Some people love solving math just for its own sake - they find beauty in mathematics. For them, math is not only used for application in physics or engineering. It's just beautiful in itself - on its own - PURE, BEAUTIFUL.
The "beauty" can be found in equations, expressions and the flow that takes place when you form a solution. Such beauty can be found in the following questions.
If you're a Calculus student or a math enthusiast - try out these 5 beautiful questions from Integral Calculus. I made these questions a long time ago and have received solutions for only two of them so far. See if you can solve them all.
Question #1
Question #2
Question #3
Question #4
Question #5
You can leave your answers as a comment here or mail them to me at dev@zarrata.com
Categories: Mathematics Tags: beautiful calculus, beautiful integral calculus, beautiful integration questions, beauty mathematical, calculus, calculus integration problems, calculus questions, difficult integration questions, hard integration questions, integral calculus, integral calculus questions, integration questions, math beauty, mathematical beauty
Mathematics is an art form. It is also the language of other art forms like music & poetry. It is the language of nature & Nature is Beautiful. Most men who have not studied the subject in depth, will find this amusing. As far as the mathematician is concerned, he is well familiar with this beauty. When you get sudden results out of nothing. Thats beauty. When new relations & expressions combine together like atoms to give an expression-a molecule. Thats beauty. When your thoughts & insights take the form of a mathematical equation. Thats beauty. Finding a relation between totally different concepts & observations. Thats beauty. The greatest example of the last point above is the Fundamental Theorm of Calculus which relates two TOTALLY different concepts-Finding instantaneous rates of change & finding areas of surfaces. The two concepts are related together by this theorm. One of the most beautiful equations in Mathematics is the Euler's Indentity.
One of the most beautiful equations in mathematics is the Euler's identity which relates the numbers e , π, & i my favourite number 0 & my second favourite number 1 . There is beauty in mathematical methods, the results of the methods, the applications of these results. Read more about the Euler's Equation on The Math Less Travelled.
If we observe closely, we find that the various branches of mathematics are all linked together in some way or the other. I show here one such observation. The binomial theorem can actually be expressed in terms of the derivatives of xn instead of the use of combinations. Lets start with the standard representation of the binomial theorm, ^{n}=x^{n}+%20^{n}C_{1}ax^{n-1}+%20^{n}C_{2}a^{2}x^{n-2}+^{n}C_{3}a^{3}x^{n-3}+...^{n}C_{n}a^{n})
We could then rewrite this as a sum, ^{n}=sum_{0}^{n}^{n}C_{r}a^{r}x^{n-r})
Another way of writing the same thing would be, ^{n}=x^{n}+nax^{n-1}+frac{n(n-1)}{1.2}a^{2}x^{n-2}+frac{n(n-1)(n-2)}{1.2.3}a^{3}x^{n-3}+...a^{n})
We observe here that the equation can be rewritten in terms of the derivatives of xn. The coefficient of a in the second terms is the first derivative of xn, similarily the coefficient of a2/2! in the second term is the second derivative of x... Lets look at the last term of the expansion, The coefficient of xn/n! should now be the nth derivative of xn. Which is very true... The following simple relation holds for all n... =n!)
Hence, the binomial expansion can now be written in terms of derivatives! We have, ^{n}=x^{n}+aD_{1}+frac{D_{2}}{2!}a^{2}+frac{D_{3}}{3!}a^{3}...frac{D_{n}}{n!}a^{n})
where Dr represents the rth derivate of xn. Hence, we can now write this as a sum, http://latex.codecogs.com/gif.latex?huge%20(x+a)^{n}=sum_{0}^{n}frac{D_{r}a^{r}}{r!} So, we now have the expansion in terms of combinations as well as in terms of derivatives! ^{n}=sum_{0}^{n}^{n}C_{r}a^{r}x^{n-r}=sum_{0}^{n}frac{D_{r}a^{r}}{r!})
n instead of the use of combinations. Lets start with the standard representation of the binomial theorm, ^{n}=x^{n}+%20^{n}C_{1}ax^{n-1}+%20^{n}C_{2}a^{2}x^{n-2}+^{n}C_{3}a^{3}x^{n-3}+...^{n}C_{n}a^{n})
We could then rewrite this as a sum, ^{n}=\sum_{0}^{n}^{n}C_{r}a^{r}x^{n-r})
Another way of writing the same thing would be, ^{n}=x^{n}+nax^{n-1}+\frac{n(n-1)}{1.2}a^{2}x^{n-2}+\frac{n(n-1)(n-2)}{1.2.3}a^{3}x^{n-3}+...a^{n})
We observe here that the equation can be rewritten in terms of the derivatives of xn. The coefficient of a in the second terms is the first derivative of xn, similarily the coefficient of a2/2! in the second term is the second derivative of x... Lets look at the last term of the expansion, The coefficient of xn/n! should now be the nth derivative of xn. Which is very true... The following simple relation holds for all n... =n!)
Hence, the binomial expansion can now be written in terms of derivatives! We have, ^{n}=x^{n}+aD_{1}+\frac{D_{2}}{2!}a^{2}+\frac{D_{3}}{3!}a^{3}...\frac{D_{n}}{n!}a^{n})
where Dr represents the rth derivate of xn.Hence, we can now write this as a sum, ^{n}=\sum_{0}^{n}\frac{D_{r}a^{r}}{r!})
Or as the sum, ^{n}=\sum_{0}^{n}\frac{D_{r}x^{r}}{r!})
So, we now have the expansion in terms of combinations as well as in terms of derivatives! ^{n}=\sum_{0}^{n}^{n}C_{r}a^{r}x^{n-r}=\sum_{0}^{n}\frac{D_{r}a^{r}}{r!})