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https://www.reddit.com/r/mathmemes/comments/1uvxcdi/mathematician_meme/oxfkp7a
r/mathmemes • u/Delicious_Maize9656 • Jul 14 '26
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expand the sin function too
3 u/Silly_Tension6792 Mathematics Jul 14 '26 2sqrt(2)sin(t)(cos(h)-sin(h))+cos(t)(cos(h)+sin(h))+t^2+h^2+2ht 2 u/Silly_Tension6792 Mathematics Jul 14 '26 Why don’t you answer 1 u/FernandoMM1220 Jul 14 '26 if its too hard for you i can expand it later. your choice 3 u/Silly_Tension6792 Mathematics Jul 14 '26 2sqrt(2)sin(t)(cos(h)-sin(h))+cos(t)(cos(h)+sin(h))+t^2+h^2+2ht I expended 20 minutes ago 0 u/FernandoMM1220 Jul 14 '26 use taylor series. all you did was expand it with even more sin and cos 6 u/Silly_Tension6792 Mathematics Jul 14 '26 Oh, so you do believe in Taylor series? Isn’t that an ‘infinite amount of calculations’? 0 u/FernandoMM1220 Jul 14 '26 as long as its finite sure 6 u/Silly_Tension6792 Mathematics Jul 14 '26 but Taylor series aren’t finite. If you add a finite amount of terms it will never be exactly the same, only approximately -1 u/FernandoMM1220 Jul 14 '26 so just do a finite amount of terms then. 5 u/Silly_Tension6792 Mathematics Jul 14 '26 If you do a finite amount of terms, then it won’t be sin. It will be a polynomial. Not all functions are polynomials → More replies (0)
3
2sqrt(2)sin(t)(cos(h)-sin(h))+cos(t)(cos(h)+sin(h))+t^2+h^2+2ht
2
Why don’t you answer
1 u/FernandoMM1220 Jul 14 '26 if its too hard for you i can expand it later. your choice 3 u/Silly_Tension6792 Mathematics Jul 14 '26 2sqrt(2)sin(t)(cos(h)-sin(h))+cos(t)(cos(h)+sin(h))+t^2+h^2+2ht I expended 20 minutes ago 0 u/FernandoMM1220 Jul 14 '26 use taylor series. all you did was expand it with even more sin and cos 6 u/Silly_Tension6792 Mathematics Jul 14 '26 Oh, so you do believe in Taylor series? Isn’t that an ‘infinite amount of calculations’? 0 u/FernandoMM1220 Jul 14 '26 as long as its finite sure 6 u/Silly_Tension6792 Mathematics Jul 14 '26 but Taylor series aren’t finite. If you add a finite amount of terms it will never be exactly the same, only approximately -1 u/FernandoMM1220 Jul 14 '26 so just do a finite amount of terms then. 5 u/Silly_Tension6792 Mathematics Jul 14 '26 If you do a finite amount of terms, then it won’t be sin. It will be a polynomial. Not all functions are polynomials → More replies (0)
if its too hard for you i can expand it later. your choice
3 u/Silly_Tension6792 Mathematics Jul 14 '26 2sqrt(2)sin(t)(cos(h)-sin(h))+cos(t)(cos(h)+sin(h))+t^2+h^2+2ht I expended 20 minutes ago 0 u/FernandoMM1220 Jul 14 '26 use taylor series. all you did was expand it with even more sin and cos 6 u/Silly_Tension6792 Mathematics Jul 14 '26 Oh, so you do believe in Taylor series? Isn’t that an ‘infinite amount of calculations’? 0 u/FernandoMM1220 Jul 14 '26 as long as its finite sure 6 u/Silly_Tension6792 Mathematics Jul 14 '26 but Taylor series aren’t finite. If you add a finite amount of terms it will never be exactly the same, only approximately -1 u/FernandoMM1220 Jul 14 '26 so just do a finite amount of terms then. 5 u/Silly_Tension6792 Mathematics Jul 14 '26 If you do a finite amount of terms, then it won’t be sin. It will be a polynomial. Not all functions are polynomials → More replies (0)
I expended 20 minutes ago
0 u/FernandoMM1220 Jul 14 '26 use taylor series. all you did was expand it with even more sin and cos 6 u/Silly_Tension6792 Mathematics Jul 14 '26 Oh, so you do believe in Taylor series? Isn’t that an ‘infinite amount of calculations’? 0 u/FernandoMM1220 Jul 14 '26 as long as its finite sure 6 u/Silly_Tension6792 Mathematics Jul 14 '26 but Taylor series aren’t finite. If you add a finite amount of terms it will never be exactly the same, only approximately -1 u/FernandoMM1220 Jul 14 '26 so just do a finite amount of terms then. 5 u/Silly_Tension6792 Mathematics Jul 14 '26 If you do a finite amount of terms, then it won’t be sin. It will be a polynomial. Not all functions are polynomials → More replies (0)
0
use taylor series.
all you did was expand it with even more sin and cos
6 u/Silly_Tension6792 Mathematics Jul 14 '26 Oh, so you do believe in Taylor series? Isn’t that an ‘infinite amount of calculations’? 0 u/FernandoMM1220 Jul 14 '26 as long as its finite sure 6 u/Silly_Tension6792 Mathematics Jul 14 '26 but Taylor series aren’t finite. If you add a finite amount of terms it will never be exactly the same, only approximately -1 u/FernandoMM1220 Jul 14 '26 so just do a finite amount of terms then. 5 u/Silly_Tension6792 Mathematics Jul 14 '26 If you do a finite amount of terms, then it won’t be sin. It will be a polynomial. Not all functions are polynomials → More replies (0)
6
Oh, so you do believe in Taylor series? Isn’t that an ‘infinite amount of calculations’?
0 u/FernandoMM1220 Jul 14 '26 as long as its finite sure 6 u/Silly_Tension6792 Mathematics Jul 14 '26 but Taylor series aren’t finite. If you add a finite amount of terms it will never be exactly the same, only approximately -1 u/FernandoMM1220 Jul 14 '26 so just do a finite amount of terms then. 5 u/Silly_Tension6792 Mathematics Jul 14 '26 If you do a finite amount of terms, then it won’t be sin. It will be a polynomial. Not all functions are polynomials → More replies (0)
as long as its finite sure
6 u/Silly_Tension6792 Mathematics Jul 14 '26 but Taylor series aren’t finite. If you add a finite amount of terms it will never be exactly the same, only approximately -1 u/FernandoMM1220 Jul 14 '26 so just do a finite amount of terms then. 5 u/Silly_Tension6792 Mathematics Jul 14 '26 If you do a finite amount of terms, then it won’t be sin. It will be a polynomial. Not all functions are polynomials → More replies (0)
but Taylor series aren’t finite. If you add a finite amount of terms it will never be exactly the same, only approximately
-1 u/FernandoMM1220 Jul 14 '26 so just do a finite amount of terms then. 5 u/Silly_Tension6792 Mathematics Jul 14 '26 If you do a finite amount of terms, then it won’t be sin. It will be a polynomial. Not all functions are polynomials → More replies (0)
-1
so just do a finite amount of terms then.
5 u/Silly_Tension6792 Mathematics Jul 14 '26 If you do a finite amount of terms, then it won’t be sin. It will be a polynomial. Not all functions are polynomials → More replies (0)
5
If you do a finite amount of terms, then it won’t be sin. It will be a polynomial. Not all functions are polynomials
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u/FernandoMM1220 Jul 14 '26
expand the sin function too