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Jul
23
comment What is the mistake in doing integration by this method?
Check the title.
Jul
23
comment What is the mistake in doing integration by this method?
-1, this doesn't answer the question which was asked.
Jul
23
comment What is the mistake in doing integration by this method?
-1, this doesn't answer the question which was asked.
Jun
5
comment Where am I wrong in $\int_0^1 x(1-x)^ndx$
@Lucas, you get another negative when you compute the integral of $(1-x)^{n+1}$.
Jun
5
comment Where am I wrong in $\int_0^1 x(1-x)^ndx$
Your first integral is incorrect: note that you are integrating a nonnegative function, so the answer should be nonnegative. Check the signs in your integration by parts.
May
31
comment Parametrization of surfaces for vector integration
What orientation should your surface have? You didn't specify one.
May
31
comment When not to use rref for finding eigenvectors?
Neither of the vectors you gave are eigenvectors of the given matrix corresponding to an eigenvalue of $50$. Is the matrix you gave supposed to be $A$ or $A- 50 I$? Even in the latter case something is wrong, please check your problem statement.
May
29
comment What are higher derivatives?
Possibly, but then it seems to be misplaced in the section it's in, which is talking about multivariable functions as studied in multivariable calculus.
May
29
comment What are higher derivatives?
Interestingly, the Wikipedia entry on the Frechet derivative outlines precisely what I say here, so this entry is at odds with the one linked to in the original post.
May
29
answered What are higher derivatives?
May
27
comment Regarding functions from R² to R: continuity and differentiability
A differentiable function need not have a continuous gradient.
May
12
comment Limit of multivariable functions Do not understand solution.
You are implicitly assuming that $x \ne 0$ when you use the inequality $1/(x^2+y^2) \le 1/x^2$, since otherwise the term on the right is undefined. This is fine as long as you also say what happens when $x=0$ separately. It would be better to use the given inequality to bound the numerator as: $|x^2||y| \le |x^2+y^2||y|$ to avoid this subtlety.
Apr
27
comment finding column vectors - linear transformations
The work you gave supporting your answer suggests that you don't quite understand what is being asked for, hence my question. You've accepted a correct answer below, but without understanding why it is correct you will likely have trouble with related concepts. Do you understand why the answer should be the first column of the matrix $A$?
Apr
26
comment finding column vectors - linear transformations
Do you understand what "matrix of $L$ with respect to $S$ and $T$" means?
Mar
31
comment Proving that $P[0, 1]$, the space of all polynomials on $[0, 1]$ is not complete.
Note that you don't necessarily have to show that the given sequence converges to $e^x$, just that it doesn't converge to a polynomial, which might be simpler depending on what you know about Taylor series.
Mar
29
comment Find all points on a surface which have a tangent plane parallel to given plane - is my method correct?
Also, when coming up with $y=x-\frac23$ it seems you divided by $x+y$ at some point, meaning that doing this ignores possible points for which $x+y=0$. So, you should consider these points separately at the end, meaning use $y=-x$ and $f_x=6$ to find precise points.
Mar
29
comment Find all points on a surface which have a tangent plane parallel to given plane - is my method correct?
Using only $f_x=-f_y$ isn't enough since you're ignoring that you have actual values for each of these, 6 for the first and -6 for the second. (Again, $k$ must be $1$ as you can determine by looking at the $z$-components of the gradients.) Your solution gives points for which $f_x=-f_y$ but which don't satisfy $f_x = 6$ and $f_y = -6$ as required. Once you have $y=x-\frac23$ you can then use $f_x=6$ for solve for $x$ to narrow down your values.
Mar
29
comment Find all points on a surface which have a tangent plane parallel to given plane - is my method correct?
@HelenByrne, see my comment to your original question.
Mar
29
comment Find all points on a surface which have a tangent plane parallel to given plane - is my method correct?
It doesn't make sense for the normal vectors to be vectors in $\mathbb R^2$ as you assume when you use $f_xe_1+f_ye_2$. You can't forget about the $z$-component, meaning that you should take the gradient of the three-variable function $f(x,y,z)=x^2-y^3-2xy-z$ of which your surface is a level surface, and similarly for the given plane. (In other words, you can't move the $z$ term to the right as you did.) The resulting gradients should be parallel yes, but by comparing the $z$-coordinates you can conclude that they must be equal, as the solution you commented on below suggests.
Mar
28
comment If $f(x)=\tilde{f}(\|x\|)$ and $f$ is continuous, is $\tilde{f}$ continuous?
@quid, fair enough.