) /2! + ((x
It’s a bit garbled, because it was copied from screenshots