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`        If y= arctan1/xx+x+1  +arctan1/xx+3x+5  +arctan1/xx+5x+7 +........upto n terms. what is dy/dx?`
2 years ago

Ashutosh Mohan Sharma
180 Points
```							The derivative of the arc tangent isddxarctan(x)=11+x2.From the formula for geometric series (see for examplethis answer for a proof) shows that1+y+y2+y3+⋯=11−yif|y|<1.Plugging in−x2fory, we get that11+x2=11−(−x2)=1+(−x2)+(−x2)2+(−x2)3+⋯+(−x2)n+⋯=1−x2+x4−x6+x8−x10+⋯provided that|−x2|<1; that is, provided|x|<1. All the computations below are done under this hypothesis (see comments at the end).So we have that:ddxarctan(x)=1−x2+x4−x6+x8−x10+⋯if|x|<1Because this is aTaylor series, it can be integrated term by term. That is, up to a constant, we have:arctan(x)=∫(ddxarctan(x))dx=∫(1−x2+x4−x6+x8−x10+⋯)dx=∫(∑n=0∞(−1)nx2n)dx=∑n=0∞(∫(−1)nx2ndx)=∑n=0∞((−1)n∫x2ndx)=∑n=0∞((−1)nx2n+12n+1)+C=C+(x−x33+x55−x77+x99−x1111+⋯).Evaluating atx=0gives0=arctan(0)=C, so we getarctan(x)=x−x33+x55−x77+x99−x1111+⋯,if|x|<1.the equality you ask about.Note however that this does not hold forallx: it certainly works if|x|<1, by the general properties of Taylor series. But the arc tangent is defined for all real numbers. The series we have here is∑n=0∞(−1)nx2n+12n+1.Using theRatio Test, we have thatlimn→∞|an+1||an|=limn→∞|x|2n+32n+3|x|2n+12n+1=limn→∞(2n+1)|x|2n+3(2n+3)|x|2n+1=limn→∞|x|2(2n+1)2n+3=|x|2limn→∞2n+12n+3=|x|2.By the Ratio Test, the series converges absolutely if|x|2<1(that is, if|x|<1) and diverges if|x|>1. Atx=1andx=−1, the series is known to converge. So the radius of convergence is1, and the equality is valid forx∈[−1,1]only (that is, if|x|≤1; we gained two points in the process).However, the arc tangent has a nice property, namely thatarctan(1x)=π2−arctan(x),So, given a value ofxwith|x|>1, you can use this identity to computearctan(x)by computingarctan(1x)instead, and for this argument the seriesisvalid.
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2 years ago
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