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Post by mondo on Sept 1, 2022 4:59:39 GMT
I try to figure out how was the last equation on page 78 calculated? There is also $r=0$ in which case the whole square brace thing should be $0$ as well. Also how is it connected to Fourier/Taylor series?
Thank you.
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Post by Admin on Sept 1, 2022 6:14:43 GMT
I try to figure out how was the last equation on page 78 calculated? There is also $r=0$ in which case the whole square brace thing should be $0$ as well. Also how is it connected to Fourier/Taylor series? Thank you. Mondo The last equation on page 78 means evaluate the derivative and then set r=0!!! Try evaluating the 1st, 2nd, 3rd, 4th etc derivatives of the third equation from the bottom and then substitute $r=0$ and $\theta=(\pi/4)$. You should then see the pattern and be able to immediately evaluate the 98th derivative. Vasco |
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Post by mondo on Sept 1, 2022 19:40:22 GMT
I try to figure out how was the last equation on page 78 calculated? There is also $r=0$ in which case the whole square brace thing should be $0$ as well. Also how is it connected to Fourier/Taylor series? Thank you. Mondo The last equation on page 78 means evaluate the derivative and then set r=0!!! Yes my mistake. It really means calculate that derivative at $0$. It will in essence just remove higher exponent terms. Right, this really means calculate that derivative $0$. It will remove all higher exponents from the derivative forumla. Try evaluating the 1st, 2nd, 3rd, 4th etc derivatives of the third equation from the bottom and then substitute $r=0$ and $\theta=(\pi/4)$. You should then see the pattern and be able to immediately evaluate the 98th derivative. Vasco Yes I see the pattern, since $98$ is even then calculating 98th derivative will zero every lower term so 97th derivative will be $98\cdot97\cdot96..\cdot 2r$ hence 98th will be just $98!$. However, we see there are even terms which are negative i.e $r^6$. How are we sure $r^98$ is positive? Thanks you. |
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Post by Admin on Sept 1, 2022 19:54:09 GMT
Mondo The last equation on page 78 means evaluate the derivative and then set r=0!!! Yes my mistake. It really means calculate that derivative at $0$. It will in essence just remove higher exponent terms. Right, this really means calculate that derivative $0$. It will remove all higher exponents from the derivative forumla. Try evaluating the 1st, 2nd, 3rd, 4th etc derivatives of the third equation from the bottom and then substitute $r=0$ and $\theta=(\pi/4)$. You should then see the pattern and be able to immediately evaluate the 98th derivative. Vasco Yes I see the pattern, since $98$ is even then calculating 98th derivative will zero every lower term so 97th derivative will be $98\cdot97\cdot96..\cdot 2r$ hence 98th will be just $98!$. However, we see there are even terms which are negative i.e $r^6$. How are we sure $r^98$ is positive? Thanks you. Mondo I don't understand your question. What do you mean "even terms which are negative"? I don't understand what you mean by "... $r^{98}$ is positive"? What about the sine part of the terms? Notice that in Latex you need to put the power in curly brackets if it is more than 1 character long. Vasco |
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Post by mondo on Sept 1, 2022 19:59:32 GMT
I meant that if you look at the Taylor series for $V_{\frac{\pi}{4}}$ you can see some powers of $r$ are have negative coefficients i.e $-\frac{1}{\sqrt{2}}r^5$ or $-r^6$. Hence the sixth derivative should be $-6!$. So my question is: how are we sure the coefficient for $r^{98}$ is positive?
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Post by Admin on Sept 1, 2022 20:20:26 GMT
I meant that if you look at the Taylor series for $V_{\frac{\pi}{4}}$ you can see some powers of $r$ are have negative coefficients i.e $-\frac{1}{\sqrt{2}}r^5$ or $-r^6$. Hence the sixth derivative should be $-6!$. So my question is: how are we sure the coefficient for $r^{98}$ is positive? Mondo Ask yourself this question: Why are some of the coefficients negative? If you can answer this question then you will see why the coefficient of $r^{98}$ is positive. Vasco |
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Post by mondo on Sept 1, 2022 20:31:13 GMT
$\sin(98*\frac{\pi}{4})$ is positive while in example $\sin(6*\frac{\pi}{4})$ is negative.
Thank you.
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Post by Admin on Sept 1, 2022 20:49:55 GMT
Mondo That's it.  I edited your last post and put \sin in place of sin because that is what it should be in Latex. Vasco |
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Post by Admin on Sept 1, 2022 20:51:38 GMT
Mondo That's it. I edited your last post and put \sin in place of sin because that is what it should be in Latex. Vasco |
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